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Question
A steel rod of cross-sectional area 4 cm2 and 2 m shrinks by 0.1 cm as the temperature decreases in night. If the rod is clamped at both ends during the day hours, find the tension developed in it during night hours. Young modulus of steel = 1.9 × 1011 N m−2.
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Solution
Given:
Cross-sectional area of steel rod A = 4 cm2 = 4 × 10−4 m2
Length of steel rod L = 2 m
Compression during night hours ΔL = 0.1 cm = 10−3 m
Young modulus of steel Y = 1.9 × 1011 N m−2
Let the tension developed at night be F.
\[Y = \frac{F}{A} \times \frac{L}{∆ L}\]
\[ \Rightarrow F = \frac{YA ∆ L}{L}\]
\[ = \frac{1 . 9 \times {10}^{11} \times 4 \times {10}^{- 4} \times {10}^{- 3}}{2}\]
\[ = 3 . 8 \times {10}^4 N\]
∴ Required tension developed in steel rod during night hours = 3.8 × 104 N.
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