Chapter – 8 : Right Circular Cylinder

Heights and Distance
WhatsApp
Book Name : Ganit Prakash
Subject : Mathematics
Class : 10 (Madhyamik)
Publisher : Prof Nabanita Chatterjee
Chapter Name : Right Circular Cylinder (8th Chapter)

Application 1.

If the length of the radius of the base of a right circular cylinder is and height is, let us write by calculating its lateral surface area.

Solution :

The length of the radius of the cylinder is \frac{12}{2}cm = 6cm.

∴ The lateral surface area of cylinder is = 2 × \pi × 6 × 21 sq.cm.

= 2 × \frac{12}{7} × 6 × 21 sq.cm. = 792sq.cm.

Application 2.

The perimeter of a right circular cylinder is 44 meter and height is 14 meter, let us write by calculating its lateral surface area. [Let me do it myself]

Solution :

Surface area of the cylinder = circumference of base x height

= (44 × 14) sqm = 616 sqm.

Application 3.

The base area of a closed cylindrical water tank is 616 sq. meters and height is 21 meters. Let us write by calculating the total surface area of that tank.

Solution :

Let the length of radius of circular base of water tank = r meter.

∴ Base area = \pi r2 sq meter

By the condition, \pi r2 = 616

or, \frac{22}{7} × r^{2} = 616

or, r2 = 616 \frac{7}{22} \quad \therefore r = \sqrt{196} = 14

Total surface area of water tank is = \left(2 \pi r^{2} + 2\pi r h\right) sq meter [where height of cylinder = h meter]

= 2 × \frac{22}{7} × r(r + h) sq meter

= 2 × \frac{22}{7} × 14(14 + 21) sq meter

= 3080 sqm.

Application 4.

If the perimeter of base of any closed cylindrical pot is 22 dm} and height is 5 dm}, let us write by calculating the area that will be coloured to paint the Outside of that pot. [Let me do it myself]

Solution :

Let radius = r dm & height = h dm.

∴ 2 \pi r = 22 \quad & h = 5

∴ 2 ×\frac{22}{7} × r = 22 \quad \therefore r = 7/2 dm.

Application 5.

(i) The total surface area of a right circular cylinder with one end Open is 1474sq.cm. If the length of diameter of the base is 14cm, let us write by calculating its height.

(ii) Again, if the pot would be closed at two ends, let us write by calculating what would be its total surface area.

Solution :

(i) The length of radius of base of right circular cylindrical pot = \frac{14}{2} cm = 7 cm

Let the height of the pot is h cm.

∴ Total surface area of pot = area of the base + area of the lateral surface

= \left(\frac{22}{7} × 7^{2} + 2 \frac{22}{7} × 7 × h\right) sq.cm.

= (154 + 44h) sq.cm.

By the condition, 154 + 44 h = 1474

∴ h = 30

∴ The height of the pot is 30 cm.

If the pot would be closed at two ends, surface area of that pot

= Area of the upper end surface + 1474sq.cm.

= \left(\frac{22}{7} × 7^{2} + i 474\right) sq.cm. = 1628 sqcm.

Application 6.

The length of the diameter of the base of a drum with lid made of steel sheet is 4.2 dcm. If 112.20 sq.dcm of steel sheet is required to make the drum let us write by calculating the height of the drum.

Again, if the price of 1 sq.dcm steel is Rs. 25 , let us calculate the production cost of the drum. [Let me do it myself]

Solution :

Let the height of the drum = h dcm.

Radius of the drum = \frac{4.2}{2} = 2.1 dcm.

According to the problem,

2 . \pi × 2.1(2.1 + h) = 112.20

or, 2 × \frac{22}{7} × \frac{22}{7}(2.1 + h) = 112.20

∴ 2.1 + h = \frac{1122 × 7 × 10}{10 × 2 × 22 × 21}

h = 8.5-2.1 = 6.4 dcm

∴ Total surface area = 112.20 sq.dcm = 1.1220 sqm.

∴ Total cost = Rs. 25 × 1.1220 = Rs. 28.05

Application 7.

