 # ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.4 in Chapter 17 - Mensuration

From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.

Solution:

As the conical cavity is drilled out from the cylinder, so the volume of the remaining solid can be calculated by subtracting the volume of cylinder and cone. Given height of the cylinder, H = 30 cm

Radius of the cylinder, r = 7 cm

Height of cone, h = 24 cm

Radius of cone, r = 7 cm

Slant height of the cone, l = √(h2+r2)

l = √(242+72)

l = √(576+49)

l = √(625)

l = 25 cm

Volume of the remaining solid = Volume of the cylinder – Volume of the cone

=π r2H – (1/3)π r2h

= π r2(H- h/3)

= (22/7)×72×(30-24/3)

= (22×7)×(30-8)

= (154)×(22)

= 3388 cm3

Volume of the remaining solid is 3388 cm3.

Total surface area of the remaining solid = Curved surface area of cylinder + surface area of top of the cylinder+ curved surface area of the cone

Total surface area of the remaining solid

= 2π rH + π r2 +π rl

= π r(2H + r + l)

=(_22/7)×7(2×30 + 7+ 25)

= 22×(60+32)

= 22×92

= 2024 cm2

Hence the total surface area of the remaining solid is 2024 cm2.

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