ML AGGARWAL CLASS 8 Chapter 18 Mensuration Exercise 18.3

  Exercise 18.3

Question 1

Sol :

Given volume of cube $=343 \mathrm{~cm}^{3}$

Let 'S' be edge of cube 

Volume of cube = $S^{3}$

$\begin{aligned} s^{3} &=343 \\ S &=\sqrt[3]{343} \\ S &=7 \mathrm{~cm} . \end{aligned}$

ஃ Length of an edge of cube = 7cm

Question 2

Sol :

 (i) Volume of Cuboid    $90 \mathrm{~cm}^{3}$    

Length 6cm   , Breadth  5cm  ,  Height  3cm

(ii) Volume of cuboid  $840 \mathrm{~cm}^{3}$ 

Length 15cm   , Breadth  15cm  ,  Height  7cm

(iii)  Volume of Cuboid   $62.5 \mathrm{~m}^{3}$ 

Length 10m   , Breadth  5m  ,  Height  1.25m

Question 3

Sol :

Given 

Volume of cuboid $=312 \mathrm{~cm}^{3}$

Base $A r e a=26 \mathrm{~cm}^{2}$

volume $=312 \mathrm{~cm}^{\circ}$

Area $\times$ height =312

$26 \times h=312$

$h=\frac{312}{26}$

h = 12cm

Question 4

Sol :

Given 

Godown  dimension $(1 \times b \times h)=$ $55 \mathrm{~m} \times 45 \mathrm{~m} \times 30 \mathrm{~m}$

Cuboidal box volume =1.2$5 m^{3}$

$\begin{aligned} \text { godown volume } &=\operatorname{l\times b} \times h \\ &=55 \times 45 \times 30 \\ &=74250 \mathrm{~m}^{3} \end{aligned}$

No. of Cuboidal boxes = $\frac{\text { godown Volume }}{\text { box volune }}$

$=\frac{74250}{1.25}$

No. of cuboidal boxes $=59400$

Question 5

Sol :

Given Dimension of rectangular pit = $1.4 \mathrm{~m} \times 90 \mathrm{~cm} \times 70 \mathrm{~cm}$

Volume of pit = $l \times b \times h$

=$140 \times 90 \times 70 \mathrm{~cm}^{3}$

Volume of pit = $882000 \mathrm{~cm}^{3}$

Given 

Brick dimension (l $\times$ b) =$2.1 \mathrm{~cm} \times 10.5 \mathrm{~cm}$

Let 'h' height of brick 

$1000 \times$ Brick volume = pit volume

$1000 \times$  $21 \times 10.5 \times h=882000$

$h=\frac{882000}{21 \times 10.5 \times 1000}$

h=4cm

$\therefore$ Height of brick =4 cm

Question 6

Sol :

Let 'a' be edge of cube 

Volume of cube = $a^{3}$

If each edge of cube is tripled = $a^{1}=3 a$

Volume of new Cube = $a^{1^{3}}$

$=(3 a)^{3}$

$=27 a^{3}$

The Volume becomes 27 times the original Volume of Cube.

Question 7

Sol :

Given 

Milk tank is in the from of cylinder 

Diameter of tank = 1.4m $\times$ 2

Height of tank = 8m 

Volume of tank = $\frac{\pi}{} d^{2} \times h$

$\frac{\pi}{} \times(1-4)^{2} \times 8$

Volume of tank = 49 .26$0 m^{3}$

∴ Volume of tank = 49.260 lit

Question 8

Sol :

Given 

Extended dimension of box = $84 \mathrm{~cm} \times 75 \mathrm{~cm} \times 64 \mathrm{~cm}$

Thickness of box = 2 cm

∴ Internal Dimension of box = ($54 -2\times 2$)cm , (75 - 2 $\times$ 2) cm ,(64- 2$\times2$) cm

$=80 \mathrm{~cm} \times 71 \mathrm{~cm} \times 60 \mathrm{cm}$

Volume of wood = External Volume - Internal Volume 

= $(84 \times 75 \times 64)-(80 \times 71 \times 60)$

$403200-340800$

 Volume of wood $=62400 \mathrm{~cm}^{3}$

Question 9

Sol :

Given 

Two Cylinder jar has same volume 

Let 

$d_{1}, d_{2}$ are diameter  jar

$h_{1,} h_{2}$ are heights of jar

Given 

$d_{1}: d_{2} \cdot 3: 4$

Volume of cylinder equal

∴ $\frac{\pi}{4} d_{1}^{2} \times h_{1}=\frac{\pi}{4} d_{2}^{2} \times h_{2}$

$d_{1}^{2} \times h_{1}=d_{2}^{2} \times h_{2}$

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{h_{2}}{h_{1}}$

$\left(\frac{3}{4}\right)^{2}=\frac{h_{2}}{h_{1}}$

$\frac{h_{1}}{h_{2}}=\frac{16}{9}$

$h_{1}: h_{2}=16: 9$

∴ Height of cylinders are in the ratio = 16: 9

Question 10

Sol :