If the length of diameter of base of glass is 11.2 cm and height is 15 cm, let us calculate the volume of water that the glass will contain.

Solution :

We understand, if the length of diameter of base of glass is 11.2 cm, height 15 cm, the glass will contain water

= \pi ×\left(\frac{11.2}{2}\right)^{2} × 15 cubic cm

= \frac{22}{7} × \frac{56}{10} × \frac{56}{10} × 15 cubic cm

= 1478.4sq.cm.

Application 8.

Height of a right circular cylinder made of iron open at two ends 42cm. If the thickness of the cylinder is 1cm and length of its external diameter is 10cm, let us calculate the volume of iron in it.

Solution :

IMG 4236

The length of external radius of the cylinder = 10cm. = 5cm.

It is of 1cm thickness. ∴ Length of internal radius of the cylinder = (5-1) cm . = 4cm. volume of iron

= \frac{22}{7}\left(5^{2}-4^{2}\right) × 42 cubic cm.

= \frac{22}{7} ×(5 + 4)(5-4) × 42 cubic cm.

\left[\because a^{2}-b^{2} = (a + b)(a-b)\right]

= \frac{22}{7} × 9 × 1 × 42 cubic cm.

= 22 × 9 × 6 cubic cm.

= 1188 cu cm.

Application 9.

If we colour inside and outside of the hollow right circular cylinder open at two ends, let us write by calculating how much area we shall colour.

Solution :

The sum of inner and outer surface area of this hollow right circular cylinder = (2 × \pi × 5 × 21 + 2 × \pi × 4 × 21)sq.cm.

= 2 × \pi × 21(5 + 4)sq.cm.

= 2 × \frac{22}{7} × 21 × 9sq.cm

= 1188sq.cm.

Application 11.

If the inner and outer radii of a hollow right circular cylindrical pipe of 6 metre length are 3.5cm and 4.2cm respectively, let us write by calculating the volume of the iron that the pipe contains.

If I cubic decimetre of iron weights 5 kilogram, let us write by calculating the weight of the pipe. [Let me do it myself]

Solution :

Height of pipe = 6 m = 60 dcm.

Interval radius of pipe = \frac{3.5}{20} = \frac{35}{200} dcm.

External radius of pipe = \frac{4.2}{20} = \frac{42}{200} dcm.

∴ Volume of Iron = \pi\left\{\left(\frac{42}{200}\right)^{2}-\left(\frac{35}{200}\right)^{2} × 60\right\} cu dcm.

= \frac{22}{7} × \frac{77}{200} × \frac{7}{200}× 60 cu dcm

= \frac{121 × 21}{1000} cu dcm . = \frac{2541}{1000} = 2.541 cu dcm.

Weight of 1 cu} dcm of iron = 5 kg.

Weight of 2.541 cu dcm of iron = (2.541 × 5) kg = 12.705 kg.

Application 12.

If the area of base of a cylinder is 13.86 sq metre and height is 8 metre, let us calculate the volume of the cylinder.

Solution : Let the length of radius of the base of the cylinder is r metre.

By the condition, \pi r2 = 13.86

or, \frac{22}{7} r^{2} = \frac{1386}{100} × \frac{7}{22} = \frac{441}{100}

∴ r = \frac{21}{10} m .

The volume of the cylinder is \frac{22}{7} × \frac{21}{10} × \frac{21}{10} × 8 cubic metre = 110.88 cu m.

Application 13.

If the perimeter of the base of a cylinder is 15.4cm and height is 10 cm, let us calculate its volume. [Let me do it myself]

Solution :

Let the radius of “base of the cylinder r cm & height = h cm = 10cm.

∴ Circumference = 2 \pi r = 2 × \frac{22}{7} × r = 15.4

r = \frac{154}{10} × \frac{7}{2 × 22} = \frac{49}{20}cm

∴ Volume of cylinder = \pi r2h

= \frac{22}{7} × \frac{49}{20} × \frac{49}{20} × 10 cu cm

= \frac{3773}{20} = 188.65 cu cm

Application 14.

