A spring balance has a scale that reads from 0 to 50 kg. the length of the scale is 20 cm. a block of mass m is suspended from this balance, when displaced annd released, it oscillates with a period 0.5 s. the value of m is (Take `g=10ms^(-2)`)

Options

A. 8 kg
B. 12 kg
C. 16 kg
D. 20 kg

Answer

C

Explanation

The 20 cm length of the scaler reads upto 50 kg.
`thereforeF=mg=(50kg)(10ms^(-2))=500N`
and `x=20cm=0.2m`
`therefore`Spring constant, `k=(F)/(x)=(500N)/(0.2m)=2500Nm^(-1)`
As `T=2pisqrt((m)/(k))`
Squaring both sides, we get
`T^(2)=(4pi^(2)m)/(k)`
`m=(T^(2)k)/(4pi^(2))=((0.5)^(2)xx(2500Nm^(-1)))/(4xx(3.14)^(2))=16kg`