The scale of a spring balance reading from $0$ to $10 \,kg$ is $0.25\, m$ long. A body suspended from the balance oscillates vertically with a period of $\pi /10$ second. The mass suspended is ..... $kg$ (neglect the mass of the spring)
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(b) Using $F = kx$
$ \Rightarrow 10g = k \times 0.25$
$\Rightarrow k = \frac{{10g}}{{0.25}} = 98 \times 4$
Now $T = 2\pi \sqrt {\frac{m}{k}}$
$\Rightarrow m = \frac{{{T^2}}}{{4{\pi ^2}}}k$
$ \Rightarrow m = \frac{{{\pi ^2}}}{{100}} \times \frac{1}{{4{\pi ^2}}} \times 98 \times 4 = 0.98\;kg$
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