Question

A steel wire of length 4 m and diameter 5 mm is stretched by 5 kg-wt. find the increase in its length, if the young’s modulus of steel is .
(A) $ 0.0041cm $
(B) $ 10.0041cm $
(C) $ 2.0041cm $
(D) $ 3.0041cm $

Answer 1

Hint: The young’s modulus given should be used to determine the strain in the wire. The strain of a body, is the ratio of the linear extension in the body to the initial length of the body.

Formula used: In this solution we will be using the following formulae;
 $ E = \dfrac{\sigma }{\gamma } $ where $ E $ is the young’s modulus of a body, $ \sigma $ is the longitudinal or normal stress on the body, $ \gamma $ is the normal strain in the body.
 $ \gamma = \dfrac{e}{l} $ where $ e $ is the longitudinal or normal extension in the body, and $ l $ is the initial length of the body.
 $ \sigma = \dfrac{F}{A} $ where $ F $ is the force acting on the body to stretch it, and $ A $ is the cross sectional area of the body.

Complete Step-by-Step solution:
To solve the above, we note that the extension of the wire is implicit in the strain of the wire. Hence, we first calculate the strain of the body, which we do from the young’s modulus.
 The young modulus of a body is given as
 $ E = \dfrac{\sigma }{\gamma } $ where $ E $ is the young’s modulus of a body, $ \sigma $ is the longitudinal or normal stress on the body, $ \gamma $ is the normal strain in the body.
Hence, strain is
 $ \gamma = \dfrac{\sigma }{E} $
And
 $ \sigma = \dfrac{F}{A} $ where $ F $ is the force acting on the body to stretch it, and $ A $ is the cross sectional area of the body.
 $ \sigma = \dfrac{{5 \times 9.8}}{{\pi {{\left( {0.0025} \right)}^2}}} = \dfrac{{49}}{{1.96 \times {{10}^{ - 5}}}} = 2.5 \times {10^6}N/{m^2} $ (all quantities are converted to SI unit)
Hence,
 $ \gamma = \dfrac{{2.5 \times {{10}^6}N/{m^2}}}{{2.4 \times {{10}^{11}}N/{m^2}}} = 1.04 \times {10^{ - 5}} $
Now, using the relation
 $ \gamma = \dfrac{e}{l} $
 $ \Rightarrow e = \gamma l = 1.04 \times {10^{ - 5}} \times 4 = 4.16 \times {10^{ - 5}}m $
Which is roughly equivalent to
 $ e = 0.0041cm $
Hence, option A is correct.

Note:
Alternatively, one can make a fairly good accurate answer without much calculation, due to a fact that the options are somewhat ridiculous. The force acting on the steel was calculated to be 49 N. This force is far too low to make a stretch a 4m long steel by 2 m let alone, 3 m or 10. Hence, we are left with only option A.