Question

The breaking strength of a rod of diameter 2cm is $2 \times {10^{ - 5}}N$. Then that for rod of same material but diameter 1cm will be
(A) $2 \times {10^{ - 5}}N$
(B) $1 \times {10^{ - 5}}N$
(C) $8 \times {10^{ - 5}}N$
(D) $0.5 \times {10^{ - 5}}N$

Answer 1

Hint: Since the material of the rod is the same in both the cases we can find the breaking stress of the material of the rod in the first case by the formula $S = \dfrac{F}{A}$ and then we can use this breaking stress to find out the force on the rod for which the radius is given.
To solve this question we will use the following formula,
$\Rightarrow S = \dfrac{F}{A}$
where $S$ is the breaking stress
$F$ is the force on the rod and $A$ is the area of cross-section of the rod.

Complete step by step answer:
In the question, we are provided that the breaking strength of a rod is $2 \times {10^5}N$. This is the force that is applied to the material of the rod that causes it to break. Now, in the first case, the diameter of the rod is given ${d_1} = 2cm = 0.02m$. So the area of cross-section of the rod in the first case is given by,
$\Rightarrow {A_1} = \pi \times {\left( {\dfrac{{{d_1}}}{2}} \right)^2}$
Now putting the value we get,
$\Rightarrow {A_1} = \pi \times {\left( {\dfrac{{0.02}}{2}} \right)^2} = \pi \times {10^{ - 4}}{m^2}$
Now the breaking stress of a material is given by the breaking force per unit area. So,
$\Rightarrow S = \dfrac{{{F_1}}}{{{A_1}}}$
By substituting the values we get
$\Rightarrow S = \dfrac{{2 \times {{10}^5}}}{{\pi \times {{10}^{ - 4}}}}$
The breaking stress is the property of the material; so in the second case, though we change the diameter, the value of breaking stress will remain the same.
So for the second case, we have the diameter ${d_2} = 1cm = 0.01m$
Therefore the area of a cross-section of the wire in the second case is,
$\Rightarrow {A_2} = \pi \times {\left( {\dfrac{{0.01}}{2}} \right)^2} = \pi \times 2.5 \times {10^{ - 5}}{m^2}$
So, using the value of the breaking stress and area of cross-section, we can find the force in the second case as,
$\Rightarrow S = \dfrac{{{F_2}}}{{{A_2}}}$
By taking the ${A_2}$ to the other side, we get
$\Rightarrow {F_2} = S \times {A_2}$
Substituting the values,
$\Rightarrow {F_2} = \dfrac{{2 \times {{10}^5}}}{{\pi \times {{10}^{ - 4}}}} \times \pi \times 2.5 \times {10^{ - 5}}$
By cancelling the $\pi $ from numerator and denominator and the ${10^5}$ and ${10^{ - 5}}$ in the numerator,
${F_2} = \dfrac{{2 \times 2.5}}{{{{10}^{ - 4}}}}$
On doing the above calculation we get,
${F_2} = 0.5 \times {10^5}$
So the breaking force will be $0.5 \times {10^5}$.
The correct option is option (D).

Note:
The breaking strength of a rod is directly proportional to the square of the radius of the rod.
$F \propto {r^2}$
So we can find the ratio of the breaking strength of the rod and find out the equation for both the cases.
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}$
Now by putting, ${r_1} = 1cm$ and ${r_2} = 0.5cm$ ${F_1} = 2 \times {10^5}N$ we get,
$\Rightarrow {F_2} = {\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)^2}{F_1}$
Substituting the values,
$\Rightarrow {F_2} = {\left( {\dfrac{{0.5}}{1}} \right)^2}2 \times {10^5}$
we get, ${F_2} = 0.5 \times {10^5}$.