Question
A bob of mass 10 kg is attached to wire 0.3 m long. Its breaking stress is 4.8 × 107 Nm2. The area of cross section of the wire is 10−6m2. The maximum angular velocity with which it can be rotated in a horizontal circle
Answer: Option B
:
B
Centripental force = breaking force
⇒ m ω2 r = Breaking stress × cross sectional area
⇒ m ω2 r = p × A ⇒ ω = √p×Amr = √4.8×107×10−610×0.3
∴ ω = 4 rad/sec
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:
B
Centripental force = breaking force
⇒ m ω2 r = Breaking stress × cross sectional area
⇒ m ω2 r = p × A ⇒ ω = √p×Amr = √4.8×107×10−610×0.3
∴ ω = 4 rad/sec
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