Question

Span of a bridge at ${10}^{\circ }\text{C}$ is $2.4\text{km}$ long. At $30{}^{\circ }\text{C}$, the cable along the span sags by $0.5\text{km}$. Change in length of the cable when temperature changes from ${10}^{\circ }\text{C}$ to ${42}^{\circ }\text{C}$ is
$[\alpha =12\times {10}^{-6}/{}^{\circ }\text{C}]$

• A. $0.758\text{m}$
• B. $0.853\text{m}$
• C. $0.998\text{m}$
• D. $0.637\text{m}$

Answer 1

The correct option is C $0.998\text{m}$

Given:
Span length of bridge,$L=2.4\text{km}=2400\text{m}$
Sag length of cable along the span at $30{}^{\circ }\text{C}$, $\Delta L=0.5\text{km}=500\text{m}$
Co-efficient of linear expansion of cable, $\alpha =12\times {10}^{-6}/{}^{\circ }\text{C}$
To find: Change in length of cable when the temperature changes from $10{}^{\circ }\text{C}$ to $42{}^{\circ }\text{C}$, $\left(\Delta L^{{}^{\prime }}\right)=?$


From figure,
Cable length at $30{}^{\circ }\text{C}$, $L=L_{PRQ}=2\sqrt{{1200}^2+{500}^2}=2600\text{m}$
Also, $L=L_0(1+\alpha \Delta T)$
[Due to linear expansion]
$\Rightarrow 2600=L_0(1+12\times {10}^{-6}\times (30-10\left)\right)$
$[L_o$ is cable length at $10{}^{\circ }\text{C}$ $]$
$\Rightarrow L_0=2599.376\text{m}$
Now, change in length of the cable due to change in temperature from $10{}^{\circ }\text{C}$ to $42{}^{\circ }\text{C}$
$\Delta L^{{}^{\prime }}=L_0\alpha \Delta T$
$=2599.376\times 12\times {10}^{-6}\times (42-10)=0.998\text{m}$