A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m. The maximum angular speed (in rad s –1 ) with which it can be rotated about its other end in space station is :
(Breaking stress of wire = 4.8 × 10⁷ Nm –2 and
area of cross-section of the wire = 10 –2 cm²) is:

Solution

Physics sir · 4 steps

In a space station there is no weight to worry about, so the wire's tension alone supplies mω² r.

1. Breaking force. Convert the area first: 10⁻² cm² = 10⁻² × 10⁻⁴ m² = 10⁻⁶ m². The largest tension the wire can bear is stress times area, 4.8 × 10⁷ × 10⁻⁶ = 48 N.
2. Only tension acts. Inside the space station the body is effectively weightless, so the tension is the only force providing the centripetal pull. It must equal mω² L.

   mωₘₐₓ² L = 48 ⇒ 10 × ωₘₐₓ² × 0.3 = 48

3. Solve. 3 ωₘₐₓ² = 48, so ωₘₐₓ² = 16.
4. So. ωₘₐₓ = 4 rad s⁻¹, so the answer is 4.

The common slip: 40. Converting 1 cm² as 10⁻² m² instead of 10⁻⁴ m² makes the breaking force 4800 N and gives 40, not 4.

The idea to keep. Breaking stress times area gives the maximum tension. Set that equal to mω² r, and add gravity only when the body actually has weight in that frame.

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