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

When bob is immersed in liquid

mg_eff = mg $$-$$ Buoyant force

mg_eff = mg $$-$$ v$$\sigma$$g ($$\sigma$$ = density of liquid)

$$ = mg - v{\rho \over 4}g$$

$$ = mg - {{mg} \over 4} = {{3mg} \over 4}$$

$$\therefore$$ $${g_{eff}} = {{3g} \over 4}$$

$${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}} $$

$${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$$

By solving

$${T_1} = {4 \over 3}2\pi \sqrt {l/g} $$

$${T_1} = {{4T} \over 3}$$

Force on each column = $${{mg} \over 4}$$

Strain = $${{mg} \over {4AY}}$$

$$ = {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$$

= 2.6 $$\times$$ 10^$$-$$7

We know,

$$\Delta l = {{WL} \over {2AY}}$$

$$\Delta l = {{10 \times 1} \over {2 \times 5}} \times 100 \times {10^{ - 4}} \times 2 \times {10^{11}}$$

$$\Delta l = {1 \over 2} \times {10^{ - 9}} = 5 \times {10^{ - 10}}$$ m

Option (d)

Viscous force = Weight

$$ = \rho \times \left( {{4 \over 3}\pi {r^3}} \right)g$$

= 3.9 $$\times$$ 10^$$-$$10