Published by:
CGP EDU Academic Team
Published on: September 12, 2026
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 θ is 60º and masses m 1 and m 2 are 2kg and 3kg respectively. Then elongation in each spring when θ is 60º is –
A
1.6 
cm
cm
B
1.6 cm
C
4.8 cm
D
None of these
Text Solution
Verified by ExpertsThe correct answer is:
A
T = 
g = 
× 10= 24 N on pulley
2k x cos30 0 = 2T
kx × 
= 24
1000 × x × 
= 24
x = 1.6 
cm
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