The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by $100^{\circ} \mathrm{C}$ is : (For steel Young's modulus is $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ and coefficient of thermal expansion is $1.1 \times 10^{-5} \mathrm{~K}^{-1}$)

(C) $2.2 \times 10^{8} \mathrm{~Pa}$

(A) $2.2 \times 10^{7} \mathrm{~Pa}$

(B) $2.2 \times 10^{6} \mathrm{~Pa}$

(D) $2.2 \times 10^{9} \mathrm{~Pa}$

JEE Main · 2014 · 06 Apr 2014 · Offline · Heat & Thermodynamics

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by

10 0^{\circ} C

is :
(For steel Young's modulus is

2 \times 1 0^{11} N m^{- 2}

and coefficient of thermal expansion is

1.1 \times 1 0^{- 5} K^{- 1}

)

(A) $2.2 \times 1 0^{7} Pa$
(B) $2.2 \times 1 0^{6} Pa$
(C) $2.2 \times 1 0^{8} Pa$ ✓ Correct
(D) $2.2 \times 1 0^{9} Pa$

Answer: (C)

Solution

If the wire is not allowed to lengthen, we must apply exactly the stress that cancels the thermal expansion. The strain the heating would have produced is

α Δ T = 1.1 \times 1 0^{- 5} \times 100 = 1.1 \times 1 0^{- 3}

. Since Young's modulus is stress divided by strain, the pressure needed is

P = Y α Δ T = 2 \times 1 0^{11} \times 1.1 \times 1 0^{- 3} = 2.2 \times 1 0^{8}

Pa.

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

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JEE Main · 2014 · 06 Apr 2014 · Offline · Heat & Thermodynamics

Solution

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