Question

Span of a bridge is 2.4 km. At 30°C a cable along the span sags by 0.5 km. Taking $\alpha = 12 \times 10^{- 6}per^{o}C$, change in length of cable for a change in temperature from 10°C to 42⁰C is

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

Answer 1

Answer: (C)

0.99 m

Answer 2

Span of bridge = 2400 m and Bridge sags by 500 m at 30⁰ (given)

From the figure L_PRQ = $2\sqrt{1200^{2} + 500^{2}} = 2600m$

But $L = L_{0}(1 + \alpha\Delta t)$ [Due to linear expansion]

⇒ $2600 = L_{0}(1 + 12 \times 10^{- 6} \times 30)$ ∴ Length of the cable $L_{0} = 2599m$

Now change in length of cable due to change in temperature from 10⁰C to 42⁰C

$\Delta L = 2599 \times 12 \times 10^{- 6} \times (42 - 10)$ = 0.99m