A $100 \,g$ mass stretches a particular spring by $9.8 \,cm$, when suspended vertically from it. ....... $g$ large a mass must be attached to the spring if the period of vibration is to be $6.28 \,s$.
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(d)
At point of equilibrium $k x=m g$
$k \times 9.8 \times 10^{-2}=100 \times 10^{-3} \times 9.8$
$k=100 \times 10^{-1}$
$k=10 \,N / m$
Period of vibration needed $=6.28 \,s$
$T=2 \pi \sqrt{\frac{m}{k}}$
$6.28=2 \times 3.14 \sqrt{\frac{m}{10}}$
$1=\frac{m}{10}$
$m=10 \,kg$ or $10^4 \,g$
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