MHT CET · Physics · Laws of Motion
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring balance reads \(49 \mathrm{~N}\), when the lift is stationary. If the lift moves downward with an acceleration of \(5 \mathrm{~m} / \mathrm{s}^2\), the reading of the spring balance will be \(\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)\)
- A \(15 \mathrm{~N}\)
- B \(24 \mathrm{~N}\)
- C \(49 \mathrm{~N}\)
- D \(74 \mathrm{~N}\)
Answer & Solution
Correct Answer
(B) \(24 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
For downward accelration of the lift,
\(\mathrm{Mg}\mathrm{T}=\mathrm{ma}\)
\(\therefore \quad \mathrm{T}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) ....(i)
\(\mathrm{W}=\mathrm{mg}\)
\(\therefore \quad \mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}=\frac{49}{9.8}\) ....(ii)
Put (ii) into (i)
\(\mathrm{T}=\frac{49}{9.8}(9.8-5)\)
\(=5(4.8)=24 \mathrm{~N}\)
\(\mathrm{Mg}\mathrm{T}=\mathrm{ma}\)
\(\therefore \quad \mathrm{T}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) ....(i)
\(\mathrm{W}=\mathrm{mg}\)
\(\therefore \quad \mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}=\frac{49}{9.8}\) ....(ii)
Put (ii) into (i)
\(\mathrm{T}=\frac{49}{9.8}(9.8-5)\)
\(=5(4.8)=24 \mathrm{~N}\)
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