Question

A simple pendulum with a solid metal bob has a period T. The metal bob is now immersed in a liquid of density one tenth that of the bob. The liquid is non viscous. Now the period of the same pendulum with its bob remaining all the time in the liquid will be

• A. t
• B. (9/10)T
• C. $\sqrt { 10 / 9 }$T
• D. $\sqrt { 9 / 10 }$ T

Answer 1

Answer: (C)

$\sqrt { 10 / 9 }$T

Answer 2

Apparent acceleration due to gravity because of force of buoyancy

g' = g $\left( 1 - \frac { \rho _ { \mathrm { i } } } { \rho _ { \mathrm { s } } } \right)$ ρ_i = ρ_s/10 ⇒ g' = g $\left( 1 - \frac { 1 } { 10 } \right)$ = $\frac { 9 } { 10 }$g T ∝ $\frac { 1 } { \sqrt { g } }$ ⇒ T' = $\sqrt { \frac { 10 } { 9 } }$ T Hence Answer is (C)