A) \[4\Omega \]
B) \[4.5\Omega \]
C) \[5\Omega \]
D) \[\frac{\sqrt{3}}{1}\]
Correct Answer: C
Solution :
The mid-point of line joining\[ds=\frac{t}{2}dt\]and\[F=ma=m\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{6{{d}^{2}}\left( \frac{{{t}^{2}}}{4} \right)}{d{{t}^{2}}}=3H\]is\[W=\int_{0}^{2}{Fds}=\int_{0}^{2}{3\frac{t}{2}}dt\], \[=\frac{3}{2}\left( \frac{{{t}^{2}}}{2} \right)_{0}^{2}=\frac{3}{2}[{{(2)}^{2}}-{{(0)}^{2}}]=3J\]Now, consider option (c) \[T=2\pi \sqrt{\frac{l}{g}}\] \[\therefore \] \[=\frac{1}{360}\] \[=\frac{24\times 60}{360}\min \]
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