First slide

Elastic behaviour of solids

Question

A 5-kg rod of square cross section 5 cm on a side and 1 m long is pulled along a smooth horizontal surface by a force applied at one end. The rod has a constant acceleration of $2 {ms}^{- 2}$ . Determine the elongation in the rod. (Young's modulus of the material of the rod is $5 \times 10^{9} N / m^{2}$ ).

Moderate

A

Zero, as for elongation to be there, equal and opposite forces must act on the rod
B

Non-zero but cannot be determine from the given situation
C

0.4 $μ$ m
D

16 $μ$ m

Solution

Let the force applied be p, then P = ma
$\Rightarrow p = 5 \times 2 = 10 N$
For the element shown in the figure, $T = \frac{m}{l} (l - x) \times a$
$\Rightarrow \Rightarrow T = \frac{5}{1} (1 - x) \times 2 = 10 (1 - x)$
Elongation in dx is $\frac{Δ (dx)}{dx} = \frac{T / A}{Y}$
$Δ (dx) = \frac{10 (1 - x)}{{(5 \times 10^{- 2})}^{2}} \times \frac{1}{5 \times 10^{9}} dx$
Total elongation, $Δ l = \int_{0}^{1} \frac{10 (1 - x)}{25 \times 10^{- 4}} \times \frac{1}{5 \times 10^{9}} d x = 0.4 \times 10^{- 6} m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App