NCERT Solutions for Class 10 Maths Chapter 13, Surface Areas and Volumes, Exercise 13.2, are available here in downloadable PDF format. The problems in this exercise have been answered by our Maths experts, to make students understand the theories in a better way.
Topics covered in these materials are according to the NCERT syllabus and guidelines. Therefore, solutions for Maths class 10, are very helpful for students who are doing the preparation for the exams and to score good marks.
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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 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 πr2h
The volume of hemisphere = ⅔πr3
So, volume of solid will be
= π cm3
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 dimensions of the model to be nearly the same.)
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 cone)
∴ Total volume = πr2h+[2×(⅓ πr2h )]
= 18 π+2(1.5 π)
= 66 cm3.
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 (see figure).
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
Now, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres
= 4.312π+(10.976/3) π
= 25.05 cm3
We know that the volume of sugar syrup = 30% of total volume
So, volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm3)
= 45×7.515 = 338.184 cm3
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.).
Volume of 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 cm3
Here, depressions are like cones and we know,
Volume of cone = (⅓)πr2h
Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of 4 cones = 4x(⅓)πr2h
= 1.46 cm2
Now, volume of wood = Volume of cuboid – 4 x volume of cone
= 525-1.46 = 523.54 cm2
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.
For the cone,
Radius = 5 cm,
Height = 8 cm
Radius of sphere = 0.5 cm
The diagram will be like
It is known that,
Volume of cone = volume of water in the cone
= ⅓πr2h = (200/3)π cm3
Total volume of water overflown= (¼)×(200/3) π =(50/3)π
Volume of lead shot
= (1/6) π
The number of lead shots = Total Volume of Water over flown/ Volume of Lead shot
= (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 cm3 of iron has approximately 8 g mass.
Given, the height of the big cylinder (H) = 220 cm
Radius of the base (R) = 24/12 = 12 cm
So, the volume of the big cylinder = πR2H
= π(12)2 × 220 cm3
= 99565.8 cm3
Now, the height of smaller cylinder (h) = 60 cm
Radius of the base (r) = 8 cm
So, the volume of the smaller cylinder = πr2h
= π(8)2×60 cm3
= 12068.5 cm3
∴ Volume of iron = Volume of the big cylinder+ Volume of the small cylinder
= 99565.8 + 12068.5
Mass = Density x volume
So, 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.
Here, the volume of water left will be = Volume of cylinder – Volume of solid
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
Total volume of solid = Volume of Cone + Volume of hemisphere
Volume of cone = π×122×103cm3 = 144×103π cm3
So, Total volume of solid = 144×103π cm3 -(⅔)×π×103 cm3
Volume of hemisphere = (⅔)×π×103 cm3
Volume of cylinder = π×602×180 = 648000 = 648×103 π cm3
Now, volume of water left will be = Volume of cylinder – Volume of solid
= (648-288) × 103×π = 1.131 m3
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 π = 3.14.
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
= 346.51 cm3
Exercise 13.2 of 10th Standard Maths have problems based on finding the volumes of;
- A solid cone in terms of Pi.
- A structure formed with a cylindrical shape object having a cone on both its sides.
- A cylindrical object immersed with some substance, occupying the volume of the object.
- A wood of cuboid shape having depressions in it.
Students can visit us anytime, to get complete solutions for chapter 13 of class 10 maths. Apart from this, they can also find other learning materials, such as notes, books, questions papers and some tips and tricks to do the preparation for the 10th Standard Maths exam.
The questions in Exercise 13.2 are prepared as per the topic, Volume of Combination of Solids, covered in the chapter before this exercise. Solutions provide generic methods for students, which they can relate to easily.