A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitudes of tensions T₁ and T₂, respectively, are (in N ) :
(Take acceleration due to gravity 10 m / s² )

Solution

Physics sir · 4 steps

Resolve both tensions along horizontal and vertical: sideways pulls cancel, upward pulls hold the 10 N weight.

1. Forces on the block. Three forces act: T₁ at 60° above the horizontal (left), T₂ at 30° above the horizontal (right), and the weight mg = 1 × 10 = 10 N downward. The block is at rest, so the net force is zero.
2. Horizontal balance. The sideways pulls must cancel, so T₁ cos 60° = T₂ cos 30°. That is T₁/2 = (√3 T₂)/2, which gives T₁ = √3 T₂.

   T₁ sin 60° + T₂ sin 30° = mg

3. Vertical balance. Put T₁ = √3 T₂ into this: √3 T₂ × √3/2 + T₂/2 = 3T₂/2 + T₂/2 = 2T₂ = 10. So T₂ = 5 N and T₁ = 5√3 N.
4. So. T₁ = 5√3 N and T₂ = 5 N, which is option (a). The steeper string carries the larger tension.

The idea to keep. For a body hanging from two strings, write one balance equation for horizontal and one for vertical. The string closer to vertical always carries more of the weight.

Answer from the NTA final answer key. Solution by PrepWiser.