Question

15. Two circular cylinders of equal volumes have their heights in the ratio 1:2. Find the
ratio of their radii.
16. The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of
its two circular faces is twice the area of the curved surface. Find the volume of the
cylinder.
17. How many cubic metres of earth must be dugout to sink a well 21 m deep and 6 m
diameter? Find the cost of plastering the inner surface of the well at 9.50 per
m?
18. The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the
trunk is 3 m. Find the volume of the timber that can be obtained from the trunk.
19. A cylindrical container with diameter of base 56 cm contains sufficient water to
submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the
rise in the level of the water when the solid is completely submerged.
20. A cylindrical tube, open at both ends, is made of metal. The internal diameter of the
tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm
everywhere. Calculate the volume of the metal.
21. From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find
the volume in litres of water delivered by the pipe in one hour.
22. A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved
surface of a right circular cylinder in two ways i.e., either by rolling the paper along​

Answer 1

15-We are given that Two circular cylinders of equal volumes have their heights in the ratio 1:2.

$Formula \: of \: volume \: of cylinder= \pi r^2 h$

$\small\bold{Let \: the \: radius \: of \: cylinder \: be \: r_1} \\ \bold{and \: r_2r }$

ATQ

$\frac{\pi (r_1)^2 h}{\pi (r_2)^2 h}=1$

$\frac{(r_1)^2}{2(r_2)^2 }=1$

$\frac{(r_1)^2}{(r_2)^2 }=\frac{1}{2}$

$\frac{(r_1)}{(r_2)}=\sqrt{\frac{1}{2}}$

Hence the ratio of their radii is 1:√2 .

16-image

17-H =21 m

D= 6m

R = 6/2m=3m

Volume of cylinder =

volumeofcylinder=πr^2h

22/7×3×3×21

= 594m^2

Cost = Rate × CSA

= > 9.50 per m^2×2πrh=>9.50perm

=>9.50×2×22/7×3×21

Cost => ₹ 3762

Hence, Solved.

18-Length of the trunk = 3m = 300 cm

Circumference of trunk of tree = 176 cm

2πr = 176

2 × 22/7 × r = 176

r = 176×7 / 2×22

= 28cm

Radius = 28cm

∴ Volume of timber can be obtained from trunk of tree = πr^2h

= 22/7 × 28 × 28 × 300

= 7392 cm^3

= 0.74 m^3

19-Diameter of base of cylindrical vessel = 56 cm

Radius of base = d/2 = 56/2 = 28cm

Dimensions of rectangular solid vessel = 32cm × 22cm × 14cm

Volume of rectangular solid vessel = 32 × 22 × 14 = 9856 cm^3

Let the rise of water level be ‘h’ cm

Volume of cylindrical container = Volume of rectangular solid vessel

πr2h = 9856

22/7 × 28 × 28 × h = 9856

h = 9856×7 / 22×28×28

= 68992 / 17248 = 4cm

∴ Rise in water level is 4cm.

20-Internal diameter of cylindrical tube = 10.4 cm

Internal radius of tube = d/2 = 10.4/2 =5.2cm

Length of tube = 25 cm

Thickness of metal = 8 mm = 0.8 cm

Outer radius of tube = R = 5.2+0.8 = 6 cm

∴ Volume of metal = π (R2 – r2) × l

= 22/7 × (62 – 5.22) × 25

= 22/7 × (36 – 27.04) × 25

= 704 cm3

21-Inner radius of tap = 0.75 cm

Find the area of cross section of pipe

Area=πr^2

Area=3.14×0.75^2

Area=1.76625 cm^2

Rate of flow of water = 7 m per second

1 m = 100 cm

7 m = 700 cm

1 hour = 3600 seconds

Rate of flow of water in 1 hour =700×3600

Find the volume in litres of water delivered by the pipe in one hour

→1.76625×700×3600=4450950 cm^3

Convert to liters

1 cm^3=0.001 liter

4450950cm^3=4450950×0.001=4450.95 liter

Thus 4450.95 liters of water delivered by the pipe in one hour

22-On rolling the paper along the length,

height (h) of cylinder = 18 cm,

radius (r) of cylinder = 15 cm.

Now,

Volume of cylinder (V) =πr^2h

V=3.14\times 15^{2} \times18V=3.14×15

2

×18 ----------(\pi =3.14π=3.14 )

V=3.14\times 15\times 15\times 18V=3.14×15×15×18

Volume of cylinder on rolling along length = 3.14\times 15\times 15\times 183.14×15×15×18 sqcm. --------(equation 1)

Now, on rolling paper along its breadth,

height (h) of cylinder = 30 cm,

radius (r) of cylinder = 9 cm.

Now,

Volume of cylinder (V) πr^2h

V=3.14×9^2×30 --------(π=3.14 )

V=3.14× 9×9× 30

Volume of cylinder on rolling along breadth=3.14×9×9×3sqcm.---------(equation 2)

Now, from equation 1 and equation 2,

Ratio=equation 1/equation 2

Ratio= 3.14×15×15×18/3.14×9×9×30

Ratio=15/9

Ratio=5/3

Therefore ratio of cylinder after rolling paper length wise and breadth wise is 5:3.