The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is 10 ~s . If the metallic bob is immersed in water, then the new time period becomes 5 \sqrtx s. The value of x will be \underline\hspace2cm .

Simple Harmonic Motion PYQ — JEE Main 2022 · 29 Jul · Shift 2 · Q67 with Solution

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Correct answer: 5

For small oscillations,

T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}}

where $g_{\text{eff}}$ is the effective acceleration acting downward.

Initially, in air:

T_1 = 2\pi \sqrt{\frac{L}{g}} = 10\text{ s}

The bob experiences buoyant force upward.

Let:

Given relative density of bob $=5$ , so

\rho_b = 5\rho_w

If volume of bob is $V$ , then

m = \rho_b V

W = mg = \rho_b V g

B = \rho_w V g

Hence the effective downward force is proportional to

W - B = (\rho_b - \rho_w)Vg

So effective acceleration is

g_{\text{eff}} = g\left(1 - \frac{\rho_w}{\rho_b}\right)

Since $\rho_b = 5\rho_w$ ,

\frac{\rho_w}{\rho_b} = \frac{1}{5}

Therefore,

g_{\text{eff}} = g\left(1-\frac15\right)=\frac{4g}{5}

Now,

T_2 = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{L}{4g/5}}

Using $T_1 = 2\pi \sqrt{L/g}$ ,

T_2 = T_1\sqrt{\frac{g}{g_{\text{eff}}}} = 10\sqrt{\frac{g}{4g/5}} = 10\sqrt{\frac{5}{4}} = 5\sqrt{5}

Given new time period is $5\sqrt{x}$ s. So,

5\sqrt{x} = 5\sqrt{5}

Hence,

x=5

Stored correct answer = $5$

Our derived answer also is $5$ . So they agree.