NCERT Solutions Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.2

NCERT Solutions for Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.2 are designed to help students apply the concepts of surface area and volume of composite 3D shapes. This exercise introduces the idea of finding the combined surface area and volume of objects formed by the combination of cylinders, cones, spheres, and hemispheres.

Exercise 13.2 focuses on:

NCERT Solutions Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.2

Surface Areas and Volumes

Medium

Q.

Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

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Surface Areas and Volumes

Medium

Q.

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

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Surface Areas and Volumes

Medium

Q.

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

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Surface Areas and Volumes

Medium

Q.

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

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Surface Areas and Volumes

Medium

Q.

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

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Surface Areas and Volumes

Medium

Q.

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see the following figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

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Surface Areas and Volumes

Medium

Q.

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 (see the following figure).

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Surface Areas and Volumes

Medium

Q.

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.

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Surface Areas and Volumes

Medium

Q.

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 8g mass. (Use π = 3.14)

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Surface Areas and Volumes

Medium

Q.

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the following figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

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Surface Areas and Volumes

Medium

Q.

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the following figure).

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• Surface area and volume of composite solids such as a cone on a cylinder or sphere in a hemisphere.

• Applying multiple formulas for the surface area and volume of different parts of the composite shape and combining them.

• Real-life applications of composite solids.

The solutions are presented step-by-step, helping students understand how to calculate the surface areas and volumes of complex shapes and solve problems efficiently for Class 10 exams.

Q1. 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 π.

Difficulty Level: Medium

Known/Given:

• The solid is made up of a cone and a hemisphere, both with radii equal to 1 cm and the height of the cone equal to its radius (1 cm).

Unknown:

• The volume of the solid.

Reasoning:

1. The total volume of the solid is the sum of the volume of the cone and the volume of the hemisphere.

2. The formulae for the volumes are:

   • Volume of the hemisphere:

     $\frac{2}{3} \pi r^3$32​πr3

   • Volume of the cone:

     $\frac{1}{3} \pi r^2 h$31​πr2h, where

     $r$r is the radius and

     $h$h is the height.

3. Volume of the solid:

    

   $$\text{Volume of solid} = \text{Volume of cone} + \text{Volume of hemisphere}$$Volume of solid=Volume of cone+Volume of hemisphere

   Since the radius

   $r = 1$r=1 cm and height of the cone

   $h = 1$h=1 cm:

    

   $$\text{Volume of the cone} = \frac{1}{3} \pi (1)^2 (1) = \frac{1}{3} \pi$$Volume of the cone=31​π(1)2(1)=31​π

   $$\text{Volume of the hemisphere} = \frac{2}{3} \pi (1)^3 = \frac{2}{3} \pi$$Volume of the hemisphere=32​π(1)3=32​π

   Therefore:

    

   $$\text{Total volume of the solid} = \frac{1}{3} \pi + \frac{2}{3} \pi = \pi \, \text{cm}^3$$Total volume of the solid=31​π+32​π=πcm3

Q2. 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 dimensions of the model to be nearly the same.)

Difficulty Level: Medium

Known/Given:

• The model is shaped like a cylinder with two cones attached to its two ends.

• Diameter of the model = 3 cm

• Length of the model = 12 cm

• Height of each cone = 2 cm

• The inner and outer dimensions are nearly the same.

Unknown:

• Volume of air contained in the model.

Reasoning:

1. The volume of the model includes the volume of the cylindrical part and both identical conical parts.

2. The formulae for the volumes are:

   • Volume of the cylinder:

     $\pi r^2 h_1$πr2h1​, where

     $r$r is the radius and

     $h_1$h1​ is the height of the cylinder.

   • Volume of the cone:

     $\frac{1}{3} \pi r^2 h_2$31​πr2h2​, where

     $r$r is the radius and

     $h_2$h2​ is the height of the cone.

3. Height of the cylindrical part:

    

   $$\text{Height of cylindrical part} = \text{Total height of the model} - 2 \times \text{Height of the cone} = 12 - 2 \times 2 = 8 \, \text{cm}$$Height of cylindrical part=Total height of the model−2×Height of the cone=12−2×2=8cm

4. Radius of the model:

    

   $$\text{Radius} = \frac{3}{2} = 1.5 \, \text{cm}$$Radius=23​=1.5cm

5. Volume of the model:

    

   $$\text{Volume of the cone} = \frac{1}{3} \pi (1.5)^2 (2) = \frac{1}{3} \pi (2.25) (2) = \frac{1}{3} \pi (4.5) = 1.5 \pi \, \text{cm}^3$$Volume of the cone=31​π(1.5)2(2)=31​π(2.25)(2)=31​π(4.5)=1.5πcm3

   $$\text{Volume of the cylinder} = \pi (1.5)^2 (8) = \pi (2.25) (8) = 18 \pi \, \text{cm}^3$$Volume of the cylinder=π(1.5)2(8)=π(2.25)(8)=18πcm3

6. Total volume of the model:

    

   $$\text{Total volume} = 2 \times \text{Volume of cone} + \text{Volume of the cylinder}$$Total volume=2×Volume of cone+Volume of the cylinder

   $$= 2 \times 1.5 \pi + 18 \pi = 3 \pi + 18 \pi = 21 \pi \, \text{cm}^3$$=2×1.5π+18π=3π+18π=21πcm3

Q3. 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.

Difficulty Level: Medium

Known/Given:

• Each Gulab jamun is shaped like a cylinder with two hemispherical ends.

• Length of each Gulab jamun = 5 cm

• Diameter of each Gulab jamun = 2.8 cm

• Syrup occupies 30% of the volume of each Gulab jamun.

• We need to find the syrup in 45 Gulab jamuns.

