A) 10%
B) 21%
C) 30%
D) 50%
Correct Answer: A
Solution :
If initial length \[{{l}_{1}}=100\] then \[{{l}_{2}}=121\] By using \[T=2\pi \sqrt{\frac{l}{g}}\Rightarrow \frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{l}_{1}}}{{{l}_{2}}}}\] Hence, \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{100}{121}}\Rightarrow {{T}_{2}}=1.1\,{{T}_{1}}\] % increase = \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}\times 100=10\,%\]
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