When a ball of copper is heated, its diameter is observed to increase by 0.01%. We can say that it's volume increases by approximately

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Consider that the ball is approximately spherical, so its volume V is proportional to the cube of its diameter d: V ∝ d^3. When the diameter increases by a small fraction Δd/d, the fractional change in volume is approximately 3 × (Δd/d), since d^3 responds to small changes as d^3 → (d + Δd)^3 ≈ d^3 + 3d^2Δd. Here, the diameter increases by 0.01%, which means Δd/d = 0.0001 in decimal form. Applying the relation, the fractional change in volume is about 3 × 0.0001 = 0.0003, i.e., 0.03%.

Why not other values? Because the approximation uses the first-order term in a Taylor expansion for small changes; higher-order terms are negligible for such a tiny percentage change. If the diameter change were larger, the linear approximation would be less precise, but for 0.01% it’s effectively exact to the first order.

In summary, a 0.01% increase in diameter yields approximately a 0.03% increase in volume for a sphere (or ball) since volume scales with the cube of the diameter.

PHYS_V 157 ALL SECTIONS 2024W1 Reading Quiz Week 2

When a ball of copper is heated, its diameter is observed to increase by 0.01%. We can say that it's volume increases by approximately

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