iit jee physics

a body is dropped from the top of the tower and falls.the distance covered by it after n second is directly proportional to.
one balloon moving upward with 5ms2 acceleration. if balloon drop the object after 2 seconds.. then find the time to reach maximum height
Disatnce b/w center of 2 stars is 10a. masses are M and 10M and radii rae A ans 2A ,resp. a body of mass m is fired straight from the surface of larger star towards smaller star. what should be its minimum speed to reach the surface of smaller star?
A cylinder of radius r full of liquid of density rho is rotated about its axis at w rad/s. What will be the increase in pressure at the centre of the cylinder ?
The initial velocity of the particle is 10m/sec and its retardation is 36m. The distance moved by the particlein 5th second of its motionis what?
a body prjected vertically upwords crosses a point twice its journey at a heigt h just after t1,t2seconds maximum height reached by the body is
A car is moving eastward .It covers 200 km in 4 hrs. Find it's velocity
a ball is thrown vertically upward covering a height of 50m with a speed of 30m/s, at the height of 50m a ball is thrown horizontally with a speed of 20m/s. calculate the minimum displacement and time taken by the ball.
For identical rods are joined end to end to form s square and each is of mass m. find moment of inertia about i)median line
ii) point of intersection of diagonals.
.a rod of length L is composed of a uniform length 1/2L of wood whose mass is m1 and a uniform length of 1/2L of brass whose mass is m2. find the moment of inertia for a perpendicular axis through the wood end.
a car moves towards north at a speed of 54km/h for 1 hr.then it moves eastward with same speed for same duration.the average speed and velocity of car for complete journey is ? how to find that? answer is 15m/s,15/root 2
Can two particles be in equilibrium under the action of their mutual gravitational force ? Can three particles
be ? Can one of the three particles be ?
A pile of looselink chain, mass per unit length ρ, lies on a rough surface with coefficient of kinetic friction μ_{k}. One end of the chain is being pulled horizontally along the surface by a constant force P. Determine the acceleration of the chain in terms of x and dx/dt=v .
 A metre stick is pivoted about its centre.A piece of wax of mass 20g travelling horizontally and perpendicular to it at 5m/s strikes and adheres to one end of the stick so that the stick starts to rotate in a horizontal circle. Given the moment of inertia of the stick and the wax about the pivot is 0.02kg m2, what is the initial angular velocity of the stick?
 What is the moment of inertia of a uniform semicircular wire of mass m and radius r, about an axis passing through its centre of mass and perpendicular to its plane?
Two point masses m and 3m are placed at a distance r.The moment of inertia of the system about an axis passing through the centre of mass of system and perpendicular to the line joining the point masses is?
Question 1: a particle is moving in a circular path of radius r at a constant speed v. draw a graphs represents its acceleration 'a' with respect to 'r' ?
Question 4 : if the motion of an object is represented by a straight line parallel to the time axis in a distance time graph, then the object undergoes
(a) An accelerated motion
(b) A decelerated motion
(c) A uniform nonzero velocity motion
(d) A zero velocity motion
Question3:Light waves projected on oil surface show seven colours due to the phenomenon of
(a) Polarisation
(b) Refraction
(c) Reflection
(d) interference
Question 2 : The working of a microwave oven involves
(a)adsorption of microwaves by matter
(b) reception of microwave by optical fibre
(c) microwave amplification by stimulated emission of radiation
(d) transmission of microwaves through a metal
Question 1: A bullet is fired vertically up from a 400m tall tower with a speed 80 m/s. If g is taken as 10 m/s^2 the time taken by the bullet to rteach the ground will be
(a) 8s (b) 16s (c) 20s (d) 24s
Sir i have score 67% in HSC class 12 and 256 in bitsat (2014)..i have took a drop do i have any chances for jee advance very tensed at this moment please help
a ball is projected vertically up with an initial speed of 20m/sec and acceleration due to gravity is 10m/s^2then how long will it take for the ball to reach a point 10m above the point of projection second line'
a ball is projected vertically up with an initial speed of 20m/sec and acceleration due to gravity is 10m/s^2then how long will it take for the ball to reach a point 10m above the point of projection second line'
a ball is projected vertically up with an initial speed of 20m/sec and acceleration due to gravity is 10m/s^2then how long will it take for the ball to reach a point 10m above the point of projection second line'
a sphere of radius 20 cm and weight 1 N is suspended against a smooth wall by a string of length 20 cm. the string joins a point in the wall and a point on the sphere . find the inclination and the tension in the string and the reaction of the wall?
a sphere of radius 20 cm and weight 1 N is suspended against a smooth wall by a string of length 20 cm. the string joins a point in the wall and a point on the sphere . find the inclination and the tension in the string and the reaction of the wall?
AN 8.0 g bullet is fired horigontally into a 9 kg cube of wood, which is at rest , and strikes in it. the cube is free to move and has a speed of 40 cm/s after impact. find the intial velocity of the bullet.
what is the tension in a rod of length L & mass M at a distance y from F when the rod is acted on by two unequal forces F & f when F>f is
a force acts for 20 seconds on a body of mass 10 kg after which force ceases and the body describes 50m in the next 10 seconds . find the magnitude of the force?
a force acts for 20 seconds on a body of mass 10 kg after which force ceases and the body describes 50m in the next 10 seconds . find the magnitude of the force?
a force acts for 20 seconds on a body of mass 10 kg after which the force ceases and the body describes 50m in the next 10 second . find the magnitude of the force?
two particles p and q start from the origin and execute simple harmonic motion alon Xaxis with the same amplitude and time periods 3s and 6s respectively. The ratio of the velocities of P and Q when they meet is
a)1:2
b)2:1
c)2:3
d)3:2
A man is walking under an inclined mirror at a constant velocity Vms^{1s.}along the X axis. If the mirror is inclined at an angle θ with the horizontal, then what is the velocity of the image?.
'for a particle travel 10m in first 5 sec and than 10m in next 3 second assume constant is acceleration what is the distance travelled in next 2 second'
a man is 48m behind a bus which is at rest.. the bus starts accelerating at the rate of 1m/s2, at the same time the man starts running with uniform velocity of 10m/s.what is the minimum time in which the man catches the bus
A ball is projected with velocity V and angle of projection is A. After what time, the ball is moving in the direction parallelto the initial direction.
a particle moving with uniform accleration from A to B with velocities v1 and v2 if c is mid point between A and B determine the velocities of particle c
a body moving with a velocity of 10m/s accelerates uniformly and covers 300m when its velocity becomes 20m/s find the time taken to achieve this velocity
a body moving with a velocity of 10 m/s accelerates uniformly and covers 300m when its velocity becomes 20m/s find the time taken to achieve this velocity
a body moving with a velocity of 10m/s accelerates uniformly and covers 300m when its velocity becomes 20m/s. find the time taken to achieve this velocity.
a body moving moving with a velocity of 10m/s accelerates uniformly and covers 300m when its velocity becomes 20 m/s.find the time taken to achieve this velocity
A boy is thrown up in alift with a velocity 'u' relative to the lift and the time of flight is found to be 't'.Find the accelaration with which the lift is moving up ?
a ball is thrown vertically upwards from the top of a tower of height'h' with velocity'v'. The ball strikes the ground after;
A stone is thrown vertically upward with an initial velocity v_{o.}The distance travelled by it in time 1.5v_{o}/g.
a particle moves in the xy plane with constant accelaration ω directive along negative yaxis.the equation of motion is y=axbx^2.find the velocity of the particle at the origin
an object is thrown vertically up from the ground passes the height 5m twice in an interval of 10s.what is time of flight answer
an object is thrown vertically up from the ground passes the height 5m twice in an interval of 10s.what is time of flight
What is kinetic energy possessed by a body with a mass of 6kg projected vertically upward with a speed of 5m/s ?
If a particle is thrown vertically upwards,then what will be its velocity so that it covers same distance in 5th & 6th second and how?
a ball falling from a height of 10m, remained in contact with the ground for 0.01s. after that ity rose to the height of 2.5m. find the acceleration of ball when it remained in contact with the ground.
A bird is flying with 20 m/s in space, calculate the velocity vector of bird in terms of i,j and k?
A bird is flying with 20 m/s in space, what is the velocity vector of bird in terms of i,j and k?
a particle is moving along the x axis so that its velocity changes with time as v^2 = 150  16x m/s . the acceleration of particle is
A resistor of resistance R is connected to an ideal battery. If the value of R is decreased, then the power dissipated in the resistor will .........???
a body dropped from top of a tower fall through 40 m during last 2 sec of its fall the height of the tower is (g=10m/s^{2} )
From the top of a tower having height of 100m a ball is dropped at the same time a ball is projected vertically upward from the ground with a velocity of 20m/s.Find when and where the two ballls meet.
A ball strikes a horizontal floor at an angle θ=45°. The coefficient of restitution between the ball and the floor is e=1/2. The fraction of its kinetic energy lost in collision is ?
A weightless rod of length l carries two equal masses m one fixed at the end and other in the middle of the rod. The rod can revolve in a vertical plane about A. Then horizontal velocity which must be imparted to end C of rod to deflect it to horizontal position...
A weightless rod of length l carries two equal masses m one fixed at the end and other in the middle of the rod. The rod can revolve in a vertical plane about A. Then horizontal velocity which must be imparted to end C of rod to deflect it to horizontal position
from a tower of height H .a partical thrown vertically upwards with speed u..the time taken by the particle to hit the ground is n times than that taken by to reach the heighesttime of its path..the relation btwn H,u,n
A body is thrown in a projectile path such dat it touches four vertices of a regular hexagon. If the side of the regular hexagon is "a", find the range of the projectile?
its like one edge is on the ground and the rest of the hexagon in air.
A person is at a distance of 10 meter from a bus. the bus acclerates with a speed on meter per second^{2. In order to catch the bus he has to move at a speed of}
^{5meter per second at what time he can catch the bus?}
The moment of inertia of the body about an axis is 1.2 kgm^{2}. Initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad/s^{2} must be applied about the axis for the duration of
The Poisson’s ratio of a material is 0.5. if a force is applied to a wire of this material, there is a decrease in the crosssectional area by 4%. The percentage increase in the length is
A capacitor of capacity 0.1 μF connected in series to a resistor of 10 MΩ is charged to a certain potential and then made to discharge through resistor. The time in which the potential will take to fall to half its original value is (Give, log_{10}2 = 0.3010)
A monatomic gas is suddenly compressed to (1/8)th of its initial volume adiabatically. The ratio of its final pressure to the initial pressure is (Given, thr ratio of the specific heats of the given gas to be 5/3)
A ball is projected from ground at an angle of 60* & initial velocity u=40m/s. At highest point , a bird of mass 500 gm., flying horizontally w.r.t. ground with constant velocity 20m/s hits the ball. How far from the point of projection will the ball fall on the ground.(Give horizontal distance).
using the expression 2dsinθ=λ,one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90°.the wavelength λ is exactly known and the error in θ is constant for all values of θ. as θ increases from 0°( JEE ADV. 2013)
a) the absolute error in d remains constant
b) the absolute error in d increases
c) the fractional error in d remains constant
d) the fractional error in d increases
A ball is projected from ground at an angle of 60* & initial velocity u=40m/s. At highest point , a bird of mass 500 gm., flying horizontally w.r.t. ground with constant velocity 20m/s hits the ball. How far from the point of projection will the ball fall on the ground.(Give horizontal distance).
A flat mirror revolves at a constant angular velocity making 2 revolution/sec. With what velocity will a light spot move along a Spherical screen with a radius of 10 m, if the mirror is at a centre of curvature of the screen ?
a bullet enters into a wall with velocity u on entering a distance x into the wall, its velocity becomes v, the bullets stop after penetrating another length y into the wall, show that v/u = root y/x+y
if mirror is rotated by angular velocity w, then find out velocity of light spot on vertical mirror
if mirror is rotated by angular velocity w, then find out velocity of light spot on vertical mirror
two particle each of mass m are moving in horizontal circle with same angular speed.if both string are of same length then the ratio of tension in string T1/T2
A particle of mass m begins to slide down a fixed smooth sphere from the top with negligible initial velocity. What is its tangetional accelaration when it breaks off the sphere?
Ans. √5 *g/3
the velocity of a particle at an instant is 10m/s and after 5sec the velocity of the particle is 20m/s the velocity in 3sec: before in m/s is?
A car A is travelling on a straight levelled road with uniform velocity 60 km/hr. It is followed by a car B with a speed of 70 km/hr. When the distance between the two cars is 2.5 km, a deceleration of 20 km/hr^{2} is given to car B. After what distance and time will B catch up with A ?
Please do answer
A car A is travelling on a straight levelled road with uniform velocity 60 km/hr. It is followed by a car B with a speed of 70 km/hr. When the distance between the two cars is 2.5 km, a deceleration of 20 km/hr^{2} is given to car B. After what distance and time will B catch up with A ?
Please do answer
A heavy uniform straight rod is suspended from a point by two strings of the same length as the rod attached to its ends.If one string is cut prove that initial angular acceleration of the rod is nine times that of the remaining string.
wat is the use of this iitbrain site where no body solve questions??
Two pendulum of time peroids 3s and 7s resp. Start oscillating simultaneously from two opposite extreme positions. After how much time they will be in same phase ?
A man walks over a rough surface. The angle between the force of friction and the instantaneous velocity of the man is
A. π
B. π/2
C. Zero
D. 2π
A man walks over a rough surface. The angle between the force of friction and the instantaneous velocity of the man is
A. π
B. π/2
C. Zero
D. 2π
A man walks over a rough surface. The angle between the force of friction and the instantaneous velocity of the man is
A. π
B. π/2
C. Zero
D. 2π
A cricketer hits a sixer.the cricket ball moves up with a velocity of 2m/s and falls down.What is it's initial velocity while falling down?
A bob executes SHM of period 20s, its velocity is 5 cm/s, 2 s after it has passed the mean position. Find the amplitude of the bob?
Two particles each of mass m are connected by a string of length 1m. The particles are kept on a smooth horizontal plane. The initial separation between the particles is 0.5 m. One of the particles is given a velocity v0 as shown in the figure. The magnitude of angular velocity of one particle with respect to the other particle just after the string becomes taut is.
Two particles each of mass m are connected by a string of length 1m. The particles are kept on a smooth horizontal plane. The initial separation between the particles is 0.5 m. One of the particles is given a velocity v0 as shown in the figure. The magnitude of angular velocity of one particle with respect to the other particle just after the string becomes taut is.
density of a substance at 0 c is 10 gm/cc and at 100 c its density is 9.7 gm/cc the coefficient of linear expansion of the substance will be ??
The speed cannot be zero since distance cannot be zero while the velocity can be zero as displacement can be zero.
I have a small question in the application of WorkEnergy theorem.
Q) A block of mass 5 Kg is attached to a spring of spring constant 4000N/m. It is stretched by a distance of 100 mm from its mean position. Then the block is released. What will be the speed of the block when it passes the mean position again? The surface has a coefficient of friction = 0.2
My answer: By Work energy theorem, Work done by friction + Work done by applied force + Work done by spring = Change in Kinetic Energy
However the answer in my book says Work done by friction + Work done by spring = Change in Kinetic Energy
Why don't we consider work done by applied force?
Thanks!
a man is suspended from ceiling by a string revolving in a horizontal circle of radius 5 cm.the tangential speed of the mass is .7 m/s what is angle between the string and the vertical?
A ball is hit 1m above ground at 35ms and 53 degrees. if there are seats with width 1m and ht 1m and first seat is 110m from crease. Which row will it hit
A uniform rod of mass m and length l rotates in a horizontal plane with angular velocity ω about a vertical axis passing through one end.The tension in the rod at a distance x from the axis is
A ball of mass M moving with speed v collides perfectly inelastically with another ball of mass m at rest. The magnitude of impulse imparted to the first ball is
A radioactive substance is decaying (converting into energy and energy does not have mass) with constant rate(i.e internal forces are acting on it).Then why the center of mass is changing?We know that internal forces cannot change center of mass?Kindly Explain?
a body is thrown up with a velocity 40 m/s. At the same time another body is dropped from a height 40m. Their relative acceleration after 1.3 is
please give me the solution for the above problem
when an object is dropped to the ground from a height 10m. what is its kinetic energy at 5 m and just before coming to ground. if the mass is doubled how its velocity changes just before reaching the ground?
when an object is dropped to the ground from a height 10m. what is its kinetic energy at 5 m and just before coming to ground. if the mass is doubled how its velocity changes just before reaching the ground?
when an object is dropped to the ground from a height 10m. what is its kinetic energy at 5 m and just before coming to ground. if the mass is doubled how its velocity changes just before reaching the ground?
there are three source of sound of equal intensities with frequency 400., 401,402 hz .Wat is the beat freq heard if all are sounded simultaneously??
