Question

Two wires of same material have lengths in the ratio $1:2$ and diameters in the ratio $2:1$. If they are stretched by the same load force, the ratio of their extensions ($\Delta l_1 : \Delta l_2$) is:

• A. 1 : 4
• B. 1 : 8
• C. 1 : 2
• D. 4 : 1 Correct Answer: (B) 1 : 8

Hint:

To solve scaling questions instantly on physics exams, write out a clean proportional statement: $\mathbf{\Delta l \propto \frac{l}{d^2}}$. Halving the length drops the extension by a factor of $2$, and doubling the diameter drops it by another factor of $2^2 = 4$. Combining these independent drops together gives a total decrease of $2 \times 4 = \mathbf{8}$, directly pointing you to $1:8$!

Answer 1

The Correct Option is B

Step 1: Write extension in terms of length and diameter.
From Young's modulus, the stretch is $\Delta l = \dfrac{Fl}{AY}$. The area of a round wire is $A = \dfrac{\pi d^2}{4}$, so \[ \Delta l = \frac{4Fl}{\pi d^2 Y} \]

Step 2: Keep only what changes.
Same material and same load mean $F$, $Y$, $\pi$ are fixed for both wires. So the stretch depends only on length over diameter squared: \[ \Delta l \propto \frac{l}{d^2} \]

Step 3: Build the ratio of the two wires.
\[ \frac{\Delta l_1}{\Delta l_2} = \frac{l_1}{l_2}\times\left(\frac{d_2}{d_1}\right)^2 \] Here $l_1:l_2 = 1:2$ and $d_1:d_2 = 2:1$, so $d_2/d_1 = 1/2$.

Step 4: Put the numbers in.
\[ \frac{\Delta l_1}{\Delta l_2} = \frac{1}{2}\times\left(\frac{1}{2}\right)^2 = \frac{1}{2}\times\frac{1}{4} = \frac{1}{8} \]
So the extensions are in the ratio $1:8$, which is option (B).
\[ \boxed{1:8} \]