Gujarat Board GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

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 it.

Solution:

Let radius be r cm.

Height of cone = Radius

h = r

r = 1 cm

h = 1 cm

Volume of solid

= Volume of cone + Volume of hemisphere

= \(\frac {1}{3}\)Ï€r^{2}h + \(\frac {2}{3}\)Ï€r^{3}

= \(\frac {1}{3}\)Ï€ x 1^{2} x 1 + \(\frac {2}{3}\)Ï€ = \(\frac {Ï€ + 2Ï€}{3}\)

= \(\frac {3Ï€}{3}\) = Ï€ cm^{3}

Question 2.

Rachel, an engineering student, was asked ta 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 dimension of the model to be nearly the same).

Solution:

Diameter of cylindrical model = 3 cm

r = \(\frac {3}{2}\)cm

Height of cylindrical part of the model

= 12 – (2 + 2)

h_{1} = 8 cm

h_{2} (height of cone) = 2 cm

and radius r = \(\frac {3}{2}\) cm (radius of both parts is same)

Volume of air contained in the model

= Volume of cylindrical part + Volume of two conical parts

= Ï€r^{2}h_{1} + 2 x \(\frac {1}{3}\)r^{2}h_{2}

= Ï€(\(\frac {3}{2}\))^{2} x 8 + \(\frac {2}{3}\)Ï€ x (\(\frac {3}{2}\))^{2} x 2

= Ï€ x \(\frac {9}{4}\) x 8 + \(\frac {2}{3}\)Ï€ x \(\frac {9}{4}\) x 2

= 18Ï€ + 3Ï€ = 21Ï€

= 21 x \(\frac {22}{7}\) = 3 x 22 = 66 cm^{3}

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.

Solution:

Total length of gulab jamun = 5 cm

Diameter = 2.8 cm

r = 1.4 cm

Height of cylindrical portion of gulab jamun

= 5 – (1.4 + 1.4)

= 5 – 2.8 = 2.2 cm

Volume of one gulab jamun = Volume of cylindrical portion + Volume of two hemispherical ends

Ï€r^{2}h + 2 x \(\frac {2}{3}\)Ï€r^{3}

= \(\frac {22}{7}\) x (1.4)^{2} x 2.2 + \(\frac {4}{3}\) x \(\frac {22}{7}\) x (1.4)^{3}

= \(\frac {22}{7}\) x (1.4) x 1.4 + 2.2 + \(\frac {4}{3}\) x \(\frac {22}{7}\) x 1.4 x 1.4 x 1.4

= 22 x 0.2 x 1.4 x 2.2 + \(\frac {4}{3}\) x 22 x 0.2 x 1.4 x 1.4

= 13.552 + 11.498

= 25.05 cm^{3}

Volume of 45 gulabjamuns

= 45 x 25.05

= 1127.25

Volume of sugar syrup in 45 gulab jamuns

= 30% of 1127.25

= \(\frac {22}{7}\) x 1127.25 = 338.175

= 338.2 cm^{3} (aprox.)

Question 4.

A pen stand made of wood is in the shape of a cuboid with four conical depression to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution:

Length of pen stand l = 15 cm

breadth b = 10 cm

and height h = 3.5 cm

Volume of cuboidal wood = l x b x h

= 15 x 10 x 3.5 = 525 cm^{3}

Radius of conical depression r = 0.5 cm

and height h = 1.4 cm

Volume of one conical depression

\(\frac {1}{3}\) Ï€r^{2}h

Volume of four conical depressions

= 4 x \(\frac {1}{3}\)Ï€r^{2}h

= \(\frac {4}{3}\) x \(\frac {22}{7}\) x (0.5)^{2} x 1.4

= \(\frac {88}{3}\) x 0.5 x 0.5 x 0.2

= \(\frac {4.4}{3}\) cm^{3} = 1.47 cm^{3}

Volume of wooden pen stand

= Volume of wood – Volume of conical depressions

= 525 – 1.47

= 523.53 cm^{3}

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:

For cone, h = 8 cm

r = 5 cm

Volume of water filled in the cone up to the brim = \(\frac {1}{3}\)Ï€r^{2}h = \(\frac {1}{3}\)Ï€ x 5^{2} x 8

Volume of 1 lead shot = \(\frac {4}{3}\)Ï€r^{3} = \(\frac {4}{3}\)Ï€ x (0.5)^{3}

No. of lead shots,

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 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 ir=3.14)

Solution:

Diameter of cylinder A = 24 cm

So radius r_{1} = 12 cm

And height h_{1} = 220cm

Radius of cylinder B = 8 cm

i.e., r_{2} = 8 cm

and height h_{2} = 60 cm

Volume of pole

= Volume of cylinder A + Volume of cylinder B

= Ï€r_{1}^{2}h_{1} + Ï€r_{1}^{2}h_{2}

= Ï€ [r_{1}^{2}h_{1} + r_{1}^{2}h_{2}]

= 3.14 [12^{2} x 220 + 8^{2} x 60]

= 3.14 [144 x 220 + 64 x 60]

= 3.14 x 35520

= 11532.8 cm^{3}

Mass of pole = Volume x Density

= 11532.8 x 8

= \(\frac{11532.8 \times 8}{1000}\)

= 892.26 kg

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 â€˜ 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 cone h = 120 cm

radius r = 60 cm

Radius of hemisphere = 60 cm

Volume of solid = Volume of cone + Volume of hemisphere

= \(\frac {1}{3}\) Ï€r^{2}h + \(\frac {2}{3}\)Ï€r^{3} = \(\frac {1}{3}\)Ï€r^{2} (h + 2r)

= \(\frac {1}{3}\) x 3.14 x 60 x 60 x (120 x 2 x 60)

= 3.14 x 60 x 60 x (120 + 2 x 60)

= 3.14 x 20 x 60 x (120 + 120)

= 904320 cm^{3}

Volume of cylindrical vessel = Ï€r^{2}h

= 3.14 x 60^{2} x 180

= 3.14 x 60 x 60 x 180 = 2034720 cm^{3
}

Volume of water left

= 2034720 – 904320 = 1130400 cm^{3}

= \(\frac{1130400}{100 \times 100 \times 100}\)m^{3} = 11304 m^{3}

Question 8.

Aspherical 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 cm^{3}. Check whether she is correct, taking the above as the inside measurements, and Ï€ = 3.14.

Solution:

Cylindrical neck of vessel has diameter = 2 cm

r = 1 cm

h = 8 cm

Volume of cylindrical portion = Ï€r^{2}h

= 3.14 x 1^{2} x 8 = 25.12 cm^{3}

Spherical part has diameter = 8.5 cm

r = \(\frac {8.5}{2}\) = 4.25 cm

Volume of spherical part = \(\frac {4}{3}\) Ï€r^{3}

= \(\frac {4}{3}\) x 3.14 x (4.25)^{3}

= \(\frac{964.176}{3}\) = 321.392 cm^{3}

Total volume of vessel

= 25.12 + 321.392

= 346.51 cm^{3}

Not correct. The correct answer is 346.51 cm^{3}.