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Karnataka Board PUCPUC Science Class 11

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,

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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.

Short/Brief Note
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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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Chapter 14: Some Mechanical Properties of Matter - Exercise [Page 300]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 14 Some Mechanical Properties of Matter
Exercise | Q 8 | Page 300

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