Question

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

• A. $30kg$
• B. $32.5kg$
• C. $35kg$
• D. $37.5kg$

Answer 1

The correct option is B $32.5kg$

Consider free body of the man

$N-250=25\left(1\right)$

$N=275N$ __(i)

Consider free body diagram of ladder

$T=N+mg$

$T=275+50=325N$

Reading of balance -

$T=m^{\prime }g$

$m^{\prime }=32.5kg$

Hence, option $\left(B\right)$ is correct answer.