A spring balance having a scale of length 10 cm is used to measure weights from 0 to 5 kg. When a small stone of mass m is suspended from the spring, the spring performs 50 vertical oscillation in 10 second. What is the mass of the stone ? [ use `g=10 m//s^(2) " and " pi^(2)=10]`.

Options

A. 100 gram
B. 250 gram
C. 500 gram
D. 750 gram

Answer

C

Explanation

When a weight of 5 k g is attached to the spring, its displacement =10 cm (on the scale)
`therefore` Force constant `(K)=(F)/(L)=(mg)/(L)`
`therefore K=(5xx10)/(0.1)=500 N//m`
and the frequency of vertical oscillations of the spring (n) `=(50)/(10)=5`
n is given by
`n=(1)/(2pi) sqrt((K)/(m)) therefore n^(2)=(1)/(4pi^(2))-(K)/(m)`
`therefore m=(K)/(4pi^(2)n^(2))=(500)/(4xx10xx25)=(1)/(2)=500` grams