MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Unless stated otherwise, take π = \(\frac{22}{7}\)

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 π.
Solution:
Here, r = 1 cm and h = 1 cm.
Volume of the conical part = \(\frac{1}{3}\) πr²h and volume of the hemispherical part = \(\frac{2}{3}\) πr³h

∴ Volume of the solid shape
= \(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr³h = \(\frac{1}{3}\) πr²h[h + 2r]
= \(\frac{1}{3}\) π(1)² [1 + 2(1)] cm³
= (\(\frac{1}{3}\) π × 1 × 3) cm³ = π cm³

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 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 to be nearly the same.)
Solution:
Here, diameter = 3 cm
⇒ Radius (r) = \(\frac{3}{2}\) cm
Total height = 12 cm
Height of a cone (h₁) = 2 cm

∴ Height of both cones = 2 × 2 = 4 cm
⇒ Height of the cylinder (h₂) = (12 – 4) cm = 8 cm.
Now, volume of the cylindrical part = πr²h₂
Volume of both conical parts = 2[\(\frac{1}{3}\) πr²h₁]
∴ Volume of the whole model

Question 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).

Solution:
Since a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.

Diameter = 2.8 cm
⇒ Radius (r) = 1.4 cm
∴ Length of the cylindrical part (h)
= 5 cm – (1.4 + 1.4) cm = 5 cm – 2.8 cm = 2.2 cm
Now, volume of the cylindrical part = πr²h and volume of both the hemispherical ends
= 2(\(\frac{2}{3}\) πr³) = \(\frac{4}{3}\) πr³
∴ Volume of a gulab jamun

Volume of 45 gulab jamuns

Question 4.
A pen stand made up 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 figure).

Solution:
Dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm.
∴ Volume of the cuboid = 15 × 10 × \(\frac{35}{10}\) cm³
= 525 cm³
Since, each depression is conical in shape with base radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of each depression
= \(\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times\left(\frac{5}{10}\right)^{2} \times \frac{14}{10} \mathrm{cm}^{3}=\frac{11}{30} \mathrm{cm}^{3}\)
Since, there are 4 depressions
∴ Total volume of 4 depressions = \(\frac{44}{30}\) cm³
Now, volume of the wood in entire stand = [Volume of the wooden cuboid] – [Volume of 4 depressions]
= 525cm³ – \(\frac{44}{30}\) cm³
= \(\frac{15750-44}{30} \mathrm{cm}^{3}=\frac{15706}{30} \mathrm{cm}^{3}\)
= 523.53 cm³

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.
Solution:
Height of the conical vessel (h) = 8 cm
Base radius (R) = 5 cm
Volume of water in conical vessel = \(\frac{1}{3}\) πR²h

Thus, the required number of lead shots = 100

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 cm³ of iron has approximately 8 g mass. (Use π = 3.14)
Solution:
Height of the big cylinder (h) = 220 cm
Base radius (r) = \(\frac{24}{2}\) cm = 12cm
∴ Volume of the big cylinder
= πr²h = (π(12)² × 220) cm³
Also, height of smaller cylinder (h₁) = 60 cm

Base radius (r₁) = 8 cm
∴ Volume of the smaller cylinder πr₁²h₁
= (π(8)² × 60) cm³
∴ Volume of iron pole = [Volume of big cylinder] + [Volume of the smaller cylinder]
= (π × 220 × 12² × π × 60 × 8²) cm³
= 3.14[220 × 12 × 12 + 60 × 8 × 8] cm³

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.
Solution:
Height of the conical part = 120 cm
Base radius of the conical part = 60 cm
Volume of the conical part
= \(\frac{1}{3}\) πr²h = [\(\frac{1}{3} \times \frac{22}{7}\) × (60)² × 120] cm³
Radius of the hemispherical part = 60 cm.

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 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Volume of the cylindrical part = πr²h
= (3.14 × 1² × 8) cm³ = \(\frac{314}{100}\) × 8 cm³

∴ Total volume of the glass vessel = [Volume of cylindrical part] + [Volume of spherical part]

⇒ Volume of water in the vessel = 346.51 cm³
Since the child finds the volume as 345 cm³
∴ The child’s answer is not correct.
The correct answer is 346.51 cm³.