Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Optional Exercise to get the best methods of solving problems.
TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise
Question 1.
A golf ball has diameter equal to 4.1cm. Its surface has 150 dimples each of radius 2mm. Calculate total surface area which is exposed to the surroundings (Assume that the dimples are all hemispherical) [π = \(\frac{22}{7}\)]
Solution:
Area exposed = surface area of the ball – total area of 150 dimples with radius 2 mm
= 4πr2 – 150 × πr2
= 4 × \(\frac{22}{7}\) × \(\frac{4.1}{2}\) × \(\frac{4.1}{2}\) – 150 × \(\frac{22}{7}\) × \(\frac{2}{10}\) × \(\frac{2}{10}\)
[∵ 2 mm = \(\frac{2}{10}\) cm]
= 52.831 – 18.85 = 33.972 cm2
Question 2.
A cylinder of radius 12 cm. contains water to a depth of 20 cm. when a spherical iron ball is dropped in to the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. [π = \(\frac{22}{7}\)]
Solution:
Rise in the water level is seen as a cylinder of radius ‘r’ = r1 = 12 cm
Height, h = 6.75 cm
Volume of the rise = Volume of the spherical iron ball dropped
πr12h = \(\frac{4}{3}\) πr23
r12h = \(\frac{4}{3}\) π23
12 × 12 × 6.75 cm3 = \(\frac{4}{3}\) × r23 cm3
r23 = \(\frac{3}{4}\) × 12 × 12 × 6.75
= 9 × 12 × 6.75
= 108 × 5.75
r23 = 729
r23 = 9 × 9 × 9
∴ 729 = (3 × 3) × (3 × 3) × (3 × 3)
∴ Radius of the ball r = r2 = 9 cm.
Question 3.
A solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at the other end. Their common diameter is 4.2 cm. and height of the cylindrical and conical portion are 12 cm. and 7 cm. respectively. Find the volume of the solid toy. [π = \(\frac{22}{7}\)]
Solution:
Volume of the toy
= volume of the hemisphere + volume of the cylinder + volume of the cone.
= \(\frac{2}{3}\) πr3 + πr2h1 + \(\frac{1}{3}\) πr2h2
= πr2(\(\frac{2}{3}\)r + h1 + \(\frac{\mathrm{h}_2}{3}\))
= \(\frac{22}{7}\) × \(\frac{4.2}{2}\) × \(\frac{4.2}{2}\)[\(\frac{2}{3}\) × \(\frac{4.2}{2}\) + 12 + \(\frac{7}{3}\)]
= 11 × 0.6 × 2.1 [1.4 + 12 + \(\frac{7}{3}\)]
= 13.86[13.4 + \(\frac{7}{3}\)]
= 13.86 × \(\left[\frac{40.2+7}{3}\right]\)
= \(\frac{13.86 \times 47.2}{3}\) = 218.064 cm3
(or)
Hemisphere :
Radius = \(\frac{\text { diameter }}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm
V = \(\frac{2}{3}\) πr3 = \(\frac{2}{3}\) × \(\frac{22}{7}\) × 2.1 × 2.1 × 2.1
= 19.404 cm3
Cylinder :
Radius, r = \(\frac{\mathrm{d}}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm
height, h = 12 cm
V = πr2h = \(\frac{22}{7}\) × 2.1 × 2.1 × 12
= 166.32 cm3
Cone :
Radius, r = \(\frac{\mathrm{d}}{2}\) = \(\frac{4.2}{2}\) = 2.1cm
Height, h = 7 cm
V = \(\frac{1}{3}\) πr2h = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 2.1 × 2.1 × 7
= 32.34 cm3
∴ Total volume = 19.404 + 166.32 + 32.34
= 218.064 cm3.
Question 4.
Three metal cubes with edges 15 cm., 12 cm. and 9 cm. respectively are melted together and formed into a single cube. Find the diagonal of this cube.
Solution:
Edges l1 = 15 cm, l2 = 12 cm, l13 = 9 cm.
Volume of the resulting cube = Sum of the volumes of the three given cubes
L3 = l31h
L3 = l13 + l23 + l33
L3 = 153 + 123 + 93
L3 = 3375 + 1728 + 729
L3 = 5832 = 18 × 18 × 18
∴ Edge of the new cube l = 18 cm
Diagonal = \(\sqrt{3 l^2}\) = \(\sqrt{3 \times 18^2}\)
= \(\sqrt{3 \times 324}\) = \(\sqrt{972}\) = 31.176 cm.
Question 5.
A hemi-spherical bowl of internal diameter 36 cm. contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm. and height 6 cm. How many bottles are required to empty the bowl ?
Solution:
Let the number of bottles required = n.
Then total valume of a ’n’ bottles = volume of the hemispherical bowl.
n. πr12h = \(\frac{2}{3}\) πr22h
Bottle :
Radius, r1 = 3 cm
Height, h = 6 cm
Volume, V = πr12h
= \(\frac{22}{7}\) × 3 × 3 × 6 = \(\frac{1188}{7}\)
∴ Total volume of n bottles = n × \(\frac{1188}{7}\) cm3.
Bowl :
∴ 72 bottles are required to empty the bowl.