Cylinder, Cone & Sphere (Surface Area & Volume) — Test Yourself

The given figure shows a solid cylinder and a solid cone on it according to the given measurement, the total surface area of the solid in terms of π and r is :

1. 4πr²

2. 5πr²

3. 6πr²

4. 7πr²

Answer

From figure,

Total surface area of solid = Curved surface area of cylinder + Curved surface area of cone + Area of cross-section

= 2πrh + πrl + πr²

= 2πr.r + πr.2r + πr²

= 2πr² + 2πr² + πr²

= 5πr².

Hence, Option 2 is the correct option.

The radii of two solid spheres are 10 cm and 20 cm. The ratio between their volumes is :

1. 3 : 8

2. 1 : 4

3. 1 : 8

4. 8 : 3

Answer

Radius of first solid sphere (r) = 10 cm

Radius of second solid sphere (R) = 20 cm

Ratio of volumes of two spheres :

$\Rightarrow \dfrac{\text{Vol. of 1st sphere}}{\text{Vol. of 2nd sphere}} = \dfrac{\dfrac{4}{3}πr^3}{\dfrac{4}{3}πR^3} \\[1em] = \dfrac{r^3}{R^3} \\[1em] = \dfrac{10^3}{20^3} \\[1em] = \dfrac{1000}{8000} \\[1em] = \dfrac{1}{8} \\[1em] = 1 : 8.$

Hence, Option 3 is the correct option.

A cone and a sphere have equal volumes. If height of the cone = radius of the sphere = 10 cm, the radius of the cone is :

1. 40 cm

2. 10 cm

3. 30 cm

4. 20 cm

Answer

Given,

Height of the cone (H) = radius of the sphere (r) = 10 cm

Volume of cone and sphere are equal.

Let radius of cone be R cm.

$\therefore \dfrac{1}{3}πR^2H = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow R^2H = 4r^3 \\[1em] \Rightarrow R^2 = \dfrac{4r^3}{H} \\[1em] \Rightarrow R^2 = \dfrac{4 \times 10^3}{10} \\[1em] \Rightarrow R^2 = 4 \times 10^2 \\[1em] \Rightarrow R^2 = 400 \\[1em] \Rightarrow R = \sqrt{400} = 20 \text{ cm}.$

Hence, Option 4 is the correct option.

A solid metal cuboid with dimension 10 cm, 11 cm and 8 cm is melted and identical solid spheres, each of radius 1 cm, are formed. The number of sphere formed is :

1. 105

2. 210

3. 315

4. 420

Answer

Given,

A solid metal cuboid with dimensions 10 cm, 11 cm and 8 cm is melted and identical solid spheres, each of radius 1 cm, are formed. Let no. of spheres formed be n.

∴ Volume of metal cuboid = n × Volume of sphere

⇒ l × b × h = n × $\dfrac{4}{3}πr^3$

⇒ 10 × 11 × 8 = n × $\dfrac{4}{3} \times \dfrac{22}{7} \times 1^3$

⇒ 880 = n × $\dfrac{88}{21}$

⇒ n = $\dfrac{880 \times 21}{88}$

⇒ n = 10 x 21 = 210.

Hence, option 2 is the correct option.

By applying force on an uniform metallic wire, the wire is extended four times along its length with same width all around. Then :

(i) The volume of metal in both the cases is same.

(ii) The surface area of solid wires in both the cases is same.

Which of the above statements is/are true?

1. only (i)

2. only (ii)

3. both (i) and (ii)

4. neither (i) nor (ii)

Answer

On transforming a solid from one shape to another by any external force, the volume of solid remains same.

Let initial length of wire be l and radius be r.

After extension,

Length of wire = 4l and Radius be R.

Since, volume is equal in both the cases :

∴ πr²h = πR²H

⇒ πr²l = πR².4l

⇒ r² = 4R²

⇒ r = $\sqrt{4R^2}$

⇒ r = 2R

⇒ R = $\dfrac{r}{2}$

Surface area in first case :

⇒ 2πrh

⇒ 2πrl

Surface area in second case :

⇒ 2πRh

⇒ $2π \times \dfrac{r}{2} \times 4l$

⇒ 4πrl.

∴ Surface area of solid wires in both the cases are different.

Hence, Option 1 is the correct option.

