A) 2 s
B) 1.2 s
C) 3s
D) \[\sqrt{3}\,s\]
Correct Answer: D
Solution :
As \[T=2\pi \sqrt{\frac{l}{g}}\] So \[2\pi \sqrt{\frac{{{l}_{1}}}{g}}=2\Rightarrow {{l}_{1}}=\frac{g}{{{\pi }^{2}}}\] and \[2\pi \sqrt{\frac{{{l}_{2}}}{g}}=1\Rightarrow {{l}_{2}}=\frac{g}{4{{\pi }^{2}}}\] Now, \[{{l}_{1}}-{{l}_{2}}=\frac{g}{{{\pi }^{2}}}\left( 1-\frac{1}{4} \right)=\frac{3}{4}\frac{g}{{{\pi }^{2}}}\] Hence, \[T=2\pi \sqrt{\frac{{{l}_{1}}-{{l}_{2}}}{g}}=2\pi \sqrt{\frac{3g}{4{{\pi }^{2}}\times g}}=\sqrt{3}s\]
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