JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Elasticity & SHM

One end of a horizontal thick copper wire of length

2 L

and radius

2 R

is welded to an end of another horizontal thin copper wire of length

L

and radius

R

. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

Answer: (C)

Solution

The elongation of a wire under tension is given by:

Δ ℓ = \frac{F ℓ}{Y A}

where

Y

is Young's modulus and

A

is the cross-sectional area.

Since the wires are in series, the tension

F

is the same in both.

For the same material,

Y

is the same. Since

A = π R^{2}

Δ ℓ \propto \frac{ℓ}{R ^{2}}

For the thin wire:

ℓ_{1} = L

R_{1} = R

For the thick wire:

ℓ_{2} = 2 L

R_{2} = 2 R

\frac{Δ ℓ _{1}}{Δ ℓ _{2}} = \frac{ℓ _{1}}{ℓ _{2}} \times \frac{R _{2}^{2}}{R _{1}^{2}} = \frac{L}{2 L} \times \frac{( 2 R ) ^{2}}{R ^{2}} = \frac{1}{2} \times 4 = 2.00

Hence the answer is (C).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option