Question:
The difference in length between two rods A and B is 60 cm at all temperatures. If $\alpha_A = 18 \times 10^{-6}/^\circ C$ and $\alpha_B = 27 \times 10^{-6}/^\circ C$, then the length of rod A and rod B at $0^\circ C$ is respectively ______.
The difference in length between two rods A and B is 60 cm at all temperatures. If $\alpha_A = 18 \times 10^{-6}/^\circ C$ and $\alpha_B = 27 \times 10^{-6}/^\circ C$, then the length of rod A and rod B at $0^\circ C$ is respectively ______.
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"Constant length difference" always means $L_1 \alpha_1 = L_2 \alpha_2$. The initial lengths are inversely proportional to their expansion coefficients. The rod with the smaller $\alpha$ must be much longer!
Updated On: Jun 19, 2026
- $l_A = 120$ cm, $l_B = 60$ cm
- $l_A = 180$ cm, $l_B = 120$ cm
- $l_A = 240$ cm, $l_B = 180$ cm
- $l_A = 270$ cm, $l_B = 210$ cm
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
Two rods maintain a constant length difference regardless of the temperature. This means they must physically expand by the exact same amount when heated. We must find their initial lengths.
Step 2: Detailed Explanation:
The formula for linear thermal expansion is:
$\Delta l = l_0 \cdot \alpha \cdot \Delta T$
If the difference in their lengths ($l_A - l_B$) is always exactly 60 cm, it means neither rod outgrows the other. Their individual expansions must be perfectly identical for any given temperature change $\Delta T$.
$\Delta l_A = \Delta l_B$
$l_A \cdot \alpha_A \cdot \Delta T = l_B \cdot \alpha_B \cdot \Delta T$
Cancel out the common temperature change $\Delta T$:
$l_A \cdot \alpha_A = l_B \cdot \alpha_B$ --- (Equation 1)
We are given the coefficients of linear expansion:
$\alpha_A = 18 \times 10^{-6} /^\circ\text{C}$
$\alpha_B = 27 \times 10^{-6} /^\circ\text{C}$
Substitute these into Equation 1:
$l_A (18 \times 10^{-6}) = l_B (27 \times 10^{-6})$
$18 l_A = 27 l_B$
Divide by 9 to simplify the ratio:
$2 l_A = 3 l_B \implies l_A = \frac{3}{2} l_B$
We are also given the constraint that their length difference is 60 cm. Since $\alpha_A < \alpha_B$, Rod A must be longer initially to compensate and match the expansion of Rod B.
$l_A - l_B = 60$
Substitute $l_A = \frac{3}{2} l_B$ into this constraint:
$\frac{3}{2} l_B - l_B = 60$
$\frac{1}{2} l_B = 60$
$l_B = 120 \text{ cm}$
Now, find $l_A$:
$l_A = l_B + 60$
$l_A = 120 + 60 = 180 \text{ cm}$
Step 3: Final Answer:
The lengths are $l_A = 180$ cm and $l_B = 120$ cm, matching option (b).
Two rods maintain a constant length difference regardless of the temperature. This means they must physically expand by the exact same amount when heated. We must find their initial lengths.
Step 2: Detailed Explanation:
The formula for linear thermal expansion is:
$\Delta l = l_0 \cdot \alpha \cdot \Delta T$
If the difference in their lengths ($l_A - l_B$) is always exactly 60 cm, it means neither rod outgrows the other. Their individual expansions must be perfectly identical for any given temperature change $\Delta T$.
$\Delta l_A = \Delta l_B$
$l_A \cdot \alpha_A \cdot \Delta T = l_B \cdot \alpha_B \cdot \Delta T$
Cancel out the common temperature change $\Delta T$:
$l_A \cdot \alpha_A = l_B \cdot \alpha_B$ --- (Equation 1)
We are given the coefficients of linear expansion:
$\alpha_A = 18 \times 10^{-6} /^\circ\text{C}$
$\alpha_B = 27 \times 10^{-6} /^\circ\text{C}$
Substitute these into Equation 1:
$l_A (18 \times 10^{-6}) = l_B (27 \times 10^{-6})$
$18 l_A = 27 l_B$
Divide by 9 to simplify the ratio:
$2 l_A = 3 l_B \implies l_A = \frac{3}{2} l_B$
We are also given the constraint that their length difference is 60 cm. Since $\alpha_A < \alpha_B$, Rod A must be longer initially to compensate and match the expansion of Rod B.
$l_A - l_B = 60$
Substitute $l_A = \frac{3}{2} l_B$ into this constraint:
$\frac{3}{2} l_B - l_B = 60$
$\frac{1}{2} l_B = 60$
$l_B = 120 \text{ cm}$
Now, find $l_A$:
$l_A = l_B + 60$
$l_A = 120 + 60 = 180 \text{ cm}$
Step 3: Final Answer:
The lengths are $l_A = 180$ cm and $l_B = 120$ cm, matching option (b).
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