The given figure shows a toy made of very thin metal sheet. The toy has the hemisphere, a cylinder and a cone all of the same radius.

Assertion(A): The minimum area of metal sheet required is:

Curved surface area of conical part + curved surface area of cylinder part - curved surface area of hemispherical part

Reason(R): Metal sheet required = curved surface area of (conical part + cylindrical part + hemispherical part)

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

The toy has the hemisphere, a cylinder and a cone all of the same radius.

Metal sheet required = minimum area of metal sheet required

= curved surface area of conical part + curved surface area of cylinder part + curved surface area of hemisphere part

= curved surface area of (conical part + cylindrical part + hemispherical part)

∴ A is false, R is true.

Hence, option 2 is the correct option.

The base radius of two right circular cone of the same height are in ratio 3 : 5.

Assertion(A): The ratio between their volume is 9 : 25.

Reason(R): As

$\dfrac{r_1}{r_2} = \dfrac{3}{5} \\[1em] \dfrac{πr_1^2h}{πr_2^2h} = \dfrac{r_1^2}{r_2^2} = \Big(\dfrac{3}{5}\Big)^2 \\[1em]$

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, the base radius of two right circular cone of the same height are in ratio 3 : 5.

As we know that volume of cone = $\dfrac{1}{3}πr^2h$

$\dfrac{\text{Volume of 1st cone}}{\text{Volume of 2nd cone}} = \dfrac{\dfrac{1}{3}πr_1^2h}{\dfrac{1}{3}πr_2^2h}\\[1em] = \dfrac{πr_1^2h}{πr_2^2h}\\[1em] = \dfrac{r_1^2}{r_2^2}\\[1em] = \dfrac{3^2}{5^2}\\[1em] = \Big(\dfrac{3}{5}\Big)^2\\[1em] = \dfrac{9}{25}.$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

A solid cone of height 3 cm and radius 3 cm is recast into solid cylinder each of height 1 cm and radius 1 cm.

Assertion(A): Number of cylinders formed = $\dfrac{1}{3}$ x 3 x 3 x 3

Reason(R): Number of cylinders formed = $\dfrac{\dfrac{1}{3}π(3)^2 \times 3}{π(1)^2 \times 1}$

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, radius of solid cone, R = 3 cm

Height of solid cone, H = 3 cm

Radius of solid cylinder, r = 1 cm

Height of solid cylinder, h = 1 cm

By formula,

Volume of cone = $\dfrac{1}{3}πR^2H$

Volume of cylinder = $πr^2h$.

Given,

Solid cone is recasted into cylinders. Let no. of cylinders formed be n.

∴ Volume of cone = n × Volume of cylinder

$\Rightarrow \dfrac{1}{3}πR^2H = n × πr^2h\\[1em] \Rightarrow \dfrac{1}{3}π(3)^2 \times 3 = n × π(1)^2 \times 1\\[1em] \Rightarrow \dfrac{1}{3} (3)^2 \times 3 = n × 1 \times 1\\[1em] \Rightarrow n = \dfrac{1}{3} \times 3 \times 3 \times 3\\[1em] \Rightarrow n = 9.$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

A solid wooden cylinder is of height h cm and radius r cm. A conical cavity of same height and the same radius is drill out of the solid cylinder.

Statement (1): The volume of the remaining wood = volume of the solid cylinder - volume of the cone drilled.

Statement (2): The volume of the remaining wood = πr²h - $\dfrac{1}{3}$ πr²h = $\dfrac{2}{3}$ πr²h

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

The volume of the remaining wood is the volume of the original solid cylinder minus the volume of the conical cavity drilled out.

Let radius and height of cylinder and conical cavity (both are equal) be r and h units.

Volume of the cylinder : πr²h

Volume of the cone : $\dfrac{1}{3}$ πr²h

The volume of the remaining wood = Volume of the solid cylinder - Volume of the cone drilled.

= πr²h - $\dfrac{1}{3}$ πr²h

= $\Big(1 - \dfrac{1}{3}\Big)$ πr²h

= $\Big(\dfrac{3 - 1}{3}\Big)$ πr²h

= $\dfrac{2}{3}$ πr²h.

∴ Both the statements are true.

Hence, option 1 is the correct option.

A solid sphere of radius 6 cm is melted and recast into solid spheres of diameter 2 cm each.

