
A steel wire can support a maximum load of W before reaching its elastic limit. How much load can another wire, made out of identical steel, but with a radius one half the radius of the first wire, support before reaching its elastic limit?
A) W
B) $\dfrac{{W}}{{2}}$
C) $\dfrac{{W}}{{4}}$
D) 4W
Answer
558.3k+ views
Hint: This question can be solved by comparing the stress of two wires made of identical material but having different radius. The relation of stress and strain can be used to compare both the wires.
Complete step by step answer:
Here we have to find the maximum load that can be held by another wire made of identical material, but radius is one half the radius of the first wire before reaching its elastic limit.
Let us first understand the meaning of stress, when a body is subjected to a deforming force, a restoring force occurs in the body which is equal in magnitude but opposite in direction to the applied force. This restoring force per unit area is known as stress. Here we can note that the elastic limit is the same for both cases.
We represent the stress in mathematical form.
Stress $=\dfrac{W}{A}$
Where W is the restoring force and A is the cross-sectional area A=\pi r^2 where r is the radius. Using this in the above equation we get,
Stress$ =\dfrac{W}{\pi r^2}$ -------(i)
Now, let us find the stress for the second wire which is made of the same material and its radius is half of the first one. i.e. $r\prime=\dfrac{r}{2} $
Then, the stress of second wire is Stress$=\dfrac{W'}{\pi{r'}^2}$
$
\Rightarrow Stress=\dfrac{W'}{\pi\left(\dfrac{r}{2}\right)^2}
\Rightarrow Stress=\dfrac{W'}{\pi\dfrac{r^2}{4}}
\Rightarrow Stress=\dfrac{4W'}{\pi r^2} ---------(ii)
$
Since the stress is same for both wires, we equate equation (i) and (ii)
$\dfrac{W}{\pi r^2}=\dfrac{4W'}{\pi r^2}|\;\;\;\;\;\; (\pi r^2)$ gets cancelled on both sides)
$
W=4W\prime
\Rightarrow W\prime=\dfrac{1}{4}W
$
Hence the second wire can hold a load of one fourth of the first wire.
Option (C) is the correct answer.
Note: Students should keep in mind that when the external force is removed the material regains its original size and shape if it’s within the elastic limit. If it is beyond the limit, material does not regain its size and shape.
Complete step by step answer:
Here we have to find the maximum load that can be held by another wire made of identical material, but radius is one half the radius of the first wire before reaching its elastic limit.
Let us first understand the meaning of stress, when a body is subjected to a deforming force, a restoring force occurs in the body which is equal in magnitude but opposite in direction to the applied force. This restoring force per unit area is known as stress. Here we can note that the elastic limit is the same for both cases.
We represent the stress in mathematical form.
Stress $=\dfrac{W}{A}$
Where W is the restoring force and A is the cross-sectional area A=\pi r^2 where r is the radius. Using this in the above equation we get,
Stress$ =\dfrac{W}{\pi r^2}$ -------(i)
Now, let us find the stress for the second wire which is made of the same material and its radius is half of the first one. i.e. $r\prime=\dfrac{r}{2} $
Then, the stress of second wire is Stress$=\dfrac{W'}{\pi{r'}^2}$
$
\Rightarrow Stress=\dfrac{W'}{\pi\left(\dfrac{r}{2}\right)^2}
\Rightarrow Stress=\dfrac{W'}{\pi\dfrac{r^2}{4}}
\Rightarrow Stress=\dfrac{4W'}{\pi r^2} ---------(ii)
$
Since the stress is same for both wires, we equate equation (i) and (ii)
$\dfrac{W}{\pi r^2}=\dfrac{4W'}{\pi r^2}|\;\;\;\;\;\; (\pi r^2)$ gets cancelled on both sides)
$
W=4W\prime
\Rightarrow W\prime=\dfrac{1}{4}W
$
Hence the second wire can hold a load of one fourth of the first wire.
Option (C) is the correct answer.
Note: Students should keep in mind that when the external force is removed the material regains its original size and shape if it’s within the elastic limit. If it is beyond the limit, material does not regain its size and shape.
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