If the glass of a tubelight is 105cm long and external circumference is 11cm and it is 0.2cm thick, let us write by calculating the volume (in cc) 0f the glass that will be required to make 5 such tube light.

Solution :

Let the length of external radius of tubelight = r1cm, length of internal radius = r2cm and height is h cm.

By the condition,

External circumference is = 2 r_{1} = 11 or, r_{1} = \frac{11 × 7}{2 × 22} = \frac{7}{4}

∴ The length of external radius is = \frac{7}{4}cm = 1.75cm.

The glass of tubelight is 0.2cm thick.

∴ Length of internal radius of tubelight = r_{2} = (1.75-0.2) cm . = 1.55cm.

External volume of the tube light – internal volume of tube light = \left(r_{1}^{2}-r_{2}^{2}\right) h

= \frac{22}{7} ×\left\{(1.75)^{2}-(1.55)^{2}\right\} × 105 cubic cm.

= \frac{22}{7} ×(1.75 + 1.55)-(1.75-1.55) × 105 cubic cm.

= \frac{22}{7} × 3.30 × 0.2 × 105 cubic cm

= 217.8 cu cm.

Application 15.

Through a hole, 110 kiloliter water enters into a ship. After closing the hole a pump is connected for the drainage of water. If the length of diameter of the pipe of the pump is 10cm and the speed of water flow is 350 metre/minute, let us write by calculating the time that will be required by the pump to clear off all water in the ship.

Solution :

Volume of water can be cleared off by pump in 1 minute

= \frac{22}{7} × \frac{10}{2} × \frac{10}{2} × \frac{1}{100} × 3500 cubic dcm.

= 2750 cubic dcm = 2750 litres [1 cubic dcm = 1 litres.]

Time required to clear off 110 kilolitres of water = \frac{110000}{2750} minutes = 40 minutes.

So, time required by the pump to clear off all water in the ship = 40 minutes.

Application 16.

A right circular cylindrical tank of 5 metre height is fixed with water. Water comes out from there through a pipe having the length of diameter 8cm at a speed of 225 metre/minute and the tank becomes empty after 45 minutes. Let us write by calculating the length of diameter of the tank. [Let me do it myself]

Solution :

Let radius of tank = r m.

∴ Radius of pipe = \frac{8}{2} = 4cm = \frac{1}{25} m.

∴ r2 × 5 = 45 × \pi\left(\frac{1}{25}\right)^{2} × 225

5 r2 = \frac{45 × 225}{25 × 25} \quad ∴ r2 = \frac{45 × 1 × 9}{5 × 25} = \frac{81}{25} \quad ∴ r = \frac{9}{5} m.

∴ Diameter of the tank = 2 × \frac{9}{5} = \frac{18}{5} = 3.6 m


LET US WORK OUT – 8

Question 1

Let us look at the picture of solid beside and answer the following :

IMG 4237

(i) Solid has _____________ surface.

Solution :

3

(ii) Number of curved surface is and number of plane surface is __________________.

Solution :

1, 2.

Question 2

Let us write the names of five solid objects of my house, the shapes of which are right circular cylinder.

Solution :

Glass, Pipe, Tube light, Pencil, Gas cylinder.

Question 3

The length of diameter of a drum made of steel covered with lid is 28cm. If 2816 \ sq cm steel sheet is required to make the drum, let us write by calculating the height of the drum.

Solution :

Let the height = h cm

& radius of the drum = \frac{28}{2} = 14cm.

According to problem,

2π(14 + h) × 14 = 2816 or, 2 × \frac{22}{7} × 14(14 + h) = 2816

or, 14 + h = 2816 × \frac{7}{2 × 22 × 14} = 32

∴ h = 32 – 14 = 18cm.

∴ Height of the drum = 18cm.

Question 4

Let us write by calculating how many cubic decimetres of concrete materials will be necessary to construct two cylindrical pillars, each of whose diameter is 5.6 decimetres and height is 2.5 metres. Let us write by calculating the cost of plastering the two pillars at Rs 125 per sq metre.