If two unit vectors are inclined at angle of 90, then magnitude of their resultant is
a2
b√2
c√ 3
dzero
A particle is executing SHM and its velocity v is related to its position (x) as v^{2} + ax^{2} = b, where a and b are positive constants. The frequency of oscillation of particle is
A simple pendulum of mass m executes SHM with total energy E. If at an instant it is at one of extreme positions, then its linear momentum after a phase shift of π/3 rad will be
A particle executes SHM and its position varies with time as x = A sin wt. Its average speed during its motion from mean position to mid point of mean and extreme position is
The SHM of a particle is given by the equation x = 2 sin wt + 4 cos wt. Its amplitude of oscillation is
In damped oscillations, damping force is directly proportional to speed of oscillator. If amplitude becomes half of its maximum value in 1s, then after 2s amplitude will be (Aο  initial amplitude).
A 100g mass stretches a particular spring by 9.8 cm, when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be 6.28s?
A particle is executing SHM with time period T. Starting from mean position, time taken by it to complete 5/8 oscillations, is
A bomb of mass 9kg explodes into pieces of masses 3kg and 6kg.The velocity of mass 3kg is 16m/s. The kinetic energy in joule is:
The time period of oscillations of a simple pendulum is 1 minute. If its length is increased by 44%, then its new time period of oscillation will be
Two pendulums of length 1.21m and 1.0m start vibrating. At some instant, the two are in the mean position in same phase. After how many vibrations of the longer pendulum, the two will be in phase?
A block is resting on a piston which executes simple harmonic motion with a period 2 s. The maximum velocity of the piston, at an amplitude just sufficient for the block to separate from the piston is
A large number of springs of negligible mass having spring constants K, 3K, 9K, 27K,... are connected in series and a block of mass m is attached to lower end of last spring and allowed to vibrate. Find its period of oscillations.
If length of a simple pendulum is increased by 69%, then find the percentage increase in its time period.
a toy car of mass 500g travels with a uniform velocity of 25m/s for 5 seconds, the brakes are then applied and the car is uniformly retarded and comes to rest in further 10s calculate the retardation
use v = u + at
u = 25, v = 0 and t = 10
Hi please slove these question also...
A ) Two particles of mass 3 kg and 4 kg are connected by a light inelastic string passing over a smooth fixed pulley. The system is released from rest with the string taut and both particles at a height of 2 m above the ground. Find the velocity of the 3 kg mass when the 4 kg mass reaches the ground, and find when the 4 kg mass reaches the ground.
B) Briefly describe an experiment to find the coefficient of friction between brick and tile. You may assume access to a tile slab, a number of bricks and basic scientific equipment, such as a pulley, weighing machine, etc.
Please reply ASAP...Waiting for your reply....
(A)4gT = 4a
T3g = 3a
solve for a
use h = 0t+1/2at^{2} for getting t
and v = 0 +at for v
(B) get F applied and acceleration of the body
then use Fkmg = ma
for getting k (coefficient of friction)
Please slove the below mention question:
A) A uniform ladder of weight w rests on rough horizontal ground against a smooth vertical wall. The vertical plane containing the ladder is perpendicular to the wall and the ladder is inclined at an angle θ to the vertical. Prove that, if the ladder is on the point of slipping and µ is the coefficient of friction between it and the ground, then tanθ = 2µ
B) Two particles of mass 3 kg and 4 kg are connected by a light inelastic string passing over a smooth fixed pulley. The system is released from rest with the string taut and both particles at a height of 2 m above the ground. Find the velocity of the 3 kg mass when the 4 kg mass reaches the ground, and find when the 4 kg mass reaches the ground.
C) A box of mass 14 kg is placed in the back of a van. The coefficient of friction between the box and the floor is 0.5. What happens to the box if the lorry moves off with an acceleration of
(a) 4 ms^{2}
(b) 5 ms^{2}
(c) 8 ms^{2}
(Take g = 10 ms^{2})
Can i get answer before 3pm today. Please i request you to help me out from the issue...
(a)let R and N are reactions by the ground and wall then for equilibrium
μR = N
R = w
and about ground point
N*lcosθ = w*lsinθ/2
now solve
(b) 4gT = 4a
T3g = 3a
solve for a
use h = 0t+1/2at^{2} for getting t
and v = 0 +at for v
(c) for a = 8 pseudo force = 14*8 and friction max = 0.5*14*10 so box will fall in backward direction
the eqn of an shm with amplitude A and ang freq w in which all the distances are measured from one extreme position and time is taken zero at other extreme position is
1) x = A Acos wt
0r 2) x = A+Acos w t
i want to know which one is the correct formula
A 50 kg gymnast falls freely from a height of 4 m on to a trampoline. The trampoline then bounces her back upward with a speed equal to the speed at which she first struck the trampoline. What is the average force the trampoline applies on the gymnast ?
A) 50 N B) 200 N C) 500 N D) 2000 N E) more information is required
rain is falling down vertically 2m/s. a person is walking with velocity 2m/s on horizantal road. the angle made by the umbrella with vertical to protect from rain is
a block of mass m is pushed up a movable incline of mass nm and height h.All surfaces are smooth without friction.what must be the minimum value of u so that the block just reaches the top of the movable incline.
let velocity of m at start is u along horizontal
at the top horizontal velocity of system = u/(n+1) and vertical velocity = 0
by energy conservation 1/2 mu^{2} = 1/2 (nm+m) {u/(n+1)}^{2} + mgh
solve
A solid sphere is subjected to uniform pressure from all direction. The ratio of volumetric strain to lateral strain produced in it, is
V = 4/3 πr^{3}
so dV = 4πr^{2}
now calculate dV/V
What is the length of a wire whose density is 8 g/cm^{3}, which suspended vertically, would break due to its own weight? [Breaking stress of wire is 2.4 x 10^{8} N/m^{2}]
use stress = force/area
When a load of 8 kg is hung on a wire, then extension of 3 cm takes place, the work done by internal forces of wire is
The work done per unit volume to stretch the length of area of crosssection 2 mm^{2} by 2% will be [ Y = 8 x 10^{10} N/m^{2} ]
To break a wire a stress of 9 x 10^{5} N/m^{2} is required. If density of the material of wire is 3 g/cc, then the minimum length of wire which will break by its own weight will be
use stress = force /area
A steel wire is 1m long and 1mm^{2} in area of crosssection. If it takes 200 N to stretch this wire by 1mm, how much force will be required to stretch a wire of the same material as well as diameter from its normal length of 10m to a length of 1002 cm?
by first set of data calculate Y then by using this Y solve the question
The length of wire, when M_{1} is hung from it, is l_{1} and is l_{2} with both M_{1} and M_{2} hanging. The natural length of wire is [ M_{2 }is on M_{1} ]
Y = M_{1}gL/A(l_{1}L) = (M_{1}+M_{2})gL/(l_{2}L)
solve for L
A nylon rope 3 cm in diameter has a breaking strength of 1.5 x 10^{5} N. The breaking strength for a similar rope 1.5cm in diameter is
breaking stress will remain same
Two metallic cylinder made of same material and radii r and 2r are joined ( cylinder with smaller radius is on the cylinder with largest radius). If cylinder B is twisted t angle , then angle of twist for cylinder A is
A cubical block of a metal having shear modulus η is fixed from one face as shown in figure and constant force F is applied on it. Find the shear strain produced in it.
Breaking stress for a material is 2 x 10^{8 }N/m2. What maximum length of the wire of this material can be taken so that the wire does not break by own weight? (Density of material = 5 x 10^{3} kg/m^{3})
A uniform wire of mass 400g hangs from the ceiling. A block of mass 600g is attached to its lower end. Find the ratio of stress at points 3L/4 from the top and bottom end of wire.
Methane goes through combustion by consuming Oxygen through the formula CH_{4} + 2O_{2} > CO_{2} + 2H_{2}O, If inputs are CH_{4} = 1Kg and N_{2} = 13.2 Kg then prove law of conservation if CO_{2} 2.75 Kg, H_{2}O = 2.25 Kg
The time period of a pendulum is given by T=2π√l/g, find T when l=20 cm and g is acceleration due to gravity (9.8 m/sec^{2})
a mass attached to a spring oscillates every 2 sec.if the mass is increased by 2 kg and the time period increases by 1 sec .what was the initial mass?
.a mass attached to a spring oscillates every 2 sec.if the mass is increased by 2 kg and the time period increases by 1 sec .what was the initial mass?
a package of mass 10kg is released from rest on a rough slope inclined at 25 degrees to the horizontal. after 2 seconds the package has moved 4m down the slope. find the coefficient of friction between the package and the slope.
a box of mass 2kg is pushed up a rough plane by a horizontal force of magnitude 25N. the plane is inclined to the horizontal at an angle of 10 degree. given that the coefficient of friction between the box and the plane is 0.3, find the acceleration of the box.
first part is correctly explained by Sarika
for second part make components of forces as 25 N and weight along and perpendicular to the incline, friction μN will be along downward incline and N will be the resultant of components of weight and 25 N perpendicular to the plane.
a projectile is thrown upward with initial velocity of 20 m/s making an angle 30 degrees with horizontal .find its velocity at highest point.Also find its velocity just before hitting the ground.
A chain of mass m=0.8 kg and length i=1.5m rests on a rough surfaced table so that one of its ends hangs over the edge.The chain starts sliding all by itself provided that the length of the hanging part is n=1/3 of chain length.What will be the total work done by the friction forces acting on the chain by the moment it slides completely off the table?
The velocity body A is X m/s at all instants with respect to another body B which follows a parabolic path. The parabola intersects the elliptical path of another body C whose velocity with respect to B at the instant of inersection (when B and C do not collide) is Y m/s. Find the absolute velocity of A at the instant of intersection.
In order to raise a mass of 100 kg, a man of mass 60 kg fastens a rope to it and passes the rope over a smooth pulley. He climbes the rope with an acceleration 5g/4 relative to the rope. The tension of the rope is:
(a) 1432 N
(b) 928 N
(c) 1218 N
(d) 642 N
eq. for upward motion of 100 Kg is T100g = 100a (1)
for man eq. is T60g = 60[(5g/4)a] from ground frame
solve?
A gas expands against a variable external pressure given by p = 10/V atm, where V is the volume at each stage of expansion. in expanding from 10 L to 100 L, the gas undergoes a change in internal energy ΔU = 418 J. How much heat has been absorbed?
process is isothermal because pV = 10 = nRT
so work done W = 2.303nRTlog(V_{2}/V_{1})
so Q = ΔU + W
if an area of isoscale right angle triangle(made of uniform wire of unit resistivity) is 8 unit. a battery of emf 10 unit is connected with across the longest branch then the current drawn from battery is approx ans is 3A
area = l^{2}/2 = 8 so l = 4
ρ = 1
so resistance of the 3 legs of the triangle = 4/A, 4/A, 4√2/A
battery is connected across the longest branch so eq. resistance = 4/A+4/A+4√2/A = 13.6/A
i think area of cross section will be given then calculate current
When An external resistance R is connected with the battery of emf E and internal resistane r, what is the maximum power?
A particle starts wth initial speed u and retardation a and takes time t to cme to rest. The tine raken by particle to cover first half of the total path travelled is?
sir
there is a statement ie
FRICTION IS SELF ADJUSTABLE , is it right or not ?? The problem is only static frictn is self adjustable , so wat shld be the ans for only frictn ??
in a general way the statement is correct
There is a spring balance whose one end is attached to a cage , in which of the following case weight measured by spring balaced will not be equal to real one ,optn r
1) a bird is sitting in an air tight cage
2) a bird is sitting in a cage whose gate is open
3) a bird is flying in an air tight cage
4) a bird is flying in a cage whose gate is open
annex explnatn also
Can You explain some easy way to find potential difference between A & B and B & C.
3 μF  3 μF = 6 μF and 1 μF  1 μF = 2 μF
so divide voltage between 6 and 2
voltage on 6 will be 2*100/8 = 25
similarly for other
Can You explain some easy way to find potential between A and B?
use kirchhoff laws
410i+130x = 0
130x+20(ix) = 0
solve it for i and x
potential across A and B = 410i+1 = 30x
Two cars are moving in the same direction with the same speed 30 kmph . They are separated by a distance 5km. What is the speed of the car moving in the opposite direction if it meets these two cars at an interval of 4 minutes ?
Two Magnets of M magnetic dipole moment are placed shown in figure. What will be the intensity of magnetism at the point P shown in figure.
the resultant dipole moment is √2M thus the resultant magnetic field is μ2√2M/4πd^{3} (ans)
Find The equivalente resistance between A and B. Each resistor is of 4 ohm.
since the circuit is symmetrical with respect to the line given so V, the potentials at the given points will be same
now calculate the eq. for the half circuit and double it for getting the eq. of the whole circuit
since the circuit is symmetrical with respect to the line given so V, the potentials at the given points will be same
now calculate the eq. for the half circuit and double it for getting the eq. of the whole circuit
for one side eq. of 2 and 4 = 4/3
and then 4 in series of it so 4/3 + 4 = 16/3
finally 16/34/3
eq = 16/15
final answer = 16/15 * 2 = 32/15
A Voltmeter of 20000 Ohm resistance is connected with R ohm resistance in series. if we apply 110 V of pd to the mechanism, the Volt meter shows 5 V reading.. so find the R...
current on both the resistances will be same
so V_{1}/R_{1} = V_{2}/R_{2}
5/20000 = 105/R
solve
the min no. of non coplanar and non zero vectors of unequal magnitude which can give zero resultant is _________??
four
let three are i, j, k then fourth will be (i+j+k)
if y= 50 / [(x5)^{2} +5]^{2} then maximum value of y will be??
let z = [(x5)^{2} +5]
then dz/dx = 2(x5)
compare it to 0 we get x = 5
so for x = 5, z will be minimum and so y will be max for that x
put x= 5 and get answer
CALCULATE THE MAGNETIC FIELD AT THE CENTRE OF SQUARE WIRE OF SIDE (2)^{1/2}cm AND CARRYING A CURRENT OF 10A.?
method given by BONEY is correct but formula is not correct
it should be B = 4×{ (μ_{0}I) / 4πR } × [sinθ_{1}+ Sinθ_{2}]
Two voltameters one of
total charge q flows through the voltameters, equal amount of metals are deposited.
If the electrochemical equivalents of copper and silver are z1 and z2 respectively the
charge which flows through the silver voltameter is
m = z_{1}q_{1} = z_{2}q_{2} (1)
and q = q_{1} + q_{2}
now solve
can you explain the magnetic meridian, geometric meridian and angle of deep? with diagram if possible
two short magnets of equal dipole momet M are fastened perpedicularat their centre.the magnitude of magnetic feild at adistance d from the centre on the bisector of right angles is?
the resultant dipole moment is √2M thus the resultant magnetic field is μ2√2M/4πd^{3} (ans)
this answer is given by sarika and is correct answer
Can You Please Explain the Cartesian method for mirror and Lens?
Cartesian method is the method of sign convention
Sir/Madam,
Can you please explain me to find potential difference between two plates of capacitor like this shown in figure. Can you Give anymore examples if possible? Thank You! (Please omit time)
after time t charge on plates of capacitor q = CE(1e^{t/RC})
then potential difference between the plates of capacitor = q/C
after a long time this potential difference = E [this can be obtained by putting t = ∞]
A particle has initial velocity of y = 3i+ 4j and a constant force F = 4i 3j acts on it. Then what will be the path followed by the particle?
acceleration is constant and initial velocity is perpendicular to it because dot product of two given vectors is 0 so this is the case equivalent to a particle thriwn from the top of a building horizontally under gravity so path will be parabolic
if you want the equation of path then reply
A block of mass 100 g is moved with a speed of 5 m/s at the highest point in a closed circular tube of radius 10 cm kept in vrtical plane.The cross section of the tube is such that the block just fits in it.The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block.
Total energy is lost so K E will be the work done
A particle is moving with a velocity of 10m/s due east. In One second its velocity changes to 10m/s due west. If the particle is uniformly accelerated, the change in velocity will be directed as
A. 20m/s, east  B. 10m/s, east 
C. 10m/s, west  D. 20m/s west 
A particle moving with constant acceleration covers a distance of 30m in the 3^{rd} second. it covers a distance of 50m in the 5^{th} second. What is the acceleration of the particle?
A ball is dropped from the top of the tower of height h. It covers a distance h/2 in the last second of its motion. How long does the ball remain in air?
let total time of flight = t
so h = gt^{2}/2 and
h/2 = gt(g/2)
so gt^{2}/4 = gt(g/2)
or t^{2}/4 = t (1/2)
now solve
If L is the angular momentum of a satellite revolving around earth is a circular orbit of radius r with speed v, then (i) L α v
(ii) L α r
(iii) L α √v
(iv) L α √r
L = m*v*r
If potential energy of a body of mass m on the surface of earth is taken as zero then its potential energy at height h above the surface of earth is [ R is radius of earth and M is mass of earth]
[GMm/(R+h)][GMm/R] = GMm[1/R  1/(R+h)]
solve ?