Statement (1): The number of solid sphere formed is 6 x 6 x 6 = 216.

Statement (2): If the smaller spheres formed are identical, the number of solid sphere formed = $\dfrac{\text{Volume of sphere melted}}{\text{Volume of each of the sphere formed}}$

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, radius of solid sphere, R = 6 cm

Diameter of new sphere, d = 2 cm

Radius of new sphere, r = $\dfrac{2}{2}$ cm = 1 cm

Volume of the sphere: $\dfrac{4}{3}$ πr³

Given,

A solid sphere of radius 6 cm is melted and recast into solid spheres of diameter 2 cm each. Let no. of spheres formed be n.

∴ Volume of sphere = n × Volume of sphere recasted.

⇒ n = $\dfrac{\text{Volume of sphere melted}}{\text{Volume of each new sphere formed}}$

∴ Statement 2 is true.

$\Rightarrow n = \dfrac{\dfrac{4}{3}πR^3}{\dfrac{4}{3}πr^3}\\[1em] \Rightarrow n = \dfrac{R^3}{r^3}\\[1em] \Rightarrow n = \dfrac{6^3}{1^3}\\[1em] \Rightarrow n = \dfrac{216}{1}\\[1em] \Rightarrow n = 216.$

∴ Statement 1 is true.

Thus, both the statements are true.

Hence, option 1 is the correct option.

Water in a canal 6 m wide and 2 m deep, is flowing with the speed of 18 km/h.

Statement (1): The volume of water that flows through the canal in 20 minutes = 6 x 2 x $\Big(18 \times \dfrac{5}{18}\Big)$ x 20 x 60 m³.

Statement (2): The volume of water that flows through the canal = 6 x 2 x 18 x 20 m³.

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, water in a canal 6 m wide and 2 m deep, is flowing with the speed of 18 km/h.

Volume of water flowing through the canal = Area of cross-section x speed x time

Area of cross-section = width × depth = (6 × 2) m²

Speed = 18 km/h = 18 × $\dfrac{1000}{3600}$ = $\Big(18 × \dfrac{5}{18}\Big)$ m/sec

Time = 20 minutes = 20 × 60 sec

∴ Volume = 6 × 2 x 18 × $\dfrac{5}{18}$ x 20 × 60 m³

Thus, Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

What is the least number of solid metallic spheres, each of 6 cm diameter, that should be melted and recast to form a solid metal cone whose height is 45 cm and diameter 12 cm ?

Answer

Given,

Diameter of metallic spheres = 6 cm

Radius of metallic spheres (r) = $\dfrac{6}{2}$ = 3 cm.

Height of cone (H) = 45 cm

Diameter of cone = 12 cm

Radius of cone (R) = $\dfrac{12}{2}$ = 6 cm.

Let no. of spheres melted to form a solid metal cone be n.

∴ n × Volume of each sphere = Volume of solid metal cone

$\Rightarrow n \times \dfrac{4}{3}πr^3 = \dfrac{1}{3}πR^2H \\[1em] \Rightarrow n = \dfrac{\dfrac{1}{3}πR^2H}{\dfrac{4}{3}πr^3} \\[1em] \Rightarrow n = \dfrac{πR^2H \times 3}{3 \times4 \times πr^3} \\[1em] \Rightarrow n = \dfrac{R^2H}{4r^3} \\[1em] \Rightarrow n = \dfrac{6^2 \times 45}{4 \times 3^3} \\[1em] \Rightarrow n = \dfrac{36 \times 45}{4 \times 27} \\[1em] \Rightarrow n = \dfrac{1620}{108} \\[1em] \Rightarrow n = 15.$

Hence, 15 solid metallic spheres need to be melted.

A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.
(Answer correct to the nearest integer)

Answer

Radius of largest sphere from a right circular cylinder will be equal to radius of circular cylinder = 7 cm.

Volume of sphere = $\dfrac{4}{3}πr^3$

$= \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \\[1em] = \dfrac{4}{3} \times 22 \times 7^2 \\[1em] = 1437 \text{ m}^3.$

Hence, volume of sphere = 1437 m³.

A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in identical cones of height 12 cm and diameter 6 cm having a hemi-spherical shape on the top. Find the number of cones required.

