JEE Advanced 2024 — Paper 1 — PHY-III solution

JEE Advanced2024Paper 1PHY-III

A thin uniform rod of length $L$ and certain mass is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point $O$ . If a horizontal impulse $P$ is imparted to the rod at a distance $x = L / n$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point $O$ , with the rod remaining aligned with the string. In such a case, the value of $n$ is ________.

Your answer

Solution

Let the rod have mass $m$ . The string is taut, so the rod-plus-string system rotates as a rigid body about $O$ immediately after the impulse, with the rod remaining colinear with the string.

Moment of inertia of the rod about $O$ : by the parallel-axis theorem,

$I_{O} = \frac{m L ^{2}}{12} + m (L + \frac{L}{2})^{2} = \frac{m L ^{2}}{12} + \frac{9 m L ^{2}}{4} = \frac{7 m L ^{2}}{3}$ .

The centre of mass of the rod is at distance $\frac{3 L}{2}$ from $O$ . For the post-impact angular velocity $ω_{0}$ :

Linear impulse on the centre of mass: $P = m v_{c} = m \cdot \frac{3 L}{2} ω_{0}$ .

Angular impulse about $O$ : $P (\frac{3 L}{2} + x) = I_{O} ω_{0}$ .

Substituting $m \frac{3 L}{2} ω_{0}$ for $P$ :

$m \frac{3 L}{2} ω_{0} (\frac{3 L}{2} + x) = \frac{7 m L ^{2}}{3} ω_{0}$

$\Rightarrow \frac{3 L}{2} + x = \frac{14 L}{9}$

$\Rightarrow x = \frac{14 L}{9} - \frac{3 L}{2} = \frac{28 L - 27 L}{18} = \frac{L}{18}$ .

Hence $n = 18$ .

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