A cylindrical tube open at both ends is made of metal. The internal diameter…
2013
A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The thickness of metal everywhere is 0.4 cm. Taking π = 22/7, the volume of the metal is:
- A.
300 cm³
- B.
306.2 cm³
- C.
312.5 cm³
- D.
322.4 cm³
Show answer & explanation
Correct answer: B
For a hollow cylindrical tube (open at both ends), the metal itself occupies the region between an outer solid cylinder of radius R and an inner hollow cylinder of radius r, both of the same length h. So the volume of metal equals the difference of the two cylinder volumes: V = π(R2 − r2)h. (As is standard for such mensuration problems, π is taken as 22/7 here; the given length, 21 cm, is itself a multiple of 7, which is why the fraction resolves to a clean value below.)
Internal radius: r = internal diameter ÷ 2 = 11.2 ÷ 2 = 5.6 cm.
External radius: R = r + thickness = 5.6 + 0.4 = 6.0 cm.
R2 − r2 = 6.02 − 5.62 = 36 − 31.36 = 4.64 cm2.
V = π(R2 − r2)h = (22/7) × 4.64 × 21 = 66 × 4.64 = 306.24 cm3 ≈ 306.2 cm3.
Cross-check using the difference-of-squares identity: R2 − r2 = (R − r)(R + r) = (0.4)(11.6) = 4.64 cm2, which matches the value used above, confirming the metal volume is 306.2 cm3 (using π = 22/7, the standard approximation for this type of problem).