A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂ (where m₂ > m₁). If the acceleration of the system is g/√(2), then the ratio of the masses (m₁)/(m₂) is: (a) (1+√(5))/(√(5)-1) (b) (√(3)+1)/(√(2)-1) (c) (√(2)-1)/(√(2)+1) (d) (1+√(5))/(√(2)-1)

Answer: (c). (m₂ - m₁)/(m₂ + m₁) = 1/√(2) Divide by m₂ with r = (m₁)/(m₂): (1 - r)/(1 + r) = 1/√(2) ⇒ √(2) - √(2)r = 1 + r. r = (√(2) - 1)/(√(2) + 1), so the answer is (c).

A light string passing over a smooth light pulley connects two blocks… | JEE Main 2024 (6 Apr, Shift 1)

Q14JEE Main 2024 · 6 APR · SHIFT 1Easy

Connected Bodies & Pulleys

A light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (where $m_2 > m_1$ ). If the acceleration of the system is $\frac{g}{\sqrt{2}}$ , then the ratio of the masses $\frac{m_1}{m_2}$ is:

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Answer: (c)

$\dfrac{m_2 - m_1}{m_2 + m_1} = \dfrac{1}{\sqrt{2}}$
Divide by $m_2$ with $r = \dfrac{m_1}{m_2}$ : $\dfrac{1 - r}{1 + r} = \dfrac{1}{\sqrt{2}} \Rightarrow \sqrt{2} - \sqrt{2}r = 1 + r$ .
$r = \dfrac{\sqrt{2} - 1}{\sqrt{2} + 1}$ , so the answer is (c).

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