Question

A thin rod having a length of \(1 m\) and area of cross-section \(3 \times 10^{-6} m ^2\) is suspended vertically from one end. The rod is cooled from \(210^{\circ} C\) to \(160^{\circ} C\) After cooling, a mass \(M\) is attached at the lower end of the rod such that the length of rod again becomes \(1m\). Young's modulus and coefficient of linear expansion of the rod are \(2 \times 10^{11} N m ^{-2}\) and \(2 \times 10^{-5} K ^{-1}\), respectively. The value of \(M\) is ____ \(kg\). (Take \(g =10 \ m s ^{-2}\))

Hint:

In such problems, the extension of the rod due to temperature change results in a force being applied, which is balanced by the weight of the mass \( M \).

Answer 1

Correct Answer: 60

 To find the mass \(M\) required to maintain the rod's original length of \(1 m\) after being cooled, we will follow these steps:

Step 1: Calculate the change in length due to temperature change.
The formula for linear expansion is:

\[\Delta L = L_0 \alpha \Delta T\\]

, where:

• \(L_0 = 1 \ m\) (initial length)
• \(\alpha = 2 \times 10^{-5} \ K^{-1}\\) (coefficient of linear expansion)
• \(\Delta T = 210^{\circ} C - 160^{\circ} C = 50 \ K\)

Substituting the values gives:

\[\Delta L = 1 \times 2 \times 10^{-5} \times 50 = 0.001 \ m\]

 

Step 2: Calculate the force required to stretch the rod by \(0.001 \ m\) back to its original length.
Young's Law for elasticity is given by:

\[\Delta L = \frac{F L_0}{A E}\]

, where:

• \(\Delta L = 0.001 \ m\)\) (change in length)
• \(\ L_0 = 1 \ m\\)(original length)
• \(\ A = 3 \times 10^{-6} \ m^2\\)(area of cross-section)
• \(\ E = 2 \times 10^{11} \ N m^{-2}\\)(Young's modulus)

Simplifying for F gives:

\[F = \frac{\Delta L \cdot A \cdot E}{L_0}\]

. By substituting, we get:

\[F = \frac{0.001 \cdot 3 \times 10^{-6} \cdot 2 \times 10^{11}}{1} = 600 \ N\]

 

Step 3: Calculate the mass M that must be hung for the force \(F\).
Using \( F = Mg\), where \( g = 10 \ m/s^2 \), solving for \( M \):

\[M = \frac{F}{g} = \frac{600}{10} = 60 \ kg\]

.

The calculated mass \(M = 60 \ kg\) is exactly within the given range (60,60).