Correct Answer is Option (C)

$\begin{array}{rl}& {\mathrm{T}}_{\text{air }}=2\pi \sqrt{\frac{\ell }{\mathrm{g}}}=\sqrt{3}\sec \\ & \ln \text{ Liquid }\to g_{\text{net }}=g(1-\frac{\rho }{\sigma })\\ & \rho =\text{ density of liquid }\\ & \sigma =\text{ density of material of bob }\\ & \text{ so }{\mathrm{T}}_{\text{Liq }}=2\pi \sqrt{\left(\frac{\ell }{g_{\text{net }}}\right)}=2\pi \sqrt{\frac{\ell }{g(1-\frac{\rho }{\sigma })}}\\ & {\mathrm{T}}_{\text{Liq }}=\frac{{\mathrm{T}}_{\text{air }}}{\sqrt{1-\frac{\rho }{\sigma }}}=\frac{\sqrt{3}}{\sqrt{1-\frac{1}{4}}}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=2\sec \end{array}$

Correct Answer is Option (A)

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

Let n₁ and n₂ be integer.

$n_1{T}_1=n_2{T}_2$

$2\pi n_1\sqrt{\frac{1.21}{g}}=2\pi n_2\sqrt{\frac{1.00}{g}}$

$\Rightarrow \frac{n_2}{n_1}=\frac{11}{10}$

$\therefore$ After completion of 11^th oscillation of shorter pendulum, it will be in phase with longer pendulum.

Correct Answer is Option (A)

In $\mathrm{\Delta }OAC,\cos \theta =A/l$

$\Rightarrow OA=lcos\theta$

$\therefore$ $AB=l\left(1-\cos \theta \right)=h$

At point, C the velocity of bob = 0. The vertical acceleration = g

$\therefore$ $v^2=2gh$

$\Rightarrow v=\sqrt{2gl\left(1-\cos \theta \right)}$

Correct Answer is Option (A)

$f_A=2f_B$

$\Rightarrow \frac{1}{2\pi }\sqrt{\frac{g}{l_A}}=2\times \frac{1}{2\pi }\sqrt{\frac{g}{l_B}}$

$\Rightarrow \frac{l}{l_A}=4\times \frac{l}{l_B}$

$\Rightarrow l_A=\frac{l_B}{4}$ , which does not depend on mass.