Question

An aluminium rod has a breaking strain 0.2%. The minimum cross-sectional area of the rod in $m^2$ in order to support a load of $10^4$N is :- (Young's modulus is 7×$10^9$ $N{m^{−2}}$)

& A)1\times {{10}^{-2}}{{m}^{2}} \\\ & B)1.4\times {{10}^{-3}}{{m}^{2}} \\\ & C)3.5\times {{10}^{-3}}{{m}^{2}} \\\ & D)7.1\times {{10}^{-4}}{{m}^{2}} \\\ \end{aligned}$$

Answer 1

Here in this type of question where Young’s modulus is given we will apply the formula for young’s modulus of a wire. Maximum breaking stress is given maximum for the wire which can only be possible if the area of wire of cross-section should be minimum. Breaking stress means required stress needed in the given formula.

Complete answer:
Young’s Modulus is defined as the ratio of normal stress to longitudinal strain within limit of proportionality. All modulus of elasticity is valid under the limit of proportionality. There is basically three types of modulus of elasticity:-
1. Young’s Modulus.
2. Bulk Modulus
3. Modulus of Rigidity.
Young’s Modulus deals with length when load is applied to it and Bulk modulus deals in volume when pressure is applied on the body and Modulus of rigidity is applicable on rigid bodies in this only shape will change. Modulus of rigidity is also called shearing modulus.
Here, it is given in the question,
Breaking strain is 0.2%.
Maximum Load of the wire is given = $10^4$ N.
Young’s Modulus of wire (Y)= 7×$10^9$ $N{m^{−2}}$.
Since formula for Young’s Modulus can be mathematically expressed as :-
$Y=\dfrac{Stress}{Strain}$
Since Stress can be expressed as Force acting per unit area of cross-section and strain can be also expressed as Change in length per unit original length .
Let us assume load on the wire can be represented by F and area of cross-section is represented by A.
So Stress can be mathematically expressed as:-
$Stress=\dfrac{F}{A}$.
Strain can be expressed as Breaking Strain . So Young’s modulus will become
$Y=\dfrac{\dfrac{F}{A}}{Breaking\\_Strain}$
$\Rightarrow Y\times Breaking\\_Strain=\dfrac{F}{A}$
$\therefore A=\dfrac{F}{Y\times Breaking\\_Strain}$
For the minimum area of cross-section breaking stress should be maximum and it is given in the question that breaking stress is 10000 N.
$\therefore A=\dfrac{{{10}^{4}}}{7\times {{10}^{9}}\times (\dfrac{2}{1000})}$
On simplifying it we get,
$\therefore A=7.1\times {{10}^{-4}}{{m}^{2}}$
So the minimum area of the cross section is $7.1\times {{10}^{-4}}{{m}^{2}}$to support the maximum load of 10000 N.

So the correct option of the solution is D.

Note:
If a rod of certain length is fixed between two rigid supports, then due to change in temperature its length will change and so it will exert a normal stress (it is compressive if temperature increases and it is tensile if temperature decreases) between the supports. This stress is called thermal stress of the linear wire.