KCET 2025 Physics Paper

The problem involves two wires, $A$ and $B$, made from the same material. The ratio of their diameters is $1: 2$, and the ratio of their lengths is $1: 3$. When the same force is applied to both wires, we need to determine the ratio of their length increases.Here's how we can solve it:The formula for the increase in length ( $\Delta \ell$ ) when a wire is stretched by a force $F$ is given by:$\Delta \ell = \;\frac{F \ell}{A Y} = \;\frac{F \ell}{\pi r^2 Y}$ Where:$\ell$ is the original length,$A$ is the cross-sectional area of the wire,$Y$ is Young's modulus,$r$ is the radius of the wire.To find the ratio of the increase in lengths of wires $A$ and $B$, we calculate: $\;\frac{\Delta \ell_1}{\Delta \ell_2} = \left(\;\frac{\ell_1}{\ell_2}\right) \times \left(\;\frac{r_2}{r_1}\right)^2$For wires A and B :The length ratio $\ell_1: \ell_2$ is $1: 3$.The diameter ratio ( $\Rightarrow$ radius ratio) is $1: 2$, so $\;\frac{r_2}{r_1} = \;\frac{2}{1}$.Substitute these values into the equation:$\;\frac{\Delta \ell_1}{\Delta \ell_2} = \;\frac{1}{3} \times \left(\;\frac{2}{1}\right)^2 = \;\frac{1}{3} \times 4 = \;\frac{4}{3}$Therefore, the ratio of the increase in their lengths is $4: 3$.