Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 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 π.

Solution:

Given:

Height of cone (h)= 1 cm

Radius of hemisphere (r) = 1 cm

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Total Volume = Volume of cone + Volume of Hemisphere

= πr²h + πr³

= πr²(h+2r)

=  × π × 1 × 1 × (1+2)

= 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 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.)

Solution:

Given:

Radius of cone and cylinder (r) =  cm

Height of cone (h) = 2 cm

Height of cylinder (H) = 12- (2+2) = 8 cm

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Total Volume = Volume of two cones + Volume of cylinder

 = πr²h + πr²h + πr²H

= πr²h + πr²H

= πr²(()h+H)

=                       (taking π=)

= 

= 

= 66 cm³

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 Fig.).

Solution:

Given, 

For 1 gulab jamun,

Height of cylindrical part (H)= 5-(2.8) = 2.2 cm

Radius of cylindrical and hemispherical part (r)=  = 1.4 cm

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Total Volume of one gulab jamun = Volume of two Hemisphere + Volume of cylinder

= πr³ + πr³ + πr²H

= πr² (r + H)

=                        (taking π=)

= 25.05 cm³

Hence, volume of 45 gulab jamun = 45 × Volume of 1 gulab jamun

= 45 × 25.05

= 1127.28 cm³

As, gulab jamun contains sugar syrup up to about 30% of its volume

Sugar syrup in 45 gulab jamun = 30% of its total volume

= × 1127.28

= 

= 338.184 cm³

Question 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:

Given:

Length of cuboid (l) = 15 cm

Breadth of cuboid (b) = 10 cm

Height of cuboid (h) = 3.5 cm

Radius of conical part (r) = 0.5 cm

Height of conical part (H) = 1.4 cm

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Total Volume wood in the entire stand = Volume of Cuboid - Volume of four conical part

= (l × b × h) - 4 × (πr²H)

= (15 × 10 × 3.5) - (4 ×  × 0.5 × 0.5 × 1.4)                           (taking π=)

= 525 - (1.466)

= 523.5333 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:

Given:

Height of cone (h) = 8 cm

Radius of cone (r) = 5 cm

Radius of sphere (R)= 0.5 cm

Let, No. of sphere in cone = n

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Volume of water = Volume of cone 

= πr²h

=  × π × 5 × 5 × 8

=  cm³

Volume of water flows out =  (total volume of water)

= 

=  cm³

Hence, the volume of n spheres = cm³

Volume of each sphere = πr³

= × π × 0.5 × 0.5 ×0.5

=  cm³

Hence, n = 

n = 

n = 100 spheres

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 8g mass. (Use π = 3.14) 

Solution:

Given:

Height of large cylinder (H)= 220 cm

Radius of large cylinder (R)=  = 12 cm

Height of small cylinder (h)= 60 cm

Radius of small cylinder (r)= 8 cm

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Total Volume = Volume of large cylinder + Volume of small cylinder 

= πR²H + πr²h

= π (R²H + r²h)

= 3.14 × ((12 × 12 × 220) + (8 × 8 × 60))                             (taking π=3.14)

= 3.14 (31680 + 3840)

= 111532.8 cm³

As given, Mass of 1cm³ = 8 g

Mass for 111532.8 cm³ = 8 × 111532.8 g

= 892,262.4 grams

= 892.2624 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 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:

Given:

Height of cylinder (H)= 180 cm

Height of cone (h)= 180 - 60 = 120 cm

Radius of cone, cylinder and hemisphere (r) = 60 cm

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Volume of water left in the cylinder = Volume of cylinder - (Volume of cone + Volume of hemisphere)

= πr²H - (πr²h + πr³)

= πr² (H - h + r)

=                          (taking π=)

=  × 60 × 60 × (100)

= 1131428.571 cm³

= 1.131 m³

Hence, the volume of water left in the cylinder = 1.131 m³

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:

Given:

Height of cylinder (h)= 8 cm

Radius of cylinder (r) = 2/2 = 1 cm

Radius of sphere (R) = cm

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Amount of water it can hold = Total Volume of this vessel

= Volume of cylinder + Volume of sphere

= πr²h + πR³

= 

= 8π + 102.35π

= 110.35π                       (taking π=3.14)

= 346.499 cm³

Volume measured by child = 345 cm³, which is INCORRECT

Correct volume = 346.5 cm³

Summary

Exercise 13.2 focuses on the surface area and volume of spheres and hemispheres. It covers calculating the surface area (curved and total) and volume of these three-dimensional shapes. Students learn to apply formulas for surface area and volume in various real-life scenarios, including composite shapes involving hemispheres combined with cylinders or cones. The problems often require conversions between different units of measurement and understanding the relationship between radius and diameter.