Two identical ideal springs of spring constant 1000 N/m are connected by an ideal pulley as shown and system is arranged in vertical plane. At equilibrium θ is 6 0 ∘ and masses m 1 and m 2 are 2 k

Question

Two identical ideal springs of spring constant 1000 N/m are connected by an ideal pulley as shown and system is arranged in vertical plane. At equilibrium $θ$ is $6 0^{\circ}$ and masses $m_{1}$ and $m_{2}$ are 2 kg and 3 kg respectively. The elongation in each spring when $θ$ is $6 0^{\circ}$ is p cm. The value of $\frac{3 p}{1.6}$ is :

TestHub · Accepted Answer

2 k x sin θ = 2 T = ( m 1 ​ + m 2 ​ 4 m 1 ​ m 2 ​ ​ ) g k x 2 3 ​ ​ 1000 × 2 x ​ × 3 ​ 2 x ​ 2 x ​ 2 P ​ 1.6 3 ​ P ​ ​ = 5 2 × 2 × 3 ​ × 10 = 8 × 3 = 125 3 ​ ​ × 100 c m = 5 3 ​ ​ × 4 = 2 x ​ = 3 ​ × 0.8 = 3 ​

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