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The minimum velocity (in ms-1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is
[AIEEE 2002]
To move a body on a circular path, centripetal force is necessary. In this question, there is an agent, friction that will provide the required centripetal force to the car.
Using the relation,
mv2r=μR,R=mg
mv2r=μmg
or, v2=μrg
or, v2=0.6×150×10
or, v=30m/s
Which of the following statements is false for a particle moving in a circle with a constant angular speed ?
[AIEEE 2004]
In uniform circular motion tangential acceleration is zero only centripetal acceleration work which is directed towards centre.
An angular ring with inner and outer radii R1 and R2 is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, F1/F2 is
F = mω2R
Since F α R, the ratio of forces will be equal to the ratio of radii.
Hence D is the correct answer.
A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out a length s = t3 + 5, where s is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t = 2s is nearly
[AIEEE 2010]
= 14m/s2
For a particle in uniform circular motion the acceleration at a point P(R,) on the circle of radius R is (here
is measured from the x-axis)
[AIEEE 2010]
In a typical power station, amid a series of energy transfers and conversions, about 70% of energy input gets wasted in form of thermal energy, efficiency of such a power station is about
Assume total input energy to be 100%
Lost energy =70%
Left energy = 100 – 70 = 30 % is the output energy
Efficiency = output energy/input energy = 30/100 = 0.3
Percentage efficiency = 0.3 × 100 = 30%
A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is
[AIEEE 2002]
The work done is stored as elastic potential energy in the spring. It is given by
W = 1/2 * k (x₂2 - x₁2)
x₁ = 5 cm
x₂ = 15 cm
Force constant, k = 800 N/m = 800/100 N/cm = 8 N/cm
W = 1/2 * 8 (15^2 - 5^2) = 1/2 * 8 (225 - 25)
W = 1/2 * 8 * 200 = 800 N cm
W = 800 N cm = 8 N m = 8 J
Therefore the work done in extending it from 5cm to 15cm is 8 J.
Hence B is the correct answer.
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to
[AIEEE 2003]
P= fv
P = mav
p = mdv/dt.v
p.dt/m = v.dv ( integration from 0-t)
pt/m = v²/2
v = √(2pt/m)
ds/dt = t½dt
s = t3/2
A spring of spring constant 5 × 103 N/m is stretched initally by 5 cm from the unstreched position. Then the work required to stretch it further by another 5 cm is
[AIEEE 2003]
work done in stretching a spring by a distance x is
W=½kx2
where k is the spring constant.
work required to stretch the string by 10cm is
W1=½kx2=½ 5×103(0.1)2=25J
work required to stretch the string by 5cm is
W2=½kx2=½ 5×103(0.05)2=6.25J
Hence the work done in stretching it from 5cm to 10cm is
W1−W2=25−6.25=18.75J
A force F = (5i + 3j +2 k) N is applied over a particle which displaces it from its origin to the point r = (2 i - j) m. The work done on the particle in joule is
[AIEEE 2004]
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?
[AIEEE 2004]
The center of mass of the hanging part is at 0.3m from table
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to
[AIEEE 2004]
The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is
[AIEEE 2005]
A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rools down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball i
[AIEEE 2005]
mgh = 1/2×mv2
v = √2gh
v = √2×10×80(actually 80 is come out of from 100-20)
v = 40 m/s
Hence A is correct.
A body of mass m is accelerated uniformly from rest to a speed V in a time T. The instantaneous power delivered to the body as function of time, is given by
[AIEEE 2005]
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion ?
[AIEEE 2005]
A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is
[AIEEE 2006]
By the Work Energy Theorem
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2
[AIEEE 2006]
The potential energy of a 1 kg particle free to move along the x-axis is given by
The total mechanical energy of the particle is 2 J. Then, the maximum speed (in ms_1) is
[AIEEE 2006]
A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10000 N/m. The spring compresses by
[AIEEE 2007]
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
[AIEEE 2008]
At time t = 0 s particle starts moving along the x-axis. If its kinetic energy increase uniformly with time t, the net force acting on it must be proportional to
[AIEEE 2011]
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