S.chand publication New Learning Composite mathematics solution of class 8 Chapter 12 Mensuration Exercise 12F

 Exercise 12F

---

Q1 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 1

Find the volume of each cylinder

(a)

Volume of cylinder=πr²h

$=\frac{22}{7}\times 3^2 \times 7$

=198cm³

(b)

Volume of cylinder=πr²h

$=\frac{22}{7}\times 7^2\times 10$

$=\frac{22}{7}\times 7\times 7\times 10$

=1540cm³

(c)

Volume of cylinder=πr²h

$=\frac{22}{7}\times 6^2\times 14$

=1584cm³

(d)

Volume of cylinder=πr²h

$=\frac{22}{7}\times (1.3)^2\times 1.4$

=7.436m³

---

Q2 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 2

Find the volume of each of the following cylinders

(a)

Radius=7cm , Height=12cm

Volume=πr²h$=\frac{22}{7}\times 7^2\times 12$=1848cm³

(b)

Radius=3cm ,Height=21cm ,

Volume=πr²h$=\frac{22}{7}\times 3^2\times 21$=594cm³

(c)

Radius=10mm ,Height=28mm ,

Volume=πr²h$=\frac{22}{7}\times (10)^2\times 28$=8800cm³

(d)

Radius=0.7 m, Height=2 m

Volume=πr²h$=\frac{22}{7}\times (0.7)^2\times 2$=3.07876 m³

---

Q3 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 3

A milk tank is in the form of a cylinder whose radius is 2 m and length is 7 m. Find the quantity of milk (in litres) that this tank can contain.

Sol :

r=2m , h=7m

The quality of milk=πr²h

$=\frac{22}{7}\times 2^2\times 7$

=88m³

=88×1000=88000L

---

Q4 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 4

Find the height of the cylinder whose volume is $7040 \mathrm{~cm}^{3}$ and radius is 8 cm

Sol :

r=8cm

Volume=7040cm

ATQ ,

πr²h=7040

$h=\frac{7040}{\pi r^2}$

$=\frac{7040}{\frac{22}{7}\times 8^2}=\frac{7040\times 7}{22\times 8\times 8}$

=35cm

---

Q5 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 5

The area of the base of a right circular cylinder is $872 \mathrm{~cm}^{2}$ and its volume is $4360 \mathrm{~cm}^{3}$. Find the height of the cylinder.

Sol :

Area=πr²=872cm²

Volume=4360

ATQ

πr²h=volume

$h=\frac{vol}{\pi r^2}=\frac{4360}{872}$

=5cm

---

Q6 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 6

The circumference of the base of a cylindrical vessel is 176 cm and its height is 30 cm. How much water (in litres) can it hold?

Sol :

Circumference of cylinder=2πr=176

∴2πr=176

$r=\frac{176}{2\times \frac{22}{7}}=\frac{176\times 7}{2\times 22}=28$cm

h=30cm

∴Volume=πr²h$=\frac{22}{7}\times (28)^2\times 30$

$=\frac{22}{7}\times 28\times 28\times 30$

=73920cm³

=73.92litres

---

Q7 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 7

How much will a metallic cylinder of radius 14 m and height 3.5 m cost if the metal used in making this cylinder is ₹ 100 per cubic metre?

Sol :

r=14m , h=3.5m

∴Volume=πr²h$=\frac{22}{7}\times (14)^2\times 3.5$

=2156 m³

1m³ =₹100

∴2156 cm³ =₹2156×100=₹ 215600

---

Q8 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 8

The radii of the bases of two cylinders are in the ratio 3: 4 and their heights are in the ratio 4: 3. Find the ratio of their volumes.

Sol :

Let, the common factor is x

∴r₁=3x

r₂=4x

Similarly,

h₁=4x

h₂=3x

∴Volume₁=πr₁²h₁

Volume₂=πr₂²h₂

∴Volume₁ : Volume₂

πr₁²h₁ : πr₂²h₂

$=\frac{\pi r_{1}^{2}h_1}{\pi r_{2}^{2}h_2}$

$=\frac{(3x)^2 \times (4x)}{(4x)^2\times (3x)}$

=3 : 4

---

Q9 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 9

The length of two cylinders of equal volumes are in the ratio 2: 3. Find the ratio of their radii.

Sol :

Let , factor common factor is x

∴h₁ : h₂

∴h₁=2x

h₂=3x

ATQ ,

πr₁²h₁ = πr₂²h₂

$\left(\frac{r_1}{r_2}\right)^2=\frac{h_2}{h_1}$

$\frac{r_1}{r_2}=\sqrt{\frac{h_2}{h_1}}$ 

$=\sqrt{\left(\frac{3x}{2x}\right)}=\frac{\sqrt{3}}{\sqrt{2}}$

r₁ : r₂ =√3 : √2

---

Q10 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 10

Find the number of coins 1.5 cm in diameter and 0.2 cm thick, which should be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Sol :

Circular cylinder volume=πr²H

$=\frac{22}{7}\times (4.5)^2 \times 10$

Coins volume=πr²h

∴Number of coins$=\frac{\pi R^2H}{\pi r^2 h}=\frac{(4.5)^2\times 10}{(1.5)^2\times 0.2}$

$=\frac{45\times 45\times 10\times 10 \times 10 \times 10}{15\times 15\times 2\times 10\times 10}$

=450 coins

---

Q11 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 11

The metal with the volume of $1496 \mathrm{~cm}^{3}$ is used to cast a pipe of length 28 cm. If internal radius of the pipe is 8 cm. Find its outer radius.

Sol :

r=8cm h=28cm

Volume of metal=1496m³

Let, outer radius=R

ATQ,

πh(R²-r²)=1496

$R^2-8^2=\frac{1496}{\pi \times h}=\frac{1496 \times 7}{22 \times 28}=17$

R²-64=17+64=81

R=√81=9cm

---

Q12 | Ex-12F |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 12

A rectangular paper of length 44 cm and width 6 cm is rolled to form a cylinder of height equal to width of the paper Find the radius of the cylinder so rolled.

Sol :

Let, radius=r

width=h=6cm

ATQ,

2πr=44

$r=\frac{44}{2\pi}=\frac{44\times 7}{2\times 22}$

=7cm