Solution :

Height of each pillar = (h) = 2.5 m = 25 dcm.

Radius of each pillar = (r) = \frac{5.6}{2} = 2.8 dcm.

Volume of plaster materials required to cover the two pillars

= 2 × π × (2.8)2 × 25 cu dcm.

= 2 × \frac{22}{7} × \frac{28}{10} × \frac{28}{10} × 25 = 56 × 22 = 1232 cu dcm.

Curved surface area of two pillars = 2 × π × 2.8 × 25 sq dcm.

= 2 × 2 × \frac{22}{7} × \frac{28}{10} = 880 Sq dcm. = 8.80 sqm

Total cost at the rate or Rs. 125 per sqm for the two pillars = Rs. 8.8 × 125 = Rs. 1100.

Question 5

If a gas cylinder for fuel purposes having a length of 7.5 dcm and the length of inner diameter of 2.8 dcm carries 15.015 kg of gas, let us write by calculating the weight of the gas per cubic dcm.

Solution :

Weight of gas in the cylinder = 15.015 kg = 15015 gm.

Radius of the cylinder = \frac{2.8}{2} = 1.4 dcm.

∴ Volume of cylinder = π(1.4)2 × 7.5 cu dcm

= \frac{22}{7} × \frac{14}{10} × \frac{14}{10} × \frac{75}{10} = \frac{462}{10} = 46.2 cu dcm.

∴ Weight of 1 cu dcm gm = \frac{15015}{46.2} gm = 325 gm.

Question 6

Out of three jars of equal diameter and height, \frac{2}{3} part of the first, \frac{5}{6} part of the second and \frac{7}{9} part of the third were filled with dilute sulphuric acid. The whole of the acid if the three jars was poured into a jar of 2.1 dcm diameter, as a result the height of acid in the jar became 4.1 dcm. If the length of the diameter of each of the three equal jars is 1.4 dcm, let us write by calculating the height of the three jars.

Solution :

Let the height of each jar = h dcm.

According to problem, πr2\left(\frac{2}{3} + \frac{5}{6} + \frac{7}{9}\right) × h =π\left(\frac{2.1}{2}\right)² × 4.1

or, \frac{12 + 15 + 14}{18} × h = \frac{22}{7} × \frac{21}{20} × \frac{21}{20} × \frac{41}{10}

∴ h = \frac{22}{7} × \frac{21}{20} × \frac{21}{20} × \frac{41}{10} × \frac{18}{41} = 4.05 dcm.

∴ Height of each jar = 4.05 dcm.

Question 7

Total surface area of a right circular pot open at one end is 2002 sq cm. If the length of diameter of base of the pot is 14cm, let us write by calculating how much liters of water the drum will contain.

Solution :

Let the height of cylinder = h cm.

& the radius of the cylinder = \frac{14}{2} = 7cm.

Accroding to the problem,

or, 2 × 7 × h + π(7)2 = 2002

or, 2 × \frac{22}{7} × 7 × h + \frac{22}{7} × 7 × 7 = 2002

or, 44h + 154 = 2002

∴ 44h = 2002 – 154 = 1848

∴ h = \frac{1848}{44} = 42

∴ Height of the cylinder = 42cm = 4.2 dcm.

Radius of the cylinder = 7cm = 0.7 dcm.

∴ Volume of water in the cylinder = π(0.7)² × 4.2 cu dcm.

= \frac{22}{7} × \frac{7}{10} × \frac{7}{10} × \frac{42}{10} = \frac{6468}{1000} = 6.468 cu dcm.

= 6.468 litre [\because 1 cu dcm = 1 litre ]

Question 8

If a pump set with a pipe of 14cm diameter can drain 2500 metres of water per minute, let us write by calculating how many kilolitres of water that pump will drain per hour. [ 1 liter = 1 cubic dcm.]

Solution :

Radius of the pipe = \frac{14}{2} = 7cm = 0.7 dcm.

Height of the pipe = 2500 m = 25000 dcm.

