The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $θ$ should be [2001]

Question

Smart Achievers · Accepted Answer

(3)

The tension in both strings will be same due to symmetry.

For equilibrium

$\sqrt{2} m g = T \cos θ + T \cos θ = 2 T \cos θ$

$∴$ $\sqrt{2} m g = 2 (m g) \cos θ$

$∴$ $\cos θ = \frac{1}{\sqrt{2}} \Rightarrow θ = 45 °$

Solution

Q.
The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $θ$ should be [2001]

1 $0^{o}$

2 $30^{o}$

3 $45^{o}$

4 $60^{o}$

Ans.
(3)

The tension in both strings will be same due to symmetry.

For equilibrium

$\sqrt{2} m g = T \cos θ + T \cos θ = 2 T \cos θ$

$∴$ $\sqrt{2} m g = 2 (m g) \cos θ$

$∴$ $\cos θ = \frac{1}{\sqrt{2}} \Rightarrow θ = 45 °$

Want Us to Email you About Special Offers & Updates?

Solution

Q. The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be [2001]

1 0o

2 30o

3 45o

4 60o

Ans. (3) The tension in both strings will be same due to symmetry. For equilibrium 2 mg=Tcosθ+Tcosθ=2Tcosθ ∴ 2 mg=2(mg)cosθ ∴ cosθ=12 ⇒ θ=45°