A student measures the internal diameter and external diameter of a hollow metal tube using a vernier caliper. She then calculates the cross-sectional area ...
Question 1 Report
A student measures the internal diameter and external diameter of a hollow metal tube using a vernier caliper. She then calculates the cross-sectional area of the metal.
The external diameter is 45 mm and the internal diameter is 28 mm.
(a) Calculate the cross-sectional area of the metal (the shaded region). [3]
(b) The tube has a length of 120 mm. Calculate the volume of metal in mm³. [1]
(c) Convert this volume to cm³. [1]
(d) State the precision of the vernier caliper. [1]
Answer Details
(a) Calculate the cross-sectional area of the metal.
The metal cross-section is the outer circle minus the inner circle (an annulus).
Outer radius = 45 / 2 = 22.5 mm
Outer area = \(\pi \times 22.5^2 = \pi \times 506.25 = 1590 \text{ mm}^2\) [1]
Inner radius = 28 / 2 = 14 mm
Inner area = \(\pi \times 14^2 = \pi \times 196 = 616 \text{ mm}^2\) [1]
Metal area = 1590 − 616 = 974 mm² [1]
(b) Calculate the volume of metal.
Volume = cross-sectional area × length = 974 × 120 = 116 880 mm³ [1]
For a uniform prism (or cylinder), the volume equals the cross-sectional area multiplied by the length.
(c) Convert this volume to cm³.
Since 1 cm = 10 mm, then 1 cm³ = 10 × 10 × 10 = 1000 mm³
116 880 mm³ / 1000 = 116.9 cm³ [1]
(d) State the precision of the vernier caliper.
The precision is 0.1 mm (or equivalently 0.01 cm). [1]
A vernier caliper typically has a main scale graduated in mm and a vernier scale that allows readings to the nearest 0.1 mm, giving one order of magnitude less precision than a micrometer screw gauge (0.01 mm).
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