Answer

Given,

Diameter of the cylinder = 12 cm

So, its radius (r) = $\dfrac{12}{2}$ = 6 cm

Height of the cylinder (h) = 15 cm

Diameter of the cone = 6 cm

So, its radius (R) = $\dfrac{6}{2}$ = 3 cm

Height of the cone (H) = 12 cm

Radius of the hemisphere = Radius of cone = R = 3 cm

Now,

Let the number of cones be n.

Volume of the cylinder = πr²h

= π × (6)² × 15

= 540π cm².

Volume of an ice-cream cone with ice-cream = Volume of cone + Volume of hemisphere

$= \dfrac{1}{3}πR^2H + \dfrac{2}{3} \times π \times R^3 \\[1em] = \dfrac{1}{3} \times π \times 3^2 \times 12 + \dfrac{2}{3} \times π \times 3^3 \\[1em] = 36 π + 18 π \\[1em] = 54 π \text{ cm}^3.$

Let no. of cones required be n.

Volume of cylinder = n × Volume of each cone

n = $\dfrac{\text{Vol. of cylinder}}{\text{Vol. of ice-cream cone}} = \dfrac{540π}{54π}$ = 10.

Hence, the number of cones required = 10.

A solid is in the form of a cone standing on a hemisphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find in terms of π, the volume of the solid.

Answer

Given,

Radius of both cone and hemisphere (r) = 8 cm

Height of cone (h) = 8 cm

From figure,

Volume of the solid = Volume of cone + Volume of hemisphere

$= \dfrac{1}{3}πr^2h + \dfrac{2}{3} πr^3 \\[1em] = \dfrac{1}{3} \times π \times 8^2 \times 8 + \dfrac{2}{3} \times π \times 8^3 \\[1em] = \dfrac{1}{3}.π.8^3 + \dfrac{2}{3}.π.8^3 \\[1em] = \Big(\dfrac{1 + 2}{3}\Big) \times π \times 512 \\[1em] = 512π \text{ cm}^3.$

Hence, volume of solid = 512π cm³.

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of wire.

Answer

Given,

Diameter of the sphere = 6 cm

So, its radius (R) = $\dfrac{6}{2}$ = 3 cm

Diameter of cylindrical wire = 0.2 cm

So, the radius of the wire (r) = $\dfrac{0.2}{2}$ = 0.1 cm

Let length of wire = h

Since, sphere is melted and recasted into a wire.

∴ Volume of sphere = Volume of wire

$\Rightarrow \dfrac{4}{3}πR^3 = πr^2h \\[1em] \Rightarrow h = \dfrac{\dfrac{4}{3}πR^3}{πr^2} \\[1em] \Rightarrow h = \dfrac{4R^3}{3r^2} \\[1em] \Rightarrow h = \dfrac{4 \times 3^3}{3 \times (0.1)^2} \\[1em] \Rightarrow h = \dfrac{4 \times 3^2}{0.01} \\[1em] \Rightarrow h = 4 \times 3^2 \times 100 \\[1em] \Rightarrow h = 3600 \text{ cm} \\[1em] \Rightarrow h = 3600 \times \dfrac{1}{100} \text{ m} \\[1em] \Rightarrow h = 36 \text{ m}.$

Hence, length of the wire = 36 m.

Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.

Answer

Let edge of the cube = a units.

Then, Volume of the cube = a x a x a = a³

The sphere that exactly fits in the cube will have diameter a units.

Radius of sphere (r) = $\dfrac{a}{2}$ units.

$\dfrac{\text{Vol. of cube}}{\text{Vol. of sphere}} = \dfrac{\text{(side)}^3}{\dfrac{4}{3}πr^3} \\[1em] = \dfrac{a^3}{\dfrac{4}{3}π \times \Big(\dfrac{a}{2}\Big)^3} \\[1em] = \dfrac{a^3 \times 3 \times 2^3}{4π \times a^3} \\[1em] = \dfrac{3 \times 8}{4 \times \dfrac{22}{7}} \\[1em] = \dfrac{3 \times 8 \times 7}{4 \times 22} \\[1em] = \dfrac{21}{11} \\[1em] = 21 : 11.$

Hence, ratio of volume of cube to volume of sphere = 21 : 11.

