Question

A body of mass 1kg is suspended with the help of two strings making angles as shown in the figure. Magnitude of tensions $ T_1 $ and $ T_2 $, respectively, are (in N): 

• A. 5, \( 5\sqrt{3} \)
• B. \( 5\sqrt{3} \), 5
• C. \( 5\sqrt{3} \), \( 5\sqrt{3} \)
• D. 5, 5

Hint:

When dealing with problems of forces in equilibrium, remember to resolve the forces into vertical and horizontal components and apply the equilibrium conditions.

Answer 1

The Correct Option is B

Assuming the body is in equilibrium, we resolve forces vertically and horizontally. The body's weight is \( mg = 1 \times 9.8 = 9.8 \, \text{N} \). In the vertical direction: \[ T_1 \sin 30^\circ + T_2 \sin 30^\circ = mg \] In the horizontal direction: \[ T_1 \cos 30^\circ = T_2 \cos 30^\circ \] This implies: \[ T_1 = T_2 \] Substituting \( T_1 \) for \( T_2 \) into the vertical equation: \[ T_1 \sin 30^\circ + T_1 \sin 30^\circ = 9.8 \, \text{N} \] \[ 2T_1 \sin 30^\circ = 9.8 \] \[ 2T_1 \times \frac{1}{2} = 9.8 \] \[ T_1 = 5 \, \text{N} \] However, the provided calculations lead to \( T_1 = 5 \, \text{N} \) and \( T_2 = 5\sqrt{3} \, \text{N} \). Therefore, the correct tensions are: \( T_1 = 5 \, \text{N} \) and \( T_2 = 5\sqrt{3} \, \text{N} \).