Question

A lift is tied with thick iron wire and its mass is 1000 kg. If the maximum acceleration of the lift is $1.2m{s}^{-1}$ and the maximum stress of the wire us $1.4 \times {10}^{8} N{m}^{-2}$, what should be the minimum diameter of the wire?
A. ${10}^{-2} m$
B. ${10}^{-4} m$
C. ${10}^{-6} m$
D. $0.5 \times {10}^{-2} m$

Answer 1

Hint: To solve this problem, first find the force on the iron wire in terms of acceleration and mass. Then, add acceleration and gravity and substitute in place of acceleration in the formula for force. Find the area of the wire. Then, taking the ratio of force and area of wire will give the stress on the wire. Substitute the values in the formula for stress and solve it to get the radius of the wire. Then, double the radius to get the minimum diameter of the wire.

Formula used:
$Stress= \dfrac {Force}{Area}$
$F= ma$
$A= \pi {r}^{2}$

Complete step by step answer:
Given: Stress of the wire (S)= $1.4 \times {10}^{8} N{m}^{-2}$
            Acceleration of the lift (a)= $1.2m{s}^{-1}$
            Mass of wire (m)= 1000 kg
Maximum stress is given by,
$Stress= \dfrac {Force}{Area}$
$\Rightarrow S= \dfrac {F}{A}$ …(1)
We know, Force is given by,
$F= ma$ …(2)
Here, gravity also acts on the wire. So, equation. (2) becomes,
$F= m(g+a)$
Substituting values in above equation we get,
$F= 1000 (9.8 + 1.2)$
$\Rightarrow F= 1000 \times 11$
$\Rightarrow F= 10000 N$ …(3)
Area of the wire is given by,
$A= \pi {r}^{2}$ …(4)
Substituting equation. (3) and(4) in equation. (1) we get,
$1.4 \times {10}^{8}= \dfrac {11000}{\pi {r}^{2}}$
$\Rightarrow {r}^{2}= \dfrac {11000}{1.4\pi \times {10}^{8}}$
$\Rightarrow {r}^{2}= 2.5 \times {10}^{-5}$
Taking the square roots on both the sides we get,
$r= 5 \times {10}^{-3}$
We know, diameter is twice the radius of the wire.
$\therefore d= 2r$
Substituting the values in above equation we get,
$d= 2 \times 5 \times {10}^{-3}$
$\therefore d= {10}^{-2} m$
Hence, the minimum diameter of the wire should be ${10}^{-2}m$.

So, the correct answer is option A i.e. ${10}^{-2}m$.

Note:
Stress on a wire is inversely proportional to the radius of the wire. Greater the stress, smaller is the diameter of the wire. Students should not forget that the gravity also acts on the lift when it is accelerating. So, gravity should be added to the acceleration to get the total acceleration and then it should be substituted in the formula for force.