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.

Answer 1

Length of the rod =L =2 m  The length increased during the night = l =0.1 cm= 0.001 m  The strain developed during the night =l/L =0.001/2 =0.0005  If the tension developed = T N  The stress developed = T/A N/m²  = T/0.0004 N/m²  But stress = strain*Y  T/0.0004 = 0.0005*1.9 ×10¹¹ N  →T = 0.0004*0.0005*1.9×10¹¹ N  →T= 2×10⁻⁷*1.9×10¹¹ N  →T= 3.8x10⁴ NRead more on Sarthaks.com - https://www.sarthaks.com/44189/steel-rod-cross-sectional-area-4cm-and-length-shrinks-1cm-the-temperature-decreases-night

Answer 2

The tension developed is $3.8\times10^{4}\ N$

Explanation:

Given that,

Area = 4 cm²

Length = 2 m

Change in length = 0.1 cm

We need to calculate the tension

Using formula of young modulus

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}$

$F=\dfrac{YA\Delta l}{l}$

Where, F = tension

A = area

l = length

$\Delta l$ = change in length

Put the value into the formula

$F=\dfrac{1.9\times10^{11}\times4\times10^{-4}\times0.1\times10^{-2}}{2}$

$F=38000\ N$

$F=3.8\times10^{4}\ N$

Hence, The tension developed is $3.8\times10^{4}\ N$

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Topic :

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