JEE Advanced 2023 — Paper 1 — PHY-III solution

JEE Advanced2023Paper 1PHY-III

Two point-like objects of masses $20$ gm and $30$ gm are fixed at the two ends of a rigid massless rod of length $10$ cm. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is $1.2 \times 1 0^{- 8} N m rad^{- 1}$ . The angular frequency of the oscillations is $n \times 1 0^{- 3} rad s^{- 1}$ . The value of $n$ is _____.

Your answer

Solution

Locate the centre of mass. Measuring from the $30$ g end, $x_{c m} = \frac{30 \cdot 0 + 20 \cdot 10}{50} = 4$ cm. So the $30$ g mass sits $4$ cm from the wire axis and the $20$ g mass sits $6$ cm from it.

Moment of inertia about the vertical wire axis. $I = 30 \cdot 4^{2} + 20 \cdot 6^{2} = 480 + 720 = 1200 g cm^{2} = 1.2 \times 1 0^{- 4} kg m^{2}$ .

Angular frequency. $ω = \frac{C}{I} = \frac{1.2 \times 1 0 ^{- 8}}{1.2 \times 1 0 ^{- 4}} = 1 0^{- 4} = 1 0^{- 2} rad s^{- 1} = 10 \times 1 0^{- 3} rad s^{- 1}$ .

Answer: $n = 10$ .

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