Question

A pendulum that beats seconds on the surface of the earth were taken to a depth of $\left(1/4\right)$th the radius of the earth. Its time period of oscillation will be

• A. $2.3s$
• B. $2.8s$
• C. $1.9s$
• D. $1.7s$

Answer 1

Answer: (A)

Gravitational acceleration at a depth $d$ from the surface of earth is given by

$g^{\prime }=g(1-\frac{d}{R})$

here $d=\frac{R}{4}$

$⟹g^{\prime }=\frac{3g}{4}$

Time period of oscillation of pendulum is given as

$T=2\pi \sqrt{\frac{l}{g}}$

Hence $\frac{T^{\prime }}{T}=\sqrt{\frac{g}{g^{\prime }}}$$=\sqrt{\frac{4}{3}}$$\approx 1.154$

$\therefore T^{\prime }=1.154\times 2=2.3s$