Question

Two strings with lengths \(l_1\) and \(l_2\), and Young's moduli \(Y_1\) and \(Y_2\) are elongated under two weights as shown. Find the ratio \(\Delta l_1 / \Delta l_2\).
Case (1): String has Young's modulus \(Y\), length \(l\), cross–sectional area \(A\) and a mass \(m\) is attached. 
Case (2): String has Young's modulus \(2Y\), length \(0.5l\), cross–sectional area \(A\) and a mass \(3m\) is attached.

• A. \(3\)
• B. \(\frac{2}{3}\)
• C. \(\frac{4}{3}\)
• D. \(\frac{1}{3}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The goal is to find the ratio of the elongations (\(\Delta l\)) of two different strings/rods supporting different masses. The properties of each string and the masses they support are given in the diagram.
Step 2: Key Formula or Approach:
The elongation \(\Delta l\) of a wire of length \(L\), cross-sectional area \(A\), and Young's modulus \(Y\) under a force \(F\) is given by:
\[ \Delta l = \frac{FL}{AY} \]
For a suspended mass \(m\), the force \(F\) is the weight \(mg\).
Step 3: Detailed Explanation:
Let's define the parameters for both setups from the image:
Setup (1):
Mass \(m_{1} = m \Rightarrow F_{1} = mg\)
Length \(l_{1} = l\)
Area \(A_{1} = A\)
Young's modulus \(Y_{1} = Y\)
Elongation \(\Delta l_{1} = \frac{mgl}{AY}\)

Setup (2):
Mass \(m_{2} = 3m \Rightarrow F_{2} = 3mg\)
Length \(l_{2} = 0.5l = \frac{l}{2}\)
Area \(A_{2} = A\)
Young's modulus \(Y_{2} = 2Y\)
Elongation \(\Delta l_{2} = \frac{(3mg)(0.5l)}{A(2Y)} = \frac{1.5 mgl}{2AY} = \frac{3 mgl}{4AY}\)

Now, we calculate the ratio \(\frac{\Delta l_{1}}{\Delta l_{2}}\):
\[ \frac{\Delta l_{1}}{\Delta l_{2}} = \frac{\frac{mgl}{AY}}{\frac{3mgl}{4AY}} \]
\[ \frac{\Delta l_{1}}{\Delta l_{2}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]
Step 4: Final Answer:
The ratio of elongations \(\Delta l_{1} / \Delta l_{2}\) is 4/3, which corresponds to option (C).