Let   'r' be the ratio's of cylinder 

h be the height of cylinder 

Volume V = $\pi r^{2} \times h$

Now radius is  $=R^{1}=\frac{r}{2}$

Height is doubled = $h^{1}=2 h$

New Volume $V^{1}=\pi r_{1}^{2} \times h^{1}$

$=\pi\left(\frac{r}{2}\right)^{2} \times(2 h)$

$=\frac{\pi r^{2}}{4} \times 2 h$

$v^{1}=\frac{\pi r^{2} \times h}{2}$

$v^{1}=\frac{v}{2}$

ஃ New Volume is half of original Volume 

Question 11

Sol :

Dimensions of tin sheet $=30 \mathrm{~cm} \times 18 \mathrm{~cm}$

When rolled along its length (30cm)

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$\begin{aligned} 2 \pi r &=30, \quad h=18 C m \\ r &=\frac{30}{2 \pi} \\ r &=4.77 \mathrm{~cm} \end{aligned}$

$\begin{aligned} \text { Volume } &=\pi r^{2} \times h \\ &=\pi \times 4.77^{2} \times 18 \\ \text { Volume } &=1289.155 \mathrm{~cm} 3 \end{aligned}$

When rolled along breadth (18cm)

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$2 \pi r=18, \quad h=30 \mathrm{~lm}$

$r=\frac{18}{2 \pi}$

r= 2.86 cm

$\begin{aligned} \text { Volume } &=\pi r^{2} \times h \\ &=\pi \times 2.86^{2} \times 30 \\ &=773.493 \mathrm{~cm} 3 \end{aligned}$

Question 12

Sol :

(i) Given dis of pipe = 7cm = 0.07m

Velocity = 5m\sec

Discharge = Area $\times$ Velocity 

$=\frac{\pi d^{2}}{4} \times V$

$=\frac{\pi \times(0.07)^{2}}{4} \times 5$

Discharge $=0.0192 \mathrm{~m}^{3} / \mathrm{sec}$

$\begin{aligned} \therefore \text { Discharge } &=19.2 \text { lits } / \text { see } \\ &=19.2 \times 60 \text { lits } / \mathrm{min} \end{aligned}$

Discharge = 1154 .53 lits\min

Discharge $\approx 1155 \operatorname{lit} / \min$

(ii) Dimension of tank = $4 m \times 3 m \times 2.31 m$

Discharge $=0.0192$ $m^{3}/sec$

$1.154 \mathrm{~m}^{3} / \mathrm{min}$

Time taken to fill the tank $=\frac{\text { Volume of tank }}{\text { Discharge }}$

$=\frac{4 \times 3 \times 2.31}{1.154}$

Time taken to fill the tank = 24min

Question 13

Given 

Vessel 1 

radius 15cm 

Height 40cm 

Volume $\pi x r^{2} \times h$

$\pi \times(15)^{2}+40$

Volume $\quad 28274.33 \mathrm{~cm}^{3}$

Vessel 2

Radius  20cm

Height  40cm

Volume  $\pi x r^{2} \times h$ 

$\pi \times(20)^{2}+45$

Volume  $56548.667 \mathrm{~cm}^{3}$

Given 

Another vessel with capcity equal to sum of vessel 1 and vessel 2

Let radius of vessel's = R

Height of vessel = 30cm

$\left(\pi R^{2}\right) \times 30=28274.33+56548.667$

$\begin{aligned} 30 \times\left(11 R^{2}\right) &=84823 \\ R^{2} &=\frac{84823}{\pi \times 30} \end{aligned}$

$R^{2}=900$

$R=\sqrt{900}$

$R=30 \mathrm{Cm}$

∴ Radius of vessel = 30cm

Question 14

Sol :

Given 

Pole height = 7m

Pole diameter = 20cm = 0.2m

Density = $225 \mathrm{~kg} / \mathrm{m}^{3}$

Volume of wood = $\frac{\pi d^{2}}{4} \times h$

$=\frac{\pi}{4} \frac{(20)^{2}}{10^{4}} \times 7$

Volume of wood $=0.219 \mathrm{~m}^{3}$

Weight of wood = 49.48kg 

Question 15

Sol :

A cylinder with diameter of 14cm and height of 30cm is the maximum Volume 

That can be cut from given Cuboid 

Volume of cylinder = $\frac{\pi}{4} d^{2} \times h$

$=\frac{\pi}{4}(14)^{2} \times 30$

Volume of cylinder = 4618.14 $\mathrm{cm}^{3}$

Volume of wood wasted = Volume of cuboid - Volume of cylinder 

= $14 \times 14 \times 30-(4618.14)$

Volume of Wood wasted = $1261.85 \mathrm{~cm}^{3}$