The negative of the distance rate of change of potential energy is equal to

force acting on the particle in the direction of displacement
acceleration of the particle, perpendicular to d is placement
power
impulse

Solution

F = - ^dU⁄_dx

The force F acting on a body moving in a circle of radius r is always perpendicular to the instantaneous velocity v.The work done by the force on the body in half rotation is

Fv
F·2πr
Fr
zero

Solution

Work done by centripetal force is zero.

A ramp is constructed in parabolic shape such that the height y of any point on its surface is given in terms of the point’s horizontal distance x from the bottom of the ramp bey = x²/2L.A block of granite is to be set on the ramp; the coefficient of static friction is 0.80. What is the maximum x coordinate at which the block can be placed on the ramp and remain at rest, if L = 10 m?

8 m
8.4 m
9 m
9.4 m

Solution

The kinetic energy of particle moving along a circle of radius R depends upon the distance and is given by K= aS where a is a constant. Then the force acting on the particle is

\(\frac{aS}{R}\)
\(\frac{2(aS)^{2}}{R}\)
\(\frac{aS^{2}}{R^{2}}\)
\(\frac{2aS}{R}\)

Solution

A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m at rest.After collision, their final velocities are V and v respectively.The value of v is

\(\frac{2uM}{m}\)
\(\frac{2um}{M}\)
\(\frac{2u}{1+\frac{m}{M}}\)
\(\frac{2u}{1+\frac{M}{m}}\)

Solution

A 10 m long iron chain of linear mass density 0.8 kg m^-1 is hanging freely from a rigid support. If g = 10 ms^-2, then the power required to left the chain upto the point of support in 10 second

10 W
20 W
30 W
40 W

Solution

A weight suspended from the free end of a vertically hanging spring produces an extension of 3 cm. The spring is cut into two parts so that the length of the longer part is ²⁄₃ of the original length, If the same weight is now suspended from the longer part of the spring, the extension produced will be

0.1 cm
0.5 cm
1 cm
2 cm

Solution

A glass marble dropped from a certain height above the horizontal surface reaches the surface in time t and then continues to bounce up and down. The time in which the marble finally comes to rest is

eⁿt
e²t
\(t\left [ \frac{1+e}{1-e} \right ]\)
\(t\left [ \frac{1-e}{1+e} \right ]\)

Solution

A motor drives a body along a straight line with a constant force. The power P developed by the motor must vary with time t according to

Solution

P = F × v ⇒ P =F at ∴ P ∝ t

A mass m is moving with velocity v collides inelastically with a bob of simple pendulum of mass m and gets embedded into it. The total height to which the masses will rise after collision is

^V²⁄_8g
^V²⁄_4g
^V²⁄_2g
^2V²⁄_g

Solution

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete