A) \[{{t}_{{}}}={{t}_{0}}\]
B) \[{{t}_{{}}}=4{{t}_{0}}\]
C) \[{{t}_{{}}}=2{{t}_{0}}\]
D) \[{{t}_{{}}}={{t}_{0}}/2\]
Correct Answer: C
Solution :
[c] in air, the time period of the bob is \[{{t}_{0}}=2\pi \sqrt{\frac{l}{g}}\] In water apparent weight \[m{g}'=\frac{4}{3}\pi {{r}^{3}}{{\rho }_{s}}g-\frac{4}{3}\pi {{r}^{3}}{{\rho }_{w}}g\] \[m{g}'=\frac{4}{3}\pi {{r}^{3}}\rho g\left( 1-\frac{{{\rho }_{w}}}{{{\rho }_{s}}} \right)=mg\left( 1-\frac{3}{4} \right)\frac{mg}{4}\Rightarrow \text{ }{g}'=g/4\] In water, the time period of the bob \[t=2\pi \sqrt{\frac{1}{g'}}=2\pi \sqrt{\frac{4l}{g}}=2{{t}_{0}}\]
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