Home
Class 12
PHYSICS
A block of weight 100 N is suspended by ...

A block of weight 100 N is suspended by copper and steel wires of same cross sectional area `0.5 cm^(2)` and, length `sqrt(3)` m and 1m, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are `30^(@) and 60^(@)`, respectively. If elongation in copper wire is `(Delta l_(C))` and elongation in steel wire is `(Delta l_(s))`, then the ratio `(Delta l_(C))/(Delta l_(s))` is -

[Young's modulus for copper and steel are `1 xx 10^(11) N//m^(2) and 2 xx 10^(11) N//m^(2)`, respectively]

Text Solution

Verified by Experts

The correct Answer is:
`2.00`

`T_(1)sin60^(@)=T_(2)sin30^(@)rArrT_(2)=sqrt(3)T_(1)`
`T_(1)cos60^(@)+T_(2)cos30^(@)=mg`
`(T_(1))/(2)+sqrt(3)T_(1)(sqrt(3))/(2)=100rArr2T_(1)=100rArrT_(1)=50N`
`T_(2)=sqrt(3)T_(1)=50sqrt(3)N`
`Deltal=(Tl)/(AY)therefore(DeltalC)/(DeltalS)=(T_(1))/(T_(2))=(l_(1))/(l_(2))(Y_(2))/(Y_(1))=((1)/(sqrt(3))(sqrt(3))/(1))(2)=2`
Promotional Banner