JAMB 2019 Physics · Question 9

A metal rod has a length of 100cm at 200°C. At what temperature will its length be 99.4cm. If the linear…

Question

A metal rod has a length of 100cm at 200°C. At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is 2 × 10−5^{-5}C−1^{-1}

  1. A.3000C300^0C
  2. B.2000C200^0C
  3. C.1000C100^0C
  4. D.−1000C-100^0C

Answer: D. −1000C-100^0C

Explanation written with AI and checked against the answer key

Analysis

To solve this problem, we need to use the formula for linear expansion, which describes how the length of a solid changes when its temperature changes.

Concept

The formula for linear expansion is:

L2=L1(1+αΔθ)L_2 = L_1(1 + \alpha\Delta\theta)

Where:

  • L1L_1 = Initial length (100 cm)
  • L2L_2 = Final length (99.4 cm)
  • α\alpha = Linear expansivity (2×10−5 ∘C−12 \times 10^{-5} \, ^\circ\text{C}^{-1})
  • Δθ\Delta\theta = Change in temperature (θ2−θ1\theta_2 - \theta_1)
  • θ1\theta_1 = Initial temperature (200∘C200^\circ\text{C})
  • θ2\theta_2 = Final temperature (what we need to find)
Step-by-Step Solution

1. Identify the change in length (ΔL\Delta L):

ΔL=L2−L1\Delta L = L_2 - L_1
ΔL=99.4−100=−0.6 cm\Delta L = 99.4 - 100 = -0.6 \text{ cm}

2. Use the linear expansivity formula:

ΔL=L1αΔθ\Delta L = L_1 \alpha \Delta\theta
−0.6=100×(2×10−5)×Δθ-0.6 = 100 \times (2 \times 10^{-5}) \times \Delta\theta

3. Solve for Δθ\Delta\theta:

−0.6=(100×0.00002)×Δθ-0.6 = (100 \times 0.00002) \times \Delta\theta
−0.6=0.002×Δθ-0.6 = 0.002 \times \Delta\theta
Δθ=−0.60.002\Delta\theta = \frac{-0.6}{0.002}
Δθ=−300∘C\Delta\theta = -300^\circ\text{C}

4. Find the final temperature (θ2\theta_2): Since Δθ=θ2−θ1\Delta\theta = \theta_2 - \theta_1:

−300=θ2−200-300 = \theta_2 - 200
θ2=−300+200\theta_2 = -300 + 200
θ2=−100∘C\theta_2 = -100^\circ\text{C}
Conclusion

The temperature at which the rod will have a length of 99.4 cm is −100∘C-100^\circ\text{C}.

Exam Tip

Always pay attention to the sign of your change in length (ΔL\Delta L). If the length decreases, ΔL\Delta L must be negative, which will lead to a negative temperature change. Don't let the negative sign confuse you—it simply means the temperature dropped!

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