A metal wire of length 0.5 m and cross-sectional area 10⁻⁴ m² has breaking stress 5 × 10⁸ Nm⁻². A block of 10 kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _________ ms⁻¹.

Solution

Physics sir · 3 steps

The wire's breaking tension is the most centripetal force it can supply, so set mv²/r equal to it.

1. Maximum tension. Breaking stress is force per unit area, so the largest tension the wire can bear is stress times area. Tₘₐₓ = 5 × 10⁸ × 10⁻⁴ = 5 × 10⁴ N.
2. Tension gives centripetal force. The block moves in a horizontal circle whose radius is the wire length, r = 0.5 m. The tension alone pulls it towards the centre, so the speed is greatest when the tension reaches Tₘₐₓ.

   Tₘₐₓ = (m vₘₐₓ²)/r ⇒ vₘₐₓ² = (Tₘₐₓ r)/m = (5 × 10⁴ × 0.5)/10 = 2500

3. So. vₘₐₓ = √2500 = 50 m s⁻¹, so the answer is 50.

The common slip: 2500. Stopping at v² gives 2500, not 50. Take the square root at the end.

The idea to keep. Breaking stress times area gives the largest force a wire or string can take. In circular motion that force is the centripetal force, so equate it to mv²/r.

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