Q.
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/s2. 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 = 10 m/s2)
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a
30 kg
b
32.5 kg
c
35 kg
d
37.5 kg
answer is B.
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Detailed Solution
If reading of spring balance is T, then applying NLM on (man + ladder) system T−(25+5)g=25aT−30g=25a⇒T−300=25(1)⇒T=325N=32.5kg
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