JEE Main 2017 · Physics · Rotational Motion · hard

A uniform thin rod of length 1 m is suspended from a horizontal axis perpendicular to the rod, passing through a point 25 cm from its centre, and…

Single correct option

A uniform thin rod of length

1\ \mathrm{m}

is suspended from a horizontal axis perpendicular to the rod, passing through a point

25\ \mathrm{cm}

from its centre, and oscillates through small angles in a vertical plane. Taking

g = \pi^2\ \mathrm{m\,s^{-2}}

, the time period of oscillation is nearly

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Answer: (A) $1.53\ \mathrm{s}$

For a compound pendulum

T = 2\pi\sqrt{\dfrac{I}{mgd}}

, where

d

is the distance of the pivot from the centre of mass. By the parallel axis theorem

I = \dfrac{mL^2}{12} + md^2 = m\left(\dfrac1{12} + \dfrac1{16}\right) = \dfrac{7m}{48}

. So the equivalent simple pendulum length is

\ell = \dfrac{I}{md} = \dfrac{7/48}{1/4} = \dfrac{7}{12}\ \mathrm{m}

. With

g = \pi^2

T = 2\pi\sqrt{\ell/\pi^2} = 2\sqrt{7/12} \approx 1.53\ \mathrm{s}

. Using

I

about the centre only gives

1.15\ \mathrm{s}

and using

\ell = L

gives

2.00\ \mathrm{s}

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake