Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

The displacement `y` of a particle executing periodic motion is given by `y = 4 cos^(2) ((1)/(2)t) sin(1000t)` This expression may be considereed to be a result of the superposition ofA. two movesB. three movesC. four movesD. five moves

Answer» If we have a term `cos kx sin omegat`, it is the superposition of two wave motions. If we have an equation of the form `cos^(2)` at `sinbt` it can be shown to be superposition of three sine waves.
Here, `y=4cos^(2)((1)/(2)t sin(1000)t)`
`=2[1+cost]sin1000t=2[sin1000t+(1)/(2)(sin100t+sin999t)]=` this comprises three waves.
Discussion
2.

The amplitude of a wave disturbance propagating in the positive x-direction is given by `y = (1)/((1 + x))^(2)` at time `t = 0` and by `y = (1)/([1+(x - 1)^(2)])` at `t = 2 seconds`, `x and y` are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is ............... m//s`.A. `1ms^(-1)`B. `0.5ms^(-1)`C. `1.5ms^(-1)`D. `2ms^(-1)`

Answer» Writing the general expression for `y` in terms of `x` as
`y=(1)/(1+(x-vt)^(2))`. At `t=0`, `y=1//(1+x^(2))`. At `t=2s`, `y=(1)/(1+[x-v(2)]^(2))`
Comparing with the given equation we get `2v=1` and `v=0.5m//s`.
Discussion
3.

A travelling wave is described by the equation `y = y_(0) sin ((ft - (x)/lambda))`. The maximum particle velocity is equal to four times the wave velocity ifA. `piy_(0)//4`B. `piy_(0)//2`C. `piy_(0)`D. `2piy_(0)`

Answer» The maximum particle velocity of a `SHM` of amplitude `y_(0)` and frequency `f` is `2pify_(0)`. The wave velocity is `f_(A)`. For `2pify_(0)` to be equal to `4flambda`, `lambda` has to be `piy_(0)//2`
(Here `lambda=a`).
Discussion
4.

When the load on a wire increased slowly from `2 kg wt.` to `4 kg wt.`, the elogation increases from `0.6 mm` to `1.00 mm`. How much work is done during the extension of the wire? `[ g = 9.8 m//s^(2)]`A. `14xx10^(-3)J`B. `0.4xx10^(-3)J`C. `8xx10^(-2)J`D. `10^(-3)J`

Answer» `W=((1)/(2)F_(2)xxDeltal_(2))-((1)/(2)F_(1)xxDeltal_(1))=20xx10^(-3)-6xx10^(-3)=14xx10^(-3)J`
Discussion