First slide

Elastic Modulie

Question

The rod AB has a cross sectional area 5 mm². Young's modulus of its material is $2 \times 10^{10} N / m^{2}$ . If mass of the rod is 10 Kg, strain produced at its mid point is

Moderate

A

0.075%
B

0.25%
C

0.75%
D

0.5%

Solution

Acceleration $a = \frac{100 - 50}{10} m / s^{2} = 5 m / s^{2}$
Tension produced at its mid section is
given by, $T - 50 = 5 \times 5 \Rightarrow T = 75 N$
$∴$ strain $= \frac{stress}{young's modulus} = \frac{75 / 5 \times 10^{- 6}}{2 \times 10^{10}} = \frac{75}{10^{5}} \times 100 %$
$\Rightarrow strain = 0 .075 %$

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