Figure shows a 5 kg ladder hanging from a string that is connected with a ceiling and is having a spring balance connected in between. A boy of mass 25 kg is climbing up the ladder at acceleration `1(m)/(s^2)`. Assuming the spring balance and the string to be massless and the spring to show a constant reading, the reading of thespring balance is: (Take`g=10(m)/(s^2)`)

Options

A. `30 kg`
B. `32.5 kg`
C. `35 kg`
D. `37.5 kg`

Answer

B

Explanation

If reading of spring balance is T, then applying NLM on (man+ ladder) system
`T-(25+5) g =25a`
`T-30 g=25a rArr T-300=25(1) rArr T=325 N=32.5 kg`.