A metal block of mass m is suspended from a rigid support through a metal wire of diameter 14 mm. The tensile stress developed in the wire under equilibrium state is 7 × 10⁵ Nm⁻². The value of mass m is _________ kg. (Take, g=9.8 ms⁻² and π=22/7 )

Solution

Physics sir · 4 steps

Stress is just weight over area, so find the wire's cross-section and multiply back to get mg.

1. Tension equals weight. In equilibrium the wire pulls up with a tension equal to the block's weight, T = mg. The tensile stress is that force spread over the wire's cross-section.

   stress = mg/(π r²) ⇒ m = (stress × π r²)/g

2. Cross-section area. Diameter 14 mm gives radius r = 7 mm = 7 × 10⁻³ m. So π r² = 22/7 × 49 × 10⁻⁶ = 1.54 × 10⁻⁴ m².
3. Find the force. mg = 7 × 10⁵ × 1.54 × 10⁻⁴ = 107.8 N.
4. So. m = 107.8/9.8 = 11 kg, so the answer is 11.

The common slip: 44. Using the 14 mm diameter as the radius makes the area four times too big and gives 44 kg instead of 11.

The idea to keep. Stress is force divided by cross-sectional area. When a wire's size is given, check whether it is a diameter or a radius before you square it.

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