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Question: A steel wire of length 4 m and diameter 5 mm is stretched by 5 kg-wt. find the increase in its lengt...

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) 0.0041cm0.0041cm
(B) 10.0041cm10.0041cm
(C) 2.0041cm2.0041cm
(D) 3.0041cm3.0041cm

Explanation

Solution

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=σγE = \dfrac{\sigma }{\gamma } where EE is the young’s modulus of a body, σ\sigma is the longitudinal or normal stress on the body, γ\gamma is the normal strain in the body.
γ=el\gamma = \dfrac{e}{l} where ee is the longitudinal or normal extension in the body, and ll is the initial length of the body.
σ=FA\sigma = \dfrac{F}{A} where FF is the force acting on the body to stretch it, and AA 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=σγE = \dfrac{\sigma }{\gamma } where EE is the young’s modulus of a body, σ\sigma is the longitudinal or normal stress on the body, γ\gamma is the normal strain in the body.
Hence, strain is
γ=σE\gamma = \dfrac{\sigma }{E}
And
σ=FA\sigma = \dfrac{F}{A} where FF is the force acting on the body to stretch it, and AA is the cross sectional area of the body.
σ=5×9.8π(0.0025)2=491.96×105=2.5×106N/m2\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,
γ=2.5×106N/m22.4×1011N/m2=1.04×105\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
γ=el\gamma = \dfrac{e}{l}
e=γl=1.04×105×4=4.16×105m\Rightarrow e = \gamma l = 1.04 \times {10^{ - 5}} \times 4 = 4.16 \times {10^{ - 5}}m
Which is roughly equivalent to
e=0.0041cme = 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.