Span of bridge is 2.4 km. At `30^@C` a cable along the span sags by 0.5 km. Taking `alpha=12xx10^(-6)//^@C`, change in length of cable for a change in temperature from `10^@C` to `42^@C` is

Options

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

Answer

C

Explanation

Span bridge`=2400m` and bridge sags by 500 m at `30^@` (given)
From Fig. `L_(PRQ)=2sqrt(1200^2+500^2)=2600m`
(due to linear expansion)
`implies2600=L_0(1+12xx10^-6xx30)`
`:.` Length of the cable `L_0=2599m`
Now change in Length of cable due to change in temperaure from `10^@C` to `42^@C`
`DeltaL=2599xx12xx10^-6xx(42-10)=0.99m`