An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm³ of iron has 8 gm of mass (approx). (Take π = $\dfrac{355}{113}$)

Answer

Given,

Diameter of cylindrical portion = 12 cm

Radius of cylindrical portion (r) = $\dfrac{12}{2}$ = 6 cm

Height of the cylindrical part (H) = 110 cm

Height of the conical part (h) = 9 cm

From figure,

Radius of conical part = Radius of cylindrical portion = r = 6 cm.

Volume of iron pole = Volume of cylindrical portion + Volume of conical portion

$= πr^2H + \dfrac{1}{3}πr^2h \\[1em] = πr^2\Big(H + \dfrac{1}{3}h\Big) \\[1em] = \dfrac{355}{113} \times 6^2 \times \Big(110 + \dfrac{1}{3} \times 9\Big) \\[1em] = \dfrac{355}{113} \times 36 \times 113 \\[1em] = 12780 \text{ cm}^3.$

Given,

Weight of 1 cm³ of iron = 8 gm.

Total weight = 12780 x 8 gm = 102240 gm

= $\dfrac{102240}{1000}$ kg = 102.24 kg.

Hence, mass of pole = 102.24 kg

A cylindrical water tank of diameter 2.8 m and height 4.2 m is being fed by a pipe of diameter 7 cm through which water flows at the rate of 4 m s^-1. Calculate, in minutes, the time it takes to fill the tank.

Answer

Given,

Diameter of cylindrical tank = 2.8 m

So, its radius (R) = $\dfrac{2.8}{2}$ = 1.4 m

Height (H) = 4.2 m

Volume of water filled in it = πr²h

= $\dfrac{22}{7}$ x 1.4 x 1.4 x 4.2

= 25.872 m³

= 25.872 × (100)³ cm³

Given,

Diameter of the pipe = 7 cm

Radius (r) = $\dfrac{7}{2}$ = 3.5 cm

Area of cross section of pipe = πr²

= $\dfrac{22}{7} \times (3.5)^2$

= 22 × 0.5 × 3.5

= 38.5 cm².

Volume of water discharged per second = 38.5 cm² × 4 m s^-1

= 38.5 × 400 cm s^-1

Let the pipe fill the tank in n seconds.

∴ n × Volume of water discharged per second = Volume of tank

$\Rightarrow n \times 38.5 × 400 = 25.872 × (100)^3 \\[1em] \Rightarrow n = \dfrac{25.872 \times 100^3}{38.5 \times 400} \\[1em] \Rightarrow n = \dfrac{25872000}{15400} \\[1em] \Rightarrow n = 1680 \text{ seconds} \\[1em] \Rightarrow n = 1680 \times \dfrac{1}{60} \text{ minutes} \\[1em] \Rightarrow n = 28\text{ minutes}.$

Hence, it takes 28 minutes to fill the tank.

Water flows, at 9 km per hour, through a cylindrical pipe of cross-sectional area 25 cm². If this water is collected into a rectangular cistern of dimensions 7.5 m by 5 m by 4 m; calculate the rise in level in the cistern in 1 hour 15 minutes.

Answer

Given,

Rate of water flow = 9 km/hr

= 9 × 10⁵ cm/hr. [As, 1 km = 10⁵ cm.]

1 hour 15 minutes = $1 + \dfrac{15}{60} = 1 + \dfrac{1}{4} = \dfrac{5}{4}$ hours.

Volume of water flowing in $\dfrac{5}{4}$ hours = Area of cross-section of pipe × Rate of water flow × $\dfrac{5}{4}$ hours

= 25 cm² × 9 km/hr × $\dfrac{5}{4}$ hr

= 25 cm² × (9 × 10⁵) cm/hr × $\dfrac{5}{4}$ hr

= $\dfrac{1125 \times 10^5}{4}$ cm³.

Let increase in level of water be h cm.

Given,

Length of rectangular cistern = 7.5 m = 750 cm

Breadth of rectangular cistern = 5 m = 500 cm

Volume of water increase in cistern = 750 × 500 × h cm³

We know that,

Volume of water increase in cistern = Volume of water flowing in $\dfrac{5}{4}$ hours

$\Rightarrow 750 × 500 × h = \dfrac{1125 \times 10^5}{4} \\[1em] \Rightarrow h = \dfrac{1125 \times 10^5}{4 \times 750 \times 500} \\[1em] \Rightarrow h = 0.00075 \times 10^5 \\[1em] \Rightarrow h = 75 \text{ cm}.$

Hence, there is an increase of 75 cm in level of water in cistern.

