First slide

Thermal expansion

Question

Span of a bridge is 2.4 km. At 30°C a cable along the span sags by 0.5 km. Taking a = 12 × $10^{- 6}$ /°C, change in length of cable for a change in temperature from 10°C to 42°C is

Easy

A

9.9 m
B

0.099 m
C

0.99 m
D

0.4 km

Solution

Span of bridge = 2400 m and bridge sags by 500 m at 30° (given)
From Fig., LPRQ = $2 \sqrt{1200^{2} + 500^{2}}$ = 2600 m
$\begin{array}{l} B u t L = L_{0} (1 + α Δ t) (D u e t o l i n e a r e x p a n s i o n) \\ \Rightarrow 2600 = L_{0} (l + 12 \times 10^{- 6} \times 30) \\ L e n g t h o f t h e c a b l e L_{0} = 2599 m \\ N o w c h a n g e i n l e n g t h o f c a b l e d u e t o c h a n g e i n t e m p e r a t u r e f r o m 10 ° C t o 42 ° C \\ Δ L = 2599 \times 12 \times 10^{- 6} \times (42 - 10) = 0.99 m \end{array}$

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