Question

A young's modulus of a 6 m rod isb186Gpa. The area of cross section of the rod is .75×10^-4. calculate the tension in the rod if the rod is streches by 0.25cm

Answer 1

Answer:

Explanation:

pls find the attachment

Answer 2

The tension in the rod when it is stretched by 0.25cm is 5.812×10³ N.

Given:

young's modulus (Y) = 186 GPa , Length (L) = 6m ,Area( A) = 0.75 × $10^{-4}$ m².

To Find:

The tension in the rod when the rod stretches' by 0.25cm.

Solution:

1GPa= $10^{9}$Pa

∴ Y= 186 ×$10^{9}$Pa

ΔL= 0.25 cm = 0.25 × $10^{-2}$ m.

Young's Modulus is given by,

$Y =\frac{F \times L}{ \triangle L\times A} \\\\\implies F = \frac{Y \times \triangle L \times A}{L}$

Substituting the given values,

$F = \frac{186 \times 10^{9} \times 0.25 \times 10^{-2} \times 0.75 \times 10^{-4} }{6} \\\\ F= \frac{34.875\times 10^{3} }{6}\\\\ F= 5.8125 \times 10^{3} N$

Therefore, The tension in the rod when it is stretched by 0.25cm is 5.812×10³ N.

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