JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 19)
Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when they are subjected to a force of 2 N. Assume that both the wires are made up of same material and the radius of wire B is 4 times that of the radius of wire A. The length of the wires A and B are in the ratio of a : b. Then a/b can be expressed as 1/x where x is _________.
Answer
32
Explanation
$${\rho _A} = {\rho _B}$$
$${r _B} = 4{r _A}$$
$$\Delta {l_a} = 2$$ mm
$$\Delta {l_B} = 4$$ mm
$$y = {{stress} \over {strain}} = {{F/A} \over {\Delta l/l}}$$
$$ \Rightarrow $$ y$${{\Delta l} \over l} = {F \over {A}}$$
$$ \Rightarrow $$ $$l = {{Ay\Delta l} \over F}$$
$$ \Rightarrow $$ $${{{I_a}} \over {{I_b}}} = {{\pi r_a^2 \times y \times \Delta {I_a} \times F} \over {\pi r_b^2 \times y \times \Delta {I_b} \times F}}$$
$$ \Rightarrow $$ $${{{I_a}} \over {{I_b}}} = {{r_a^2 \times \Delta {I_a}} \over {r_b^2 \times \Delta {I_b}}} = {{r_a^2 \times 2} \over {{{(4{r_a})}^2} \times 4}} = {{r_a^2} \over {16r_a^2 \times 2}}$$
$$ \Rightarrow $$ $${{{I_a}} \over {{I_b}}} = {1 \over {32}}$$
$$ \therefore $$ $$x = 32$$
$${r _B} = 4{r _A}$$
$$\Delta {l_a} = 2$$ mm
$$\Delta {l_B} = 4$$ mm
$$y = {{stress} \over {strain}} = {{F/A} \over {\Delta l/l}}$$
$$ \Rightarrow $$ y$${{\Delta l} \over l} = {F \over {A}}$$
$$ \Rightarrow $$ $$l = {{Ay\Delta l} \over F}$$
$$ \Rightarrow $$ $${{{I_a}} \over {{I_b}}} = {{\pi r_a^2 \times y \times \Delta {I_a} \times F} \over {\pi r_b^2 \times y \times \Delta {I_b} \times F}}$$
$$ \Rightarrow $$ $${{{I_a}} \over {{I_b}}} = {{r_a^2 \times \Delta {I_a}} \over {r_b^2 \times \Delta {I_b}}} = {{r_a^2 \times 2} \over {{{(4{r_a})}^2} \times 4}} = {{r_a^2} \over {16r_a^2 \times 2}}$$
$$ \Rightarrow $$ $${{{I_a}} \over {{I_b}}} = {1 \over {32}}$$
$$ \therefore $$ $$x = 32$$
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