Cross-sectional area of the mouth of the pipe

= \frac{22}{7} × 7² sq cm. = 154 sq cm . = 1.54 sq dm

The volume of water irrigated by the pipe in one minute

= (1.54 × 25000) cu dcm = 38500 cu dcm

∴ Volume of water irrigated by the pipe in one hour

= (38500 × 60) cu dcm = 2310000 cu dcm.

Now, 1 cu dcm = 1 litre

∴ The pump set can irrigate 2310000 litres

= 2310 Kilolitre in 1 hour.

Question 9

There are some water in a long gas jar of 7cm diameter. If a solid right circular cylindrical pipe of iron having 5cm length and 5.6cm diameter be immersed completely in that water, let us write by calculating how much the level of water will rise.

Solution :

Let the water level will rise by h cm.

Accroding to the problem,

\pi\left(\frac{7}{2}\right)^{2} × h = π\left(\frac{5.6}{2}\right)² × 5

\frac{49}{4} h= \frac{28}{10} × \frac{28}{10} × 5

∴ h = \frac{28 × 28 × 5}{10 × 10 × 49} × 4 = \frac{32}{10} = 3.2

∴ The water level will rise by 3.2cm.

Question 10

If the surface area of a right circular cylindrical pillar is 264 sq metre and the volume is 924 cubic metres, let us write by calculating the height of the diameter of this pillar.

Solution :

Let the radius of the pillar = r m & height = h m.

According to 1st condition, 2rh = 264 – – – – – – – – – – – – (i)

According to 2nd condition, r² h = 924 – – – – – – – – – – – – – (ii)

\frac{\pi r² h}{2r h} = \frac{924}{264} \quad or,  r = 7

∴ 2r h = 2 × \frac{22}{7} × 7 × h = 264 \quad \therefore h = \frac{264}{44} = 6

∴ Diameter of the pillar = 2 × 7 = 14 m

& height of the pillar = 6 m.

Question 11

A right circular cylindrical tank of 9 metre height is filled with water. Water comes out from there through a pipe having length of 6cm diameter with a speed of 225 metre per minute and the tank becomes empty after 2 hr 24 minutes, let us write by calculating the length of diameter of the tank.

Solution :

Let the radius of the tank = R m

& radius of the pipe = \frac{6}{2} = 3cm . = 0.03 m.

According to the problem,

πR² × 9 = 36(0.03)² × 225 (Where radius of pipe = R)

R² = 36 × \frac{3}{100} × \frac{3}{100} × 225 × \frac{1}{9} = \frac{81}{100}

∴ R = \frac{9}{10}

∴ Diameter of the tank = 2 × \frac{9}{10} = 1.8 m.

Question 12

The curved surface area of the right circular cylindrical log of wood of uniform density is 440 sq dcm. If 1 cubic dcm of wood weighs 1.5 kg and the weight of the log is 9.24 quintals. Let us write by calculating the length of the diameter of the log and its height.

Solution :

Weight of the wooden log = 9.24 quintal = 924 kg.

∴ Volume of the log = \frac{924}{1.5} cu dcm = 6.16 cu dcm.

Let the radius of the log = r dcm & height of the log = h dcm.

According to 1st condition, 2πrh = 264 – – – – – – – – – – – – (i)

According to 2nd condition, π2rh = 264 – – – – – – – – – – – – (ii)

& ∴ \frac{\pi r² h}{2r h} = \frac{616}{440} \quad \text { or, } \frac{r}{2} = \frac{616}{440} \quad \therefore r = \frac{2 × 616}{440} = \frac{28}{10}

& ∴ 2r h = 440

or, 2 \times \frac{22}{7} \times \frac{28}{10} \text{ h } = 440 \quad \text{ h } = \frac{100}{4} = 25

∴ Diameter of the log = 2 × \frac{28}{10} = \frac{56}{10} = 5.6 dcm.

Height of the log = 25 dcm.

Question 13

The lengths of inner and outer diameter of a right circular cylindrical pipe open at two ends are 30cm and 26cm respectively and length of the pipe is 14.7 metre. Let us write by calculating the cost of painting its all surfaces with coal tar at Rs. 2.25 per dcm2.

