GSEB Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.2

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: Let the radius be \( r \) cm. The height of the cone is equal to its radius, so \( h = r \). Given that the radius \( r = 1 \) cm, this means the height \( h = 1 \) cm as well. The volume of the solid is calculated by adding the volume of the cone and the volume of the hemisphere.
Volume of solid \( = \) Volume of cone \( + \) Volume of hemisphere
\( = \frac {1}{3}\pi r^2 h + \frac {2}{3}\pi r^3 \)
Substitute the values \( r=1 \) cm and \( h=1 \) cm:
\( = \frac {1}{3}\pi (1)^2 (1) + \frac {2}{3}\pi (1)^3 \)
\( = \frac {1}{3}\pi + \frac {2}{3}\pi \)
\( = \frac { \pi + 2\pi }{3} \)
\( = \frac {3\pi}{3} \)
\( = \pi \) cm\( ^3 \)
The volume of the solid is \( \pi \) cm\( ^3 \).
In simple words: We find the volume by adding the cone's volume to the hemisphere's volume. Since the radius and cone's height are both 1 cm, we plug these values into the formulas and simplify to get the final volume in terms of pi.

Exam Tip: Remember to clearly state the dimensions of each part (radius, height) and use the correct formulas for the volume of a cone and a hemisphere. Combine the terms carefully to get the final answer in terms of \( \pi \).

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 aluminum 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 dimension of the model to be nearly the same).
Answer: First, we determine the dimensions of each part of the model. The diameter of the cylindrical model is 3 cm, so its radius \( r = \frac {3}{2} \) cm. The total length of the model is 12 cm. Since each cone has a height of 2 cm, the combined height of the two cones is \( 2 + 2 = 4 \) cm. Therefore, the height of the cylindrical part \( h_1 = 12 - 4 = 8 \) cm. The height of each cone \( h_2 = 2 \) cm, and the radius of both the cylindrical and conical parts is \( r = \frac {3}{2} \) cm.
To find the total volume of air, we add the volume of the cylindrical part and the volume of the two conical parts.
Volume of air contained in the model \( = \) Volume of cylindrical part \( + \) Volume of two conical parts
\( = \pi r^2 h_1 + 2 \times \frac {1}{3}\pi r^2 h_2 \)
Substitute the values: \( r = \frac {3}{2} \) cm, \( h_1 = 8 \) cm, and \( h_2 = 2 \) cm.
\( = \pi \left( \frac{3}{2} \right)^2 \times 8 + 2 \times \frac {1}{3}\pi \left( \frac{3}{2} \right)^2 \times 2 \)
\( = \pi \times \frac {9}{4} \times 8 + \frac {2}{3}\pi \times \frac {9}{4} \times 2 \)
\( = 18\pi + 3\pi \)
\( = 21\pi \)
Using \( \pi = \frac {22}{7} \):
\( = 21 \times \frac {22}{7} = 3 \times 22 = 66 \) cm\( ^3 \)
The total volume of air contained in the model is 66 cm\( ^3 \).
In simple words: To find the total air inside the model, we calculate the volume of the central cylinder and add it to the volumes of the two cones at each end. First, find the radius from the diameter, then the height of the cylinder by subtracting the cone heights from the total length. Add all the individual volumes.

Exam Tip: For composite shapes, always break the model into simpler geometric figures (cylinder, cone, hemisphere) and calculate their individual volumes. Pay close attention to how the dimensions (radius, height) relate to each other and to the overall shape.

Question 3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately what volume of 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: We begin by finding the dimensions of a single gulab jamun. The total length is 5 cm and the diameter is 2.8 cm, so the radius \( r = \frac {2.8}{2} = 1.4 \) cm. Since there are hemispherical ends, the height of the cylindrical portion \( h = 5 - (1.4 + 1.4) = 5 - 2.8 = 2.2 \) cm.
The volume of one gulab jamun is the sum of the volume of the cylindrical part and the volume of two hemispherical ends.
Volume of one gulab jamun \( = \) Volume of cylindrical portion \( + \) Volume of two hemispherical ends
\( = \pi r^2 h + 2 \times \frac {2}{3}\pi r^3 \)
\( = \pi r^2 h + \frac {4}{3}\pi r^3 \)
Substitute \( \pi = \frac {22}{7} \), \( r = 1.4 \) cm, \( h = 2.2 \) cm:
\( = \frac {22}{7} \times (1.4)^2 \times 2.2 + \frac {4}{3} \times \frac {22}{7} \times (1.4)^3 \)
\( = \frac {22}{7} \times 1.96 \times 2.2 + \frac {4}{3} \times \frac {22}{7} \times 2.744 \)
\( = 13.552 + 11.498 \) (approximately)
\( = 25.05 \) cm\( ^3 \)
Next, we calculate the total volume for 45 gulab jamuns.
Volume of 45 gulab jamuns \( = 45 \times 25.05 = 1127.25 \) cm\( ^3 \).
Finally, we find the volume of sugar syrup, which is 30% of the total volume.
Volume of sugar syrup in 45 gulab jamuns \( = 30\% \) of \( 1127.25 \)
\( = \frac {30}{100} \times 1127.25 = 338.175 \)
\( \approx 338.2 \) cm\( ^3 \)
Approximately 338.2 cm\( ^3 \) of sugar syrup would be found in 45 gulab jamuns.
In simple words: First, calculate the volume of one gulab jamun by combining a cylinder and two hemispheres. Then, multiply this by 45 to get the total volume of all gulab jamuns. Finally, take 30% of that total volume to find the amount of sugar syrup.

Exam Tip: Remember to account for the two hemispherical ends when determining the cylindrical height. Make sure to calculate the total volume for all items before finding the percentage of syrup, and round to an appropriate decimal place if specified.

Question 4. A pen stand made of wood is in the shape of a cuboid with four conical depression to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Answer: We first determine the volume of the cuboidal pen stand. The length \( l = 15 \) cm, breadth \( b = 10 \) cm, and height \( h = 3.5 \) cm. The volume of the cuboidal wood is given by \( l \times b \times h \).
Volume of cuboidal wood \( = 15 \times 10 \times 3.5 = 525 \) cm\( ^3 \).
Next, we find the volume of each conical depression. The radius of each depression \( r = 0.5 \) cm and the depth (height) \( h = 1.4 \) cm. The volume of one conical depression is \( \frac {1}{3}\pi r^2 h \).
Volume of one conical depression \( = \frac {1}{3}\pi (0.5)^2 (1.4) \).
Since there are four such depressions, we calculate their total volume.
Volume of four conical depressions \( = 4 \times \frac {1}{3}\pi r^2 h \)
\( = 4 \times \frac {1}{3} \times \frac {22}{7} \times (0.5)^2 \times 1.4 \)
\( = \frac {4}{3} \times \frac {22}{7} \times 0.25 \times 1.4 \)
\( = \frac {88}{21} \times 0.35 \)
\( = \frac {4.4}{3} \) cm\( ^3 \)
\( \approx 1.47 \) cm\( ^3 \)
To find the volume of wood in the stand, we subtract the total volume of the conical depressions from the volume of the cuboidal wood.
Volume of wooden pen stand \( = \) Volume of cuboidal wood \( - \) Volume of conical depressions
\( = 525 - 1.47 \)
\( = 523.53 \) cm\( ^3 \)
The volume of wood in the entire stand is approximately 523.53 cm\( ^3 \).
In simple words: First, figure out the total volume of the wooden cuboid. Then, calculate the volume of one pen-holding cone and multiply it by four for all the depressions. Finally, subtract the total volume of the cone holes from the cuboid's volume to get the amount of wood remaining.

Exam Tip: When dealing with objects with carved-out portions, remember to subtract the volume of the removed parts from the total volume of the original solid. Keep track of units and use the correct value of \( \pi \) if specified.

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: First, we find the volume of water in the conical vessel. For the cone, the height \( h = 8 \) cm and the radius \( r = 5 \) cm.
Volume of water filled in the cone \( = \frac {1}{3}\pi r^2 h = \frac {1}{3}\pi \times 5^2 \times 8 \)
\( = \frac {1}{3}\pi \times 25 \times 8 = \frac {200}{3}\pi \) cm\( ^3 \).
Next, we find the volume of water that flows out, which is one-fourth of the total water.
Volume of water that flows out \( = \frac {1}{4} \times \frac {200}{3}\pi = \frac {50}{3}\pi \) cm\( ^3 \).
Now, we calculate the volume of a single lead shot. Each lead shot is a sphere with radius \( r_{shot} = 0.5 \) cm.
Volume of 1 lead shot \( = \frac {4}{3}\pi r_{shot}^3 = \frac {4}{3}\pi \times (0.5)^3 \)
\( = \frac {4}{3}\pi \times 0.125 = \frac {0.5}{3}\pi \) cm\( ^3 \).
To find the number of lead shots, we divide the volume of displaced water by the volume of one lead shot.
Number of lead shots \( = \frac{\text{Volume of water flows out}}{\text{Volume of 1 lead shot}} \)
\( = \frac{\frac{50}{3}\pi}{\frac{0.5}{3}\pi} \)
\( = \frac{50}{0.5} \)
\( = 100 \)
Therefore, 100 lead shots were dropped into the vessel.
In simple words: First, calculate how much water is in the cone. Then, find out how much water overflowed, which is a quarter of the total. Next, calculate the volume of one small lead ball. Finally, divide the volume of overflowed water by the volume of one lead ball to find how many balls were dropped.

Exam Tip: This problem involves the principle of water displacement. The volume of water that flows out is equal to the total volume of the objects dropped into the vessel. Make sure to use the correct volume formulas for both the cone and the sphere, and to simplify \( \pi \) terms if they cancel out.

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 cm\( ^3 \) of iron has approximately 8 g mass. (Use \( \pi = 3.14 \))
Answer: We need to calculate the total volume of the iron pole, which consists of two cylinders. For the lower cylinder (cylinder A), the diameter is 24 cm, so the radius \( r_1 = \frac{24}{2} = 12 \) cm. Its height \( h_1 = 220 \) cm.
For the upper cylinder (cylinder B), the radius \( r_2 = 8 \) cm and its height \( h_2 = 60 \) cm.
The total volume of the pole is the sum of the volumes of these two cylinders.
Volume of pole \( = \) Volume of cylinder A \( + \) Volume of cylinder B
\( = \pi r_1^2 h_1 + \pi r_2^2 h_2 \)
\( = \pi (r_1^2 h_1 + r_2^2 h_2) \)
Substitute \( \pi = 3.14 \), \( r_1 = 12 \) cm, \( h_1 = 220 \) cm, \( r_2 = 8 \) cm, \( h_2 = 60 \) cm:
\( = 3.14 (12^2 \times 220 + 8^2 \times 60) \)
\( = 3.14 (144 \times 220 + 64 \times 60) \)
\( = 3.14 (31680 + 3840) \)
\( = 3.14 \times 35520 \)
\( = 111532.8 \) cm\( ^3 \)
Now, we calculate the mass of the pole using the given density: 1 cm\( ^3 \) of iron has approximately 8 g mass.
Mass of pole \( = \) Volume \( \times \) Density
\( = 111532.8 \times 8 \) g
\( = 892262.4 \) g
To convert this mass to kilograms, we divide by 1000.
\( = \frac{892262.4}{1000} \) kg
\( = 892.2624 \) kg
The mass of the pole is approximately 892.26 kg.
In simple words: First, calculate the volume of each cylinder that makes up the pole. Add these volumes together to get the pole's total volume. Then, multiply this total volume by the iron's density (8 grams per cubic centimeter) to find the mass in grams. Finally, convert this mass into kilograms.

Exam Tip: For problems involving mass and density, remember that Mass = Volume \( \times \) Density. Ensure that all dimensions are in consistent units before calculating volume. If the final answer requires a different unit (like kg from g), perform the conversion at the end.

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: We need to find the volume of water left in the cylinder after the composite solid is placed inside. This is done by subtracting the volume of the solid from the volume of the cylinder.
First, let's calculate the volume of the composite solid (cone on a hemisphere).
For the cone: height \( h_c = 120 \) cm, radius \( r = 60 \) cm.
For the hemisphere: radius \( r = 60 \) cm.
Volume of solid \( = \) Volume of cone \( + \) Volume of hemisphere
\( = \frac {1}{3}\pi r^2 h_c + \frac {2}{3}\pi r^3 \)
\( = \frac {1}{3}\pi r^2 (h_c + 2r) \)
Using \( \pi = 3.14 \):
\( = \frac {1}{3} \times 3.14 \times (60)^2 \times (120 + 2 \times 60) \)
\( = \frac {1}{3} \times 3.14 \times 3600 \times (120 + 120) \)
\( = \frac {1}{3} \times 3.14 \times 3600 \times 240 \)
\( = 3.14 \times 1200 \times 240 \)
\( = 3.14 \times 288000 \)
\( = 904320 \) cm\( ^3 \)
Next, we calculate the volume of the cylindrical vessel.
For the cylinder: radius \( r_{cyl} = 60 \) cm, height \( h_{cyl} = 180 \) cm.
Volume of cylindrical vessel \( = \pi r_{cyl}^2 h_{cyl} \)
\( = 3.14 \times (60)^2 \times 180 \)
\( = 3.14 \times 3600 \times 180 \)
\( = 3.14 \times 648000 \)
\( = 2034720 \) cm\( ^3 \)
Finally, the volume of water left in the cylinder is the difference between the cylinder's volume and the solid's volume.
Volume of water left \( = \) Volume of cylindrical vessel \( - \) Volume of solid
\( = 2034720 - 904320 \)
\( = 1130400 \) cm\( ^3 \)
Converting to cubic meters (since \( 1 \text{ m} = 100 \text{ cm} \), \( 1 \text{ m}^3 = 100^3 \text{ cm}^3 = 1,000,000 \text{ cm}^3 \)):
\( = \frac{1130400}{1000000} \) m\( ^3 \)
\( = 1.1304 \) m\( ^3 \)
The volume of water left in the cylinder is \( 1.1304 \) m\( ^3 \).
In simple words: First, find the volume of the solid object by adding the cone's volume to the hemisphere's volume. Then, calculate the total volume of the water-filled cylinder. Subtract the solid's volume from the cylinder's volume to find the remaining water. Finally, convert the answer from cubic centimeters to cubic meters.

Exam Tip: Problems involving water displacement in a container require calculating the volume of the container and the volume of the immersed object. The volume of water displaced (or remaining) is determined by subtraction. Always ensure units are consistent or convert them as needed at the end.

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 cm\( ^3 \). Check whether she is correct, taking the above as the inside measurements, and \( \pi = 3.14 \).
Answer: We need to calculate the actual volume of the glass vessel and compare it to the child's measurement.
First, consider the cylindrical neck:
Diameter \( = 2 \) cm, so radius \( r_{cyl} = \frac {2}{2} = 1 \) cm.
Length (height) \( h_{cyl} = 8 \) cm.
Volume of cylindrical portion \( = \pi r_{cyl}^2 h_{cyl} \)
Using \( \pi = 3.14 \):
\( = 3.14 \times (1)^2 \times 8 \)
\( = 3.14 \times 1 \times 8 = 25.12 \) cm\( ^3 \).
Next, consider the spherical part:
Diameter \( = 8.5 \) cm, so radius \( r_{sph} = \frac {8.5}{2} = 4.25 \) cm.
Volume of spherical part \( = \frac {4}{3}\pi r_{sph}^3 \)
Using \( \pi = 3.14 \):
\( = \frac {4}{3} \times 3.14 \times (4.25)^3 \)
\( = \frac {4}{3} \times 3.14 \times 76.765625 \)
\( = \frac {964.176}{3} \) (approximately)
\( = 321.392 \) cm\( ^3 \) (approximately).
The total volume of the vessel is the sum of the cylindrical and spherical parts.
Total volume of vessel \( = \) Volume of cylindrical portion \( + \) Volume of spherical part
\( = 25.12 + 321.392 \)
\( = 346.512 \) cm\( ^3 \)
The child's measurement for the volume was 345 cm\( ^3 \).
Comparing the calculated volume to the child's measurement, \( 346.512 \text{ cm}^3 \neq 345 \text{ cm}^3 \).
Therefore, the child's measurement is not correct. The accurate volume is approximately \( 346.51 \) cm\( ^3 \).
In simple words: Calculate the volume of the cylindrical neck and the spherical body separately. Add these two volumes together to get the total volume of the glass vessel. Compare this calculated total volume with the child's measured volume of 345 cubic centimeters to see if her measurement is correct.

Exam Tip: When verifying a measurement, ensure you calculate the theoretical volume precisely by breaking down the object into its basic geometric shapes (cylinder, sphere). Use the given value of \( \pi \) and round off only at the final step to maintain accuracy. Always state your conclusion clearly.