Question

The coefficient of linear expansion of steel and brass are $$11\times {10}^{-6}/{}^{\circ }C$$ and $$19\times {10}^{-6}/{}^{\circ }C$$ respectively. If their difference in lengths at all temperature has to kept constant at 30 cm, their lengths at $$0{}^{\circ }C$$ should be
A. 71.25 cm and 41.25 cm
B. 82 cm and 52 cm
C. 92 cm and 62 cm
D. 62.25 cm and 32.25 cm

Answer 1

Hint: Steel and the brass rod will expand according to the temperature rise. So from the equation of linear expansion coefficient, we can find out the new lengths of the rods. Since we are dealing with the same conditions, we can find the ratio of new lengths of each rod.
Formula used: $${l_1}^{\prime }=l_1(1+{\alpha }_1\mathrm{\Delta }T)$$, where $${l_1}^{\prime }$$ is the new length of the steel rod, $$l_1$$ is the initial length of the steel rod, $${\alpha }_1$$ is the coefficient of linear expansion of steel rod and $$\mathrm{\Delta }T$$ is the temperature change.
$${l_2}^{\prime }=l_2(1+{\alpha }_2\mathrm{\Delta }T)$$, where $${l_2}^{\prime }$$ is the new length of the brass rod, $$l_2$$ is the initial length of the brass rod, $${\alpha }_2$$ is the coefficient of linear expansion of brass rod and $$\mathrm{\Delta }T$$ is the temperature change.

Complete step by step answer:
 Coefficient of linear expansion of steel can be written as $${\alpha }_1$$ and coefficient of linear expansion of brass can be written as $${\alpha }_2$$. We can assume the length of the steel rod is $$l_1$$ and the length of brass rod is $$l_2$$. Here the difference in their length kept constant as 30 cm at all temperatures. So we can write this as,
$$l_2-l_1=30$$
If we are increasing temperature, suppose the lengths of the rods may vary. We can find this by using the following equation.
$${l_1}^{\prime }=l_1(1+{\alpha }_1\mathrm{\Delta }T)$$, where $${l_1}^{\prime }$$ is the new length of the steel rod, $$l_1$$ is the initial length of the steel rod, $${\alpha }_1$$ is the coefficient of linear expansion of steel rod and $$\mathrm{\Delta }T$$ is the temperature change.
$${l_2}^{\prime }=l_2(1+{\alpha }_2\mathrm{\Delta }T)$$, where $${l_2}^{\prime }$$ is the new length of the brass rod, $$l_2$$ is the initial length of the brass rod, $${\alpha }_2$$ is the coefficient of linear expansion of brass rod and $$\mathrm{\Delta }T$$ is the temperature change.
According to the question even if they are changing their initial lengths of the rod, the difference between the lengths of the rods will be the same. So we can write as,
$${l_2}^{\prime }-{l_1}^{\prime }=30$$
$${l_2}^{\prime }-{l_1}^{\prime }=l_2(1+{\alpha }_2\mathrm{\Delta }T)-l_1(1+{\alpha }_1\mathrm{\Delta }T)$$
We can simplify this equation as,
$${l_2}^{\prime }-{l_1}^{\prime }=l_2-l_1+(l_2{\alpha }_2\mathrm{\Delta }T-l_1{\alpha }_1\mathrm{\Delta }T)$$
Here the difference between the lengths of steel and brass rod will be the same at any temperature.
So,
$${l_2}^{\prime }-{l_1}^{\prime }=l_2-l_1$$
So the remaining terms of the above equation will be equal to zero. Thus we can equate that term to zero.
$$(l_2{\alpha }_2\mathrm{\Delta }T-l_1{\alpha }_1\mathrm{\Delta }T)=0$$
Here the temperature change is the same for both rods. So we can take it outside and equate the remaining terms to zero.
$$(l_2{\alpha }_2-l_1{\alpha }_1)=0$$
$$l_2{\alpha }_2=l_1{\alpha }_1$$
From this, we can find out the ratio of lengths of steel and brass rod.
$${\displaystyle \frac{l_2}{l_1}}={\displaystyle \frac{{\alpha }_1}{{\alpha }_2}}$$
We can assign the coefficient of linear expansion of steel and brass rod in respective places.
$${\displaystyle \frac{l_2}{l_1}}={\displaystyle \frac{11\times {10}^{-6}/{}^{\circ }C}{19\times {10}^{-6}/{}^{\circ }C}}$$
$${\displaystyle \frac{l_2}{l_1}}={\displaystyle \frac{11}{19}}$$
We have to divide all the options that are already given in the question. So we can find out, one option will obey this ratio.
Option A is correct.
$${\displaystyle \frac{41.25}{71.25}}={\displaystyle \frac{11}{19}}$$, all other options are in a different ratio.
Note: Since we are dealing with 2 values of each quantity, it is advised to do the calculations with care. Otherwise, the ratio will be different. So you will get the wrong answer. Even if you are getting the right answer from the first option itself, it is better to check other options also to confirm your result. Here we are dealing with the temperature at $$0{}^{\circ }C$$ for both rods. If the temperature is varying differently for each rod, we have to consider that also in the ratio.