Answer:
(a) When two wires of same size are suspended in parallel, a force F equal to the breaking force will act on each wire if a breaking force of 2F is applied on the parallel combination. (b) \[F=\frac{\Upsilon A\Delta l}{l}=\frac{\Upsilon .\pi {{r}^{2}}\Delta l}{l}\] i.e., \[F\propto {{r}^{2}}\] Thus for a single wire of double the thickness, the breaking force will be 4F.
You need to login to perform this action.
You will be redirected in
3 sec