
$\frac{F}{A_{2}}=Y \frac{\triangle l_{2}}{\lim }$ plies $\triangle l-2=\frac{F l}{A_{2} Y}=\frac{F l}{4 A_{1}}=8 m m$
$\Delta l-1+\triangle l_{2}=10 m m$
$\frac{F l}{A_{1} y}+\frac{F l}{4 A_{1} Y}=10 m m \Rightarrow \frac{F l}{A_{1} Y}=8 m m$

Assertion $(A)$:The stretching of a spring is determined by the shear modulus of the material of the spring.
Reason $(R)$:A coil spring of copper has more tensile strength than a steel spring of same dimensions.
In the light of the above statements, choose the most appropriate answer from the options given below:
