from a solid cylinder of height 14 cm and base diameter 7 cm 2 equal conical halls each of radius 2.1 cm and height 4 cm are cut off find the volume of the remaining solid
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Height of the cylinder (H) = 14 cm
Diameter of the base of cylinder = 7 cm
Radius of base of cylinder =7/2cm
⇒ Volume of cylinder = πR2H
=22/7*(7/2)²*14cm
= 539 cm3
Now Radius of base of cone (r) = 2.1 cm
Therefore, Height of cone (h) = 4 cm
=>Volume of cone=1/3πr²h
=>Volume of 2 cone =2/3πr²h
=>2/3*22/7*(2.1)²*4cm³
= 36.96 cm3
Hence the volume of remaining solid = Volume of cylinder – Volume of 2 cones
= 539 cm3 – 36.96 cm3
= 502.04 cm3
Diameter of the base of cylinder = 7 cm
Radius of base of cylinder =7/2cm
⇒ Volume of cylinder = πR2H
=22/7*(7/2)²*14cm
= 539 cm3
Now Radius of base of cone (r) = 2.1 cm
Therefore, Height of cone (h) = 4 cm
=>Volume of cone=1/3πr²h
=>Volume of 2 cone =2/3πr²h
=>2/3*22/7*(2.1)²*4cm³
= 36.96 cm3
Hence the volume of remaining solid = Volume of cylinder – Volume of 2 cones
= 539 cm3 – 36.96 cm3
= 502.04 cm3
Answered by
1
volume of remaining solid = volume of cylinder - 2×volume of cone
= 22/7×r×r×h - 1/3×22/7×r'×r'×h'×2
=22/7×3.5×3.5×14 - 22/7×1/3×2.1×2.1×4×2
=539-36.96
=502.04cm^3
= 22/7×r×r×h - 1/3×22/7×r'×r'×h'×2
=22/7×3.5×3.5×14 - 22/7×1/3×2.1×2.1×4×2
=539-36.96
=502.04cm^3
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