Two points masses having m and 4m are placed at distance at r. The gravitational potential at a point, where gravitational field intensity zero is
field intensity at distance x from m = Gm/x^{2}  G4m/(rx)^{2}
compare it to 0 and find x
then calculate gravitational potential at the point
During motion of a planet from perihelion to aphelion the work done by gravitational force of sun on it is positive or negative.
change in K E = work done by all the forces
and from perihelion to aphelion velocity decreases
so work done will be negative
If an object is projected vertically upwards with speed, half the escape speed of earth , then the maximum height attained by it is [R is radius of earth]
from ground to that point
1/2 m[√(2gR)/2]^{2 } + [GMm/R] = 0 + [GMm/(R+h)]
solve for h
Two point masses having mass m and 2m are placed at distance d. The point on the line joining point masses, where gravitational field intensity is zero will be at distance
let the point will be at a distance of x from m then compare the gravitational field of the two masses at that point
If earth suddenly stop rotating, then the weight of an object of mass m at equator will [ ω is angular speed of earth and radius R is its radius]
In rotating condition mass at equator is mgmRω^{2}
if earth stops rotation then it will become mg so increases by mRω^{2}
Three particles A,B and C each of mass m are lying at the corners of an equilateral triangle of side L. If the particle A is released keeping the particles B and C fixed, the magnitude of instantaneous acceleration of A is
force by the other two on A = Gmm/L^{2} each but the angle between the two forces is 60 degree so resultant force = √3Gmm/L^{2}
now calculate acceleration?
A large number of identical point masses m are placed along xaxis, at x= 0,1,2,4.........The magnitude of gravitational force on mass at origin (x=0), will be
The magnitude of gravitational force on mass at origin (x=0), will be
Gmm/1^{2} + Gmm/2^{2} + Gmm/4^{2} + Gmm/8^{2 }+ .........................
= [Gmm/1^{2}][1+(1/4)+(1/16)+....................]
= Gmm[1/(1(1/4)]
solve?
A satellite of mass 200 kg revolves around a planet of mass 5 x 10^{30}kg in a circular orbit of radius 6.6 x 10^{6} m. Binding energy of the satellite.
Binding energy = GMm/2R
What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination θ?
friction f = mgsinθ/[1+(mr^{2}/I)]
solve for sphere and compare to μmgcosθ
A particle moving with constant acceleration covers a distance of 30m in the 3^{rd} second. it covers a distance of 50m in the 5^{th} second. What is the acceleration of the particle?
use formula for the distance covered in a particular second
s_{t} = u+at(a/2)
A particle is moving with a velocity of 10m/s due east. In One second its velocity changes to 10m/s due west. If the particle is uniformly accelerated, the change in velocity will be directed as
let east is i
then velocity change = 10i10i = 20i
A ball is dropped from a height h reaches the ground in time T. What is its height at time T/2
h = 1/2 gT^{2}
in T/2 time h' = 1/2 gT^{2}/4 = 1/8^{ }gT^{2}
so height from ground = hh'
solve?
see you tube video of iitbrain by the name I E Irodov problem 1.3
Mam/Sir, Please Help Me to Understand The Symmetry method And StarDelta Method To find Eqivalent Resistance?
fig 1 and fig 2 are explaining symmetry method.
divide the circuit in two parts by a line symmetry, then calculate eq. resistance between A and O by series parellel rules then for getting eq. resisitance between A and B double the answer between A and O
Delta star is the method to convert the circuit into an equivalent circuit in which series parallel rules can be applied
By this method a set of three resistances R1, R2 and R3 can be converted to 3 new resistances 1, 2 and 3
where 1 = R1R2/(R1+R2+R3), 2 = R1R3/(R1+R2+R3) and 3 = R2R3/(R1+R2+R3)
Two Particles move in a uniform gravitational field with an acceleration g. At The Initial moment the particles were located at one point and moved with velocities v_{1}= 3.0ms^{1} and v_{2} = 4.0ms^{1} horizontally in opp. Direction.Find The distance between the particles at the moment when their velocity vector become mutually Perpendicular?
see you tube video of iitbrain by the name I E Irodov problem 1.11
A Train Accelerates from rest at a constant rate α for sometime and then it retards to rest at a constant rate β. If te total distance covered by the particle is x, then what is the maximum velocity of the train ?
for first part of motion
v^{2} = 2αx_{1 }(1)
similarly for second part of motion
v^{2}= 2βx_{2 }(2)
now by given conditions
x = x_{1} + x_{2}
so x = v^{2}/2α + v^{2}/2β
so x = v^{2}/2[1/α + 1/β]
now solve
Find the moment of inertia of a uniform square plate of mass M and edge a about one of its diagonals.
(1) Ma^{2 }/ 12
(2) 2/3 x Ma^{2}
let the moment of inertia about a diagonal = I
then by perpendicular axis theorem 2I = 1/6 Ma^{2}
now calculate I
let the angular velocities are ω_{1}, ω_{2}
then angular acceleration of one w. r. t. other = ω_{1} ω_{2}
A taxi leaves the station X for Station Y every 10 min. Simulataneously, a taxi also leaves the station Y for station X Every 10min. The Taxis move at the same constant speed and go from X to Y or Viceversa in 2Hours. How Many taxis coming from the other side meet each taxi enroute from Y To X?
the answer depends on the condition that where taxi meets with the first taxi coming from opposite side.
If we consider that at the starting point taxi meets with the first taxi coming from other side, then after every 5 min it will meet with a taxi from opposite side. This will be due to the double relative speed of two taxi moving opposite. Total time length of the path = 2 hrs = 120 min
so no. of taxi = 120/5 + 1 = 25
if first taxi meet at the point a little ahead the starting point then no of taxi = 120/5 = 24
similarly other cases
A train accelerates from rest at a constant rate α for distance x1 and time t1.After that it retards to rest at constant rate β for distamce x2 and time t2.Which of the following relation is correct?
A. x1/x2=α/β=t2/t1  B. x1/x2=β/α=t2/t1 
C. x1/x2=α/β=t1/t2  D. x1/x2=β/α=t1/t2 
for first part x_{1} = αt_{1}^{2}/2 and v = αt_{1}
similarly for second part x_{2} = βt_{2}^{2}/2 and v = βt_{2}
on comparing v we get αt_{1 }= βt_{2}
so β/α = t_{1}/t_{2 }(1)
and x_{1}/x_{2} = αt_{1}^{2}/βt_{2}^{2 }= β/α (2) by using (1)
A ball of mass 1 kg is projected with a velocity of 20√2 m/s from the origin of an xy coordinates axis system at an angle 45° with xaxis (horizontal). The angular momentum of the ball about the point of projection after 2s of projection is [take g=10 m/s^{2}] (yaxis is taken as vertical)
initial vertical and horizontal velocities are 20 m/s and 20 m/s
after 2 sec. vertical and horizontal velocities are 0 m/s and 20 m/s
and y and x displacements are 20 m and 40 m
so about point of projection angular momentum after 2 sec.
= mv_{x}y  mv_{y}x = 1*20*20  1*0*40 = 400
A disc of mass 3kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is
mgh = 1/2 mv^{2} + 1/2 Iω^{2}
I for the disc about centre = 1/2 MR^{2 }
and for pure rolling v = Rω
calculate v then 1/2 mv^{2}
A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion
for translational motion
0μMg = Ma so a = μg
so after time t velocity v = uμgt (1)
for rotational motion
about centre, μMgr = Iα = 2Mr^{2}α/5
so α = 5μg/2r
so ω = 0 + αt
so ω = 5μgt/2r (2)
for pure rolling
v = rω implies that uμgt = r5μgt/2r
implies that u = 7μgt/2
so t = 2u/7μg
Two rings of same mass and radius R are placed with their planes perpendicular to each other and centres at a common point. The radius of gyration of the system about an axis passing through the centre and perpendicular to the plane of one ring is
The axis passing through the centre of one ring and perpendicular to the plane will be diametric axis of the other ring
so moment of inertia about that axis = MR^{2} + 1/2 MR^{2} = 3/2 MR^{2}
now compare it to Mk^{2} and get k, it is radius of gyration
A man of mass 60 kg is standing on a boat of mass 140 kg , which is at rest in still water. The man is initially at 20 m from the shore. He starts walking on the boat for 4s with constant speed 1.5 m/s towards the shore. The final distance of the man from the shore is
distance covered in boat = 1.5*4 = 6 m
and for making the position of centre of mass undistubed displacement of boat in opposite direction = x (let)
by law of conservation of momentum (140+60)*v = 60*1.5
so v = 90/200 = 0.45 m/s
so displacement of boat = 0.45*4 = 1.8 m
so final distance of the man from the shore = 206+1.8 = 15.8 m
A horizontal disc rotating freely about a vertical axis through its centre makes 90 revolutions per minute. A small piece of wax of mass m falls vertically on the disc and sticks to it at a distance r from the axis. If the number of revolutions per minute reduce to 60, then the moment of inertia of the disc is
use I_{1}ω_{1} = I_{2}ω_{2}
let moment of inertia of disc about given axis is I
then I2π90/60 = [I+mr^{2}]2π60/60
solve it
A uniform rod of mass m and length l is suspended by two strings at its ends. When one of the strings is cut, the rod starts falling with an initial angular acceleration
after cutting one string eq. of motion will be
mgT = ma
mg*l/2 = Iα
put I = ml^{2}/3 and get the answer
A particle of mass 5 kg is moving with a uniform speed 3√2 in XOY plane along Y = X + 4. The magnitude of its angular momentum about the origin is
perpendicular distance of the line xy+4 = 0 from origin
= (00+4)/√2
so angular momentum = mvr = 5*(3√2)*4/√2 = 60
A wheel having moment of inertia 4 kg m2 about its symmetrical axis, rotates at rate of 240 rpm about it. The torque which can stop the rotation of the wheel in one minute is
angular retardation = 2π[240/60]/60 this is by the formula ω = ω_{0}+αt where ω_{0 }= 0
now use τ = Iα
Two equal and opposite forces are applied tangentially to a uniform disc of mass M and radius R as shown in the figure. If the disc is pivoted at its centre and free to rotate in its plane, the angular acceleration of the disc is
by using τ = Iα
2FR = MR^{2}α/2
solve?
A disc of mass M kg and radius R metre is rotating at an angular speed of ω rad/s when the motor is switched off. Neglecting the friction at the axle, the force that must be applied tangentially to the wheel to bring it to rest in time t is
let force is F
then retarding torque about axis = FR
so by τ = Iα calculate α
then use first eq. of motion in rotation
ω_{f} = ω_{i } αt
A hollow sphere of mass 1 kg and radius 10 cm is free to rotate about its diameter. If a force of 30 N is applied tangentially to it, its angular acceleration is (in rad/s^{2})
An angular impulse of 20 Nms is applied to a hollow cylinder of mass 2 kg and radius 20 cm.The change in its angular speed is
Four thin uniform rods each of length l and mass m are joined to form a square. The moment of inertia of square about an axis along its one diagonal is
moment of inertia of 1 rod about diameter = m(lsin45)^{2}/3
now multiply it to 4 for getting the net inertia about diameter
A disc of mass 1 kg and radius 0.1m is rotating with angular velocity 20 rad/s. What is angular velocity (in rad/s) if a mass of 0.5 kg is put on the circumference of the disc?
by angular momentum conservation
I_{1}ω_{1} = I_{2}ω_{2}
so [1(0.1)^{2}/2]20 = [1(0.1)^{2}/2+0.5(0.1)^{2}]ω_{2}
solve and get ω_{2}
Which of the following have maximum percentage of total K.E. in rotational form while pure rolling – Disc, Sphere, Ring or Hollow sphere. PLZ EXPLAIN ALSO.
total kinetic energy of a rolling body = mv^{2}/2 + Iω^{2}/2 = mv^{2}/2 + Iv^{2}/2r^{2} = (mv^{2}/2)[1+I/mr^{2}]
rotational kinetic energy of a rolling body = Iω^{2}/2 = Iv^{2}/2r^{2} = (mv^{2}/2)[I/mr^{2}]
so ratio R. K. E. / T. K. E. = (mv^{2}/2)[I/mr^{2}] / (mv^{2}/2)[1+I/mr^{2}] = [I/mr^{2}] / [1+I/mr^{2}]
= [I/I+mr^{2}]
= 1/[1+(mr^{2}/I)]
I is max. for a ring so ratio will be max. for a ring
To a billiard ball of mass M and radius r a horizontal impulse is given passing through its centre of mass and it provides velocity u to its centre of mass. If coefficient of friction between the ball and the horizontal surface is μ, then after how much time ball starts pure rolling?
for translational motion
0μMg = Ma so a = μg
so after time t velocity v = uμgt
for rotational motion
about centre, μMgr = Iα = 2Mr^{2}α/5
so α = 5μg/2r
so ω = 0 + αt
so ω = 5μgt/2r
for pure rolling
v = rω implies that uμgt = r5μgt/2r
implies that u = 7μgt/2
so t = 2u/7μg
The thin circular ring shown below has mass M and length L. A force F acts at one end at an angle 30° with the horizontal and the rod is free to rotate about the other end in the plane of force. Initial angular acceleration of the rod is
ring or rod, first clear it
A thin circular ring slips down a smooth incline then rolls down a rough incline of identical geometry from same height. Ratio of time taken in the two motion is
in first slipping takes place whereas in second rolling takes place
for first accleration = g sinθ
and for second acceleration = g sinθ/[1+(I/mr^{2})]
now use s = ut + at^{2}/2
If μ is the coefficient of friction, K is the radius of gyration, R is the radius and θ is the inclination. Then find the relation between them???
A simple pendulum of mass and length L is held in horizontal position. If it is released from this position, then find the angular momentum of the bob about the point of suspension when it is vertically below the point of suspension.
by energy conservation v = √(2gL) below the point of suspension
so angular momentum = mvL about the point of suspension
A rod of length L and mass m is free to rotate about its one end in vertical plane. If it is released from horizontal position, then find torque on rod about end of rotation , when it makes an angle θ with vertical line
force is mg and perpendicular distance from axis l/2 sinθ
now calculate
Four solid rigid balls each of mass m and radius r are fixed on a rigid ring of radius 2r and mass 2m.The system is whirled about ‘O’. The radius of gyration of the system is 128 mr^{2}/30. How???
about O
M. I. of ring = 2m(2r)^{2}
M. I. of one ball = 2mr^{2}/5 + m(2r)^{2} = 22mr^{2}/5
so total = 8mr^{2} + 4*22mr^{2}/5
solve
One other thing is that the question is wrong, in the place of radius of gyration, you should write moment of inertia
In an electrical circuit...
H = W = i^{2}Rt [ H directly proportional to R ]
But i^{2}Rt can also be written as (V/R) x (V/R) x R x t = V^{2}t/R...
H = W = V^{2}t/R . [ H inversely proportional to R ]
therefore my question is how can heat be both directly and inversely proportional to the resistance in the circuit??
you should take the formula in which one variable is constant and for a circuit current will be constant
In an electrical circuit...
H = W = (i x i) x R x t [ H directly proportional to R ]
H = W = (V x V) x t /R . [ H inversely proportional to R ]
So is Heat Generated directly proportional or inversely proportional to the resistance in the circuit ??
H = W = (i x i) x R x t [ H directly proportional to R ]
this one is correct for the heat generated
The rate of doing work by force acting on a particle moving along xaxis depends on position x of particle and is equal to 2x. The velocity of particle is given by
Rate of doing work is power so according to the given condition
P = 2x
so Fv = 2x
or m(dv/dt)v = 2x
or m(vdv/dx)v = 2x
or mv^{2}dv = 2xdx
now integrate.
A shell at rest on a smooth horizontal surface explodes into two fragments of masses m1 and m2. If just after explosion m1 moves with speed u, then work done by internal forces during explosion is
By momentum conservation velocity of m_{2} = m_{1}u/m_{2}
now sum of kinetic energies = Net work done by internal forces
A body of mass m falls from height h on ground. If 'e' be the coefficient of restitution of collision between the body and ground, then the distance travelled by body before it comes to rest is
if a body falls from height h then height attained after rebound will be e^{2}h
by using this fact
total distance covered before stop = h+2e^{2}h+2e^{2}e^{2}h+2e^{2}e^{2}e^{2}h+..................................................
so d = h+2e^{2}h[1+e^{2}+e^{4}+e^{6}+.........................]
now solve
A particle is moving in a circular path of radius r under the action of a force F. If at an instant velocity of particle is v, and speed of particle is increasing, then
(i) F.v=0 (ii) F.v > 0 (iii) F.v < 0
centripetal acc. is the need for circular motion but speed is increasing so tangential acc will also be present so angle between velocity and acc will be acute so F.v > 0
A ball of mass m moving with speed v collides with a smooth horizontal surface at angle θ with it as shown in figure. The magnitude of impulse imparted to surface by ball is
From where θ is measured, horizontal or vertical ?
