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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{	ext{eff}}}}.$ In air,the effective acceleration due to gravity is $

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$4T\sqrt{\frac{1}{15}}$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{	ext{eff}}}}.$ In air,the effective acceleration due to gravity is $g_{	ext{eff}} = g.$ So,$T = 2\pi \sqrt{\frac{L}{g}}.$ When the bob is immersed in a liquid of density $ho_l$ and the bob has density $ho_b,$ the effective acceleration due to gravity $g'$ is given by $g' = g \left( 1 - \frac{ho_l}{ho_b} ight).$ Given $ho_l = \frac{1}{16} ho_b,$ we have $g' = g \left( 1 - \frac{1}{16} ight) = g \left( \frac{15}{16} ight).$ The new time period $T'$ is $T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g \cdot \frac{15}{16}}}.$ Simplifying this,$T' = 2\pi \sqrt{\frac{L}{g}} \cdot \sqrt{\frac{16}{15}} = T \cdot \frac{4}{\sqrt{15}} = 4T \sqrt{\frac{1}{15}}.$

$A$ simple pendulum oscillating in air has period $T.$ The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac{1}{16}$ of the material of the bob. If the bob is inside the liquid all the time,its period of oscillation in this liquid is:

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