The given figure shows the cross-section of a cone, a cylinder and a hemisphere all with the same diameter 10 cm, and the other dimensions are as shown.

Calculate :

(i) the total surface area,

(ii) the total volume of the solid and

(iii) the density of the material if its total weight is 1.7 kg.

Answer

(i) Given,

Diameter of cone, cylinder and hemisphere = 10 cm

Radius of cone, cylinder and hemisphere (r) = $\dfrac{10}{2}$ = 5 cm.

From figure,

Height of cone (h) = 12 cm

By formula,

⇒ l² = r² + h²

⇒ l² = 5² + 12²

⇒ l² = 25 + 144

⇒ l² = 169

⇒ l = $\sqrt{169}$ = 13 cm.

Total surface area = Surface area of cone + Surface area of cylinder + Surface area of hemisphere

= πrl + 2πrh + 2πr²

= πr(l + 2h + 2r)

= $\dfrac{22}{7} \times 5 \times (13 + 2 \times 12 + 2 \times 5)$

= $\dfrac{110}{7} \times (13 + 24 + 10)$

= $\dfrac{110 \times 47}{7}$

= 738.57 cm².

Hence, surface area of figure = 738.57 cm².

(ii) From figure,

Volume of figure = Volume of cone + Volume of cylinder + Volume of hemisphere

$= \dfrac{1}{3}πr^2h + πr^2h + \dfrac{2}{3}πr^3 \\[1em] = πr^2\Big(\dfrac{1}{3}h + h + \dfrac{2}{3}r\Big) \\[1em] = \dfrac{22}{7} \times 5^2 \times \Big(\dfrac{1}{3} \times 12 + 12 + \dfrac{2}{3} \times 5\Big) \\[1em] = \dfrac{550}{7} \times \Big(4 + 12 + \dfrac{10}{3}\Big) \\[1em] = \dfrac{550}{7} \times \dfrac{58}{3} \\[1em] = \dfrac{31900}{21} \\[1em] = 1519.05 \text{ cm}^3.$

Hence, volume of figure = 1519.05 cm³.

(iii) By formula,

Density = $\dfrac{\text{Mass}}{\text{Volume}}$

Given,

Mass = 1.7 kg = 1700 gm

Substituting values we get,

Density = $\dfrac{1700}{1519.05}$ = 1.12 g/cm³.

Hence, density of material = 1.12 g/cm³.

A solid, consisting of a right circular cone standing on a hemisphere, is placed upright in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of cone is 4 cm. Give your answer to the nearest cubic centimeter.

Answer

Given,

Height of cylinder (H) = 6 cm

Radius of cylinder (R) = 3 cm

Height of cone (h) = 4 cm

Radius of cone = Radius of hemisphere = r = 2 cm

Volume of water left in cylinder = Volume of cylinder - Volume of cone - Volume of hemisphere

$= πR^2H - \dfrac{1}{3}πr^2h - \dfrac{2}{3}πr^3 \\[1em] = π\Big(R^2H - \dfrac{1}{3}r^2h - \dfrac{2}{3}r^3\Big) \\[1em] = \dfrac{22}{7} \times \Big(3^2 \times 6 - \dfrac{1}{3} \times 2^2 \times 4 - \dfrac{2}{3} \times 2^3\Big) \\[1em] = \dfrac{22}{7} \times \Big(54 - \dfrac{16}{3} - \dfrac{16}{3}\Big) \\[1em] = \dfrac{22}{7} \times \Big(54 - \dfrac{32}{3}\Big) \\[1em] = \dfrac{22}{7} \times \Big(\dfrac{162 - 32}{3}\Big) \\[1em] = \dfrac{22}{7} \times \dfrac{130}{3} \\[1em] = \dfrac{2860}{21} \\[1em] = 136 \text{ cm}^3.$

Hence, volume of water left in cylinder = 136 cm³.

A metal container in the form of a cylinder is surmounted by a hemisphere of the same radius. The internal height of the cylinder is 7 m and the internal radius is 3.5 m. Calculate :

(i) the total area of the internal surface, excluding the base;

(ii) the internal volume of the container in m³.

Answer

Given,

Radius of cylindrical portion = Radius of hemispherical portion = r = 3.5 m.

