MHT CET · Physics · Oscillations

The bob of a simple pendulum performs SHM with period $T$ in air and with period $T_1$ in water. Relation between $T$ and $T_1$ is (neglect friction due to water, density of the material of the bob is $\frac{9}{8} \times 10^3 kg / m ^3$, density of water is $1 g / cc$ )

A $T_{1} = 3 T$
B $T_{1} = 2 T$
C $T_{1} = T$
D $T_{1} = \frac{T}{2}$

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Answer & Solution

Correct Answer

(A) $T_{1} = 3 T$

Step-by-step Solution

Detailed explanation

T = 2 π \sqrt{\frac{l}{g_{e f f}}}

In air

g_{e f f} = g

where as in water

g_{e f f} = g (\frac{σ - ρ}{σ}) = 9.8 (\frac{\frac{9}{8} \times 10^{3} - 10^{3}}{\frac{9}{8} \times 10^{3}})

= 9.8 (\frac{\frac{9}{8} - 1}{\frac{9}{8}}) = 9.8 \times (\frac{1}{8} \times \frac{8}{9}) = \frac{9.8}{9}

∴ T_{1} = 2 π \sqrt{\frac{l}{g_{e f f}}} = 2 π \sqrt{\frac{9 l}{g}}

\Rightarrow T_{1} = \sqrt{9} T = 3 T

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The bob of a simple pendulum performs SHM with period \(T\) in air and with period \(T_1\) in water. Relation between \(T\) and \(T_1\) is (neglect friction due to water, density of the material of the bob is \(\frac{9}{8} \times 10^3 kg / m ^3\), density of water is \(1 g / cc\) )

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