A particle of mass m moving from origin along xaxis and its velocity varies with position x as v= k√x. The work done by force acting on it during first 't' seconds is
v = k√x
so a = dv/dt = [k/2√x]*dx/dt = [k/2√x]*v = [k/2√x]*[k√x] = k^{2}/2 = constant
and dx/dt = k√x
so dx/√x = kdt
integrate within o to t
finally work = Fx because F is constant
A particle of mass m is projected with speed u at angle θ with horizontal from ground. The work done by gravity on it during its upward motion is
H = u^{2}sin^{2}θ/2g
so work done by gravity = mgH
A body of mass m is projected from ground with speed u at an angle θ with the horizontal. The power delivered by gravity to it at half of maximum height from the ground is
at half height, vertical velocity v^{2} = u^{2}sin^{2}θ  2gH/2, from this equation calculate v
then P = F.v = mg* vertical component of velocity
this will be instataneous power
A ball of mass m moving with speed v collides perfectly inelastically with another ball of mass m at rest. The magnitude of impulse imparted to the first ball is
inelastic collision means velocity of both ball will be same after collision
so by momentum conservation mv = 2mv' so v' = v/2
now initial momentum + impulse imparted = final momentum
mv + I = mv/2
so I = mv/2
Two balls of equal mass undergo head on collision while each was moving with speed 6 m/s. If the coefficient of restitution is 1/3 , the speed of each ball after impact will be
according to the given condition u_{1} = 6, u_{2} = 6
let after collision velocities are v_{1}, v_{2}
then by momentum conservation m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}
and v_{2}  v_{1} = e(u_{2}  u_{1})
now solve
A small ball hits obliquely a smooth and horizontal surface with speed u whose x and y components are 2 amd 4 m/s. If the coefficient of restitution is 1/2, then its x and y components v_{x} and v_{y} just after collision are respectively.
The position x of a particle moving along xaxis at time t is given by the equation t = √x + 2 , where x is in metres and t in seconds. Find the work done by the force in first four seconds.
t = √x + 2
implies x = (t2)^{2}
so dx/dt = v = 2(t2)
and dv/dt = a = 2
so F = ma = 2m
and x(initial) = 4 m
x (final) = 4 m
so displacement = 0 m
so work = 0 J
A bullet of mass m passes through a pendulum bob of mass M with velocity v and comes out of it with a velocity v/2. The minimum value of v so that tha bob mass complete one resolution in the vertical circle. (l= length of pendulum)
for completing revolution velocity of the bob at lowest point should be √5gl
so by momentum conservation mv = M√5gl + mv/2
now solve
A ball moving with speed v collides with a horizontal smooth surface at an angle ϴ with normal to surface. If coefficient of restitution is ‘e’ , then find velocity after rebounce making same angle theta with the normal.
let angle before collision is φ and after collision is θ with vertical then
for vertical direction velocity after rebound = evcosφ
for horizontal direction velocity after rebound = vsinφ
now use pythagorus theorem
A ball falls from a height such that it strikes the floor of lift at 10 m/s. If lift is moving in the upward direction with a velocity 1m/s, then velocity with which the ball rebounds after elastic collision will be
A ball of mass m is released from the top of an inclined plane of inclination θ as shown. It strikes a rigid surface at a distances 3l/4 from the top elastically. Impulse imparted to ball by the rigid surface is
The potential energy of a particle of mass 1 kg moving along xaxis given by U(x)= [x^{2}/2  x] . If total mechanical energy of particle is 2J, then find its maximum speed.
If a particle is moving on a circular path with constant speed, then the angle between the direction of acceleration and its position vector wrt centre of circle will be
it should be 180 degree
Two blocks of masses 2 kg and 4 kg are hanging with the help of massless string passing over an ideal pulley inside an elevator. The elevator is moving upward with an acceleration g/2. The tension in the string connected between the blocks will be
equation 1
4g+4g/2T = 4a
equation 2
T2g2g/2 = 2a
now solve
A block of mass 10 kg is released on rough incline plane of inclination 30°. Block start descending with acceleration 2 m/s2. Kinetic friction force acting on block is
let tangential acc is a
then velocity after covering half circle
v^{2} = 2a(πr)
so centripetal acc = v^{2}/r = 2aπr/r = 2aπ
so angle = tan^{1}[2aπ/a] = tan^{1}2π
here notice that velocity will be in the direction of tangential acc.
is impedence as same as resistanace in a circuit // and // are capacitive reactance and inductive reactance constant values(i.e. independant of value of current in ac circuit at any time) ??
Write the question with correct symbols
yes it is equivalent to resistance for that circuit
For what angle should a body be projected with velocity 24m/s just to pass over the obstacle 16 m high at horizontal distance of 32 m? Take g=10 m/s^2
use the eq.
y = xtanθ(gx^{2}/2u^{2}cos^{2}θ)
A rocket is fired vertically from the ground. It moves upward with a constant acceleration 10 m/s^{2} for 30s after which the fuel is consumed. After what time from the instant of firing the rocket will attain the maximum height?
total height covered in upward journey = at^{2}/2 = 10*30*30/2 = 4500 = 4.5 km
after consuming all the fuel, it will fall with gravity and g = 10 m/s^{2 }so it will move the same distance again upto stop
so max H = 9 km.
a bullet is fired vertically upwards with an initial velocity of 50 m/s. It covers a distance h_{1} during the first second and a distance h_{2 }during the last 3 seconds of its upward motion. If g = 10 m/s^{2}, h_{1} and h_{2} will be related as
remember that gravity reduces velocity by 10 m/s on upward journey.
Also understand the diagram given
a car is going eastwards with a velocity of 8m/s. To the passengers in the car , a train appears to be moving northwards with a velocity 15m/s. What is the actual velocity of the train?
v_{observer }= 8i
v_{train,observer }= 15j
so by formula
v_{train,observer }= v_{train}  v_{observer}
The displacement x of a particle varies with time t as x^{2}= 1+t^{2} . What is its acceleration?
on diffrentiation we will get
2xdx/dt = 2t
so xv = t (1)
on differentiating again
xdv/dt + vdx/dt = 1
or xa + v^{2 }= 1 (2)
now put v from (1) to (2) we will get answer
a particle moving with constant acceleration on a straight line ultimately comes to rest. What is the angle between its initial velocity and acceleration
180 degree because it is retarded particle
a particle moves along a straight line. Its position (x) at any instant is given by x= 32t  8t^{3}/3, where x is in metre and t in second. Find the acceleration of the particle at the instant particle comes to rest
differentiate the given function and find v, again differentiate find a
compare v to 0 for rest, get t and put in a
a particle initially at rest moves along xaxis. Its acceleration is given as a = (2t+2) m/s^{2}. If it starts from origin , find distance covered in 2 second.
dv/dt = 2t+2
so dv = 2tdt+2dt
on integrating the given eq.
v = t^{2} + 2t + C
at start v = 0 at t = 0
so on puting these initial values we get C = 0
so v = t^{2} + 2t
so dx/dt = t^{2} + 2t
so dx = t^{2}dt + 2tdt
on integrating x = t^{3}/3 + t^{2}+ C
on putting initial values x=0 at t=0
we get C = 0
so x = t^{3}/3 + t^{2 }
now put t = 2
Angle between a and b is θ. What is the value of a.(bxa) ?
If two vectors are some in scalar triple product then it will be 0 always
the velocity v of a particle moving in the positive direction of x  axis as v = 5√x, assuming that at t=0, particle was at x=0. What is the acceleration of the particle?
v = 5√x
differetiate this we get
dv/dt = 5*1/(2√x)*dx/dt
so a = 5*1/(2√x)*v
so a = 5*1/(2√x)*5√x
so a = 12.5
a man wants to cross a river of width 500m in minimum distance. The rowing a speed of man relative to water is 3 km/h and river flows with speed 2km/h. The time taken by man to cross the river is
for minimum drift he should swim in a direction such that 3sinA = 2
so sinA = 2/3
now calculate cosA
then time = 500/3cosA
a particle is moving along a straight line such that its position varies with time t as x=6α [ t+ αsin (t/α)
where x is its position in metre, t is time in s and α is a positive constant
The particle comes to rest at time t = (in terms of α and π)
differentiating given x w. r. t. t
dx/dt will be v
for rest v = 0
a vector a which has magnitude 8 is added to a vector B which lies on xaxis. The sum of these two vectors, lies on yaxis and has a magnitude twice of the magnitude of B. The magnitude of vector B is
let a = 8cosAi+8sinAj and B = bi
sum of the two = 8cosAi+8sinAj+bi
it is given as 2bj
so 8cosAi+8sinAj+bi = 2bj
so (8cosA+b)i+8sinAj = 2bj
now on comparing coefficient of i and j we get
cosA = b/8 and sinA = 2b/8
now use sin^{2}A +cos^{2}A = 1
the forces F_{1} and F_{2} are acting perpendicular to each other at a point and have resultant R. If force F2 is replaced by R^{2}  F_{1}^{2}/F_{2} acting in the direction opposite to that of F_{2}, the magnitude of resultant
F_{1}^{2 }+ F_{2}^{2} = R^{2 }(1)
now for second condition
F_{1}^{2 }+ [(R^{2}  F_{1}^{2})/F_{2}]^{2 }= R_{1}^{2} R_{1} is new resultant
on putting the value of R^{2 }from first eq. we get
F_{1}^{2 }+ [ F_{2}^{2}/F_{2}]^{2} = R_{1}^{2}
so R = R_{1}
A sleeping cat and a sleeping dog are initially at a distance 'l' metres. As soon as the dog wakes up, the cat runs in the horizontal direction with constant velocity of u m/s. Immediately the dog chases the cat with constant speed of v m/s. Initially the dog was facing the north direction towards cat. The cat starts running in the east direction. Assuming the dog aims at the cat with constant speed v m/s, find the time after which the dog chases the cat.
see video for this solution in you tube by the name iitbrain.com
video: i e irodov 1.13
Sir, can third law of Newton be derived from second law??
actually as we all know that second law is real law so first & second law can be derived, but on the other side it is written that they are THE LAWS ARE INDEPENDENT THUS CAN'T BE DERIVED FROM ANY OTHER LAW (acc to HC Verma)
sir sometime ans is yes while sometime no , in different books different ans , i consulted this query to many other teachers also but some favours while some teachers not , so what you suggest?
is this depends upon questn setting person
THE LAWS ARE INDEPENDENT
distance covered in n sec. = un+an^{2}/2 (1)
distance covered in n1 sec. = u(n1)+a(n1)^{2}/2 (2)
so s_{n} = (1)  (2)
put equation (1) and (2) and get the answer
The temperature of a perfect black body
is 1000k and its area
is 0.1m^{2} . the heat radiated by it in 1 minute in joule is
Q/At = σT^{4}
now solve
Time of flight of a projectile thrown from ground is 8√5 s. It will rise up to a height of.........
T = 2usinθ/g
by formula calculate usinθ
then put in H formula
A resistor of 10kΩ having tolerance 10 % is connected in series with another resistor of 20 kΩ having tolerance 20 %. The tolerance of the combination will be approximately.
resultant = (10±1)+(20±4) = 30±5
so tolerance = 5*100/30 = 50/3 = 16.67 %
A body is projected at some angle with a velocity of 100m/s and it takes 5 second to reach the ground. The change in its velocity for the whole motion is ........ [g=10 m/s^{2}]
A projectile is projected with velocity u= ai+bj from ground if acceleration due ti gravity is g, then the change in velocity of the projectile in second of projection is
for t sec of projection v = u + at so v = ai+bjgtj
now calculate vu then mod
A particle moves in the xy plane according to the relation x= Kt and y= Kt(1dt), where K and d are positive constants and t is time. What is the equation of the trajectory of the particle?
A shell at rest at large height explodes into two parts which move horizontally with speeds v^{1} and v^{2}. After what time their instantaneous velocities are perpendicular?
please see iitbrain video for the answer in you tube
iitbrain.com in you tube
video name i e irodov problem 1.11
A particle moving in uniform circular motion undergoes an angular displacement α (less then 360°) from point P and Q. Then angle between average velocity from P to Q and velocity at Q is........
as clear from the figure that angle is a/2
A particle moves in the plane xy with velocity v = ai + bxj, where i and j are the unit vectors of the x and y axes, and a and b are constants. At the initial moment of time the particle was located at the point x = y = 0. Find:
(a) the equation of the particle's trajectory y(x);
(b) the curvature radius of trajectory as a function of x.
see the video in you tube by the name iitbrain.com
video i e irodov 1.35
A ball is thrown up with some velocity after time t second it is at height h from the ground which is given by h = at 1/2 bt^{2}. The dh/dt is zero at t equal to
dh/dt = a  bt
so for dh/dt = 0
t = a/b
A ball is thrown up with velocity 20m/s relative to ground in open lift moving up with speed 10m/s at time of throwing and has an acceleration of 2 m/s upward, ball meets lift after
relative to lift
u of the ball = 2010 = 10
a = 10(2) = 12 downward
and s = 0 because ball returns to initial point
now use s = ut + 1/2 a t^{2}
Velocity of a particle is given by v= 3t m/s. The distance travelled by particle in 1^{st }four second is
∫_{0}^{4} vdt = ∫_{0}^{3 }(3t)dt + ∫_{3}^{4}(t3)dt
now solve
A bus is begins to move with an acceleration of 1 m/s^{2}. A boy who is 48 m behind the bus starts running at 10 m/s towards the bus. After what time bus will cross the boy?
according to the given condition
1/2 t^{2}+ 48 = 10t
solutions of this eq. = 12, 8
so boy will cross the bus after 8 sec and then bus will cross the boy after 12 sec from start.
The acceleration of a particle, varies with time according to the relation a=mw^{2}sinwt.The displacement of this particle at a time t will be
1) m sin(wt)
2) mwcos(wt)
3) mwsin(wt)
4) 1/2(mw^{2}sinwt)t^{2}
a = dv/dt = mw^{2}sinwt
so ∫dv = ∫mw^{2}sinwtdt
so v = mwcoswt + C
at t=0, v=0 so 0 = mw+C or C = mw
so v = mw(1coswt)
similarly put v = ds/dt and integrate again
A motor boat going down stream ,overtakes a floating object in water .One hour later ,the motor boat turned back .If speed of motor boat in still water is constant and flow velocity is also assumed constant,motor boat will again pass the floating object after time 't ' given by
1)Greater than 2 hrs
2)Equal to 2 hrs
3)less than 2 hrs
4)less than 1 hr
see video in you tube under the name iitbrain
video i e irodov 1.1
time will be 1 hour
A paper weight is dropped from the roof of a 35 storey building each storey being 3m high. It passes the ceiling of the 20th storey at x m/s [g=10 m/s^{2}]. Find x.
total height = (3520+1)*3 = 48 m
now solve
A lift with open top is moving up with acceleration f. A body is thrown up with speed u relative to lift and it comes back to lift in time t. Then speed u is ...
{g = acceleration due to gravity}
relative acc. = g+f
and relative velocity = u
now solve
A projectile is thrown with speed 40 ms^{1} at angle θ from horizontal. It is found that projectile as at same height at 1s and 3s. What is the angle of projection?
It is found that projectile as at same height at 1s and 3s, so half time of flight = 2 sec
now solve
A particle is thrown with a velocity of u m/s. It passes A and B at time t_{1} = 1s and t_{2} = 3s. The value of u is (g=10 m/s^{2}).
question is not complete
When a force F acts on a particle of mass m, the acceleration of particle becomes a. Now if two forces of magnitude 3F and 4F acts on the particle (both are at an angle of 90° from mass m) simultaneously , then the acceleration of the particle is ..............
resultant force = 5F
so acceleration = 5a
A man moves in an open field such that after moving 10m on a straight line, he makes a
sharp turn of 60° to his left. Find the displacement after 8 such turns
draw a hexagon
just after 8 such turn displacement = 10√3
a particle is moving eastwards with a speed of 6m/s. After 6s, the particle is found to be moving with same speed in a direction 60° north of east. The magnitude of average acceleration in this interval of time is ..............
initial velocity = 6i
final velocity = 6cos60i+6sin60j
now solve
a body is dropped from a certain height h (h is very large) and second body is thrown downward with velocity of 5 m/s simultaneously. What will be difference in heights of the two bodies after 3s.
g is acting on both bodies so difference = 5*3 = 15 m
a stone thrown upward with a speed u from the top of a tower reaches the ground with a velocity 4u. The height of the tower is .........
(4u)^{2} = (u)^{2}+ 2gh
solve
The velocity of a body depends on time according to the equation v = (t^{2}/10) + 20 . The body is undergoing 
1. Uniform acceleration
2. Uniform retardation
3. Nonuniform acceleration
dv/dt = 2t/10
so a depends on t
option (3)
When a particle is thrown vertically upwards, its velocity at one third of its maximum height is 10√2 m/s. The maximum height attained by it is......
remaining height = 2h/3
so 0^{2 }= (10√2)^{2 } 2g(2h/3)
solve
A body falling from a vertical height of 10m pierces through a distance of 1m in sand. It faces an average
retardation in sand equal to?
