A thin brass sheet at `10^(@)C` and a thin steel sheet at `20^(@)C` have the same surface area. The common temperature at which both would have the same area is (Coefficient of linear expansion for brass and steel are respectively, `19 xx 10^(-6//@)C` are `11 xx 10^(-6//@)C)`

Options

A. `-3.75^(@)C`
B. `-2.75^(@)C`
C. `+2.75^(@)C`
D. `+3.75^(@)C`

Answer

A

Explanation

The area of brass sheet at `10^(@)C = l_(1)b_(1)`
The area of steel sheet at `20^(@)C = l_(2)b_(2)`
Given that `l_(1) b_(1) = l_(2) b_(2)`
Let at temperature `T^(@)C` the areas of brass and steel sheets be same.
Length of brass sheet at `T^(@)C =l_(1) [1 + alpha_(b) Delta T_(b)]`
Length of steel sheet at `T^(@)C = l_(2) [1 + alpha_(s) Delta T_(s)]`
Breadth of brass sheet at `T^(@)C = b_(1) [1 + alpha_(b) Delta T_(b)]`
Breadth of steel sheet at `T^(@)C = b_(2) [1 + alpha_(s) Delta T_(s)]`
According to problem, `l_(1) [1 + alpha_(b) Delta T_(b)] xx b_(1) [1 + alpha_(b) Delta T_(b)] = l_(2) [1 + alpha_(3) Delta T_(s)] xx b_(2) [1 + alpha_(s) Delta T_(s)]`
or `l_(1) b_(1) [1 + alpha_(b) Delta T_(b)]^(2) = l_(2) b_(2) [1 + alpha_(s) Delta T_(s)]^(2)`
or `l_(1)b_(1) [1 + 19 xx 10^(-6) xx (10-T)]^(2) = l_(2)b_(2) [1 + 11 xx 10^(-6) (20-T)]^(2)`
or `19 xx 10^(-6) (10-T) = 11 xx 10^(-6) (20-T)`
`:. T= - (30)/(8) = -3.75^(@)C`