CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 16

NCERT Exercise

Question 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of \( \pi \).
Answer:

Radius \( r = 1 \) cm and height of the cone \( h = 1 \) cm.
Volume \( = \) volume of cone \( + \) volume of hemisphere \( = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi + \frac{2}{3}\pi = \pi \) cm³.
Hence, the volume of the solid is \( \pi \) cm³.
In simple words: Add the volumes of the cone and the hemisphere.

Teacher's Note:
1. Keep the answer in terms of pi as asked.
2. Both parts have radius 1 cm.

Question 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Answer:

Radius \( r = \frac{3}{2} \) cm. Height of each cone \( = 2 \) cm, so the height of the cylinder \( = 12 - 2 \times 2 = 8 \) cm.
Volume of air \( = \) volume of cylinder \( + 2 \times \) volume of a cone \( = \pi r^2 \times 8 + 2 \times \frac{1}{3}\pi r^2 \times 2 = \pi\left(\frac{3}{2}\right)^2\left(8 + \frac{4}{3}\right) = 21\pi \) cm³.
\( = 21 \times \frac{22}{7} = 66 \) cm³.
Hence, the volume of air in the model is 66 cm³.
In simple words: Cylinder plus two identical cones.

Teacher's Note:
1. Subtract the heights of both cones from the total length.
2. The air fills the whole model.

Question 3. A Gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 Gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.
Answer:

Radius \( r = 1.4 \) cm. Height of the cylindrical part \( = 5 - 2 \times 1.4 = 2.2 \) cm.
Volume of one Gulab jamun \( = \pi r^2 h + 2 \times \frac{2}{3}\pi r^3 = \frac{22}{7} \times 1.96 \times 2.2 + \frac{4}{3} \times \frac{22}{7} \times 2.744 \approx 13.55 + 11.50 = 25.05 \) cm³.
Syrup in one Gulab jamun \( = 30\% \) of 25.05 \( \approx 7.515 \) cm³. In 45 Gulab jamuns \( = 45 \times 7.515 \approx 338 \) cm³.
Hence, the syrup in 45 Gulab jamuns is about 338 cm³.
In simple words: Volume of one piece first, then take 30 percent and multiply by 45.

Teacher's Note:
1. Two hemispherical ends make one full sphere of the same radius.
2. Use the cylinder length 5 - 2.8 = 2.2 cm.

Question 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Answer:
Volume of the cuboid \( = 15 \times 10 \times 3.5 = 525 \) cm³.
Volume of one conical depression \( = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 \) cm³, so four depressions \( = 4 \times \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4 \approx 1.47 \) cm³.
Volume of wood \( = 525 - 1.47 = 523.53 \) cm³.
Hence, the volume of wood in the stand is 523.53 cm³.
In simple words: Subtract the four cone-shaped hollows from the cuboid.

Teacher's Note:
1. Volume of wood = cuboid minus four cones.
2. Keep two decimal places.

Question 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Answer:

Volume of the cone \( = \frac{1}{3}\pi(5)^2(8) = \frac{200}{3}\pi \) cm³. Water that flows out \( = \frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi \) cm³.
Volume of one lead shot \( = \frac{4}{3}\pi(0.5)^3 = \frac{1}{6}\pi \) cm³.
Number of lead shots \( = \frac{\frac{50}{3}\pi}{\frac{1}{6}\pi} = 100 \).
Hence, 100 lead shots were dropped.
In simple words: Water pushed out equals the total volume of the lead shots.

Teacher's Note:
1. The volume displaced equals the volume of the shots.
2. Pi cancels.

Question 6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use \( \pi = 3.14 \))
Answer:

Bigger cylinder: \( R = 12 \) cm, \( H = 220 \) cm, volume \( = 3.14 \times 144 \times 220 = 99475.2 \) cm³. Smaller cylinder: \( r = 8 \) cm, \( h = 60 \) cm, volume \( = 3.14 \times 64 \times 60 = 12057.6 \) cm³.
Volume of the pole \( = 99475.2 + 12057.6 = 111532.8 \) cm³.
Mass \( = 111532.8 \times 8 = 892262.4 \) g \( \approx 892.26 \) kg.
Hence, the mass of the pole is about 892.26 kg.
In simple words: Add the volumes and multiply by the density.

Teacher's Note:
1. Convert grams to kilograms at the end.
2. The base diameter 24 cm means radius 12 cm.

Question 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Answer:

Volume of the cylinder \( = \pi r^2 h = \frac{22}{7} \times 60^2 \times 180 \) cm³.
Volume of the solid \( = \frac{1}{3}\pi r^2 \times 120 + \frac{2}{3}\pi r^3 = \pi \times 60^2\left(40 + 40\right) \) cm³.
Water left \( = \pi \times 60^2 \times (180 - 40 - 40) = \frac{22}{7} \times 3600 \times 100 \approx 1131428.57 \) cm³ \( \approx 1.13 \) m³.
Hence, the volume of water left is about 1.13 m³.
In simple words: Water left is the cylinder's volume minus the volume of the immersed solid.

Teacher's Note:
1. Cone plus hemisphere together take up 80 cm of the height in effect.
2. 1 m³ = 1000000 cm³.

Question 8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and \( \pi = 3.14 \).
Answer:

Cylindrical neck: \( r = 1 \) cm, \( h = 8 \) cm. Sphere: radius \( = \frac{8.5}{2} = 4.25 \) cm.
Volume \( = \pi r^2 h + \frac{4}{3}\pi R^3 = 3.14 \times 1 \times 8 + \frac{4}{3} \times 3.14 \times (4.25)^3 \approx 25.12 + 321.39 = 346.51 \) cm³.
The child measured 345 cm³, which is not correct. The correct volume is 346.51 cm³.
In simple words: Add the volumes of the neck and the spherical part.

Teacher's Note:
1. Use the inside measurements as given.
2. Compare the computed volume with the child's value.