A) \[T\]
B) \[\frac{10}{9}T\]
C) \[\sqrt{\left( \frac{9}{10} \right)}T\]
D) \[\sqrt{\left( \frac{10}{9} \right)}T\]
Correct Answer: D
Solution :
The time period pendulum in air \[T=2\pi \sqrt{\frac{l}{g}}\] ?(i) \[l\]being the length of simple pendulum. In liquid, effective weight of sphere \[w'=\]weight of bob in air - up thrust \[\Rightarrow \] \[\rho v{{g}_{ff}}=mg-m'g\] \[=\rho vg=\rho 'vg=(\rho -\rho ')vg\] where\[\rho '=\]density of sphere \[\rho =\]density of liquid \[\therefore \] \[{{g}_{eff}}=\left( \frac{\rho -\rho /10}{\rho } \right)g=\frac{9}{10}g\] Thus, \[T'=2\pi \sqrt{\frac{l}{{{g}_{eff}}}}=2\pi \sqrt{\frac{l}{\frac{9}{10}g}}\] \[\frac{T'}{T}=\sqrt{\frac{10}{9}}\] or \[T'=\sqrt{\frac{10}{9}}\,T\]
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