First slide

Thermal expansion

Question

A metal rod has a length of 1m at 30°C. 'α' of metal is 2.5 ×10^–5/°C. The temperature at which it will be shortened by 1mm is

Easy

A

–30°C
B

–40°C
C

–10° C
D

10°C

Solution

$\large l=1m,\;t_1=30^0C,\;t_2=?$
$\large \alpha =2.5\times 10^{-5}/^0C,\;\Delta l=-1mm=-1\times 10^{-3}m$
we have $\large \alpha =\frac {\Delta l}{l \Delta t}\Rightarrow 2.5\times 10^{-5}=\frac {-1\times 10^{-3}}{(1)(t_2-30)}$
$\large t_2=-10^0C$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App