Question

At 40 \[{}^\circ C\], a brass wire of 1mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40 \[{}^\circ C\] to 20 \[{}^\circ C\] it regains its original length of 0.2m. The value of M is close to: (Coefficient of linear expansion and Young's modulus of brass are \[{{10}^{-5}}/{}^\circ C\]and \[{{10}^{11}}N/{{m}^{2}}\] respectively; g= \[10m/{{s}^{2}}\]
A. 1.5 kg
B. 9 kg
C. 0.9 kg
D. 0.5 kg

Answer 1

Hint: In this problem, we are given the expansion in length due to application of force on account of load hung at the end and then that length extension is neutralized by a change in temperature. Thus, we have to use the concept of extension of length involving Young modulus and the change in length due to temperature change.

Complete step by step answer:
Radius = 1 mm
=0.001 m
Initial length, l = 0.2 m
Using the formula of Young’s modulus, \[Y=\dfrac{Fl}{A\Delta l}\]
\[\begin{align}
  & \dfrac{\Delta l}{l}=\dfrac{F}{AY} \\
 & \Delta l=\dfrac{Fl}{AY} \\
\end{align}\]
But on heating, the change in length is given as \[\Delta l=l\alpha \Delta T\]
Thus, we get,
\[\begin{align}
  & l\alpha \Delta T=\dfrac{Fl}{AY} \\
 & \alpha \Delta T=\dfrac{F}{AY} \\
\end{align}\]
Also, since it was hung from the ceiling, force is due to gravity, hence, \[\alpha \Delta T=\dfrac{mg}{AY}\]
\[m=\dfrac{\alpha AY\Delta T}{g}\]---------(1)
Now given values are, \[\alpha ={{10}^{-5}}/{}^\circ C\], Y= \[{{10}^{11}}N/{{m}^{2}}\], \[\Delta T=40-20=20{}^\circ C\], g= \[10m/{{s}^{2}}\], A=
\[\begin{align}
  & A=\pi {{(0.001)}^{2}} \\
 & =0.00314{{m}^{2}} \\
\end{align}\]
Putting all the values in eq (1) we get 6.28 kg which is closest to 9kg.

So, the correct answer is “Option B”.

Note:
Young’s modulus is dependent on the nature of the material, it does not depend on other parameters such as length or area of the cross-section or the temperature.