Question

A brass rod and a lead rod each $80\text{cm}$ long at $0{}^{\circ }\text{C}$ are clamped together at one end with their free ends coinciding. The separation the of free ends of the rods, if the system is placed in a steam bath is
$[{\alpha }_{\text{brass}}=18\times {10}^{-6}/{}^{\circ }\text{C}$ and ${\alpha }_{\text{lead}}=28\times {10}^{-6}/{}^{\circ }\text{C}\text{and temperature of steam}={100}^{\circ }\text{C}]$

• A. $0.2\text{mm}$
• B. $0.8\text{mm}$
• C. $1.4\text{mm}$
• D. $1.6\text{mm}$

Answer 1

The correct option is B $0.8\text{mm}$

Given:
Length of each rod, $L=80\text{cm}$
Change in temperature, $\Delta T=100-0=100{}^{\circ }\text{C}$
Coefficient of expansion of brass, ${\alpha }_b=18\times {10}^{-6}/{}^{\circ }\text{C}$
Coefficient of expansion of lead, ${\alpha }_l=28\times {10}^{-6}/{}^{\circ }\text{C}$
To find:
Separation between the free ends of the rod when they are kept in steam bath, $\Delta L=?$



We know that,
Separation between the free ends $=$ linear expansion of lead $-$ linear expansion of brass
$\Rightarrow \Delta L=L{\alpha }_l\Delta T-L{\alpha }_b\Delta T$
[ from formula of linear expansion ]
$\Rightarrow \Delta L=80\times 28\times {10}^{-6}\times 100-80\times 18\times {10}^{-6}\times 100$
$\Rightarrow \Delta L=0.08\text{cm}=0.8\text{mm}$