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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$1.$ The ratio of $x_1 / x_2$ is
$(A)$ $2$ $(B)$ $\frac{1}{2}$ $(C)$ $\sqrt{2}$ $(D)$ $\frac{1}{\sqrt{2}}$
$2.$ When disc $\mathrm{B}$ is brought in contact with disc $\mathrm{A}$, they acquire a common angular velocity in time $\mathrm{t}$. The average frictional torque on one disc by the other during this period is
$(A)$ $\frac{2 \mathrm{I} \omega}{3 \mathrm{t}}$ $(B)$ $\frac{9 \mathrm{I} \omega}{2 \mathrm{t}}$ $(C)$ $\frac{9 \mathrm{I} \omega}{4 \mathrm{t}}$ $(D)$ $\frac{3 \mathrm{I} \omega}{2 \mathrm{t}}$
$3.$ The loss of kinetic energy during the above process is
$(A)$ $\frac{\mathrm{I} \omega^2}{2}$ $(B)$ $\frac{\mathrm{I} \omega^2}{3}$ $(C)$ $\frac{\mathrm{I} \omega^2}{4}$ $(D)$ $\frac{\mathrm{I} \omega^2}{6}$
Give the answer question $1,2$ and $3.$
