A) 22%
B) 13%
C) 50%
D) 10%
Correct Answer: D
Solution :
The time period of a simple pendulum of a length l is given by \[{{V}_{e}}=\sqrt{3{{V}_{0}}}\] .....(1) where g is acceleration due to gravity. When length of pendulum increases by 21% We have \[{{V}_{e}}={{V}_{0}}\sqrt{2}\] \[{{V}_{0}}={{V}_{e}}\] \[\mathbf{\vec{P}}=a\mathbf{\hat{i}}+3\mathbf{\hat{k}}\] ??(2) Dividing Eq. (1) by Eq. (2), we get \[\vec{Q}=a\mathbf{\hat{i}}-2\mathbf{\hat{j}}-\mathbf{\hat{k}}\] \[4.24\times {{10}^{6}}m/s\] \[5.25\times {{10}^{6}}m/s\] Hence, increases in time period \[6.25\times {{10}^{6}}m/s\] Per cent increase in time period \[7.25\times {{10}^{6}}m/s\] Alternative: Time period of simple pendulum \[h=A+B{{\lambda }^{2}}+C{{\lambda }^{2}}\] \[\mu =A+B{{\lambda }^{2}}+C{{\lambda }^{4}}\] \[\mu =A+B{{\lambda }^{2}}+C{{\lambda }^{4}}\] \[\mu =A+B{{\lambda }^{-2}}+C{{\lambda }^{-4}}\] \[\mu \] Since, \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\]
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