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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Mass of the body, m = 10\,\text{kg}
• Length of the wire, l = 0.3\,\text{m}
• Breaking stress of wire, \sigma = 4.8 \times 10^7\,\text{N}\,\text{m}^{-2}
• Cross-sectional area of wire, A = 10^{-2}\,\text{cm}^2 = 10^{-2} \times 10^{-4} = 10^{-6}\,\text{m}^2
Step 2: Write the Expression for Tension During Circular Motion
When the mass is rotated in a circle of radius l , the necessary centripetal force is provided by the tension T in the wire. The formula for tension is:
T = m \omega^2 \, l
where m is the mass, \omega is the angular speed, and l is the length (radius of circular path).
Step 3: Relate Tension to Breaking Stress
The maximum tension the wire can sustain is determined by its breaking stress. The breaking stress \sigma is given by:
\sigma = \frac{T}{A} \quad \Longrightarrow \quad T = \sigma \cdot A
Step 4: Find the Maximum Tension and Equate to Centripetal Force
The wire will break if the tension exceeds \sigma \cdot A . Therefore,
m \omega^2 l = \sigma A
Substitute the known values:
m = 10\,\text{kg}, \quad l = 0.3\,\text{m}, \quad \sigma = 4.8 \times 10^7\,\text{N}\,\text{m}^{-2}, \quad A = 10^{-6}\,\text{m}^2
So we have:
10 \,\omega^2 \,(0.3) = (4.8 \times 10^7)\,(10^{-6})
Step 5: Solve for Angular Speed
First compute the right-hand side:
(4.8 \times 10^7) \times (10^{-6}) = 48\,\text{N}
Therefore, the equation becomes:
3 \,\omega^2 = 48 \quad \Longrightarrow \quad \omega^2 = 16 \quad \Longrightarrow \quad \omega = 4 \,\text{rad\,s}^{-1}
Answer
The maximum angular speed with which the body can be rotated without breaking the wire is \boxed{4\text{ rad s}^{-1}} .