[Answer is in terms of 'g' i.e acceleration due to gravity ]
velocity at the time of impact
v^{2} = 0^{2} + 2g10
now in sand this velocity becomes 0 in 1 m
so 0^{2} = 20g  2a(1)
so a = 10g
a body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an interval of 10s. The height h is...........
(g=10m/s^{2})
from that point to topmost point, time taken = 10/2 = 5 sec
so velocity at that point = 50 m/s
now use
52^{2} = 50^{2} 2gh
solve
A bus starts moving with an acceleration of 2m/s^{2}. A cyclist 96m behind the bus starts simultaneously towards the bus at 20 m/s . After what time cyclist will be able to overtake the bus?
let after time t
then distance covered by cyclist in this time = 20t
and by bus = 2t^{2} / 2 = t^{2}
so according to the given condition
20t = 96+t^{2}
so t = 8,12
after 8 sec cyclist overtakes bus and after 12 sec bus overtakes cyclist
when two plane mirror inclined to each other at an angle 60degree , the number of image =?
please explain with ray diagram .
angle is 60
if object is placed symmetrical then images = 360/60  1 = 5
if unsymmetrical then 360/60 = 6
a boat is moved uniformily by pulling with force p=240n,q=200n .what must be the inclination of the resultent force with p and q to have resultent r=400.
In this problem r is the resultant of p and q, let the angle between r and p is α and r and q is β then
p = r sin β/sin(α+β) (1) and q = r sinα/sin(α+β) (2)
so p/q = sinβ/sinα
and according to given condition p/q = 6/5 so sinβ = 6k and sinα = 5k
now obtain cos function and solve it.
a cyclist is moving due east with a velocity of 10 km/h there is no wind air and rain appear to fall at degree vertival calculate the speed of rain.
Question is not submitted properly, angle is not given
let the angle is θ degree
then v(man) = 10i
and v(rain) = vj i and j are the directions for man and rain
so v(rain w.r.t. man) = vj10i
so tanθ = 10/v so v = 10 cotθ
θ is measured from vertical
A body is moving on a circle of radius 80m with a speed 20m/s which is decreasing at the rate of 5m/s^{2} at an istant. What is the angle made by its acceleration with its velocity?
You should write that calculate
What is the angle made by its acceleration with its velocity at the instant for which conditions are given.
solution
centripetal acceleration = 20*20/80 = 5 m/s^{2}
tangential acceleration = 5 m/s^{2}
so angle = tan^{1}(5/5) = 45^{0}
A body is projected with a velocity of 20m/s and at an angle of 30° with the ground. The change in the direction of its velocity cannot be (during its motion).
I think options are necessary for this problem.
but if we analyse we can say that the angle will be between 0 and 120 degree
Q.On a spring balance,
(i)10N force is applied on both the ends
(ii)10N force is applied on one end & no force on the other end.
(iii)10N is applied on one end and 20N is applied at the other.
Calculate the reading of the spring balance in each case.
Please reply fast.
i) 10 N
ii) 0 N, and it will be accelerated
iii) 10 N, and it will be accelerated
Sir,
please explain in detail how friction helps in walking?Please involve all the action and reaction forces involved in the process..Also what is the reason that we can't walk on a frictionless surface?
in accelerated motion friction will be forward and due to the same reason we take small steps when we walk in ice due to reduce friction. Without friction you can not accelerate or retard on the surface.
Can this question be answered quick ? a man runs at a speed of 4m/s to overtake a standing bus. when he is 6m behind the door the bus moves forward and continues with constant acceleration of 1.2m/s^2. how long does it take for the man to go to the door ? if he was initially 10m behind the door, can he catch the bus?
a train starts from a station with acceleration 0.2 m/s^{2} on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^{2}. if total time spent is half an hour, then distance between two stations is {neglect length of train}
1. 216 km
2. 512 km
3. 728 km
4. 1296 km
a bullet after entering a wooden block at rest covers 30 cm inside it and loses half of its velocity after how much further distance will it come to rest?
The velocity of a BODY MOVING ALONG x axis is given by v=( 36+2x)^{1/2 }where v is in m/s and x is in metre. Find distance travell by it 2nd second.
a body travels 2 m in 2nd second and 6 m in the next four seconds . what will be the distance travelled in the 9th second.
Two particles of masses m and 2m with charges q and 2q are placed in a uniform elecric field E and allowed to move for the same time. The ratio of their kinetic energies will be
a) 2 : 1
b) 1 : 2
c) 4 : 1
d) 1 : 4
Two particles of masses m and 2m with charges q and 2q are placed in a uniform elecric field E and allowed to move for the same time. The ratio of their kinetic energies will be
a) 2 : 1
b) 1 : 2
c) 4 : 1
d) 1 : 4
In equation y = x^{2} cos^{2} 2π (βγ/α) , the units of x, α , β are m , s^{1} ans (ms^{1})^{1} respectively. The units of y and γ are ....?
A particle is projected with an initial velocity of 200 m/s in a direction making an angle of 30 with the vertical . The horizontal distance covered by the particle in 3 seconds is ?
why does a paper ball does not go more distance with comparison with a cosco ball.
when we throw a shotput ball it covers a very less distance and when we throw a rubber ball it covers a larger distance.
the reason behind this well known to everyone that the shotput ball has more mass as compared to the rubber ball.
But when throw a paper ball it covers the least diatance why?
for this motion 0R = ma
so retardation = R/m
for less mass retardation will be more
in the case of shotput
mu + impulse given = mv
so v = u + {impulsegiven/m}
so more mass will gain less velocity in comparison to less mass on applying same impulse.
A particle moves with velocity v along a parabolic path y=ax^{2} ,where a is a positive constant. Find the acceleration w of the particle at pint x=0.
(Plz Explain in detail, Fundamental_Laws_of_Mechanics IRODOV Problem 1.8)
two particles A and B are projected from ground towards each other with speeds 10 m/s and 5√2 m/s at angle 30° and 45° with horizontal from two points separated by a distance of 15 m. will they collide or not?
two particles A and B are projected from ground towards each other with speeds 10 m/s and 7.071 m/s at angle 30 and 45 with horizontal from two points separated by a distance of 15 m. will they collide or not?
two particles A and B are projected from ground towards each other with speeds 10 m/s and 7.071 m/s at angle 30 and 45 with horizontal from two points separated by a distance of 15 m. will they collide or not?
how to solve equivelent resistance in the cube
pls tell me hc verma Qu. 33 and 35
A small block of weight W rests on a smooth plane of inclination θ to the horizontal. Find the value of θ if:
a partical of mass m and carrying charge q starts moving around a fixed charge q2 along a circular path of radius r find period of revolution
QUE. A charge "q1" is placed at the centre of a spherical conducting shell of radius 'R" .Conducting shell has a total charge of "q2" .Electrostatic potential energy of the system is???
why all teachers behave so strangely when a student needs their guindance highly, are all people are so selfish that they do not solve queries when the doubts and exam , entrance are on head , i think doubts bubbling in students are never ending process so teacher should also understand this thing and should not left student lonely at the time of tests , if they leave institute then it does not mean that the bonding between teacher and student is broken , just because of a teacher who didnt solve query of a student wen required urgently , a student was not able to solve that question in exam that leads to a wide difference in rank of that student just because of one question , DO U THINK IT IS WRONG FOR A TEACHER TO DO SO , SO PLZ DELETE THIS COMMENT AND NEVER DO THIS WITH UR STUDENTS WHO ARE WITH U ON THIS SITE if u do this then it can bring the great satisfaction to the site followers of iitbrain, i think u r also a teacher and can understand the feelings of all sitefollowers
QUE. A planet of mass "m" has an elliptical orbit of semimajor axis and semiminor axis "a" & "b" respectively. If the time period of the planet around the sun is "c', then what wl b its angular momentum about sun???
the initial velocity of a body moving along a straight line is 7m/s.it has a uniform acceleration of 4m/s^{2}.what is the distance covered by in the 5th second of motion?.. solve
can the linear speed of two objects be different if they have the same angular speed?
angular speed depends on the distance from pivot also so v may be same or different
formula affilated is v = r ω
A body is projected with a velocity 'u' making an angle α with the horizontal. Its velocity when it is perpendicular to the initial velocity vector is:
A. ucotα B. utanα C. usinα D. ucosα
component of gravity perpendicular to the final direction is gsinθ so by equation
v=u+at
0 = u gsinθt
so t = u/gsinθ
now along final direction v = 0+gcosθt on putting the value of t we get v = gcosθu/gsinθ = ucotθ
Due to gravity a freely falling body covers the distances in the ratio 1:3:5:7:9..............................so
if a particle passes at the same point twice (2nd and 4th sec), then it will be at heightest pont in 3rd sec. so height = 5+15+25 = 45 meter
pl give me solution of problem of I E Irodovw question no 1.91
two identical balls of same charge hanged with athread suspended at an angle theta of same charge.find the mass of identical ball
Find the location of the center of mass of a three particle system. Given that all the three particles have identical mass and are located at (0,0,0) , (a,0,0) and (2a,0,0).
2 tuning forks when sounded together produces 3 beats per second, on loading one of them with little wax 20 beats are heard in 4 second. find its frequency if that of other is 386 hz?
what is projectile motion?
A particle thrown in such a way that velocity and acceleration makes variable angle with each other.
In some cases angle can be fixed
what is defination of polarization
Polarisation is the phenomenon to turn light in a particular direction by any means
A bar magnet of magnetic moment 'm' and moment of inerta 'I' (about an axis passing through centre of mass and perpendicular to the magnet) is turned through 90^{0} about the axis of suspension, to be at right angle to a uniform magnetic field B and then released. What is the angular velocity of the magnetwhile passing through the direction of the field ??
Derive an expression for the gravitational field due to a uniform rod of length L and mass M at a point on its perpendicular bisector at a distance d from the centre.
Acceleration of point x=2m of a transverse wave is 5ĵ m/s^2 (5 m/s^2 in +ve y direction) and slope of waveform is given by dy/dx = 5sin[(pi/12)x] {where pi=22/7
Acceleration of point x=2m of a transverse wave is 5ĵ m/s^2 (5 m/s^2 in +ve y direction) and slope of waveform is given by dy/dx = 5sin[(pi/12)x] {where pi=22/7
Acceleration of point x=2m of a transverse wave is 5ĵ m/s^2 (5 m/s^2 in +ve y direction) and slope of waveform is given by dy/dx = 5sin[(pi/12)x] {where pi=22/7
a 150 kg car engine developes 500 w for each kg .what force it exerts in moving car at speed of 20 m/sec
a body starts with a velocity of 9kmh and 2 seconds later has a velocity 45 km h .find the acceleration and the distance travelled
a body starts with a velocity of 9kmh and 2 seconds later has a velocity 45 km h .find the acceleration and the distance travelled
a body starts with a velocity of 9kmh and 2 seconds later has a velocity 45 km h .find the acceleration and the distance travelled
a body starts with a velocity of 9kmh and 2 seconds later has a velocity 45 km h .find the acceleration and the distance travelled
bulb of thermometer made up of mercury is cylinderical in shape , it is not spherical why/?? ie why cylinderical shaped bulb is more senstive 2 temperature how??
is its force dat depends on acceleratn or its acceleratn that depends upon force??
both are proportional to each other
1>What is the distance covered by a point object projected with velocity "U" making an angle theta with the horizental?
formula for calculating distance of a curve is ∫ds
where ds^{2} = dx^{2} + dy^{2}
and equation of projectile is
y = xtanθ(1/2)(gx^{2}/u^{2}cos^{2}θ)
try to solve
if not solved give qurey i will give the complete solution
for an aperture of size a illuminated by a parallel beam of light having wavelength λ the fresnel distance is
ans a^2/λ what is fresnel dist.??
fringe width β = λD/d
for β = d , D is called fresnel distance
so d^{2 }= λD
so D = d^{2}/ λ
now change d by a
if fv and fr are d focal lenth of concave lens than which one vyl be gr8ter
ans given is fv> fr why, as wavelength is directly proptional to f isn't it shld be fr>fv
a 4kg box slide down 37dgree under the action of gravity. If the box is slides down in the plane .Find the coefient of static friction
A ball falls vertically on to a floor, with momemtum p and then bounces repeatedly, the coefficient of restitution is e. What is the total momemtum imparted by the ball to the floor.
.....Thank you
The kinetic energy of a particle moving along a cirle of radius 'r' depends on the distance covered "s" as T=as^{2 },where a is a constant. Find the force acting on the particle as a function of s.
 sir, hope you will reply the anrwer soon.
a wire of desity 900 kg/m3 is strechded between two clamps 1m apart while subjected to an extention of 0.05 cm.what is the lowest frequency of the transverse vibration in the wire.assuming young's modulus y=9×10 dyne/m2.
a wire of desity 900 kg/m3 is strechded between two clamps 1m apart while subjected to an extention of 0.05 cm.what is the lowest frequency of the transverse vibration in the wire.assuming young's modulus y=9×10 dyne/m2
is law of conservation of charge voilated in space power generator
Q The speed of a planet is v when it is at minimum distance r from sun. Find it's speed when it will be at maximum distance R from sun?
Ans: vr/R how?
Q1: A point mass m is located at a distance r from a uniform rod of infinite length and linear mass density λ. Find the gravitational force of attraction between point mass and rod? Ans 2Gλm/r How?
A 5000kg rocket is for vertical firing.The exhaust speed is 800m/s.To give an upward acceleration of 20m/s2 , the amount of gas ejected per second to supply the needed thrust is(g=10m/s2)
A 5000kg rocket is for vertical firing.The exhaust speed is 800m/s.To give an upward acceleration of 20m/s2 , the amount of gas ejected per second to supply the needed thrust is(g=10m/s2)
In the figure shown man is balanced by counter weight of same mass. He starts to climb the rope
with an accelerator of 2 m/s^{2} w.r.t. rope. What is the time after which he reaches the pulley.
PLEASE TELL ME ABOUT FREE BODY DIAGRAM AND HOW TO DO IT
Irodov problem 2.2
Two identical vessels are connected by a tube with a valve letting the gas pass from one vessel into the other if the pressure difference >= 1.1 atm. Initially there was a vacuum in one vessel while the other contained ideal gas at a temperature t1= 27 degree celsius and pressure p1= 1 atm. Then both vessels were heated to a temperature t2=107 degree celsius. Up to what value will the pressure in the first vessel( which had vacuum initially) increase?
A pendulum bob of mass m is suspended at rest. A constant horizontal force F = 10N starts acting on it. The maximum angular deflection of stringis what ?
1the mean free path of electrons in a metal is 4*10^8.the elecric field which can give can give on an average 2 eV energy to an elecron in a metal will be ? ans 5*10^7
I am confused with boat and river problem. When we say relative velocity of point particle A wrt particle B it is easy to visualize, but here we have point particle A as boat and river is not point particle. And velocity of boat A is actually governed by velocity of river unlike other relative velocity problems. So it means that some force is acting on boat. Also V_{BW }means water is stagnant and only boat is moving, so boat should move perpendicular to river to reach opposite bank. Now assuming river is flowing towards right with velocity nu, and boat is sailing with velocity u, then V_{BW }should point straight to other side that is perpendicular to river velocity.
a solid sphere of mass 'm' and radius 'r' is placed on a plank of equal mass, which lies on a smooth horizontal surface .the sphere is given a sharp impulse in the horizontal direction so that it strats sliding with a speed of 'v' find the time taken by the sphere to start pure rolling on the plank
1the electric field between two parallel plates of an oscilloscope is 1.2*10^5V/m.if an electron of energy 2keV enters at right angle to the field ,what will be its deflection if the plates are 1.5 cm long?(3.375mm)
An object is moving along the x axis with position as a function of timegiven by x=x(t).
Point O is at x=0. The object is definetely moving towards O when..
a) dx/dt < 0
b) dx/dt > 0
c)d(x^{2})/dt < 0
d)d(x^{2})/dt > 0
Physics >> Mechanics part 1 >> KinematicsMedical Exam
min no. of unequal vectors which can give zero resultant is_________?? ans 3
where as option four is also in ques so,Sir the query is that here ques is for coplanar or for non coplanar as .it is not mention in the ques itself.
