CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 08

Mathematics Assignment No. 07 - Volumes And Surface Area

Q1. A cylindrical jar of r = 6 cm contains water. Iron spheres, each of radius 1.5 cm, are immersed in the water. How many spheres are necessary to raise the level of the water by 2 cm.
Answer: Radius of the cylindrical jar (\(R\)) = 6 cm
Rise in water level (\(h\)) = 2 cm
Volume of water raised in the cylinder = \(\pi R^2 h = \pi \times (6)^2 \times 2 = 72\pi \, \text{cm}^3\)
Radius of each iron sphere (\(r\)) = 1.5 cm = \(\frac{3}{2} \, \text{cm}\)
Volume of one iron sphere = \(\frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times \left(\frac{3}{2}\right)^3 = \frac{4}{3} \times \pi \times \frac{27}{8} = \frac{9}{2}\pi \, \text{cm}^3\)
Number of spheres required = \(\frac{\text{Volume of water raised}}{\text{Volume of one sphere}} = \frac{72\pi}{\frac{9}{2}\pi} = \frac{72 \times 2}{9} = 16\)
In simple words: Divide the volume of the water pushed up in the cylinder by the volume of a single small sphere to find the total number of spheres needed.

Exam Tip: Do not substitute the value of \(\pi\) early in the calculation, as it usually cancels out in ratio problems.

Q2. 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 rim. 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: Radius of the conical vessel (\(R\)) = 5 cm
Height of the conical vessel (\(H\)) = 8 cm
Volume of the conical vessel = \(\frac{1}{3} \pi R^2 H = \frac{1}{3} \pi (5)^2 (8) = \frac{200}{3}\pi \, \text{cm}^3\)
Volume of water that flows out = \(\frac{1}{4} \times \text{Volume of cone} = \frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi \, \text{cm}^3\)
Radius of each spherical lead shot (\(r\)) = 0.5 cm = \(\frac{1}{2} \, \text{cm}\)
Volume of one lead shot = \(\frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{1}{2}\right)^3 = \frac{4}{3} \pi \times \frac{1}{8} = \frac{1}{6}\pi \, \text{cm}^3\)
Number of lead shots = \(\frac{\text{Volume of water flowed out}}{\text{Volume of one lead shot}} = \frac{\frac{50}{3}\pi}{\frac{1}{6}\pi} = \frac{50}{3} \times 6 = 100\)
In simple words: Find the total volume of the cone, calculate one-fourth of that volume, and then divide it by the volume of one lead shot.

Exam Tip: Carefully read fractions such as "one fourth of the water flows out" to determine the exact volume overflowed.

Q3. A hemispherical bowl of internal radius 15 cm contains a liquid. The liquid is to be filled into cylindrical shaped bottles of diameter 5 cm and height 6 cm. How many bottles are necessary to empty the bowl?
Answer: Radius of the hemispherical bowl (\(R\)) = 15 cm
Volume of the liquid in the bowl = \(\frac{2}{3} \pi R^3 = \frac{2}{3} \pi (15)^3 = \frac{2}{3} \pi \times 3375 = 2250\pi \, \text{cm}^3\)
Diameter of the cylindrical bottle = 5 cm, so radius (\(r\)) = \(\frac{5}{2} \, \text{cm}\)
Height of the cylindrical bottle (\(h\)) = 6 cm
Volume of one cylindrical bottle = \(\pi r^2 h = \pi \left(\frac{5}{2}\right)^2 \times 6 = \pi \times \frac{25}{4} \times 6 = \frac{75}{2}\pi \, \text{cm}^3\)
Number of bottles required = \(\frac{\text{Volume of hemispherical bowl}}{\text{Volume of one cylindrical bottle}} = \frac{2250\pi}{\frac{75}{2}\pi} = \frac{2250 \times 2}{75} = 60\)
In simple words: Divide the total volume of liquid in the large bowl by the capacity of one small bottle.

Exam Tip: Always convert diameter to radius before applying cylinder and sphere volume formulas.

Q4. The largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.
Answer: Radius of the cylinder (\(R\)) = 7 cm
Height of the cylinder (\(H\)) = 14 cm
For the largest sphere to be carved out of the cylinder, the diameter of the sphere must equal the diameter of the cylinder (or height, whichever is smaller). Here, diameter = \(2 \times 7 = 14\) cm, which matches the height of 14 cm.
Therefore, diameter of the sphere = 14 cm, and its radius (\(r\)) = 7 cm
Volume of the sphere = \(\frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (7)^3 = \frac{4}{3} \times \frac{22}{7} \times 343 = \frac{30184}{21} = 1437.33 \, \text{cm}^3 \approx 1437 \, \text{cm}^3\)
In simple words: The biggest sphere that fits inside a cylinder will have a radius equal to the cylinder's radius.

Exam Tip: Check whether the height or the diameter of the cylinder limits the sphere's dimensions.

Q5. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the Volume and S.A. of the solid.
Answer: Diameter of the cylinder = 7 cm, so radius (\(r\)) = \(\frac{7}{2} = 3.5 \, \text{cm}\)
Total height of the solid (\(H\)) = 19 cm
Height of the cylindrical part (\(h\)) = Total height - Radius of top hemisphere - Radius of bottom hemisphere = \(19 - 3.5 - 3.5 = 12 \, \text{cm}\)
1. Volume of the solid = Volume of cylinder + Volume of two hemispheres
Volume = \(\pi r^2 h + 2 \left(\frac{2}{3} \pi r^3\right) = \pi r^2 \left(h + \frac{4}{3}r\right)\)
Volume = \(\frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times \left(12 + \frac{4}{3} \times \frac{7}{2}\right)\)
Volume = \(\frac{22}{7} \times \frac{49}{4} \times \left(12 + \frac{14}{3}\right) = \frac{77}{2} \times \frac{50}{3} = \frac{3850}{6} = 641.67 \, \text{cm}^3\)
2. Total Surface Area of the solid = Curved surface area of cylinder + Curved surface area of two hemispheres
S.A. = \(2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)\)
S.A. = \(2 \times \frac{22}{7} \times \frac{7}{2} \times \left(12 + 2 \times \frac{7}{2}\right) = 22 \times (12 + 7) = 22 \times 19 = 418 \, \text{cm}^2\)
In simple words: Combine the formulas for a cylinder and two hemispheres to find both total volume and surface area.

Exam Tip: Subtract the radii of the hemispherical ends from the total height to find the exact height of the middle cylindrical section.

Q6. A solid toy is in the form of a right circular cylinder with hemispherical shape at one end and a cone at the other end. Their common diameter is 4.2 cm and the height of the cylindrical and conical portions are 12 cm and 7 cm respectively. Find the volume of the solid toy.
Answer: Common diameter = 4.2 cm, so radius (\(r\)) = \(\frac{4.2}{2} = 2.1 \, \text{cm} = \frac{21}{10} \, \text{cm}\)
Height of the cylindrical portion (\(h_1\)) = 12 cm
Height of the conical portion (\(h_2\)) = 7 cm
Radius of the hemisphere = \(r = 2.1 \, \text{cm}\)
Total volume of the toy = Volume of hemisphere + Volume of cylinder + Volume of cone
Volume = \(\frac{2}{3} \pi r^3 + \pi r^2 h_1 + \frac{1}{3} \pi r^2 h_2\)
Volume = \(\pi r^2 \left(\frac{2}{3}r + h_1 + \frac{1}{3}h_2\right)\)
Volume = \(\frac{22}{7} \times \left(\frac{21}{10}\right)^2 \times \left(\frac{2}{3} \times \frac{21}{10} + 12 + \frac{1}{3} \times 7\right)\)
Volume = \(\frac{22}{7} \times \frac{441}{100} \times \left(\frac{14}{10} + 12 + \frac{7}{3}\right)\)
Volume = \(\frac{22 \times 63}{100} \times \left(\frac{7}{5} + 12 + \frac{7}{3}\right) = \frac{1386}{100} \times \left(\frac{21 + 180 + 35}{15}\right)\)
Volume = \(13.86 \times \frac{236}{15} = 13.86 \times 15.7333 = 218.06 \, \text{cm}^3\)
In simple words: Add the individual volumes of the hemisphere, cylinder, and cone together.

Q7. A cylindrical tub of r = 5 cm and length 9.8 cm is full of water. A solid in the shape of a cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of water left in the tub.
Answer: Radius of the cylindrical tub (\(R\)) = 5 cm, length/height (\(H\)) = 9.8 cm
Volume of the cylindrical tub = \(\pi R^2 H = \frac{22}{7} \times (5)^2 \times 9.8 = \frac{22}{7} \times 25 \times 9.8 = 22 \times 25 \times 1.4 = 770 \, \text{cm}^3\)
Radius of the hemisphere (\(r\)) = 3.5 cm = \(\frac{7}{2} \, \text{cm}\)
Height of the cone (\(h\)) = 5 cm
Volume of the solid immersed = Volume of hemisphere + Volume of cone
Volume = \(\frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r + h)\)
Volume = \(\frac{1}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times \left(2 \times \frac{7}{2} + 5\right)\)
Volume = \(\frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} \times (7 + 5) = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} \times 12 = \frac{22 \times 7 \times 12}{4} = 154 \, \text{cm}^3\)
Volume of water left in the tub = Total volume of tub - Volume of solid immersed
Volume left = \(770 - 154 = 616 \, \text{cm}^3\)
In simple words: Subtract the volume of the immersed toy from the total volume of the cylindrical tub to find the remaining water.

Exam Tip: The volume of water displaced equals the exact volume of the solid submerged inside the liquid.

Q8. The diameters of the internal and external surfaces of a hollow hemispherical shell are 6 cm and 10 cm respectively. If it is melted and recasted into a solid cylinder of diameter 14 cm, find the height of the cylinder.
Answer: Internal diameter = 6 cm, internal radius (\(r\)) = 3 cm
External diameter = 10 cm, external radius (\(R\)) = 5 cm
Volume of the hollow hemispherical shell = \(\frac{2}{3} \pi (R^3 - r^3) = \frac{2}{3} \pi (5^3 - 3^3) = \frac{2}{3} \pi (125 - 27) = \frac{2}{3} \pi (98) = \frac{196}{3}\pi \, \text{cm}^3\)
Diameter of the solid cylinder = 14 cm, so radius (\(r_1\)) = 7 cm
Let the height of the cylinder be \(h\).
Volume of the solid cylinder = \(\pi r_1^2 h = \pi (7)^2 h = 49\pi h\)
Since the shell is melted and recasted, their volumes are equal:
\(49\pi h = \frac{196}{3}\pi\)
\(h = \frac{196}{3 \times 49} = \frac{4}{3} = 1.33 \, \text{cm}\)
In simple words: Equate the volume of the hollow shell to the volume of the new cylinder to solve for height.

Exam Tip: When a solid is melted and recasted, equate the volume of the initial shape to the final shape.

Q9. A well with diameter 3 m is dug 14 m deep. Earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Answer: Diameter of the well = 3 m, radius (\(r\)) = 1.5 m = \(\frac{3}{2} \, \text{m}\)
Depth of the well (\(h\)) = 14 m
Volume of earth dug out = \(\pi r^2 h = \pi \left(\frac{3}{2}\right)^2 \times 14 = \pi \times \frac{9}{4} \times 14 = \frac{63}{2}\pi \, \text{m}^3\)
Width of embankment = 4 m
Inner radius of embankment (\(r\)) = 1.5 m
Outer radius of embankment (\(R\)) = Inner radius + Width = \(1.5 + 4 = 5.5 \, \text{m} = \frac{11}{2} \, \text{m}\)
Area of the circular ring (embankment base) = \(\pi (R^2 - r^2) = \pi \left[\left(\frac{11}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right] = \pi \left(\frac{121 - 9}{4}\right) = \frac{112}{4}\pi = 28\pi \, \text{m}^2\)
Height of the embankment (\(H\)) = \(\frac{\text{Volume of earth}}{\text{Area of embankment base}} = \frac{\frac{63}{2}\pi}{28\pi} = \frac{63}{56} = \frac{9}{8} \, \text{m}\)
In simple words: Divide the total volume of dug-out soil by the base area of the circular embankment ring to find its height.

Q10. The slant height of the frustum of a cone is 4 cm and the perimeters of circular ends are 18 cm and 6 cm. Find the C.S.A. of the frustum.
Answer: Slant height of the frustum (\(l\)) = 4 cm
Perimeter of the upper circular end (\(2\pi r_1\)) = 18 cm
Perimeter of the lower circular end (\(2\pi r_2\)) = 6 cm
Curved Surface Area (C.S.A.) of the frustum = \(\pi (r_1 + r_2) l = \frac{1}{2} (2\pi r_1 + 2\pi r_2) l\)
C.S.A. = \(\frac{1}{2} \times (18 + 6) \times 4 = \frac{1}{2} \times 24 \times 4 = 12 \times 4 = 48 \, \text{cm}^2\)
In simple words: Multiply half the sum of the boundary perimeters by the slant height to find the curved surface area.

Exam Tip: Remember the alternative formula for the C.S.A. of a frustum: \(\pi(r_1 + r_2)l = \frac{1}{2}(2\pi r_1 + 2\pi r_2)l\).