JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Elasticity & SHM

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

\frac{a}{b}

can be expressed as

\frac{1}{x}

where

x

is ⋯------.

Answer: 32

Solution

Young's modulus gives

Y = \frac{F L}{A Δ L}

, so for the same material and the same stretching force

L = \frac{Y A Δ L}{F} \Rightarrow \frac{L _{A}}{L _{B}} = \frac{A _{A} Δ L _{A}}{A _{B} Δ L _{B}}

Since

r_{B} = 4 r_{A}

, the cross-sections satisfy

A_{B} = 16 A_{A}

, and the extensions are

Δ L_{A} = 2

mm and

Δ L_{B} = 4

mm, so

\frac{L _{A}}{L _{B}} = \frac{1 \times 2}{16 \times 4} = \frac{1}{32}

Hence

x = 32

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Elasticity & SHM

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