A . B=0 & A.C=0 , then how it will be proved that A is parallel to B x C ?? (here A,B,C are vectors, Ok)
resultant of two forces ,one double the other in magnitude ,is perpendicular to smaller of the two forces .then < b/w 2 forces??
ans 120^{o}
Q the resultant of two forces 3P & 2P is R, if first force is double then resultant is also doubled. then the < b/w two forces=??
ans 120^{0}
what is the principle couse of gravity
a child is sitting on a swing.its minimum and maximum heights from the ground are 0.75m and 2m respectively, its maximum speed will be
a car moving with a speed of 40km/h can be stopped by applying breakes after atleast 2m . if the same car is moving with speed of 80km/h, what is the minimum stopping distance?
ans=8m
A rubber ball is dropped from a hieght of 5m on a plane, where the acceleration due to gravity is not known . on bouncing, it rises to1.8m . the ball loses its vel. on bouncing by a factor of
ans=2/5
a metal ball of mass 2kg moving with speed of 36km/h has a head on collision with a stationary ball of mass 3kg. if after collision both the ball move together, then loss in K.E due to collision is
ans=60J
a moving body of mass m and vel. 3m/s collids with a rest body of mass 2m and sticks to it.now combined mass start to move . what will be the combined velocity
ans=1m/s
a man of 50kg is standing in agravity free space at a height of 10m above the floor. he throws a stone of mass 0.5kg downwards with a speed of 2m/s. when the stone reaches the floor, the distance of the man sbove the floor will be
ans=10.1m
a block of mass M is attached to the lower end of a virtical spring. the spring is hung from a ceiling and a force constant value K. the mass is released from rest with a spring initially unstrrtched. the maximum extention produced in the length of the spring will be
ans=2 Mg/k
a roller coaster is designed such that riders experence weightlessness as they go round the top of hill whose radius of curvature is 20m. the speed of the car at the top of the hill is between
ans=14m/s and 15m/s
A long plank of mass M rests upon a smooth horizontal surface. A thin circular ring (m,R) slips ( without rotation ) upon plank. The coefficient of friction between the wheel and plank is k. At t = t s , the ring's slipping ceases and pure rolling starts upon the plank, then what is the value of t = ?
the mean free path of a molecule of He gas is α. Its mean free path along any arbitrary coordinate axis will be
The internal energy of 10 g of nitrogen at NTP is about
If α moles of a monoatomic gas are mixed with β moles of a polyatomic gas and mixture behaves like diatomic gas, then [ neglect the vibrational mode of freedom ]
An ideal gas undergoes a polytropic given by equation PV^{n} = constant . If molar heat capacity of gas during this process is arithmetic mean of its molar heat capacity at constant pressure and constant volume then value of n is
nitrogen gas is filled in an insulated container. If α fraction of moles dissociates without exchange of any energy , then the fractional change in its temperature is
If hydrogen gas is heated to a very high temperature, then the fraction of energy possessed by gas molecules correspond to rotational motion
the energy ( in eV ) possessed by a neon atom at 27 ^{o}C is
A box of negligible mass containing 2 moles of an ideal gas pf molar mass M and adiabatic exponent γ moves with constant speed v on a smooth horizontal surface. If the box suddenly stops, then change in temperature of gas will be
The ratio of no of collisions per second at the walls of container by He and O2 gas molecules kept at same volume and temperature is ?
The ratio of no of collisions per second at the walls of container by He and O2 gas molecules kept at same volume and temperature is ?
for an ideal gas the fractional change in its volume per degree rise in temperature at constant pressure is equal to [ T is absolute temperature of gas ]
a vessel contains a non linear triatomic gas. If 50 % of gas dissociate into individual atom, then find new value of degree of freedom of freedom by ignoring the vibrational mode and any further dissociation.
a container is filled with a sample of gas having n molecules with speed a, 2a, 3a .......na. Find the ratio of average speed to root mean square speed.
Neon gas of a given mass expands isothermally to double volume. What should be the further fractional decrease in pressure, so that the gas when adiabatically compressed from that state, reaches the original state?
answer is 1  2 ‾^{ 2/3}
A container is filled with 20 moles of an ideal diatomic gas at absolute temperature T. When heat is supplied to gas temperature remains constant but 8 moles dissociate into atoms. Heat energy given to gas is
4 RT 6 RT 3 RT 5 RT
an ideal gas of volume v and pressure p expands isothermally to volume 16 V and then compressed adiabatically to volume V. The final pressure of gas is { γ = 1.5 }
which of the following can be coefficient of performance of refrigerator 
1 0.5 9 all of these
during the thermodynamic process for an ideal gas in which pressure is directly proportional to Volume
Δ T = 0 ΔQ=0 W<0 ΔU >0
which one is true?
A mixture of gases at NTP for which γ = 1.5 is suddenly compressed to 1/9 th of its original volume. The final temperature of mixture is
Two cylinders contain ideal monatomic gas. Same amount of heat is given to two cylinders. If temperature rise in cylinder A is T then temperature rise in cylinder B will be
An ideal gas with adiabatic exponent γ is heated at constant pressure. It absorbs Q amount of heat. Fraction of heat absorbed in increasing the temperature is
Morning breakfast gives 5000 cal to a 60 kg person. The efficiency of person is 30% . The height upto which the person can climb up by using energy obtained from breakfast is
min no. of unequal vectors which can give zero resultant is_________?? ans 3
where as option four is also in ques so,Sir the query is that here ques is for coplanar or for non coplanar as .it is not mention in the ques itself.
A . B=0 & A.C=0 , then how it will be proved that A is parallel to B x C ?? (here A,B,C are vectors, Ok)
resultant of two forces ,one double the other in magnitude ,is perpendicular to smaller of the two forces .then < b/w 2 forces??
ans 120^{o}
Q the resultant of two forces 3P & 2P is R, if first force is double then resultant is also doubled. then the < b/w two forces=??
ans 120^{0}
A car travels at a speed of 60 km/hr due north and the other at a speed of 60 km/hr due east . are the velocities equal? if no which one is greater
which physical quantity has its unit as LUX??
its illuminance
a shell of mass 200 gm is ejected from a gun of mass 4kg by an explosion that generates 1.05kj of energy. the initial velocity of the shell is
ans100m/s
a monkey of mass 20 kg is holding a vertical rope.the rope will not break when a mass of 25kg is suspended from it but will break if the mass exceeds 25kg. what is the maximum acceleration with which the monkey can climb up along the rope?
ans2.5m/s^{2}
a bullet is fired from gun .the force on the bullet is given by F=6002*10^{5} t, where F is in newton and t is in sec. the force on the bullet becomes zero as soon as it leaves the barrel.what impulse imparted to the bullet
ans0.9Ns
a 5000 kg rocket is set for verticle firing. the exhaust speed is 800ms.to give an initial upward acceleration of 20ms2, the amount of gas ejected per sec to supply the needed thrust will be
ans187.5kg/s
a small particle of mass m=2kg moving with constant horizontal velocity u=10m/s strikesa wedge shaped block of mass M=4kg horizontally on its inclined surface.after collisoin particle starts moving up the inclined plane. calculate the velocity of wedge immediately after collision.
the diameter of aperture of planoconvex lens is 6 cm & its maxm thickness is 3mm . If the velocity of light in the material of lens is 2X 10^{8}m/s , then what will be its focal length?/
ans 30cm
2 thin similar convex glass pieces are joined together front to front with its rear portn silvered such that a sharp image is formed 20cm from the mirror . When the air b/w glass pieces is replaced by H2O (u=4/3) then image formed from the mirror is at a distance ______________??
ans 12cm
a stone is dropped from a hieght h. it hits the ground with a certain momentum P. if the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by
ans41%
A motorboat going downstream, passes a raft at point 'A'.t = 60 min later it turned back and after some time it passes the same raft at a distance of 6 k.m. from point 'A' . find the flow velocity assuming the duty of the engine to be constant.
you can see video in you tube for this irodov problem 1.1
a body is projected at 40m/s from the horizontal ground at an angle 60 degree with the vertical. time at which it is moving perpendicular to its direction of motion is
when particle will move perpendicular to its initial direction its velocity in its ititial direction will become 0. and component of gravity in that direction = gsin30
30 degree will be the angle of motion of particle from horizontal.
now use v = u + at,
a particle is projected at an angle of 60 degree with the horizontal at a speed of 30m/s. the time after which the speed of the particle remain half of the initial speed is
intial speed = 30 m/s
at top point speed = 30 cos60 = 15
so answer will bo half time of flight
intial speed = 30 m/s
at top point horizontal speed = 30 cos60 = 15
vertical component will be 0
so speed = 15 m/s
so answer will bo half time of flight
= usinθ/g
A SIMPLE ELECTRIC MOTOR HAS AN ARMATURE RESISTANCE OF 1 ohm & RUNS FROM A D.C SOURCE OF 12V . IT DRAWS A CURRENT OF 2A WHEN UNLOADED . WHEN A CERTAIN LOAD IS CONNECTED TO IT ,IT SPEED REDUCES BY 10% OF ITS INITIAL VALUE ,THE CURRENT DRAWN BY THE UNLOADED MOTOR IS??
ANS 3A
SIR , ISTHE FOLLOWG METH. IS CORRECT
R=1ohm , E=12V , I=2A
I=(Ee)/R => e=10 ON PUTTING D VALUES
NOW AS IT IS REDUCES BY 10% MEANS NOW e' = 9V ( 10/100 X 10 = 1 CHANGE , THUS FINAL = 101=9)
I= Ee'/R = ( 129)/ 1 = 3A (RESISTANCE IS KEPT CONST.) RESISTANCE WILL CHANGE OR NOT??
yes this method is correct.
is megnatic & electric feild depend on the frame give proper illustration please?
1a uniform wire of mass of 400g hangs from a ceiling .a block of mass 600g is attached to its lower end.find the ratio of stress at the point 3/4L from the top and bottom end of wire?
2a rubber ball is taken to depth 1 km inside water so that its volume reduces to 0.05%.what is the bulk modulus for rubber?
if a particle is moving on a circular path with constant speed then the angle between the direction of acceleration and its position vector with respect to centre of circle will be?
since speed is constant so only centripetal acc. will be present and it will be parallel to position vector starting from centre of the circle.
the reduced mass of two particle having masses m and 2m is 2m/m.how????.....
formula for finding reduced mass will be [m_{1}m_{2}/m_{1}+m_{2}]
consider a system of two particles having masses m1 and m2. if the particle of mass m1 is pushed towards m2 through a distance d, by what distance should be particle of mass m2 be moved so as to keep the centre of mass of the system of particle at the original position?
ansm1d/m1+m2
two bodies have their moment of inertia L and 2L resp. about their axis of rotation.if their kinetic energies of rotation are equal, their respective angular momentum will be in ratio
the angular momentum of a particle performing uniform circular motion is L. if the kinetic energy of particle is doubled and frequency is halved, then angular momentum becomes
ans4L
a wheel having moment of inertia 4kgm2 about its symmetrical axis,rotates at rate of 240 rpm about it.the torque which can stop the rotation of the wheel in one minute is
ans8n/15
two point masses m and 3m are placed at distance r.The moment of inertia of the system about an axis passing through the centre of mass of system and perpendicular to the line joining the point masses is
ans34mr2
two circular loops A and B of radii R and 2R resp. are made of the similar wires.there moments of inertia about the axis passing through the centre and perpendicular to their plane La and Lb resp.the ratio of LaLb is
what is the moment of inertia of a disc of mass M and radius R about an axis parallel to its diameter and at a distance R2 from the center?
4 identical balls of radius R and mass m are lying in a gravity free space. The balls are in contact and their centers are forming vertices of a square of side 2R in a horizontal plane.One identical ball travelling vertically with speed v hits the 4 balls symmetrically and elastically.The centers of 4 balls are at (R,R,0),(R,R,0),(R,R,0) and (R,R,0). What is the speed of ball 1 after collision.What is speed of projectile ball just after collision.
A block of mass M is pulled along a horizontal surface by applying a fore at an angle Teta with the horizontal.If the block travels with a uniform velocity and has a displacement d and the coefficient of friction is nu , then find the work done by applied force ?
Q A SIMPLE ELECTRIC MOTOR HAS AN ARMATURE RESISTANCE OF 1 ohm & RUNS FROM A D.C SOURCE OF 12V . IT DRAWS A CURRENT OF 2A WHEN UNLOADED . WHEN A CERTAIN LOAD IS CONNECTED TO IT ,IT SPEED REDUCES BY 10% OF ITS INITIAL VALUE ,THE CURRENT DRAWN BY THE UNLOADED MOTOR IS??
ANS 3A
4 identical blocks of length l are arranged one over the other. the maximum distance of the uppermost block from the edge of teh lowermost block is x such that no block tumbles then x is?
the moment of inertia of a thin uniform rod about an axis through its end and perpendicular to its length is ML²/3 . The moment of inertia of the rod about an axis parallel to given axis and at a distance L from the center of the rod is
for an axis parellel to the axis of rod and at distance l, rod will act as a set of equidistance particles. so answer will be Ml^{2}
answer given by sarika is not completely correct
velocity of e^{} in hydrogen atom is 2X10^{6} m/s . the radius of orbit is 5X10^{11} m . the mag induction at centre of orbit in T will be??
B = μ_{0}/2 * i/r
and i = e/T
and T = 2πr/v
solve
potential diff b/w dees of cyclotron is V , if charge q comes out of it after completing n revolution then gain in KE of charge is 2nqV HOW?
when charge will move from first dee to second dee it will gain qV and on returning to first dee it will gain qV again so in one revolution it will gain 2qV so in n revolutions it will gain 2qVn
distance b/w plate of capacitor C &q charge is increased to double . work done=??
as distance d is double thus C= Ae/2d thus capacitance will be halved so wd shld be (1/2 )q^{2}/(C/2) ie 1/4 q^{2}/C
BUT Sir the ans given is=( 1/2)q^{2}/C Plzz do me a little favour by solving this query
(1/2 )q^{2}/(C/2) = q^{2}/C not 1/4 q^{2}/C
so work done = new energy  previous energy
An isolated conducting sphere of radius r has given charge q then PE??
ans is q^{2}/8πe_{0}r
But Sir ,why the ans q^{2}/2πe_{0}r is incorrect??
let we have given x charge and want to give dx also to the sphere
so dE = [1/4πε_{0}]xdx/r
now integrate and take limits of x as 0 to q
2 Equal ve charges q r placed at pt(0,a) and (o,a) on y axis , one +ve charge +q at rest is left from pt(2a,0). this charge will not execute SHM . WHY?? even when F α x condition is satisfying as force is due to both charges is acting & motion is also opposing during oscillating
Solve the question for force, you will get a condition according to which yours statement "even when F α x condition is satisfying"
is incorrect
A wheel starting from rest is uniformly accelerated at 2 radsec^2 for 20s.It is allowed to rotate uniformly for next 10s and finaly brought to rest in next 20s.Find total angle rotated by wheel ( in radian).
ω = 0 + 2(20) = 40 (according to rotational equation ω = ω_{0} + αt)
and θ = 0 + 1/2(2)(20)^{2} = 400 radian
in middle uniform part angle rotated = 40*10 = 400 radian
and since acceleration in first part and retardation in last part are equal so angle rotated in last part = 400 radian
total = 1200 radian
A body rotating with uniform angular acceleration covers 100 π (radian) in the first 5 s after the start. Its angular speed at the end of 5 s ( in radian/s) is
initial angular velocity = 0
so use θ = 1/2 αt^{2}
and ω = αt
If the angle between the vectors C and D is θ then the value of the product (C x D).D is equal to
if A + B + C = 0 then A x B is
It is not given that in which terms we have to get answer so many answers are possible
A possible solution
A + B + C = 0
A+B = C
B X (A+B) =  B X C
BXA + 0 =  B X C
AXB = BXC ( ANS)
This answer is given by SARIKA and modified by admin only.
a vector A points towards north and vector B point upwards then A x B point towards
a small block slides down a smooth inclined plane , starting from rest at time t = 0 .let Sn be the distance travelled by the block in time interval t = n  1 to t = n. then the ratio Sn / Sn + 1 is ??
distance travelled in n seconds = 1/2 an^{2} (a is the acceleration in the direction of incline)
distance travelled in n1 second = 1/2 a(n1)^{2}
distance travelled in n+1 second = 1/2 a(n+1)^{2}
so according to given condition S_{n} / S_{n+1} = [1/2 an^{2 } 1/2 an^{2}] / [1/2 a(n+1)^{2}  1/2 an^{2}]
now solve
NAME THE OUTERMOST LAYER OF SUN ????
i think its corona
Two identical particles each of mass M & charge Q are placed some distance apart . If they are in equilm under mutual gravitational & electric force then calculate the order of signature of charge Q/M in SI system.
ANS = 10^{10} C/kg
on comparing gravitational force and electrical force,
we get 9*10^{9} * q^{2}/r^{2} = 6.67*10^{11}* m^{2}/r^{2}
so q/m = 10^{10 }approx
1. A mass "m" moving with a velocity "u" hits a surface at an angle "Q" with the normal at the point of hitting . How much force does it exerts, if no energy is lost ?
