JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Elasticity & SHM

Steel wire of length 'L' at

4 0^{\circ} C

is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is

1 0^{- 5} /^{\circ} C

, Young's modulus of steel is

1 0^{11} N / m^{2}

and radius of the wire is 1 mm. Assume that

L ≫

diameter of the wire. Then the value of 'm' in kg is nearly

Answer: 3

Solution

When mass m is hung, the wire stretches by:

Δ L_{1} = \frac{m g L}{Y A}

When cooled by

Δ T = 10° C

, the wire contracts by:

Δ L_{2} = Lα Δ T

For the wire to regain original length:

Δ L_{1} = Δ L_{2}

\frac{m g L}{Y A} = Lα Δ T

m = \frac{Y A α Δ T}{g}

With

A = π r^{2} = π \times (1 0^{- 3})^{2} = π \times 1 0^{- 6}

m²:

m = \frac{1 0 ^{11} \times π \times 1 0 ^{- 6} \times 1 0 ^{- 5} \times 10}{10} = π \approx 3.14 kg

Therefore, m ≈ 3 kg.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Integer answer

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JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Elasticity & SHM

Solution

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