Question

Q.1 A wire can be broken by suspending a minimum load of 30 kg. The minimum mass required to break the wire of twice the diameter is

Answer 1

Answer:

60 kg is the answer

I hope this may help you

Answer 2

The minimum mass required to break the wire of twice the diameter is 120kg.

For, a wire radius 'r', the minimum breaking load is 30kgf (F). The breaking stress is represented as F/r².

The breaking stress is the same for the same material. So, it is equal to the wire with double the radius (2r) and with minimum breaking load with F'.

So, F/r² = F'/(2r)²

⇒ F/r² = F'/4r²

Replacing the value of F as 30.

⇒ F' = 30×4 = 120kgf

So, the minimum mass required is 120kg