Let T0=2π√(L/g) be the small-oscillati…

Let T0=2π√(L/g) be the small-oscillation period of a simple pendulum of length L at rest in a uniform gravitational field g. List-I describes four…

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Let

T_0=2\pi\sqrt{L/g}

be the small-oscillation period of a simple pendulum of length

L

at rest in a uniform gravitational field

g

. List-I describes four oscillators, each undergoing small oscillations. List-II gives possible values of

(T/T_0)^2

, where

T

is the period of the oscillator. Match List-I with List-II.

List-I
(P) A simple pendulum of length

L

inside a lift accelerating upward at

3g

(Q) A uniform thin rod of length

L

pivoted at one end and swinging in a vertical plane
(R) A simple pendulum of length

L

in a vehicle accelerating horizontally at

\sqrt3\,g

(oscillating about its new equilibrium)
(S) A uniform thin disc of radius

L

pivoted at a point on its rim, swinging in its own vertical plane

List-II
(1)

1/4

(2)

2/3

(3)

4

(4)

3/2

(5)

1/2

Show answer and solution

Answer: (A) P→1, Q→2, R→5, S→4

For a physical pendulum

T=2\pi\sqrt{I/(mgd\,)}

, with

g\to g_{eff}

if the frame accelerates, and

T_0^2\propto L/g

. (P)

g_{eff}=g+3g=4g

, so

(T/T_0)^2=1/4

, which is (1). (Q) For the rod

I=\tfrac13mL^2

d=L/2

, so

T^2=4\pi^2\dfrac{(1/3)L^2}{g\,L/2}=4\pi^2\dfrac{2L}{3g}

, giving ratio

2/3

, which is (2). (R) The effective gravity is

g_{eff}=\sqrt{g^2+3g^2}=2g

, so the ratio is

1/2

, which is (5). (S) For a disc about a rim point,

I=\tfrac12mL^2+mL^2=\tfrac32mL^2

and

d=L

, so

T^2=4\pi^2\dfrac{3L}{2g}

, giving ratio

3/2

, which is (4). The value

4

in (3) corresponds to

g_{eff}=g/4

and is unused. Hence P→1, Q→2, R→5, S→4.

Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake