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Chapter 12 · Question 6

A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 55 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 1010 cm, find the apparent capacity of the glass and its actual capacity. (Use π=3.14\pi = 3.14)

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Q6

A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 55 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 1010 cm, find the apparent capacity of the glass and its actual capacity. (Use π=3.14\pi = 3.14)

Answer Revealed
Direct Answer:
The glass is a cylinder with a hemispherical bump at the bottom. Radius r=52=2.5r = \frac{5}{2} = 2.5 cm, height of cylinder h=10h = 10 cm. Apparent capacity (volume of the full cylinder) =πr2h=3.14×(2.5)2×10=3.14×6.25×10=196.25= \pi r^2 h = 3.14 \times (2.5)^2 \times 10 = 3.14 \times 6.25 \times 10 = 196.25 cm3^3. Volume of the hemispherical raised portion =23πr3=23×3.14×(2.5)3=23×3.14×15.625=98.1253=32.71= \frac{2}{3}\pi r^3 = \frac{2}{3} \times 3.14 \times (2.5)^3 = \frac{2}{3} \times 3.14 \times 15.625 = \frac{98.125}{3} = 32.71 cm3^3 (approx.). Actual capacity =Apparent capacity−Volume of hemispherical portion=196.25−32.71=163.54= \text{Apparent capacity} - \text{Volume of hemispherical portion} = 196.25 - 32.71 = 163.54 cm3^3.

Simple Explanation

The glass looks like a regular cylinder — that is the 'apparent' capacity: π(2.5)2×10=196.25\pi (2.5)^2 \times 10 = 196.25 cm3^3. But the bottom has a dome-shaped bump sticking up into the glass, which takes up space. That bump is a hemisphere with volume 23π(2.5)3≈32.71\frac{2}{3}\pi (2.5)^3 \approx 32.71 cm3^3. Subtract it: 196.25−32.71=163.54196.25 - 32.71 = 163.54 cm3^3 of juice actually fits in the glass. The juice seller seems generous with a tall glass, but the raised bottom tricks you into thinking it holds more.

Exam-Ready Structure

This problem illustrates the difference between the nominal (apparent) volume and the actual usable volume when a solid intrudes into the cavity: 1. The glass is a cylinder (r=52=2.5r = \frac{5}{2} = 2.5 cm, h=10h = 10 cm) with a hemispherical raised portion at the bottom that points inward (upward) into the cylinder. This hemispherical bump occupies space that would otherwise hold liquid. 2. Apparent capacity: This is the volume the glass seems to have if you ignore the raised bottom — it is simply the volume of the full cylinder. Vapparent=πr2h=3.14×2.5×2.5×10=3.14×6.25×10=196.25V_{\text{apparent}} = \pi r^2 h = 3.14 \times 2.5 \times 2.5 \times 10 = 3.14 \times 6.25 \times 10 = 196.25 cm3^3. 3. Volume of the hemispherical raised portion: The hemisphere at the bottom has the same radius as the glass interior. Its flat face sits on the bottom of the cylinder and the curved surface bulges into the glass. Volume Vhemisphere=23πr3=23×3.14×(2.5)3=23×3.14×15.625=98.1253≈32.71V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3} \times 3.14 \times (2.5)^3 = \frac{2}{3} \times 3.14 \times 15.625 = \frac{98.125}{3} \approx 32.71 cm3^3. 4. Actual capacity: The volume of liquid the glass can actually hold =Vapparent−Vhemisphere=196.25−32.71=163.54= V_{\text{apparent}} - V_{\text{hemisphere}} = 196.25 - 32.71 = 163.54 cm3^3. 5. The difference 32.7132.71 cm3^3 is the volume of juice that cannot be poured into the glass because the hemispherical bump occupies that space. 6. Concept note: This is the reverse of a 'combination of solids' volume problem — here a solid is REMOVED (or rather, a solid intrusion reduces the usable volume) rather than added.

Key Points

  • Glass dimensions: r=2.5r = 2.5 cm, h=10h = 10 cm
  • Apparent capacity =πr2h=196.25= \pi r^2 h = 196.25 cm3^3 (full cylinder volume)
  • Hemispherical bump volume =23πr3≈32.71= \frac{2}{3}\pi r^3 \approx 32.71 cm3^3
  • Actual capacity =196.25−32.71=163.54= 196.25 - 32.71 = 163.54 cm3^3
  • This is volume subtraction: an intrusion reduces available space, unlike combination where volumes add

Common Mistakes

  • Adding the hemisphere volume instead of subtracting it — the bump takes space AWAY from the glass, reducing capacity
  • Using the height of the glass as 1010 cm for the cylinder but forgetting the height of the hemisphere is not subtracted from the cylinder height — the hemisphere sits at the bottom, inside the 1010 cm height

Related Questions

Q5

A vessel is in the form of an inverted cone. Its height is 88 cm and the radius of its top, which is open, is 55 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.50.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Q4

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 22 cm and the diameter of the base is 44 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π=3.14\pi = 3.14)

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