Unknown:

• Volume of syrup in 45 Gulab jamuns.

Reasoning:

1. The length of the cylindrical part is:

    

   $$\text{Length of cylindrical part} = \text{Length of Gulab jamun} - 2 \times \text{Radius of hemispherical part}$$Length of cylindrical part=Length of Gulab jamun−2×Radius of hemispherical part

   $$\text{Radius of cylindrical and hemispherical parts} = \frac{2.8}{2} = 1.4 \, \text{cm}$$Radius of cylindrical and hemispherical parts=22.8​=1.4cm

   $$\text{Length of cylindrical part} = 5 - 2 \times 1.4 = 5 - 2.8 = 2.2 \, \text{cm}$$Length of cylindrical part=5−2×1.4=5−2.8=2.2cm

2. Volume of 1 Gulab jamun:

   • Volume of the cylindrical part:

      

     $$V_{\text{cylinder}} = \pi r^2 h = \pi (1.4)^2 (2.2) = \pi (1.96)(2.2) = 4.312 \pi \, \text{cm}^3$$Vcylinder​=πr2h=π(1.4)2(2.2)=π(1.96)(2.2)=4.312πcm3

   • Volume of the two hemispherical parts:

      

     $$V_{\text{hemispheres}} = 2 \times \left( \frac{2}{3} \pi r^3 \right) = \frac{4}{3} \pi (1.4)^3 = \frac{4}{3} \pi (2.744) = 3.659 \pi \, \text{cm}^3$$Vhemispheres​=2×(32​πr3)=34​π(1.4)3=34​π(2.744)=3.659πcm3

3. Total volume of 1 Gulab jamun:

    

   $$V_{\text{Gulab jamun}} = 4.312 \pi + 3.659 \pi = 7.971 \pi \, \text{cm}^3$$VGulab jamun​=4.312π+3.659π=7.971πcm3

4. Total volume of syrup in 45 Gulab jamuns:

    

   $$V_{\text{syrup in 45}} = 0.30 \times 45 \times 7.971 \pi = 0.30 \times 358.695 \pi = 107.609 \pi \, \text{cm}^3$$Vsyrup in 45​=0.30×45×7.971π=0.30×358.695π=107.609πcm3

   $$\approx 337.76 \, \text{cm}^3$$≈337.76cm3

Q4. 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.

Difficulty Level: Medium

Known/Given:

• The pen stand is in the shape of a cuboid with four conical depressions.

• Dimensions of the cuboid: 15 cm × 10 cm × 3.5 cm

• Radius of each conical depression = 0.5 cm

• Depth of each conical depression = 1.4 cm

Unknown:

• Volume of wood in the entire pen stand.

Reasoning:

1. The volume of the stand is the volume of the cuboid minus the volume of the four conical depressions.

2. Volume of the cuboid:

    

   $$V_{\text{cuboid}} = l \times b \times h = 15 \times 10 \times 3.5 = 525 \, \text{cm}^3$$Vcuboid​=l×b×h=15×10×3.5=525cm3

3. Volume of one conical depression:

    

   $$V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (0.5)^2 (1.4) = \frac{1}{3} \pi (0.25)(1.4) = 0.1167 \pi \, \text{cm}^3$$Vcone​=31​πr2h=31​π(0.5)2(1.4)=31​π(0.25)(1.4)=0.1167πcm3

4. Total volume of the four conical depressions:

    

   $$V_{\text{4 cones}} = 4 \times 0.1167 \pi = 0.4668 \pi \, \text{cm}^3 \approx 1.465 \, \text{cm}^3$$V4 cones​=4×0.1167π=0.4668πcm3≈1.465cm3

5. Volume of wood in the stand:

    

   $$V_{\text{wood}} = V_{\text{cuboid}} - V_{\text{4 cones}} = 525 - 1.465 = 523.535 \, \text{cm}^3$$Vwood​=Vcuboid​−V4 cones​=525−1.465=523.535cm3

FAQs: Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.2

Q1. What is the focus of Exercise 13.2?
Answer:
Exercise 13.2 focuses on solving problems related to the surface area and volume of composite 3D solids, such as combinations of cylinders, cones, spheres, and hemispheres.

Q2. How do I calculate the surface area of a composite solid?
Answer:
To calculate the surface area of a composite solid:

1. Break down the solid into its individual shapes (cylinder, cone, sphere, etc.).

2. Calculate the surface area of each part using the appropriate formula.

3. Add or subtract areas where parts overlap or are internal.

Q3. How do I calculate the volume of a composite solid?
Answer:
To calculate the volume of a composite solid:

1. Divide the solid into simpler shapes (e.g., a cylinder and a hemisphere).

2. Calculate the volume of each part using the volume formulas for the respective shapes.

3. Add or subtract the volumes based on the given conditions.

Q4. What are the formulas for surface area and volume for the shapes in this exercise?
Answer:
Here are the key formulas:

• Surface area of a cylinder:

  $A = 2\pi r(h + r)$A=2πr(h+r)

• Volume of a cylinder:

  $V = \pi r^2 h$V=πr2h

• Surface area of a cone:

  $A = \pi r(l + r)$A=πr(l+r)

• Volume of a cone:

  $V = \frac{1}{3} \pi r^2 h$V=31​πr2h

• Surface area of a sphere:

  $A = 4\pi r^2$A=4πr2

• Volume of a sphere:

  $V = \frac{4}{3} \pi r^3$V=34​πr3

• Surface area of a hemisphere:

  $A = 3\pi r^2$A=3πr2

• Volume of a hemisphere:

  $V = \frac{2}{3} \pi r^3$V=32​πr3

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide step-by-step methods for solving problems involving composite solids. By practicing these solutions, students can build confidence and improve their problem-solving skills for Class 10 exams.