Question

At \[{0^0}C\]a steel rod of square cross-section of side \[\sqrt 2 cm\] is rigidly clamped at both ends so that its length cannot change. If the Young’s modulus of steel is \[20 \times {10^{10}}N/{m^2}\] and the linear coefficient is \[12 \times {10^{ - 6}}{/^0}C\], then the force exerted on the clamps if the temperature is raised to \[{20^0}C\] is:
A) \[2400N\]
B) \[4800N\]
C) \[9600N\]
D) \[7200N\]

Answer 1

Hint: The rod is rigidly clamped hence its length cannot change. Use formula for Young’s modulus and then substitute the values to solve accordingly. Convert the value of the cross section in \[{m^2}\] to avoid calculation mistakes. The force exerted on both the ends will be the same. Recall the formula and solve.

Complete step by step solution:
Since the steel rod is clamped on both the ends hence its length cannot change. Since the rod is clamped and cannot increase in length but still due to change in temperature the length of rod must take place.
In the process of expansion of the rod, it will exert force on both ends of its clamp. To find this force we will use Young’s modulus.
Young’ modulus is defined as the ratio of Stress upon Strain. Where Stress is the force exerted per unit area of the cross section. Strain is the ratio of change in property to its original property, in this case Strain will be changed in length divided by original length. Mathematically,
\[E = \dfrac{\sigma }{\varepsilon } = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}\]
Where \[E\] is the Young’s modulus,
\[\sigma \] Is the stress per unit area
\[\varepsilon \] Is the Strain
\[F\] is the force applied
And \[A\] is the area of cross section
\[\dfrac{{\Delta l}}{l}\] is strain

Now from above formula we can calculate the force exerted :
\[F = (E)(A)(\dfrac{{\Delta l}}{l})\]
As we know that \[\Delta l = l\alpha \Delta T\] hence the above formula becomes
\[F = (E)(A)(\alpha \Delta T)\]
Where \[\alpha \] is the linear coefficient
And \[\Delta T\] is the change in temperature
Here the given values are \[E = 20 \times {10^{10}}N/{m^2}\], \[A = {(\sqrt 2 cm)^2} = 2 \times {10^{ - 4}}c{m^2}\], \[\alpha = 12 \times {10^6}{/^0}C\], \[\Delta T = {20^0}C\](the change in temperature is from zero to twenty)
Now substituting the values we get:
\[ \Rightarrow F = (2 \times {10^{10}})(2 \times {10^{ - 4}})(12 \times {10^{ - 6}})(20)\]
\[F = 9600N\]

Therefore option (C) is the correct answer.

Note: Don’t forget to convert the unit of cross section to \[{m^2}\]. Remember all the formulas. It is to be noted that the change in length of rod due to change in temperature is responsible for the force. Also Strain is calculated not only for change in length divided by original length but also many other physical quantities.