Solution :

Internal radius of the pipe = \frac{26}{20} = 1.3 dcm.

External radius of the pipe = \frac{30}{20} = 1.5 dcm.

Height of the pipe = 14.7 m = 147 dcm.

Total surface area of the pipe

= \left[2π(1.5 + 1.3) 147 + 2π\left\{(1.5)² - (1.3)²\right\}\right] sq dcm.

= [2× 2.8 × 147 + 2× 2.8 × 0.2] sq dcm.

= 2 × 2.8(147 + 0.2)

= 2 × \frac{22}{7} × \frac{28}{10} × 147.2 sq dcm.

= 17.6 × 147.2 = 2590.72 sq dcm.

Total cost for paintaing with coal tar

= Rs. (2.25 × 2590.72) = Rs. 5829.12 .

Question 14

Height of a hollow right circular cylinder, open at both ends, is 2.8 metre. If the length of the inner diametre of the cylinder is 4.6 dcm and the cylinder is made up of 84.48 cubic dcm of iron, let us calculate the length of outer diameter of the cylinder.

Solution :

Let the external radius = r dcm. of the cylinder & the height of the cylinder = 2.8 m = 28 dcm.

The internal radius of the cylinder = \frac{4.6}{2} = 2.3 dcm.

According to the problem,

\pi\left\{r² - (2.3)²\right\} × 28 = 84.48

or, \frac{22}{2}\left(r² - 5.29\right) × 28 = 84.48

r² – 5.29 = \frac{84.48}{88} = 0.96

∴ r² = 5.29 = 5.29 + 0.96 = 6.25

Question 15

Height of a right circular cylinder is twice of its radius. If the height would be 6 times of its radius, then the volume of the cylinder would be greater by 539 cubic dcm, let us write by calculating the height of the cylinder.

Solution :

Let the height of the cylinder = h dcm & radius = r dcm.

∴ h = 2 r

Volume of the cylinder = π r² × 2 r = 2r³

If height h = 6 r, volume = π r² × 6 r = 6r³

According to the problem, 6π r³ – 2π r³ = 539

4πr³ = 539 or, 4 × \frac{22}{7} r³ = 539 \quad \therefore= \frac{539 × 7}{88} = \left(\frac{7}{2}\right)³

∴ r = \frac{7}{2} \quad \therefore 2 r = 7

∴ Height = 2 r = 7 dcm.

Question 16

A group of firebrigade personnel carried a right circular cylindrical tank filled with water and pumped out water at a speed of 420 metre per minute to put out the fire in 40 minutes by three pipes of 2cm diameter each. If the diameter of tank is 2.8 meter and its length is 6 metre, then let us calculate (i) what volume of water has been spend in putting out the fire and (ii) the volume of water that still remains in the tank.

Solution :

Radius of the tanker = \frac{2.8}{2} = 1.4 = 14 dcm.

Length of the tanker = 6 ~m} = 60 dcm.

Volume of the tanker = π(14)² × 60 cu dcm.

Radius of each pipe = \frac{2}{2} = 1cm = \frac{1}{10} dcm.

Length of the pipe = 420 m = 4200 dcm.

Quantity of water ejected in 40 minutes by 3 pipes

= 40 × 3 × π(\frac{1}{10})^{2} × 4200 cu dcm.

= 40 × 3 × \frac{22}{7} × \frac{1}{10} × \frac{1}{10} × 4200 cu dcm.

= 15840 cu dcm.

∴ Quantity of water left in the tanker

= (36960 – 15840) cu dcm.

= 21120 cu dcm.

Question 17

It is required to make a plastering of sand and cement with 3.5cm thick, surrounding four cylindrical pillars, each of whose diameter is 17.5cm.

(i) If each pillar is of 3 metre height, let us write by calculating how many cubic dcm of plaster materials will be needed.

(ii) If the ratio of sand and cement in the plaster material be 4 : 1, let us write how many cubic dcm of cement will be needed.

Solution :

Length of each pillar = 3 m = 30 dcm.