Height of cylinder (h) = 7 m.

(i) Area of internal surface = Surface area of cylinder + Surface area of hemisphere

= 2πrh + 2πr²

= 2πr(h + r)

= $2 \times \dfrac{22}{7} \times 3.5 \times (7 + 3.5)$

= 2 × 22 × 0.5 × 10.5

= 231 m².

Hence, the total area of the internal surface = 231 m².

(ii) Internal volume of container = Volume of hemisphere + Volume of cylinder

$= \dfrac{2}{3}πr^3 + πr^2h \\[1em] = πr^2\Big(\dfrac{2r}{3} + h\Big) \\[1em] = \dfrac{22}{7} \times (3.5)^2 \times \Big(\dfrac{2 \times 3.5}{3} + 7\Big) \\[1em] = 22 \times 0.5 \times 3.5 \times \Big(\dfrac{7}{3} + 7\Big) \\[1em] = 38.5 \times \dfrac{28}{3} \\[1em] = 359.33 \text{ m}^3.$

Hence, volume of container = 359.33 m³.

An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above ground is 85 m and height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for fold and for stitching. Give your answer to the nearest m².

Answer

Given,

Diameter of base = 168 m

From figure,

Diameter of cylindrical base = Diameter of cone

Radius of cylindrical base = Radius of conical base = r = $\dfrac{168}{2}$ = 84 m.

Height of cylindrical part (H) = 50 m

Height of conical part (h) = 35 m

By formula,

⇒ l² = r² + h²

⇒ l² = 84² + 35²

⇒ l² = 7056 + 1225

⇒ l² = 8281

⇒ l = $\sqrt{8281}$ = 91 m.

Surface area of exhibition tent = Surface area of cylindrical part + Surface area of conical part

= 2πrH + πrl

$= 2 \times \dfrac{22}{7} \times 84 \times 50 + \dfrac{22}{7} \times 84 \times 91 \\[1em] = 26400 + 24024 \\[1em] = 50424 \text{ m}^2.$

Area required for folds and stitching = $\dfrac{20}{100} \times 50424$ = 10084.8 m².

Total area of canvas required = 50424 + 10084.8 = 60508.8 m² ≈ 60509 m² (to the nearest m²).

Hence, area of canvas required = 60509 m².

A test tube consists of a hemisphere and a cylinder of the same radius. The volume of the water required to fill the whole tube is $\dfrac{5159}{6}$ cm³, and $\dfrac{4235}{6}$ cm³ of water is required to fill the tube to a level which is 4 cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.

Answer

Given,

Let radius of hemisphere be r cm and height of cylindrical portion be h cm.

Total volume of test tube = Volume of hemisphere + Volume of cylinder

$\Rightarrow \dfrac{5159}{6} = \dfrac{2}{3}πr^3 + πr^2h \\[1em] \Rightarrow \dfrac{5159}{6} = πr^2\Big(\dfrac{2}{3}r + h\Big) \\[1em] \Rightarrow \dfrac{5159}{6} = πr^2\Big(\dfrac{2r + 3h}{3}\Big) \\[1em] \Rightarrow πr^2(2r + 3h) = \dfrac{5159}{6} \times 3 \\[1em] \Rightarrow πr^2(2r + 3h) = \dfrac{5159}{2} ..............(1)$

Given,

$\dfrac{4235}{6}$ cm³ of water is required to fill the tube to a level which is 4 cm below the top of the tube.

$\therefore \dfrac{4235}{6} = \text{Vol. of hemisphere + Vol. of cylinder[upto (h - 4) cm]} \\[1em] \Rightarrow \dfrac{4235}{6} = \dfrac{2}{3}πr^3 + πr^2(h - 4) \\[1em] \Rightarrow \dfrac{4235}{6} = πr^2\Big(\dfrac{2}{3}r + h - 4\Big) \\[1em] \Rightarrow \dfrac{4235}{6} = πr^2\Big(\dfrac{2r + 3h - 12}{3}\Big) \\[1em] \Rightarrow \dfrac{4235 \times 3}{6} = πr^2(2r + 3h) - 12πr^2 \\[1em] \Rightarrow \dfrac{4235}{2} = πr^2(2r + 3h) - 12πr^2$

Substituting value of πr²(2r + 3h) from equation 1 in above equation.