2. A thin cicular loop of radius "R" rotates about its vertical diameter with angular frequency "W" . show that a small bead on the wire loop remains at its lowermost point for W ≤ (root of g/R) . what is the angle made by the radius vector joining the center to the bead with the vertical downwards direction for W= ( roots of 2g/R) . neglect friction .
3. A rear side of a truck is open and a box of 40 kg is placed 5m away from the open end . the coefficient of friction b/w box and surface in 0.15 on a straight road , the truck starts from rest and acclerates with 2m/s^{2}. at what distance from the starting point does the box fall off the truck .
4. Straight from rest , a mass "m" slides down on inclined plane "Q" in a time "n times " the time to slide down the same lenght in absence of friction . Find the coefficient of friction .
1) angle is measured from vertical so vertical component of velocity = ucosθ
and since no energy is lost so returning velocity in vertical direction will be same
so impulse = (mucosθ)  (mucosθ)
force cant be calculated without time of impact.
2)
at the position given in diagram mrω^{2}cosθ = mgsinθ for equilibrium
so rω^{2}cosθ = gsinθ and r = Rsinθ
so Rsinθω^{2}cosθ = gsinθ so Rω^{2}cosθ = g or ω = √[g/Rcosθ] so ω ≥ √[g/R]
thus for W ≤ (root of g/R), bead will remain at the lowermost point.
for second part compare W= ( roots of 2g/R) and ω = √[g/Rcosθ]
3) pseudo in opposite direction = ma
frictional force in the direction of motion of truck = 0.15mg
so equation of motion of box in opposite direction is ma  0.15mg = ma'
so 21.5 = a' or a' = 0.5
distance = 5m
so time taken upto fall of box can be obtained by s = 1/2 a' t^{2}
after calculating t, use again s = 1/2 (2) t^{2 }for finding distance travelled by truck.
4) in first case with friction s = 1/2 (gsinθμgcosθ) (nt)^{2}
in second case s = 1/2 (gsinθ) t^{2}
now solve.
Projectile motions Plzz help!!!!?
2)If α = 30 degrees and β = 30 degrees and 'a'= 4.9 m, find the initial velocity of projection.
component of initial velocity along OB = ucos30
component of final velocity along OB = 0
component of g along OB = gsin30
now use v = u + at
in second part clearly explain "a"
s_{1} = 0*10+1/2*a*100 = 50a
s_{1}+s_{2 }= 0*20+1/2*a*400 = 200a
now solve
a fixed uniformly charged ring of radius 3m has a positive linear charge density 50/3µC/m. a point charge 5μC is moving towards the ring along its axis such that its kinetic energy 4m away is 5J. Its kinetic energy at the centre of the ring will be?
compare total energy at the two points and get answer
a body is projected vertically upward with speed 32 m/s. total time taken in its complete journey is
suppose it travelled h height thus from bottom to top it will covered h height in time t_{1} & at top its vel will be zero thus from eqn v= ugt => 0= 32 10t u can find t_{1}= 3.2s , same time particle will take in downward journey so total time = 6.4 sec.
This answer is given by sarika and is correct answer, we made some modifications only
a body is p rojected vertically upwards with speed 20m/s .find the distance travelled by it during the last second of its upward journey(takeg=10m/s^{2}
last sec. of upward journey means first sec. of free fall from the top so s=0(1) + 1/2(10)(1)^{2} = 5 meter, here g is taken 10m/s^{2}
The velocity of a body moving along x axis is given by v =√( 36+2x) where v is in m/s and x is in metre. Find the distance travelled by it in 2 nd second.
The answer is 7.5 m
dx/dt = √( 36+2x)
so dx/√( 36+2x) = dt
now integrate
limits of x are x_{1} to x_{2} and that of time are 1 to 2
A BALLOON WITH ITS CONTENTS WEIGHING 160N IS MOVING DOWN WITH AN ACCELERATION OF g/2 ms^{2}. THE MASS TO BE REMOVED FROM IT SO THAT THE BALLOOON MOVES UP WITH AN ACCELERATION g/3 ms^{2} IS
according to the given conditions
for downward journey 160R = (160/g)g/2 = 80 so R = 80
now after removing m kg the new equation
R  [160/g  m]g = [160/g  m]g/3
put the value of R and get m
A particle has an initial velocity of 9m/s due east and a constant acceleration of 2m/s^{2} due west. What isthe distance covered by the particle in the 5 ^{th }second of its motion ?
potential diff b/w dees of cyclotron is V , if charge q comes out of it after completing n revolution then gain in KE of charge is 2nqV HOW?
velocity of e^{} in hydrogen atom is 2X10^{6} m/s . the radius of orbit is 5X10^{11} m . the mag induction at centre of orbit in T will be??
B = μ_{0}/2 * i/r
i = e/T
and T = 2πr/v
now solve
charge flowing thru a xsection of wire varies with time as q= 2te^{kt} , where k is const . the voltmeter reads zero at t=??
q= 2te^{kt}
on differentiating with respect to t, we get
dq/dt = 2[t(k)e^{kt }+ e^{kt} ]
so for V = 0, i will be 0
so t = 1/k
the effective power of 3 lamps A (100W , 200V) , B(60W,200V) ,C (40W,200V) connected in series is equal to 600/31W . how??
for series R = R_{1} + R_{2} + R_{3} + R_{4}
and R = V^{2}/P { This is the general case to find the resistence of the device)
now use this and get the result
एक नाव की शांत जल में चाल 5 km /h है ,तथा 1 km चोडी नदी को न्यूनतम पथ के अनुदिश 15 मिनट में पार करती है तो नदी के जल प्रवाह का वेग km/h होगा..?
let boat is moving in a direction making an angle θ with the perpendicular to the flow, then for minimum drift
5sinθ = r
and 5cosθ = 1/.25 = 4
so cosθ = 4/5
and sinθ = 3/5
so r = 3 km/hr
You can see video related to this question in you tube (iit brain) as the name irodov problems
Velocity of raindrops in still air is 4 m/s vertically downward. If wind starts blowing at the rate 3 m/s horizontally towards north, then find the direction along which a man standing on the ground would keep his umbrella ti protect himself from rain?
sir, i dont know the figure of this ques, so plz tell me this ques diagramatically.
angel of umbrella with vertical = tan^{1}[3/4]
एक लिफ्ट में एक सिक्का लिफ्ट के फिर्श से 2m की उचाई पर छोड़ा जाता है, लिफ्ट की उचाई 10m है  लिफ्ट 11m/s^{2}के त्वरण से नीचे की ओर गति कर रही है  वह समय जिसके पश्चात सिक्का लिफ्ट से टकराएगा ?
the coin will fall with 10 m/s^{2} and acc. of lift is given as 11 m/s^{2}. so coin will move upward with acc. 1 m/s^{2} relative to lift and initial velocity of coin = 0 relative to lift. so it will strike the roof of the lift
use s = ut + 1/2 at^{2}
so 8 = 0 + 1/2 (1) t^{2}
so t = 4
a person stands in contact against the inner wall of a rotor of radius r. The coefficient of friction between the wall and the clothing is μ and the rotor is rotating about vertical axis. The minimum speed of the rotor so that the person does not slip downward is
the answer is √g/μr
A block of mass m is at rest on a rough inclined plane of angle of inclination θ . If coefficient of friction between the block and the inclined plane is μ, then the minimum value of force along the plane required to move the block on the plane is mg [sin θ μ cos θ ] . How?
A monkey of mass 40 kg climbs up a rope, of breaking load 600 N hanging from a ceiling. If it climbs up the rope with the maximum possible acceleration, then the time taken by monkey to climb up is [ Length of rope is 10 m ]
1. 2s 2. 1s 3. 4s
A block of mass 10 kg is released on rough incline plane. Block start decending with acceleration 2 m/s² . Kinetic friction force acting on block is ( take g= 10m/s² )
1. 10N 2. 30N
question is not correct
A bomb of mass M @ rest explodes into 2 fragments of masses m1 &m2 . The total energy released in the explosn is E if E1 &E2 represent the energies carried by m1 &m2. then which of the followg is correct
a) E1=(m2/M)E
B) E1= m1/m2 E
C) E1= m1/M E
D) E1= m2/m1 E
(1) is correct
the ratio of energy will be E_{1}/E_{2} = m_{2}/m_{1}
Three forces F1 = ( 2i+ 4j ) N ; F2 = ( 2jk ) N and F3 = (k  4i  2j ) N are applied on an object of mass 1 kg at rest at origin. The position of the object at t=2s will be
1. (2m, 6m) 2. (4m , 8m)
a person stands in contact against the inner wall of a rotor of radius r. The coefficient of friction between the wall and the clothing is μ and the rotor is rotating about vertical axis. The minimum speed of the rotor so that the person does not slip downward is
1. √μg/r 2. √g/μr
A block of mass m takes t to slide down on a smooth inclined plane of angle of inclination and height h. If same block slids down on a rough inclined plane of same angle of inclination and some height and takes time n times of initial value, then coefficient friction between block and inclined plane is
the answer is [1 1/n² ] tanθ
in first case acceleration is gsinθ
and in second case acceleration is gsinθμgcosθ
now use s = 1/2 a t^{2 }in both the case and solve
A block of mass m is at rest on a rough inclined plane of angle of inclination θ . If coefficient of friction between the block and the inclined plane is required to move the block on the plane is
how the answer is mg[sinθ  μ cosθ] .
i think one line is missing in the question.
A small metallic sphere of mass m is suspended from the ceiling of a car accelerating on a horizontal road with constant acceleration a. The tension in the string attached with metallic sphere is
1. mg 2. m√g² + a²
A block of weight 1 N rests on an inclined plane of inclination θ with the horizontal. The coefficient of friction between the block and the inclined plane is μ. The minimum force that has to be applied parallel to the inclined plane to make the body just move up the plane is
1. μ sinθ 2. μ cosθ + sinθ
a point charge q is placed at a distance 2r from the centre O of a conducting charged sphere of radius r. due to induced charges on the sphere , find the elecric potential at point P on the surface of sphere in volt if kq/r =18V.
in the above ques, last time u found out potential at center but according to ques we hav eto find out on the surface and ans. is 3
if the answer is 3 V then 2r should be the distance of q from surface not from centre.
first confirm
In the ques previously submitted by someone which was as follow
the electric field vector is given by E= a(x)^{1/2} i^ . find the φ thru a cube bounded by surface x=l, x=2l, y=0, y=l , z=l, z=0.
Sir , my query is that in the solutn submitted the leaving flux at x=2l is given as( 2l)^{1/2} X l^{2} . but as the electric field is itself along x then why 2l will be used as along x axis it will be zero because cosθ used is perpendicular to surface with electric field axis thus cos90 =0??
θ should be measured between E and perpendicular drawn on the surface
a small body starts sliding from a height h down an inclined groove passing into a half circle of radius h/2. find the speed of the bodywhen it reaches the highest point of its trajectory
since body starts from height h at rest and ends in the same height so it will again be in rest
A BLOCK OF WOOD IS KEPT ON THE FLOOR OF A STATIONARY ELEVATOR. D ELEVATOR BEGINS TO DESCEND WITH AN ACCLN 12m/s^{2} . IF g =10 , THEN DISPLACEMENT OF BLOCK W.R.T. GROUND DURING FIRST 0.2 SEC. AFTER D START WILL BE ??
ANS 0.2m
s = ut + 1/2 at^{2}
so s = 0 + 1/2 10(0.2)^{2}
so s = 5(0.04) = 0.2
since elevator is moving with an acc. greater than g so we should take it a free fall
The energy required to accelerate a car from rest to 10m/s is E . What energy will be required to accelerate the car from 10m/s to 20m/s??
(ANS 3E)
4 particles each of mass m and charge q are held at the corners of a square of side a . they are released at t=0 and move under mutual repulsice force. find the speed of any particle when its distance from the center is double.
potential energy in initial condition = 4*k*q*q/a + 2*k*q*q/a√2
potential energy in final condition = 4*k*q*q/2a + 2*k*q*q/2a√2
and kinetic energy in the final condition = 4*mv^{2}/2
now use energy conservation
a point charge q is placed at a distance 2r from the centre O of a conducting charged sphere of radius r. due to induced charges on the sphere , find the elecric potential at point P on the surface of sphere in volt if kq/r =18V.
potential at the centre due to q = kq/2r = 9 V
potential at centre due to surface charge = 0 V
total potential = 9 V
now at surface potential due to q = kq/r = 18 V
but it should be 9 V totally so due to surface induction, potential will be 9V
the electric field vector is given by E=a(x)^{½ }iˆ.find the flux through a cube bounded by surfaces x=l ,x=2l, y=0 ,y=l, z=0, z=l
entering flux at x=l will be a(l)^{1/2} * l^{2}
leaving flux at x=2l will be a(2l)^{1/2} * l^{2}
so net flux will be the difference of these 2 flux
A body of mass 32kg is suspended by a spring balance from the roof of a vertically operating lift going downward from rest . At that instant the lift covered 20m and 50m , the spring balance showed 30 kg & 36 kg respt. the vel. is increasing at 20m & decreasing at 50m. Plzz sir explain this.
for downward journey
first case
32g30g = 2g = 32a so a = g/16
second case
32g36g = 4g = 32a so a = g/8
so it is acceleration upto 20 m and retardation after that
d benches of a gallery in a cricket stadium r 1m wide & 1m high . a boy strikes d ball at a level of 1m above d ground & hits a sixer. d ball starts at 35m/s at an < of 53^{0} with d horizontal . d benches r perpendicular to plane of motion & d first bench is 110m from d boy. ON WHICH BENCH WILL THE BALL HIT??
ANS 6TH
we know that y = xtanθ  gx^{2}/2u^{2}cos^{2}θ
according to the given conditions
x = 110+n
y = n here n is the number of bench in which ball strikes
θ, u and g are given
A 6 kg bomb at rest explodes into three equal pieces P, Q and R. If P flies with speed 30 m/s and Q with speed 40 m/s making an angle 90° with the direction of P. The angle between the direction of motion of P and R is about
1. 143° 2. 127° 3. 120°
since initially bomb is in rest so initially momentum of the bomb = 0i+0j+0k
after explosion total momentum = 30mi+40mj+mv here v is a vector
now by momentum conservation 0 = 30mi+40mj+mv so v = 30i40j
so vector along P is i and along R is 3i4j
so by A.B = ABcosθ
we get i.[3i4j] = 1*5*cosθ
so 3/5 = cosθ so θ = 127
If n balls hit elastically and normally on a surface per unit time and all the balls of mass m are moving with same velocity u, then force on surface is
2 mun . How ?
momentum change in elastic collision = mu(mu) = 2mu here after collision partical will retrace its direction
for n balls it will be 2mun
so force = rate of change of momentum = 2mun/1 = 2mun
A man of mass 50 kg carries a bag of weight 40 N on his shoulder. The force with which the floor pushes up his feet will be
1. 882 N 2. 530 N 3. 90 N 4. 600 N
reaction by ground is due to both bag and man so 50g+40 = 50*9.8+40 = 490+40 = 530 N
When a 4 kg rifle is fired, the 10 g bullet receives an acceleration of 3 x 10^6 cm/s^2 . The magnitude of the force acting on the rifle (in newton) is
1. 300 2. 200 3. 3000
force on the bullet = [10/1000]*3*10^{4 }= 300 N
so that will be same for rifle
A body of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The force required to keep the body moving with the same velocity is
1. 8 N 2. 0 N 3. 1/2 N
no force is required
Two particles each of mass m are moving in horizontal circle with same angular speed. If both string are of same length then the ratio of tension in string T1 / T2 is
1. 3/2 2. 3 3. 2 4. 1/3
it will be 1:1
A particle is moving on a circular path of radius 1 m with 2 m/s. If speed starts increasing at a rate of 2 m/s² , then acceleration of particle is
centripetal acceleration = v^{2}/r = 2*2/1 = 4
and tangential acceleration = 2 [given]
so acceleration = [4^{2}+2^{2}]^{1/2} = √ 20
A particle is revolving in a circular path of radius 2 m with constant angular speed 4 rad/s. The angular acceleration of particle is
1. zero 2. 8π² 3. 16π²
since angular velocity is constant, so zero will be the answer
If a particle is moving on a circular path with constant speed, then the angle between the direction of acceleration and its position vector w.r.t. centre of circle will be
1. zero 2. π 3. π/2 4. 2π
position will be radically outward and acceleration will be radically inward, so angle = π
If a particle of mass 2 kg is moving in circular path with constant speed 20 m/s. The magnitude of change in velocity when particle travels from A to P will be
1. 20√2 m/s 2. 40 m/s
in this question you should define A and P first then we will give you an answer.
A block of mass m kg on a weighing machine in an elevator is retarding upward by a m/s^2, the reading of weighing machine is (in kgf)
How the answer is m(1a/g).
A book of mass 5 kg is placed on a table and it is pressed by 10 N force then normal force exerted by the table on the book is
My answer is 60 N but the correct answer is 59 N. How?
