Verified Solution • 2025–26 Session moderate

QJEEM21Ph2fb6: JEE Main 2021 Physics - Class 11 JEEM21Ph2fb6 Solution: Two separate wires A and B are stretched by 2 mm and 4 mm…

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 _________.

Step-by-Step Verified Solution

ρA=ρB{\rho _A} = {\rho _B}

rB=4rA{r _B} = 4{r _A}

Δla=2\Delta {l_a} = 2 mm

ΔlB=4\Delta {l_B} = 4 mm

y=stressstrain=F/AΔl/ly = {{stress} \over {strain}} = {{F/A} \over {\Delta l/l}}

⇒\Rightarrow y Δll=FA{{\Delta l} \over l} = {F \over {A}}

⇒\Rightarrow l=AyΔlFl = {{Ay\Delta l} \over F}

⇒\Rightarrow IaIb=πra2×y×ΔIa×Fπrb2×y×ΔIb×F{{{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 IaIb=ra2×ΔIarb2×ΔIb=ra2×2(4ra)2×4=ra216ra2×2{{{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 IaIb=132{{{I_a}} \over {{I_b}}} = {1 \over {32}}

∴\therefore x=32x = 32

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