Internal radius of each pillar = \frac{17.5}{2} = \frac{175}{20}cm . = \frac{7}{8} dcm.

External radius of each pillar = \left(\frac{7}{8} + \frac{75}{10}\right) dcm.

= \left(\frac{7}{8} + \frac{35}{100}\right) = \left(\frac{7}{8} + \frac{7}{20}\right) \text{ dcm } = \frac{35 + 14}{40} = \frac{49}{40} dcm.

(i) Plaster material is required for 4 pillars

= 4π\left\{\left(\frac{49}{40}\right)² - \left(\frac{7}{8}\right)²\right\} × 30 cu dcm.

= 4 × \frac{22}{7}\left\{\left(\frac{49}{40} + \frac{7}{8}\right)\left(\frac{49}{40} - \frac{7}{8}\right)\right\} × 30 cu dcm.

= 4 × \frac{22}{7} × \frac{84}{40} × \frac{14}{40} × 30 \text{ cu dcm } = \frac{2772}{10} = 277.2 cu dcm.

(ii) Volume of cement required = \frac{1}{5} × 277.2 cu dcm . = 55.44 cu dcm.

Question 18

The length of outer and inner diameter of a hollow right circular cylinder are 16 cm and 12cm respectively. Height of the cylinder is 36cm. Let us calculate how many solid cylinders of 2cm radius and 6cm length may be obtained by melting this cylinder.

Solution :

External radius of the cylinder = \frac{16}{2} = 8cm.

Internal radius of the cylinder = \frac{12}{2} = 6cm.

Height of the cylinder = 36cm.

Radius of the solid cylinder = \frac{2}{2} = 1cm.

Height of the cylinder = 6cm.

Volume of hollow cylinders = π\left\{(8)² - (6)²\right\} × 36 cu cm.

Volume of x no. of solid cylinders = x : π(1)² . 6 cu cm.

According to the problem,

x \cdotπ(1)² \cdot 6 =π\left\{(8)² - (6)²\right\} × 36

or, x . 1 . 6 = π(64 – 36) × 36

or, x = \frac{28 × 36}{6} = 168

∴ No. of solid cylinders = 168.


Very short answer type questions. (V.S.A.)

M.C.Q. :

Question 1

If the lengths of radii of two solid right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, then ratio of their lateral surface areas is

  1. 2 : 5
  2. 7 : 7
  3. 10 : 9
  4. 16 : 9

Solution :

Ratio of area of curved surfaces

= 2 × 2 × 5 : 2 × 3 × 3 = 10 : 9

Ans. (c) 10 : 9

Question 2

If the lengths of radii of two solid right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, then the ratio of their volumes is

  1. 27 : 20
  2. 20 : 27
  3. 40 : 9
  4. 9 : 4

Solution :

= π(2)² × 5 : π(3)² × 3 = 20 : 27

Ans. (b) 20 : 27

Question 3

If volumes of two solid right circular cylinders are same and their heights are in the ratio 1 : 2, then the ratio of lengths of radii is

  1. 1 : \sqrt{2}
  2. \sqrt{2} : 1
  3. 1 : 2
  4. 2 : 1

Solution :

Ratio of radius of the cylinder = \sqrt{2} : 1

Ans. (b) \sqrt{2} : 1

Question 4

In a right circular cylinder, if the length of radius is halved and height is doubled, volume of the cylinder will be

  1. Equal
  2. Double
  3. Half
  4. 4 times

Solution :

Let radius & height of cylinder are r unit & h unit respectively, volume \left(v_{1}\right) = r² h cu unit.

Now, if radius & height of cylinder \frac{r}{2} unit & 2 h unit respectively volume \left(v_{2}\right)

= π\left(\frac{r}{2}\right)^{2} × 2 h = \frac{1}{2}r² h = \frac{1}{2} v_{1}

Ans. (c) Half

Question 5

If the length of radius of a right circular cyclinder is doubled and height is halved, the lateral surface area will be

  1. Equal
  2. Double
  3. Half
  4. 4 times

Solution :

Let radius & height of cylinder are r unit & h unit.

