Question

Two identical ideal springs of spring constant 1000 N/m as connected by an ideal pulley as shown and system is arranged in vertical plane. At equilibrium q is 60ŗ and masses m₁ and m₂ are 2kg and 3kg respectively. Then elongation in each spring when q is 60ŗ is –

• A. 1.6 $\sqrt{3}$ cm
• B. 1.6 cm
• C. 4.8 cm
• D. None of these

Answer 1

Answer: (A)

1.6 $\sqrt{3}$ cm

Answer 2

T = $\frac{2m_{1}m_{2}}{m_{1} + m_{2}}$g = $\frac{2 \times 2 \times 3}{5}$× 10= 24 N on pulley

2k x cos30⁰ = 2T

kx × $\frac{\sqrt{3}}{2}$= 24

1000 × x × $\frac{\sqrt{3}}{2}$= 24

x = 1.6 $\sqrt{3}$cm