When a person stands on a weighing balance, working on the principle of Hooke's law, it shows a reading of `60kg` after a long time and the spring gets compressed by `2.5cm`. If the person jumps on the balance from a height of `10cm`, the maximum reading of the balance will be

Options

A. (a) `60kg`
B. (b) `120kg`
C. (c) `180kg`
D. (d) `240kg`

Answer

D

Explanation

Initially, `60g=kx=k(2.5)`
Let `x^'` be the maximum compression when
the person jumps on the balance, then `1/2kx^('2)=60g(x^'+10)`
`implies1/2[(60g)/(2.5)]x^('2)=60g(x^'+10)`
`impliesx^('2)=5x^'+50`
`impliesx^('2)-5x^'-50=0`
Solving for `x^'`, we get `x^'=10cm`
If `mkg` is the reading, then
`mg=k(10)` (ii)
From Eqs. (i) and (ii), we get `m=240kg`