Question

A steel tape is calibrated at ${20}^{\circ }C$. It is used to measure a length of an object on a hot summer day when the temperature is ${40}^{\circ }C$. If measured length is 5 m, the actual length is($a=1\times {10}^{-5}{/}^{\circ }C$ )

• A. 5.001m
• B. 5.01 m
• C. 5.1 m
• D. 4.99 m

Answer 1

The correct option is A 5.001m

The scale is calibrated at ${20}^0$C
At ${40}^0$C , the measuring gaps will be more. So, it will measure lesser than the actual.
Using, $l_2=l_1(1+\alpha dt)$ where $\alpha ={10}^{-5}$, dt = 20 and $l_1$ = 5, we get $l_2$ = 5.001m