A) \[t={{t}_{0}}\]
B) \[t={{t}_{0}}/2\]
C) \[t=2{{t}_{0}}\]
D) \[t=4{{t}_{0}}\]
Correct Answer: C
Solution :
The time period of simple pendulum in air \[T={{t}_{0}}=2\pi \sqrt{\left( \frac{l}{g} \right)}\] ?.(i) \[l,\]being the length of simple pendulum. In water, effective weight of bob \[w'=\]weight of bob in air - upthrust \[\Rightarrow \] \[\rho V{{g}_{eff}}=mg-m'g\] \[=\rho Vg-\rho 'Vg=(\rho -\rho ')Vg\] where\[\rho '=\]density of bob, \[\rho =\]density of water. \[\therefore \] \[{{g}_{eff}}=\left( \frac{\rho -\rho '}{\rho } \right)g=\left( 1-\frac{\rho '}{\rho } \right)g\] \[\therefore \] \[t=2\pi \sqrt{\left[ \frac{l}{\left( 1-\frac{\rho '}{\rho } \right)g} \right]}\] ?..(ii) Thus, \[\frac{t}{{{t}_{0}}}=\sqrt{\left[ \frac{1}{\left( 1-\frac{\rho '}{\rho } \right)} \right]}\] [dividing Eq. (ii) by Eq. (i)] \[=\sqrt{\left( \frac{1}{1-\frac{1000}{(4/3)\times 1000}} \right)}=\sqrt{\left( \frac{4}{4-3} \right)}\] \[=2\Rightarrow t=2{{t}_{0}}\]
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