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Question

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 the spring balance is : $$\left( take\ g=10{ { m }/{ s } }^{ 2 } \right)$$
1262820_26b91aed785d4f4591d2a5f745bb60f7.png


A
30 kg
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B
32.5 kg
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C
35 kg
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D
37.5 kg
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Solution

The correct option is B $$32.5\ kg$$
Consider free body of the man 
$$N - 250 = 25(1)$$
$$N = 275 N $$ __(i)
Consider free body diagram of ladder
$$T = N + mg$$
$$T = 275 + 50 = 325 N$$
Reading of balance -
$$T = m'g$$
$$m' = 32.5 kg$$
Hence, option $$(B)$$ is correct answer.
1380606_1262820_ans_1c51749ada1444b9aac482733ef79000.png

Physics

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