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A simple pendulum of mass $200\, gm$ and length $100\, cm$ is moved aside till the string makes an angle of $60^o$ with the vertical. The kinetic and potential energies of the bob, when the string is inclined at $30^o$ to the vertical, are
$7.174 \times 10^6\, erg,\, 2.626 \times 10^6\, erg$
$8.2 \times 10^6\, erg,\, 2.2 \times 10^6 \,erg$
$2.6 \times 10^6 \,erg,\, 5.6 \times 10^6\, erg$
$3.6 \times 10^6\, erg,\, 6.2 \times 10^6\, erg$
Solution
Reference level
(Here $\mathrm{m}=200 \mathrm{gm}$
$\mathrm{L}=100 \mathrm{cm}=1 \mathrm{m}$
$\left.g=10 m / \sec ^{2}\right)$
$\mathrm{W}=\Delta \mathrm{k}$
$mgL\left( {\cos {{30}^\circ } – \cos {{60}^\circ }} \right) = K{E_f} – K{E_i}$
${\rm{K}}.{{\rm{E}}_f} = {\rm{mgL}}\left( {\cos {{30}^\circ } – \cos {{60}^\circ }} \right)$
$=0.732 \mathrm{J}$
$=7.32 \times 10^{6} \mathrm{erg}$
$\mathrm{PE}=\mathrm{mgL}\left(1-\cos 30^{\circ}\right)$
$=0.268 \mathrm{J}$
$=2.68 \times 10^{6} \mathrm{erg}$
