Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2 NCERT Solutions

Exercise 12.2 of Class 10 Maths Chapter 12, Surface Areas and Volumes, focuses on applying formulas to calculate the volumes and surface areas of different solid shapes. This includes problems based on real-life situations, helping you understand how these concepts are used practically, as covered in the CBSE Class 10th syllabus.

These questions will help you connect different formulas and apply them correctly depending on the shape and given conditions. With regular practice, you can improve your problem-solving approach and handle numerical questions more efficiently in exams.

NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2

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 being equal to its radius. Find the volume of the solid in terms of π.

Here r = 1 cm and h = 1 cm. The diagram is as follows.

Now, Volume of solid = Volume of conical part + Volume of hemispherical part We know the volume of cone = ⅓ πr 2 h And, The volume of the hemisphere = ⅔πr 3 

So, the volume of the solid will be= π cm 3

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 are nearly the same.)

Solution:

Given, Height of cylinder = 12–4 = 8 cm 

Radius = 1.5 cm 

Height of cone = 2 cm 

Now, the total volume of the air contained will be = Volume of cylinder+2×(Volume of the cone) 

∴ Total volume = πr 2 h+[2×(⅓ πr 2 h )] 

= 18 π+2(1.5 π) 

= 66 cm 3 .

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 a length of 5 cm and a diameter of 2.8 cm (see figure).

Solution:

It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends. So, the total height of a gulab jamun = 5 cm. 

Diameter = 2.8 cm 

So, radius = 1.4 cm 

∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm =2.2 cm 

Now, the total volume of one gulab jamun = Volume of cylinder + Volume of two hemispheres 

= πr 2 h+(4/3)πr 3 

= 4.312π+(10.976/3) π 

= 25.05 cm 3 

We know that the volume of sugar syrup = 30% of the total volume.

So, the volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm 3 ) 

= 45×7.515 

= 338.184 cm 3

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 (see Fig.).

Solution:

The volume of the cuboid = length x width x height 

We know the cuboid’s dimensions as 15 cmx10 cmx3.5 cm 

So, the volume of the cuboid = 15x10x3.5 = 525 cm 3 

Here, depressions are like cones, and we know, Volume of cone = (⅓)πr 2 h 

Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm 

∴ Volume of 4 cones = 4x(⅓)πr 2 h = 1.46 cm 2 

Now, the volume of wood = Volume of the cuboid – 4 x volume of the cone 

= 525-1.46 

= 523.54 cm 2

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.

Solution:

For the cone, Radius = 5 cm, Height = 8 cm Also, Radius of sphere = 0.5 cm The diagram will be like

It is known that, 

The volume of cone = volume of water in the cone = ⅓πr 2 h = (200/3)π cm 3 

Now, Total volume of water overflown= (¼)×(200/3) π =(50/3)π 

The volume of lead shot = (4/3)πr 3 = (1/6) π 

Now, The number of lead shots = Total volume of water overflown/Volume of lead shot = (50/3)π/(⅙)π = (50/3)×6 = 100

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.

Solution:

Given the height of the big cylinder (H) = 220 cm 

The radius of the base (R) = 24/2 = 12 cm 

So, the volume of the big cylinder = πR 2 H = π(12) 2 × 220 cm 3 = 99565.8 cm 3 

Now, the height of the smaller cylinder (h) = 60 cm 

The radius of the base (r) = 8 cm 

So, the volume of the smaller cylinder = πr 2 h 

= π(8) 2 ×60 cm 3 

= 12068.5 cm 3 

∴ The volume of iron = Volume of the big cylinder+ Volume of the small cylinder 

= 99565.8 + 12068.5 =111634.5 cm 3 

We know, Mass = Density x volume 

So, the mass of the pole = 8×111634.5 = 893 Kg (approx.)

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.

Solution:

Here, the volume of water left will be = Volume of the cylinder – Volume of solid 

Given, Radius of cone = 60 cm

Height of cone = 120 cm

Radius of cylinder = 60 cm 

Height of cylinder = 180 cm 

Radius of hemisphere = 60 cm 

Now, The total volume of solid = Volume of Cone + Volume of the hemisphere 

Volume of cone = 1/3πr 2 h 

= 1/3 × π×60 2 ×120cm 3 

= 144×10 3 π cm 3

 Volume of hemisphere = (⅔)×π×60 3 cm 3 

= 144×10 3 π cm 3 

So, total volume of solid =  144×10 3 π cm 3 + 144×10 3 π cm 3 

= 288 ×10 3 π cm 3 

Volume of cylinder = π×60 2 ×180 

= 648000 = 648×10 3 π cm 3 

Now, the volume of water left will be 

= Volume of the cylinder – Volume of solid 

= (648-288) × 10 3 ×π 

= 1.131 m 3

8. A spherical glass vessel has a cylindrical neck 8 cm long and 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 π = 3.14.

Solution:

Given, For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm 

For the spherical part, Radius (r) = (8.5/2) = 4.25 cm

Now, volume of this vessel = Volume of cylinder + Volume of sphere 

= π×(1) 2 ×8+(4/3)π(4.25) 3 

= 346.51 cm 3 

Hence, the child’s calculation is not correct.

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

• Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

• Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

• Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

• Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

• Solve Previous Year Papers:

Practising CBSE Class 10 Previous Year Question Papers (PYQs) helps you understand question patterns and important topics.

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