Question

A thin metallic square plate has length l when it is heated from 0C to 100C it's length increases by 1%what is the%increase in the area...

Answer 1

Given,

$\bold{\frac{\Delta l}{l}\times100 = 1}$

$\bold{\frac{\Delta l}{l}=\alpha\Delta t=\frac{1}{100}=0.01}$

$\bold{\frac{\Delta A}{A}=2\alpha\Delta t=2\times0.01=0.02}}$

$\bold{\frac{\Delta A}{A}\times100=2\%}$

Area,

$\bold{A=l\times b=2l^2}$

$\bold{\frac{\Delta A}{A}=\frac{2\Delta l}{l} = 2\%}$

$\boxed{\bold{\therefore{Increase\:in\:Area=2\%}}}$