A solid consisting of a right circular cone o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Radius $(r)$ of the hemispherical part = Radius $(r)$ of the conical part = $60\, cm$.Height $(h_2)$ of the conical part of the solid = $120\, cm$

Vedclass · Accepted Answer

$1.131$

Vedclass · Answer

(C) Radius $(r)$ of the hemispherical part = Radius $(r)$ of the conical part = $60\, cm$.Height $(h_2)$ of the conical part of the solid = $120\, cm$.Height $(h_1)$ of the cylinder = $180\, cm$.Radius $(r)$ of the cylinder = $60\, cm$.The volume of water left in the cylinder = Volume of the cylinder - Volume of the solid.Volume of the solid = Volume of the cone + Volume of the hemisphere = $\frac{1}{3}\pi r^2 h_2 + \frac{2}{3}\pi r^3$.Volume of water left = $\pi r^2 h_1 - (\frac{1}{3}\pi r^2 h_2 + \frac{2}{3}\pi r^3)$.Substituting the values: $\pi(60)^2(180) - (\frac{1}{3}\pi(60)^2 	imes 120 + \frac{2}{3}\pi(60)^3)$.$= \pi(60)^2 [180 - (40 + 40)] = \pi(3600)(100) = 360,000\pi\, cm^3$.Using $\pi = \frac{22}{7}$,Volume $\approx 360,000 	imes 3.14159 = 1,130,973.36\, cm^3$.Since $1\, m^3 = 1,000,000\, cm^3$,the volume in $m^3$ is $\approx 1.131\, m^3$.

Similar Questions

$A$ medicine capsule is in the shape of a cylinder of diameter $0.5\, cm$ with two hemispheres stuck to each of its ends. The length of the entire capsule is $2\, cm$. The capacity of the capsule is (in $cm^3$):

$1 \ m^{3} = \dots \dots \dots \text{ kilolitre}$

Derive the formula for the volume of the frustum of a cone.

$A$ metallic sphere of radius $21 \,cm$ is melted and recast into a wire with radius $0.5 \,cm$. Find the length of the wire in metres.

Vedclass Test Series

Exam Paper Generator

Online Exam Module