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Question: The pressure that has to be applied to the ends of a steel wire of length \( 10cm \) to keep its len...

The pressure that has to be applied to the ends of a steel wire of length 10cm10cm to keep its length constant when its temperature is raised by 100C100^\circ C is:
(For steel Young’s modulus is 2×1011Nm22 \times {10^{11}}N{m^{ - 2}} and coefficient of thermal expansion is 1.1×105K11.1 \times {10^{ - 5}}{K^{ - 1}} )
(A) 2.2×107Pa2.2 \times {10^7}Pa
(B) 2.2×106Pa2.2 \times {10^6}Pa
(C) 2.2×108Pa2.2 \times {10^8}Pa
(D) 2.2×109Pa2.2 \times {10^9}Pa

Explanation

Solution

Hint We need to find the change in length in the steel wire due to the change in the temperature. Then by using the formula for Young’s modulus, we need to find the pressure by using the given values.

Formula Used: In the solution we will be using the following formula,
ΔL=αLΔT\Rightarrow \Delta L = \alpha L\Delta T where ΔL\Delta L is the change in length,
α\alpha is the coefficient of thermal expansion
LL is the length and ΔT\Delta T is the change in temperature.
Y=F/AΔL/L\Rightarrow Y = \dfrac{{F/A}}{{\Delta L/L}}
where YY is the Young’s modulus,
FF is the force and AA is the area of cross section

Complete step by step answer
In this question we are told that the temperature of the steel wire is to be raised by the amount of 100C100^\circ C . Now due to the increase in the temperature, the length of the steel wire will also increase. Now the value of this increase in length will be given by the formula,
ΔL=αLΔT\Rightarrow \Delta L = \alpha L\Delta T
In the problem we are given, L=10cmL = 10cm . So in SI units it will be, L=0.1mL = 0.1m . Now the thermal coefficient is given as, α=1.1×105K1\alpha = 1.1 \times {10^{ - 5}}{K^{ - 1}} and the change in temperature is given as, ΔT=100C\Delta T = 100^\circ C , which is equal to ΔT=100K\Delta T = 100K . Now substituting the values we get,
ΔL=1.1×105×0.1×100\Rightarrow \Delta L = 1.1 \times {10^{ - 5}} \times 0.1 \times 100
On calculating we have,
ΔL=1.1×104m\Rightarrow \Delta L = 1.1 \times {10^{ - 4}}m
So pressure has to be applied to prevent this change in length.
Now from the formula for the Young’s modulus we have,
Y=F/AΔL/L\Rightarrow Y = \dfrac{{F/A}}{{\Delta L/L}}
Here the term FA\dfrac{F}{A} is the force per unit area, which is the pressure. Therefore we can write this formula for the pressure as,
P=FA=YLΔL\Rightarrow P = \dfrac{F}{A} = \dfrac{Y}{L}\Delta L
Now we are given Y=2×1011Nm2Y = 2 \times {10^{11}}N{m^{ - 2}}
So substituting we get the value,
P=2×10110.1×1.1×104\Rightarrow P = \dfrac{{2 \times {{10}^{11}}}}{{0.1}} \times 1.1 \times {10^{ - 4}}
Hence on calculating we get,
P=2.2×108Pa\Rightarrow P = 2.2 \times {10^8}Pa
So the correct answer is option C.

Note
The Young’s modulus is the modulus of elasticity of a substance. It is the mechanical property that measures the tensile stiffness of a solid. It can be calculated by the ratio of the tensile stress to the axial strain of the body.