JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 26)
A steel rod with y = 2.0 $$\times$$ 1011 Nm$$-$$2 and $$\alpha$$ = 10$$-$$5 $$^\circ$$C$$-$$1 of length 4 m and area of cross-section 10 cm2 is heated from 0$$^\circ$$C to 400$$^\circ$$C without being allowed to extend. The tension produced in the rod is x $$\times$$ 105 N where the value of x is ____________.
Answer
8
Explanation
Given, the Young's modulus of the steel rod, Y = 2 $$\times$$ 1011 Pa
Thermal coefficient of the steel rod, $$\alpha$$ = 10$$-$$5$$^\circ$$C
The length of the steel rod, l = 4 m
The area of the cross-section, A = 10 cm2
The temperature difference, $$\Delta$$T = 400$$^\circ$$C
As we know that,
Thermal strain = $$\alpha$$ $$\Delta$$T
Using the Hooke's law
Young's modulus (Y) = $${{Thermal\,stress} \over {Thermal\,strain}} = {{F/A} \over {\alpha \,\Delta \,T}}$$
Thermal stress, $$F = YA\,\alpha \,\Delta \,T$$
Substitute the values in the above equation, we get
$$F = 2 \times {10^{11}} \times 10 \times {10^{ - 4}} \times {10^{ - 5}} \times (400)$$
$$ = 8 \times {10^5}N$$
Comparing with, $$F = x \times {10^5}N$$
The value of the x = 8.
Thermal coefficient of the steel rod, $$\alpha$$ = 10$$-$$5$$^\circ$$C
The length of the steel rod, l = 4 m
The area of the cross-section, A = 10 cm2
The temperature difference, $$\Delta$$T = 400$$^\circ$$C
As we know that,
Thermal strain = $$\alpha$$ $$\Delta$$T
Using the Hooke's law
Young's modulus (Y) = $${{Thermal\,stress} \over {Thermal\,strain}} = {{F/A} \over {\alpha \,\Delta \,T}}$$
Thermal stress, $$F = YA\,\alpha \,\Delta \,T$$
Substitute the values in the above equation, we get
$$F = 2 \times {10^{11}} \times 10 \times {10^{ - 4}} \times {10^{ - 5}} \times (400)$$
$$ = 8 \times {10^5}N$$
Comparing with, $$F = x \times {10^5}N$$
The value of the x = 8.
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