A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $\mathrm{y} \pi \times 10^{-2} \mathrm{~s}$, where the value of $y$ is (Acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$, density of water $=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ )

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Elasticity & SHM

A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is

y π \times 1 0^{- 2} s

, where the value of

y

is
(Acceleration due to gravity,

g = 10 m / s^{2}

, density of water

= 1 0^{3} kg / m^{3}

)

(A)2✓ Correct
(B)6
(C)4
(D)1

Answer: (A)

Solution

The cube of side

L = 0.1 m

and mass

m = 10 g = 1 0^{- 2} kg

floats in water. When pushed down by a small depth

x

, the extra buoyant force acts as a restoring force:

F = - ρ g L^{2} x

so the equation of motion is

m \overset{x}{¨} = - ρ g L^{2} x

, which is SHM with

ω^{2} = \frac{ρ g L ^{2}}{m}

Substituting

ρ = 1 0^{3} kg/m^{3}

g = 10 m/s^{2}

L^{2} = 1 0^{- 2} m^{2}

m = 1 0^{- 2} kg

ω^{2} = \frac{1 0 ^{3} \times 10 \times 1 0 ^{- 2}}{1 0 ^{- 2}} = 1 0^{4} \Rightarrow ω = 100 rad/s

T = \frac{2 π}{ω} = 2 π \times 1 0^{- 2} s

Comparing with

y π \times 1 0^{- 2} s

y = 2

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Elasticity & SHM

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