9 Views

Question : A hemisphere of lead of radius 4 cm is cast into a right circular cone of height 72 cm. What is the radius of the base of the cone?

Option 1: 1.63 cm

Option 2: 1.35 cm

Option 3: 1.33 cm

Option 4: 1.45 cm


Team Careers360 21st Jan, 2024
Answer (1)
Team Careers360 23rd Jan, 2024

Correct Answer: 1.33 cm


Solution : Volume of hemisphere = $\frac{ 2 \pi \times \text{Radius}^3}{3}$
Volume of cone = $\frac{\pi \times \text{Radius}^2 \times \text{Height}}{3}$
Given, Radius = 4 cm
Volume of hemisphere = $\frac{ 2 \pi \times 4^3}{3}$ = $\frac{ 128 \pi}{3}$ cm 3
Height of the cone = 72 cm
Let the radius of the base of the right circular cone be $R$ cm.
Volume of the cone = $\frac{\pi \times R^2 \times 72}{3}$
According to the question,
$\frac{ 128 \pi}{3} = \frac{\pi \times R^2 \times 72}{3}$
$⇒R^2 = \frac{16}{9}$
$⇒R = \frac{4}{3}$
$\therefore R= 1.33$
$\therefore$ The radius of the base of the cone is 1.33 cm.
Hence, the correct answer is 1.33 cm.

SSC CGL Complete Guide

Candidates can download this ebook to know all about SSC CGL.

Download EBook

Know More About

Related Questions

TOEFL ® Registrations 2024
Apply
Accepted by more than 11,000 universities in over 150 countries worldwide
Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books