Area\left(A_{1}\right) = 2rh

Now, if radius & height be 2 r unit & \frac{h}{2} unit respectively

Area \left(A_{2}\right) = 2π(2 r) \frac{h}{2} = 2r h = A_{1}.

Ans. (a) Equal


Let us write true or false for the following statements :

Question 1

The length of a right circular drum is rcm and height is h cm. If half part of the drum is filled with water then the volume of water will ber² h cubic cm.

Solution :

FALSE

Question 2

If the length of radius of a right circular cylinder is z unit, the numerical value of volume and surface area of cylinder will be equal for any height.

Solution :

TRUE


Let us fill in the blanks :

Question 1

The length of a rectangular paper is ℓ units and breadth is b units. The rectangular paper is rolled and a cylinder is formed whose perimeter is equal to the length of the paper. The lateral surface are of the cyclinder is __________ sq unit.

Solution :

ℓb

Question 2

The longest rod that can be kept in a right circular cylinder having the diameter of 3cm and height of 4cm, then the length of rod is ______________ cm.

Solution :

5.

Question 3

If the numerical values of volume and Iateral cylinder are equal then the length of diameter of the surface area of a right circular cylinder are equal then the lenght of diameter of the cylinder is _____________ unit.

Solution :

4.


Short answer type question (S.A.) :

Question 1

If the lateral surface area of a right circular cylindrical pillar is 264 sq metres and volume is 924 cubic metres, let us write the length of radius of the base of the cylinder.

Solution :

Let the radius & height of the pillar = r m & h m respectively.

According to 1st condition, 2r h = 264 ————— (i)

According to 2nd condition, πr2h = 924 ————— (ii)

(ii) ÷ (i), \frac{\pi r² h}{2r h} = \frac{924}{264} \text{ or, } \frac{r}{2} = \frac{7}{2}

∴ r = 7

\because Radius = 7cm.

Question 2

If the lateral surface area of a right circular cylinder is c square units, length of radius of base is r unit and volume is v cubic unit, let us write the value of \frac{c r}{v}.

Ans. Let height = h unit.

\therefore \frac{C r}{v} = \frac{2r h × r}{\pi r² h} = 2

Question 3

If the height of a right circular cylinder is 14cm and lateral surface area is 264 sq cm, let us write the volume of the cylinder.

Solution :

Let height of the cylinder = rcm.

∴ 2rh = 264 or, 2 × \frac{22}{7} × r × 14 = 264 \quad \therefore r = 3

∴ Volume = π(3)² × 14 cu cm = \frac{22}{7} × 9 × 14 cu cm = 396 cu cm.

Question 4

If the heights of two right circular cylinders are in the ratio of 1 : 2 and perimeters are in the ratio of 3 : 4, let us write the ratio of their volumes.

Solution :

Radius & height of the two cylinders are r unit & R unit & heights are h unit & 2 h unit respectively.

∴ 2r : 2R = 3 : 4 \quad \therefore r : R = 3 : 4 \quad \therefore h = \frac{3 R}{r}.

Ratio of volume = π(r)² h : (R)² 2 h = \left(\frac{3 R}{4}\right)² : 2 R².

or, \frac{9 R}{16} : 2 R² = 9 : 32

Question 5

The length of radius of a right circular cylinder is decreased by 50% and height is increased by 50%, let us write by how much percent of the volume will be changed.

Solution :

Let radius & height are r unit & h unit respectively.

Volume \left(v_{1}\right) = r² h cu. unit.

It is reduced by 50%, \frac{50 r}{100} = \frac{r}{2} unit

& height is increased by 50% \frac{150 ~h}{100} = \frac{3 h}{2} unit.

∴ Volume \left(v_{2}\right) =π\left(\frac{r}{2}\right)² × \frac{3 h}{2} = \frac{3}{8}r²h cu unit.

∴ Volume reduced = \left[\frac{\pi r² h - \frac{3}{8}r²}{\pi r² h} × 100\right] \%

= \frac{5}{8} × 100 \% = 62 \frac{1}{2}%

Was This helpful ?

Subscribe
Notify of