A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of 4 ~ms^-1 , is \underline\hspace2cm cm. (take g = 10 ~ms^-2 )

Rotational Motion PYQ — JEE Main 2022 · 27 Jul · Shift 2 · Q64 with Solution

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Correct answer: 120

A solid cylinder is suspended symmetrically by two light strings wound around it. As it descends, the strings unwind without slipping.

We need the downward distance $h$ from the initial rest position so that the cylinder attains speed $v=4\ \text{m s}^{-1}.$

Since the strings are massless and unwind without slipping, loss of gravitational potential energy becomes:

So, $mgh=\frac12 mv^2+\frac12 I\omega^2.$

For a solid cylinder about its central axis, $I=\frac12 mR^2.$

Also, because the string unwinds without slipping, $v=R\omega \quad \Rightarrow \quad \omega=\frac{v}{R}.$

Substitute into rotational KE:

=\frac14 mv^2.$$ Thus total kinetic energy is $$\frac12 mv^2+\frac14 mv^2=\frac34 mv^2.$$ Hence, $$mgh=\frac34 mv^2.$$ Cancelling $m$, $$gh=\frac34 v^2.$$ --- 4. **Substitute values** Given: $$g=10\ \text{m s}^{-2}, \qquad v=4\ \text{m s}^{-1}.$$ So, $$10h=\frac34(4^2)=\frac34(16)=12.$$ Therefore, $$h=\frac{12}{10}=1.2\ \text{m}.$$ Convert to cm: $$1.2\ \text{m}=120\ \text{cm}.$$ --- 5. **Final answer** The required distance is $$\boxed{120\ \text{cm}}.$$ --- 6. **Comparison with stored answer** Stored correct answer = $120$ Our derived answer also = $120$. So the answer agrees.