Question

A brass rod at ${30^ \circ}\,C$ is observed to be $100\,cm$ long when measured by a steel scale which is correct at ${0^ \circ}\,C$. $\alpha$ for steel is $12 \times {10^{ - 6}}{\,^ \circ}{C^{ - 1}}$ and $\alpha$ for brass is $19 \times {10^{ - 6}}{\,^ \circ }{C^{ - 1}}$. The correct length of brass rod at ${0^ \circ}\,C$ is:
A) $100.021\,cm$
B) $99.979\,cm$
C) $100.042\,cm$
D) $99.958\,cm$

Answer 1

Hint:- In this problem, first we have to determine the actual length of the brass rod measured at ${30^ \circ}\,C$ with the help of the $\alpha$ value given in the problem and by the help of the actual length of the brass rod, the length of the brass rod at ${0^ \circ}\,C$ can be determined.

Useful formula:
The length of the rod can be determined by,
$\Rightarrow L\left( {1 + t \times \alpha } \right)$
Where, $L$ is the length of the rod, $t$ is the temperature and $\alpha$ is the coefficient of linear expansion.

Complete step by step solution:
Given that,
The length of brass rod at ${30^ \circ}\,C$ is $100\,cm$,
$\alpha$ for steel is $12 \times {10^{ - 6}}{\,^ \circ }{C^{ - 1}}$,
$\alpha$ for brass is $19 \times {10^{ - 6}}{\,^ \circ }{C^{ - 1}}$.
Now the actual length measured by steel scale at the given temperature of ${30^ \circ}\,C$ is,
$L = 100\left( {1 + t \times \alpha } \right)$
On substituting the temperature and the coefficient of linear expansion of steel scale in the above equation, then
$L = 100\left( {1 + 30 \times 12 \times {{10}^{ - 6}}} \right)$
On multiplying the above equation, then the above equation is written as,
$L = 100\left( {1 + 3.6 \times {{10}^{ - 4}}} \right)$
Now adding the terms inside the bracket, then the above equation is written as,
$L = 100\left( {1.00036} \right)$
Now multiplying, then the above equation is written as,
$L = 100.036\,cm$
The above equation shows the actual length measured by the steel scale.
So, the brass rod has the length of $100.036\,cm$ at the temperature of ${30^ \circ}\,C$ and assume the length of the brass rod at ${0^ \circ}\,C$ as $L'$, then,
$L = L'\left( {1 + t \times \alpha } \right)$
Now substituting the known values, then
$100.036 = L'\left( {1 + 30 \times 19 \times {{10}^{ - 6}}} \right)$
Now solving the terms inside the bracket, then
$100.036 = L'\left( {1.00057} \right)$
By keeping the term $L'$ in one side and the other terms in other side, then
$L' = \dfrac{{100.036}}{{1.00057}}$
On dividing the above equation,
$L' = 99.979\,cm$
Thus, the above equation shows the length of the brass rod at ${0^ \circ}\,C$.

Hence, the option (B) is correct.

Note: The temperature of the brass is transferred to the steel scale through conduction, so in first the length of the steel scale is determined, and then the brass rod length at ${0^ \circ}\,C$ is equated with the length of the steel scale, then the length of the brass rod at ${0^ \circ}\,C$ is determined.