First slide

Stress and strain

Question

The length of an elastic string is ‘a’ meter when the tension is 4 N and ‘b’ meter when the tension is 5 N. The length, (in meter), when the tension is 9 N, is

Difficult

A

(a + b)
B

$(4 b - 5 a)$
C

$(5 b - 4 a)$
D

$(9 b - 9 a)$

Solution

Let l be the natural length and $k (= \frac{YA}{I})$ be the force constant of wire. Then, $a = l + \frac{4}{k} \dots (i) (∵ F = kΔL or Δl = \frac{F}{k})$
and $b = l + \frac{5}{k}$
On solving Eqs. (i) and (ii), we get
or $\frac{1}{k} = (b - a) and l = (5 a - 4 b)$
Now, when T = 9N
$L^{'} = l + \frac{9}{k} = (5 a - 4 b) + 9 (b - a) = (5 b - 4 a)$

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