Question

the pulleys and strings shown in figure are smooth and of negligible mass . for the system to remain in equilibrium , the angle theta should be

Answer 1

45°The Tension in the string is mg as no acceleration T-mg=m(0) =>T=mg. Next the component of T i.e. Tcos(theta)Balance √2mg since both sides of the system are symmetrical then 2Tcos(theta)=√2mg =>2mgcos(theta)=√2mgCos(theta )=1/√2=>theta =45°

Answer 2

Answer:

45°

The Tension in the string is mg as there is no acceleration

T-mg=m(0)

=>T=mg. - ( 1 )

Next take the component of T along the string i.e. Tcos(theta) Balances √2mg since both sides of the system are symmetrical then

=>2Tcos(theta)=√2mg - ( 2 )

=>2mgcos(theta)=√2mg from (1) and ( 2 )

=>Cos(theta )=1/√2

=>theta =45°

Fyi - This question came in jee 2001