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 (Take $g=10m/s^2$ )

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

Answer 1

Answer: (B)

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