A particle of mass m strikes elastically on a wall with velocity v, at an angle 60° from the wall the magnitude of change in momentum of ball along the wall is
(1) zero (2) mv
since momentum of the ball before strike and after strike are same along the wall so change will be 0
sir i had asked a ques and its ans was also replied by you, but still i have a doubt that in 1st part you take g as ve because motion is against gravity and it is clear to me but in 2nd why g is taken as ve? here, the motion is not against gravity, so plz tell me the concept of g? wten it is ve or +ve.
b) in second part we have to calculate the time of journey from top of 25 m high building to ground.
for this we should use
h = ut + gt^{2}/2 [in this case we are using upward direction as  ve and downward as + ve]
so 25 = 20t + 10t^{2}/2
or 5t^{2}  20t  25 = 0
or t^{2} 4t  5 = 0
or (t+1)(t5) = 0
or t = 5 , 1
so t = 5 sec.
This is answer of the same problem but here i am using g +ve, I want to say only that it will always be your opinion to consider the sign of direction. You can consider any of the direction as positive and negative.
A smooth sphere of radius R is made to translate in a straight line with constnt acceralation a . A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. find the speed of the particle with respect to the sphere as a function of the angle θ it slide
use energy conservation to solve it
if not possible then send a reply we will give you complete answer.
a ball is thrown vertically upwars with a velocity 20m/s from the top of a multistorey building. the height of the point from where the ball is thrown is 25m from the ground.
a) how high will the ball rise?
b) how long will it be before the ball hits the ground? take g=10m/s square
a) At heighest poing velocity will be 0
so by formula v^{2} = u^{2}  2gh (here  is because you are moving against gravity)
so h = 20*20/2*10 = 20 m
so ball will rise upto 25+20 m from the ground.
b) in second part we have to calculate the time of journey from top of 25 m high building to ground.
for this we should use
h = ut  gt^{2}/2 [in this case we are using upward direction as + ve and downward as  ve]
so 25 = 20t  10t^{2}/2
or 5t^{2}  20t  25 = 0
or t^{2} 4t  5 = 0
or (t+1)(t5) = 0
or t = 5 , 1
so t = 5 sec.
A particle starts from rest and moves with constant acceleration. The magnitude of its displacement is
1) more than its distance traversed
2) less '' '' '' ''
3) equal to '' '' ''
4) zero
ans given is (3) but I want to know why it vyl not be less than dist. travelled & why (4) optn is wrong as in case of uniform circular motion displacement is 0 but have const. acc.??
I think the particle is moving in a straight line
How fast a car can pull 100kg Iron? the energy of car is 10hp
E= ∫Fds
= ∫ma ds
= ∫m( dv / dt) ds
= ∫m dv (ds/dt)= ∫mvdv
= 1/2 mv^{2}
so v =(2E/m) ^{1/2 }now use mass 100kg & given energy after converting in SI units
This answer is given by SARIKA SHARMA and is correct, we made some modifications only
IF 2sin^{2}22^{0} = p+qcosθ, FIND SMALLEST POSITIVE VALUE OF θ, WHEN p=1 AND q = 1.
2sin^{2}22.5 = 1 cos45 = 1 + (1)cos45
so θ = 45
A car moves towards north at a speed of 54 km/h for 1 h. Then it moves eastward with same speed for same duration. The average speed and velocity of car for complete journey is
avg speed = total distance / total time = 54+54/1+1 = 54
avg velocity = total displacement / total time = 54sqrt(2) / 2 = 27sqrt(2)
If P & Q are two vectors such that PQ = P+Q , then find angle between P and Q.
The acceleration (a) of a body moving along xaxis is given by a=4x^{3} ,where a is in m/s^{2} and x is in metre.if at x=0 the velocity of body is 2m/s,then find its velocity at x=4m?
a = 4x^{3}
so dv/dt = 4x^{3}
or dv/dx * dx/dt = 4x^{3}
or dv/dx * v = 4x^{3}
or vdv = 4x^{3}dx
or ∫vdv = 4∫x^{3}dx
or v^{2}/2 = 4x^{4}/4 + C
at x = 0, v = 2
so 4/2 = 0+C
so C = 2
on putting the value of C
we get v^{2}/2 = x^{4}+ C
now put x = 4 and get v
The position x of a particle moving along xaxis at time t is given by equation t=√x +4 , where x is in meter and t is in seconds. Find the position of particle when its velocity is zero.
given that x = (t4)^{2}
so dx/dt = 2(t4)
or v = 2t8
now for v=0, t=4
put t=4 and get x
A body projected upwards is at same height form ground at t=3s and t=7s . Find the maximum height attained by it. ( take g= 10 m/s^2 ).
A body projected upwards is at same height form ground at t=3s and t=7s. It will gain max. height after 5 second
so maximum height = gt^{2}/2 = 10*5*5/2 = 125
another direct method for finding total distance = 5+15+25+35+45 = 125
these are the distances in 1st, 2nd, 3rd and etc sec.
A particle starts moving with acceleration 2 m/s^2 . Distance travelled by it in 5 th half second is
(a) 1.25 m (b) 2.25 m (c) 6.25 m (d)30.25 m
What would be the Potential energy of two points charges q &q which r separated by a dist d ( IF POTENTIAL AT MID PT IS TAKEN TO BE ZERO)
potential energy = kqq/d
a particle , with an inital velocity v_{0} in a plane , is subjected to a constant acceleration in the same plane . then in general ,
the path of the particle would be
1) circle
2) ellipse
3) parabola
4) hyperbola
parabola
for understanding this make components of v parallel and perpendicular to a
a train is moving with a velocity of 30m/s. When brakes are applied , it is found that the velocity reduces to 10m/s in 240m.
When the velocity of the train becomes zero , the total distance travelled is??
270m
by v^{2} = u^{2}2as
100 = 9002*a*240
or 800 = 480a
or a = 80/48 = 10/6 = 5/3
so total distance upto rest is u^{2}/2a = 900*3/2*5 = 270m
A CARNOT ENGINE WORKING B/W 27^{0} C & 123^{0}C HAS EFFICIENCY η . IF TEMP OF SINK IS DECREASED BY 50K THE EFFICIENCY BECOMES??
ANS 4η/3
according to the given condition η = 1(150/300) = 0.5
now after decreasing temperature of sink by 50K
we get the new efficiency = 1(100/300) = 2/3 = 4*0.5/3 = 4η/3
in a thermodynamic process on 2 moles of a monoatomic gas work done on gas is 100J & change in temp is 30 C . The change in internal energy of the gas in this process is??
change in internal energy = nC_{v}dT = 2*[R/(γ1)]*30 = 60*8.31/[(5/3)1]
while measuring the speed of sound by performing a resonance column expt , a student gets d first resonance conditn at a column length of 18 cm during winter . Repeating d same expt during summer , she measure d column length to be x cm for the second resonance. THEN
1)18>X
2)X>54
3) 54>X>36
4) 36>X>18
ANS (2)
AIEEE 2008
since l α λ α v/n n is frequency
on increasing temperature, v increases so λ increases
for first resonance l = λ/4
and for third resonance L = 3λ^{'}/4
and λ^{' }> λ so x will be more than 54
angle b/w vectors (i^+j^)and(j^+k^) is
1) 60^{o}
2) 90^{o}
3) 30^{0}
4) 45^{0}
use the formula A.B = ABcosθ
what is definition of time period?
what is the definition of velocity?
chock coil is used to
1) increase current
2) increase voltage
3) decrease current
4) decrease voltage
here we know power= IV & chock coil is used to reduce the power loss , so here what vyl be the answer
chock coil in ac is the device equivalent to rheostat in dc, so it is used to increase current.
If a particle (strong enough to survive in space) is released in space from earth. Where it will go? What about its motion? Will it revolve around the sun or remain stationary?
it depends on the speed with which it is thrown
a large open tank has two holes in the wall. one is a square hole of side L at a depth Y from the top and the other is a circular hole of radius R At a depth 4Y from the top . when the tank is completely filled with H2O , the quantities of H2O flowing / sec from both holes are same . then R IS equal to??
ans L / [√(2π)]
speed of water from any hole = √2gh
so volume of water per sec. = A√2gh
for square hole volume/sec = L^{2}√2gY
for circular hole volume/sec = πR^{2}√2g4Y
compare and get answer
when a tuning fork vibrates with 1.0 m or 1.05m long wire of a sonometer, 5 beats per second are produced in each case. what will be the frequency of the tuning fork ??
for sonometer wire
l = λ/2
or 2l = v/n here n represents frequency
or n = v/2l
for first wire of 1 m length n= v/2
for second wire of 1.05 m length n=v/2.1
let frequency of tuning fork = n_{1}
so according to the given condition v/2 = n_{1}+ 5
and v/2.1 = n_{1}  5
solve now
if ω_{1 &}ω_{2} are dispersive powers of lenses of focal length f1 & f2 respectively . then the condition of chromatic aberration for two lenses in contact is
1) ω1/ω2 = f2/f1
2) ω1/ω2 = f1/f2
3) ω1/ω2=  f1/f2
4) ω2/ω1= f1/f2
3 is the true condition and can be proved
the half life of radium is 1620 year & its atomic weight is 226gm/mole. the number of atoms that will decay from its 1g sample per second is
1) 3.16 x 10^{10}
2)3.6 x 10^{12}
3) 3.1 x 10^{15}
4) 31.1 x 10^{15}
for radioactivity we know that it is a first order process
so dN/dt = λN
so dN/1 = [0.6932/(1620*3.1*10^{7})]*(1/226)*6.023*10^{23}
solve now
An open knife edge of mass m is dropped from a height h on a wooden floor. if the blade penetrates a dist s into the wood , the avg. resistance offered by wood to blade. ANS mg(1+h/s)
velocity at the time of strike of the knife at the wood = √(2gh)
now equation for penetration on the wood is Rmg = ma here R is reaction by the wood
so R = m(g+a)
here it is given that s is the distance penetrated by the knife on the wood so 0^{2} = (√2gh)  2as so a = 2gh/2s = gh/s
on taking this value of a
we get R = m(g+gh/s) = mg(1+h/s)
A at road of 200km & B at 100km road. A with vel 20m/s & B with 7.5m/s . Then wat will b d time when B overtake A.
1) 24s
2) 40s
3) 80s
4)nothing can be predicted
data are irrelevent
A LIFT IS MOVING UPWARD WITH ACC 2m/s^{2} .WHEN IT GAIN THE VELOCITY 4m/s THEN AT THAT MOMENT A BALL IS PROJECTED BY 30^{0} (wrt floor to lift) WITH VEL. 4m/s RELATIVE TO LIFT . FIND HOW MUCH TIME WILL IT TAKE TO 1) TO REACH GROUND & 2) TO THE FLOOR OF LIFT AGAIN
since velocity of the ball is 4 m/s relative to lift so
time taken by the ball to reach the floor of lift T = 2usinθ/g = (2*4*1/2)/10 = 0.4 sec
height of the lift at the time of dropping the ball = (v^{2}u^{2})/2g = 16/20 = 8/10 = 0.8 m = h
and vertical component of the velocity of ball with respect to the ground = 6 m/s = u
so use h = ut  gt^{2}/2
A STONE WEIGHS 9N AT A PLACE ON THE SURFACE OF EARTH HAVING LATITUDE π/2 . ITS MASS AT CENTRE OF EARTH WILL BE (kg)
1) 10/ 37√2
2 ) 10√2
3) 5
4) 15
π/2 means it is pole so mg = 9 implies that m = 9/9.8
same will be the mass at centre of earth
A CHARGED PARTICLE IS PROJECTED PARALLEL TO A UNIFORM FIELD . THE PATH OF D PARTICLE WILL BE
1) PARABOLIC
2) ELLIPTICAL
3) CIRCULAR
4) LINEAR
linear and accelerated.
CHARGES ARE PLACED ON VERTICES OF A SQ. WITH a, b HAS +q & c,d has q charge . LET E ELECTRIC FIELD & V IS POT. AT CENTRE . IF CHARGES ON a &b R INTERCHANGED WITH THOSE OF c& d RESPTLY THEN E CHANGES WHY??
in this case magnitude will remain same. Only direction of E will change.
a II plate capacitor has an electric field of 10^{5} V/m b/w d plates . if charge on capacitor plate is 1uC d force on each capacitor is??
ans 0.05N
F = σq/2ε_{0} = Eq/2 = 10^{5}*10^{6}/2 = 0.05
A charge of +2.0 x 10^8 C is placed on the positive plate and a charge of 1.0 x 10^8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10^3 uF. Calculate the potential difference developed between the plates. Please add explanation!
due to induction the charges on the four faces will be
so V = q/C = 1.5 * 10^{8}/1.2 * 10^{9}
solve now
2 MOLES OF MONOATOMIC GAS IS MIXED WITH 1 MOLE OF A DIATOMIC GAS. THEN γ FOR THE MIXTURE??
ANS 1.55
total no of freedom for this mixture = (2*3+1*5)/(2+1) = 11/3
so γ = 1+[2/f] = 1 + 2/[11/3] = 1 + [6/11] = 17/11 = 1.55
a thin square plate with each side equal to 10cm , is heated by a blacksmith . the rate radiated energy by the heated plate is 1134W the temp of hot square plate is
( σ = 5.67 X 10^{8} Wm^{2}/k^{4} emissivity of plate =1)
ANS 1000K
according to the Stefans law P = AeσT^{4}
so 1134 = 0.02*1*5.67*10^{8}*T^{4 }(area of both the faces are considered)
so 1134*10^{10}/11.34 = T^{4}
so T = 1000
TWO PENDULUM HAVE TIME PERIODS T &5T/4. THEY STARTS SHM AT THE SAME TIME FROM THE MEAN POSITION . WHAT WILL BE THE PHASE DIFFERENCE B/W THEM AFTER D BIGGER PENDULUM COMPLETED ONE OSCILLATn ??
ans 90^{0}
When bigger pendulum will complete its one oscillation, the smaller one will complete [1+(1/4)] oscillation. so phase difference = 2π*(T/4)/T = π/2
QUES A MAN RUNS AT A SPEED OF 4M/s TO OVERTAKE A STANDING BUS . WHEN HE IS 6m BEHIND THE DOOR (AT t=0) , THE BUS MOVES FORWARD & CONTINUES WITH A CONSTANT ACCELERATn OF 1.2m/s^{2} . the MAN SHALL GAIN THE DOOR AT TIME t EQUAL TO
A) 5.2s
b) 4.3s
c)2.3s
d) the man shell never gain the door.
ans 4.3s , Sir ,on solving I 'm getting both 4.3 & 2.3 s ( when we solve sq root) so here i want to know why not the ans is 2.3s. AMU 2012
After 2.3 sec. the man overtake the door first, in this time velocity of bus will be less than that of man so it is not possible to catch the bus then after 4.3 sec. the bus door will again reach to the man, in this time velocity of bus and man are same, so relative velocity is zero and catching of bus is possible.
A RADIOACTIVE SUBSTANCE EMITS n BETA PARTICLES IN THE FIRST 2 SECONDS & 0.5 n BETA PARTICLES IN THE NEXT 2 SECONDS . THE MEAN LIFE OF THE SAMPLE IS
ANS 2/(ln2) s
A RADIOACTIVE SUBSTANCE EMITS n BETA PARTICLES IN THE FIRST 2 SECONDS & 0.5 n BETA PARTICLES IN THE NEXT 2 SECONDS. Definitely by definition its half life will be 2 sec.
so mean life = half life / ln2 This is the relation between mean life and half life.
A NUCLEUS OF MASS NO. 220, INITIALLY AT REST , EMITS AN .α PARTICLE . IF THE Q VALUE OF THE Rxn IS 5.5MeV THE ENERGY OF THE EMITTED α PARTICLE WILL BE ??
ANS 5.4 MeV
5.5*216/220 = 216/40 = 5.4 MeV here 216 is the mass of nucleus after ejecting α particle.
THE TERMINAL SPEED ATTAINED BY AN Al SPERE OF RADIUS 1mm FALLING THROUGH H2O AT 20^{0}C WILL BE CLOSE TO ??
ANS 4.6 m/s
(assume laminar flow , specific gravity of Al=2.7 & η H2O =8X 10^{4 }PL.)
v = 2r^{2}g(Dd)/9η
so v = 2(.001)^{2}*9.8(27001000)/9*8*10^{5}
solve it
according to the diagram
tanθ = qE/mg = q(σ/2ε_{0})/mg
solve this eq. for σ.
Remember that length of the thread is not applicable in the question.
what is a short circuit?
to connect  and + of the cell together
IN THE SOLUTION GIVEN BY AMIT TO MY PREVIOUS QUERY , THE FORMULA USED IS E=kq/r ,SIR I WANT TO KNOW IS IT IS CORRECT , DON'T YOU THINK IT SHOULD BE E=kq/r^{2 }??
yes you are right