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 (neglect the mass of the spring)

Options

A. 20 kg
B. 10 kg
C. 0.98 kg
D. 5 kg

Answer

C

Explanation

The extension is 0.25 m for a load of 10 kg wt
`therefore K=(F)/(x)=(10xx g)/(0.25) =40xx9.8=98xx 4 N//m`
for the oscillating spring,
`T=2pi sqrt((m)/(K)) therefore T^(2)=(4pi^(2)m)/(K)`
`therefore m =(T^(2)K)/(4pi^(2))=(pi^(2))/(100)xx(98xx4)/(4 pi^(2))=(98)/(100) =0.98 kg `