A bob of mass m suspended by a thread of length ℓ undergoes simple...

A bob of mass

m

suspended by a thread of length

ℓ

undergoes simple harmonic oscillations with time period

T .

If the bob is immersed in a liquid that has density

\dfrac{1}{4}

times that of the bob and the length of the thread is increased by ${\left(\dfrac{1}{3}\right)}^{rd} of the original length, then the time period of the simple harmonic oscillations will be:

Options

A $\dfrac{3}{4}T$
B $\dfrac{4}{3}T$
C $T$
D $\dfrac{3}{2}T$

Correct answer

B. $\dfrac{4}{3}T$

Step-by-step solution

T = 2\pi\sqrt{\dfrac{ℓ}{g}}

{g}_{eff} = g - \dfrac{g}{4} = \dfrac{3g}{4} ∴ T ' = 2\pi\sqrt{\dfrac{4ℓ / 3}{3g / 4}} = 2\pi\sqrt{\dfrac{16ℓ}{9g}}

T ' = \dfrac{4}{3}T

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