Question

22. Three bars having length I, 21 and 31 and area of
cross-section A, 2A and 3A are joined rigidly end
to end. Compound rod is subjected to a stretching
force F. The increase in length of rod is (Young's
modulus of material is Y and bars are massless)​

Answer 1

The increase in length of rod is  $\bf{ \frac{3Fl}{AY}}$

Explanation:

Given that

Three bars with length I, 21 and 31 and area of  cross-section A, 2A and 3A.

Force F is exerted to rod to stretch.

Young’s Modulus Formula , $Y = \frac{FL}{A \triangle L}$---------------(1)

where,

Y= the young modulus in pascals(Pa)

F = force in newtons(N)

L = original length in metres(m)

A = area in square metres$(m^2)$

$\triangle L$ = change in length in metres (m)

From equation (1)

We get, $\triangle L = \frac{Fl}{A Y}$

Change in length, $\triangle l_1 = \frac{Fl}{A Y}$ -----------(1)

$\triangle l_2 = \frac{F2l}{2A Y}$--------------(2)

$\triangle l_3 = \frac{F3l}{3A Y}$--------------(3)

Adding 1, 2 and 3

Extension in three rods is given by :

$\triangle l = \frac{Fl}{AY} + \frac{F2l}{2AY} + \frac{F3l}{3AY}$

$= \frac{Fl}{AY} + \frac{Fl}{AY} + \frac{Fl}{AY}$

$= \frac{3Fl}{AY}$

Thus, increase in length of rod is $= \frac{3Fl}{AY}$