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201.

The centre of mass of a body is a point at which the entire mass of the body is supposed to be concentrated. The position vector `overset rarr(r )` of c.m of the system of tow particles of masses `m_(1)` and `m_(2)` with position vectors `overset rarr(r_(1))` and `overset rarr(r_(2))` is given by `overset rarr(r ) = (m_(1)overset rarr(r_(1)) + m_(2)overset rarr(r_(2)))/(m_(1) + m_(2))` For an isolated system, where no external force is acting, `overset rarr(v_(cm)) =` constant Under no circumstances, the velocity of centre of mass of an isolated system can undergo a change With the help of the comprehension given above, choose the most appropriate alternative for each of the following questions : A bomb dropped from an aeroplane in level flight explodes in the middle. the centre of mass of the fragmentsA. is a restB. moves vertically downwardsC. moves vertically upwardsD. continues to follow the same parabolic path which it would have followed if there was no exposion.

Answer» Correct Answer - D
As explosion is due to inertal forces only, the centre of mass of fragments continues to follow the same parabolic path, which it would have followed, if there was no explosion.
Discussion
202.

Find the components along the `x,y,z` axes of the angular momentum `overset rarr(L)` of a particle, whose position vector is `overset rarr(r )` with components `x,y,z` and momentum is `overset rarr(p)` with components `p_(x), p_(y) and p_(z)`. Show that if the particle moves only in the `x-y` plane, the angular momentum has only a z-component.

Answer» We know that angular momentum `vec(l)` of a particle having position vector `vec (r )` and momentum `vec(p)` is given by
`vec(l)=vec(r )xx vec(p)`
But `vec(r )=[xhat(i)+yhat(j)+zhat(k)]` where x, y, z are the components of
`vec(r ) " and " vec(p)=[p_(x)vec(i)+p_(y)hat(j)+p_(z)hat(k)]`
`therefore " " vec (l)=vec(r )xxvec(p)=[xhat(i)+yhat(j)+zhat(k)]xx[p_(x)hat(i)+p_(y)hat(j)+p_(z)hat(k)]`
or ` (l_(x)hat(i)+l_(y)hat(j)+l_(z)hat(k))= |{:(hat(i), hat(j), hat(k)),(x,y,z),(p_(x),p_(y),p_(z)):}|`
`=(yp_(z)-zp_(y))hat(i)+(zp_(x)-xp_(z))hat(j)+(xp_(y)-yp_(x))hat(k)`
From this releation, we conclude that
`l_(x)=yp_(z)-zp_(y) l_(y)=zp_(x)-xp_(z) " and " l_(z)=xp_(y)-yp_(x)`
If the given particle moves only in the x-y plane, then z=0 and `p_(z)=0` and hence,
`vec(l)=(xp_(y)-yp_(z))hat(k)`, which is only the z-comonent of `vec(l)`.
It means that for a particle moving only in the x-y plane, the angular momentum has only the z-component.
Discussion