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Exercises · 10.8

Q.A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm4.24\ \text{cm} at 27.0 ∘C27.0\ ^\circ\text{C}. What is the change in the diameter of the hole when the sheet is heated to 227 ∘C227\ ^\circ\text{C}? Coefficient of linear expansion of copper =1.70×10−5 K−1= 1.70 \times 10^{-5}\ \text{K}^{-1}.

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The hole expands exactly as if it were made of copper, so its diameter change follows linear expansion: Δd=d0αΔT\Delta d = d_0 \alpha \Delta T. Substituting gives Δd=4.24×1.70×10−5×200=1.44×10−2 cm\Delta d = 4.24 \times 1.70\times10^{-5} \times 200 = 1.44\times10^{-2}\ \text{cm}.

When you heat a metal sheet with a hole, a common worry is whether the hole shrinks because the material around it expands inward. That’s a natural guess, but it’s wrong. The correct picture: the hole expands as if it were made of the same metal. Think of it this way — if you drew a circle on the sheet before heating, that circle would expand with the sheet. The hole is just the absence of material inside that circle. So its boundary moves outward exactly as the drawn circle would.

This is why we use the linear expansion formula directly on the hole’s diameter:

ΔL=L0 α ΔT\Delta L = L_0 \, \alpha \, \Delta T

where L0L_0 is the original length (here, the diameter), α\alpha is the coefficient of linear expansion, and ΔT\Delta T is the temperature change.

Now let’s work it through.

  1. Identify the given data

    Original diameter: d0=4.24 cmd_0 = 4.24\ \text{cm}

    Initial temperature: T1=27.0 ∘CT_1 = 27.0\ ^\circ\text{C}

    Final temperature: T2=227 ∘CT_2 = 227\ ^\circ\text{C}

    Coefficient: α=1.70×10−5 K−1\alpha = 1.70 \times 10^{-5}\ \text{K}^{-1}

  2. Find the temperature change

ΔT=T2−T1=227−27.0=200 ∘C\Delta T = T_2 - T_1 = 227 - 27.0 = 200\ ^\circ\text{C}

Since a change of 1 ∘C1\ ^\circ\text{C} equals a change of 1 K1\ \text{K}, we can use ΔT=200 K\Delta T = 200\ \text{K} directly.

  1. Apply the linear expansion formula The change in diameter is:

Δd=d0 α ΔT\Delta d = d_0 \, \alpha \, \Delta T

Substitute:

Δd=4.24×(1.70×10−5)×200\Delta d = 4.24 \times (1.70 \times 10^{-5}) \times 200

  1. Calculate step by step

    First, 1.70×10−5×200=1.70×10−5×2×102=3.40×10−31.70 \times 10^{-5} \times 200 = 1.70 \times 10^{-5} \times 2 \times 10^2 = 3.40 \times 10^{-3}

    Then, 4.24×3.40×10−3=(4.24×3.40)×10−34.24 \times 3.40 \times 10^{-3} = (4.24 \times 3.40) \times 10^{-3}

    4.24×3.40=4.24×(3+0.4)=12.72+1.696=14.4164.24 \times 3.40 = 4.24 \times (3 + 0.4) = 12.72 + 1.696 = 14.416

    So Δd=14.416×10−3=1.4416×10−2 cm\Delta d = 14.416 \times 10^{-3} = 1.4416 \times 10^{-2}\ \text{cm}

  2. Round to appropriate significant figures …

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