Question

Two wires made of same material are subjected to forces in the ratio 1:4.Their lengths are in ratio 2:1 and diameters in the ratio 1:3.what is the ratio of their ex-tensions ?​

Answer 1

Given :

• Their forces are in the ratio =1:4

•Their lengths are in the ratio =2:1

•Their diameters in the ratio =1:3

To Find :

•The ratio of their extension =?

Formulas :

$\bullet\bf{A= \pi r^{2} }$

$\\ \bullet\bf{Young \: Modulus (Y) =\dfrac{Fl }{A \delta l} }$

Solution :

By using formulas :

$\\ \implies\sf{Young \: Modulus (Y) =\dfrac{Fl}{\pi r^{2} \delta l} }$

$\\ \implies\sf{or \: \delta l=\dfrac{Fl}{\pi r^{2} Y} }$

$\\ \implies\sf{or \: \delta l \: is \: proportional \: to \: \dfrac{Fl}{r^{2} } }$

>> (Y is same for two wires)

$\\ \implies\sf{\therefore \dfrac{\delta l_{1} }{\delta l_{2} }=\dfrac{F_{1} }{F_{2} } \times \dfrac{l_{1} }{l_{2} } \times \bigg(\dfrac{r_{2} }{r_{1} } \bigg)^{2} }$

$\\ \implies\sf{\dfrac{1}{4} \times \dfrac{2}{1} \times \bigg(\dfrac{3}{1} \bigg) ^{2} }$

$\\ \implies\boxed{\sf{\dfrac{9}{2} }}$

$\\ \implies\sf{ \dfrac{\delta l_{1} }{\delta l_{2} }=\dfrac{9}{2} }$

$\\ \implies\boxed{\bf{\dfrac{\delta l_{1} }{\delta l_{2} }=\dfrac{9}{2} }}$

Their extension are in the ratio 9:2 .