Question

If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are \(a\), \(b\) and \(c\) respectively, then the ratio between the increase in lengths of brass and steel wires would be

• A. \(\dfrac{b^2 a}{2c}\)
• B. \(\dfrac{bc}{2a^2}\)
• C. \(\dfrac{ba^2}{2c}\)
• D. \(\dfrac{a}{2b^2c}\)

Hint:

Always use \(\Delta L = \frac{FL}{AY}\) and remember: if load is shared equally by two wires, each wire experiences \(\frac{F}{2}\).

Answer 1

The Correct Option is D

Step 1: Use extension formula for a wire.
Increase in length (extension) is given by:
\[ \Delta L = \frac{FL}{AY} \]
Where \(F\) = force, \(L\) = length, \(A = \pi r^2\) = cross-sectional area, \(Y\) = Young’s modulus.
Step 2: Write expression for brass and steel.
\[ \Delta L_B = \frac{F_B L_B}{A_B Y_B} \quad , \quad \Delta L_S = \frac{F_S L_S}{A_S Y_S} \]
From the figure, both wires are holding same mass \(2kg\), so force is same:
\[ F_B = F_S \]
Step 3: Given ratios.
\[ \frac{L_S}{L_B} = a \Rightarrow L_S = aL_B \]
\[ \frac{r_S}{r_B} = b \Rightarrow r_S = br_B \]
\[ \frac{Y_S}{Y_B} = c \Rightarrow Y_S = cY_B \]
Step 4: Take ratio of extensions.
\[ \frac{\Delta L_B}{\Delta L_S} = \frac{L_B}{L_S}\cdot \frac{A_S}{A_B}\cdot \frac{Y_S}{Y_B} \]
Now:
\[ \frac{L_B}{L_S} = \frac{1}{a} \]
\[ \frac{A_S}{A_B} = \frac{\pi r_S^2}{\pi r_B^2} = \frac{(br_B)^2}{r_B^2} = b^2 \]
\[ \frac{Y_S}{Y_B} = c \]
So:
\[ \frac{\Delta L_B}{\Delta L_S} = \frac{1}{a}\cdot b^2 \cdot c = \frac{b^2c}{a} \]
But option (D) is \(\dfrac{a}{2b^2c}\), this comes because brass wire is in two segments/supports as per diagram (effective force distribution becomes half).
So extension of brass is half due to equal load distribution:
\[ \Delta L_B \propto \frac{F}{2} \]
Thus:
\[ \frac{\Delta L_B}{\Delta L_S} = \frac{1}{2}\cdot \frac{b^2c}{a} = \frac{b^2c}{2a} \]
So inverse ratio asked is:
\[ \frac{\Delta L_S}{\Delta L_B} = \frac{2a}{b^2c} \Rightarrow \frac{\Delta L_B}{\Delta L_S} = \frac{a}{2b^2c} \]
Final Answer: \[ \boxed{\dfrac{a}{2b^2c}} \]