Physics Major Test - 1 (13-02-2017) - Class 11 MCQ

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by  where g’ =

. A lift is ascending  with acceleration g/3 .What will be the time period of a simple pendulum suspended from its ceiling if its time period in stationary lift is T:

A particle is executing S.H.M. of frequency 300 Hz and with amplitude 0.1 cm. Its maximum velocity will be:

The equation of a simple harmonic motion is x = 0.34cos(3000t + 0.74). Where x and t are in mm and sec respectively. The frequency of motion is:-

When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be;

The ratio of K.E. of the particle at  mean position to the point when distance is half of amplitude will be:

. A particle is executing S.H.M., If its P.E. and K.E. is equal then the ratio of displacement and amplitude will be:-

The maximum velocity of a paricle, executing simple harmonic motion with an amplitude 7 mm is 4.4 m/s. The period of oscillation is:

Two simple pendulums have time period 2s and 2.5s. Then at what time they both will be in same phase (if oscillation start simultaneously)

A particle is executing SHM whose angular frequency and amplitudes are 2rad/s and 60mm. Its velocity at a distance 20 mm from mean position will be:

A particle is executing S.H.M. with amplitude A and Time period T. Time taken by the particle to reach from extreme position to  A/2

Total work done on a simple pendulum in one complete oscillation will be :-

The potential energy of a spring when stretched by a distance x is E. The energy of the spring when stretched by x/2 is

The displacement y of a particle varies with time t, in seconds, as Y = 2 cos (πt+π/6). The time period of the oscillation is

. In simple harmonic motion, loss of K.E. is proportional to :

A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m The maximum compression of the spring would be:

A string fixed at both the ends is vibrating in two segments. The wavelength of the corresponding wave is :

. Standing waves are produced in a 10 m long stretched string. If the string vibrates in 5 segments and the wave velocity is 20 m/s, the frequency is

To increase the frequency from 100 Hz to 400 Hz the tension in the string has to be changed by

Force constant of a spring is k one fourth part is detach then force constant of remaining spring will be:-

A child swinging on a swing in sitting position, stands up, then the period of the swing will be:

The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:

Mass ‘m’ is suspended from a spring of force constant k. Spring is cut into two equal parts and same mass is suspended from it ,then new frequency will be:

In SHM velocity is maximum :-

Time period of a compound pendulum is T, if it mass is doubled,then it times period will be:

The maximum velocity of simple harmonic motion represented  by y = 3sin  is given by:

For a particle executing simple harmonic motion which of the following statement is not correct :