Question

Span of a bridge is 2.4 km. At ${30}^{\circ }C$ a cable along the span sags by 0.5 km. Taking $\alpha =12\times {10}^{-6}pe{r}^{\circ }C,$ change in length of cable for a change in temperature from ${10}^{\circ }Cto{42}^{\circ }C$ is

• A. 9.9 m
• B. 0.099 m
• C. 0.99 m
• D. 0.4 km

Answer 1

The correct option is C 0.99 m

Span of bridge = 2400 m and Bridge sags by 500 m at ${30}^{\circ }$ (given)
From the figure $L_{PRO}=2\sqrt{{1200}^2+{500}^2}=2600mButL=L_0(1+\alpha \Delta t)\left[Duetolinearexpansion\right]\Rightarrow 2600=L_0(1+12\times {10}^{-6}\times 30)\therefore Lengthofthecable{L}_0=2599m$
Now change in length of cable due to change in temperature from ${10}^{\circ }Cto{42}^{\circ }C\Delta L=2599\times 12\times {10}^{-6}\times (42-10)=0.99m$