# Mensuration – Quiz 8

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# Mensuration – Quiz 8

### Introduction

Mensuration is one of important topic in Quantitative Aptitude Section. In Mensuration – Quiz 8 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Mensuration – Quiz 8 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

A cone and sphere have the same radius of 12 cm. Find the height of the cone if the cone and sphere have the same volume.
A. 18 cm B. 24 cm C. 36 cm D. 48 cm

D
Let the height of the cone be h $\frac {}{}$
Volume of the cone
= $\frac {1}{3} \times \pi \times {(12)}^{2} \times h$
= 48πh ${cm}^{3}$
Volume of the sphere
= $\frac {4}{3} \times \pi \times {(r)}^{3}$
= $\frac {4}{3} \times \pi \times {(12)}^{3}$
= 2304 ${cm}^{3}$
Since the volumes are equal
48πh = 2304π
Solving for h
h = $\frac {2304π}{48π}$
= 48 cm

### Q2

A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:
A. 12π ${cm}^{3}$ B. 15π ${cm}^{3}$ C. 16π ${cm}^{3}$ D. 20π ${cm}^{3}$

A

Clearly, we have r = 3 cm and h = 4 cm.
Volume = $\frac {1}{3} \times \pi \times {r}^{2} \times h$
= Volume of the cone
= $\frac {1}{3} \times \pi \times {(3)}^{2} \times 4$${cm}^{3}$ = 12π ${cm}^{3}$

### Q3

In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
A. 75 cu. m B. 750 cu. m C. 7500 cu. m D. 75000 cu. m

B
1 hectare = 10,000 ${m}^{2}$
So, Area = (1.5 x 10000) ${m}^{2}$ = 15000 ${m}^{2}$.
Depth = $\frac {5}{100} m = \frac {1}{20}m$
Volume = (Area x Depth) = $(15000 \times \frac {1}{20}) {m}^{3} = 750 {m}^{3}$

### Q4

A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
A. 720 B. 900 C. 1200 D. 1800

C
2(15 + 12) x h = 2(15 x 12)
$\Rightarrow h = \frac {180}{27} m = \frac {20}{3} m$
i.e,Volume = $(15 \times 12 \times \frac {20}{3}){m}^{3} = 1200 {m}^{3}$

### Q5

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/${cm}^{3}$, then the weight of the pipe is:
A. 3.6 kg B. 3.696 kg C. 36 kg D. 36.9 kg

B
External radius = 4 cm,
Internal radius = 3 cm.
Volume of iron = $(\frac {22}{7} \times [{(4)}^{2}) - {(3)}^{2}]\times 21) {cm}^{3}$
= $(\frac {22}{7} \times 7 \times 1 \times 21) {cm}^{3}$
= 462 ${cm}^{3}$

### Study Guide

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### Exams

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