First slide

Thermal expansion

Question

A metal rod has a length of 1m at $30^{0} C$ , $' α'$ of metal is $2.5 \times 10^{- 5} /^{0} C$ , the temperature at which it will be shortened by 1mm is

Moderate

A

$- 30^{0} C$
B

$- 40^{0} C$
C

$- 10^{0} C$
D

$10^{0} C$

Solution

$\begin{array}{rcl} c o e f f i c i e n t o f l i n e a r e x p a n s i o n o f s o l i d i s \\ α & = & \frac{Δ l}{l Δ t} \\ h e r e Δ l & = & - 1 m m = - 10^{- 3} m = d e c r e a s e i n l e n g t h \\ l i s i n i t i a l l e n g t h & = & 1 m; Δ t i s c h a n g e i n t e m p e r a t u r e; \\ i n i t i a l t e m p e r a t u r e & = & 30^{0} C; α = 2.5 x 10^{- 5} / {}^{0}C \\ s u b s t i t u t e g i v e n v a l u e s i n a b o v e e q u a t i o n \\ 2.5 \times 10^{- 5} & = & \frac{- 10^{- 3}}{1 \times (t_{2} - 30)} \\ t_{2} - 30 & = & \frac{- 1}{2.5 \times 10^{- 2}} = \frac{- 100}{2.5} \\ t_{2} - 30 & = & - 40 \\ t_{2} & = & - 40 + 30 = - 10^{0} C = t e m p e r a t u r e a t w h i c h r o d w i l l b e s h o r t e n e d \end{array}$

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