$\Rightarrow \dfrac{4235}{2} = \dfrac{5159}{2} - 12πr^2 \\[1em] \Rightarrow 12πr^2 = \dfrac{5159}{2} - \dfrac{4235}{2} \\[1em] \Rightarrow 12πr^2 = \dfrac{924}{2} \\[1em] \Rightarrow r^2 = \dfrac{924}{2 \times 12π} \\[1em] \Rightarrow r^2 = \dfrac{924}{24 \times \dfrac{22}{7}} \\[1em] \Rightarrow r^2 = \dfrac{924 \times 7}{24 \times 22} \\[1em] \Rightarrow r^2 = \dfrac{6468}{528} \\[1em] \Rightarrow r^2 = 12.25 \\[1em] \Rightarrow r = \sqrt{12.25} \\[1em] \Rightarrow r = 3.5 \text{ cm}.$

Substituting value of r in equation 1, we get :

$\Rightarrow πr^2(2r + 3h) = \dfrac{5159}{2} \\[1em] \Rightarrow \dfrac{22}{7} \times (3.5)^2 \times (2 \times 3.5 + 3h) = \dfrac{5159}{2} \\[1em] \Rightarrow 22 \times 0.5 \times 3.5 \times (7 + 3h) = \dfrac{5159}{2} \\[1em] \Rightarrow 38.5 \times (7 + 3h) = \dfrac{5159}{2} \\[1em] \Rightarrow 7 + 3h = \dfrac{5159}{2 \times 38.5} \\[1em] \Rightarrow 7 + 3h = \dfrac{5159}{77} \\[1em] \Rightarrow 7 + 3h = 67 \\[1em] \Rightarrow 3h = 67 - 7 \\[1em] \Rightarrow 3h = 60 \\[1em] \Rightarrow h = \dfrac{60}{3} \\[1em] \Rightarrow h = 20 \text{ cm}.$

Hence, radius of cylindrical part = 3.5 cm and height = 20 cm.

A solid is in the form of a right circular cone mounted on a hemisphere. The diameter of the base of the cone, which exactly coincides with hemisphere, is 7 cm and its height is 8 cm. The solid is placed in a cylindrical vessel of internal radius 7 cm and height 15 cm. How much water, in cm³, will be required to fill the vessel completely?

Answer

Given,

Diameter of cone = 7 cm

Radius of cone (r) = $\dfrac{7}{2}$ = 3.5 cm

Height of cone (h) = 8 cm

Radius of hemisphere = radius of cone = r = 3.5 cm

Radius of cylindrical vessel (R) = 7 cm

Height of cylindrical vessel (H) = 15 cm.

Volume of solid = Volume of cone + Volume of hemisphere

$= \dfrac{1}{3}πr^2h + \dfrac{2}{3}πr^3 = πr^2\Big(\dfrac{1}{3}h + \dfrac{2}{3}r) \\[1em] = \dfrac{22}{7} \times (3.5)^2 \times \Big(\dfrac{1}{3} \times 8 + \dfrac{2}{3} \times 3.5\Big) \\[1em] = 22 \times 0.5 \times 3.5 \times \Big(\dfrac{8}{3} + \dfrac{7}{3}\Big) \\[1em] = 38.5 \times \dfrac{15}{3} \\[1em] = 38.5 \times 5 \\[1em] = 192.5 \text{ cm}^3.$

By formula,

Volume of cylindrical vessel = πR²H

$= \dfrac{22}{7} \times 7^2 \times 15 \\[1em] = 22 \times 7 \times 15 \\[1em] = 2310 \text{ cm}^3.$

Volume of water required to fill the vessel completely = Volume of cylindrical vessel - Volume of solid

= 2310 - 192.5

= 2117.5 cm³.

Hence, volume of water required to fill the vessel completely = 2117.5 cm³.

A certain number of metallic cones, each of radius 2 cm and height 3 cm, are melted and recast into a solid sphere of radius 6 cm. Find the number of cones used.

Answer

Given,

Radius of cone (r) = 2 cm,

Height of cone (h) = 3 cm,

Radius of sphere (R) = 6 cm.

Let n cones be melted to recast into a solid sphere.

∴ n × Volume of each cone = Volume of sphere

$\Rightarrow n \times \dfrac{1}{3}πr^2h = \dfrac{4}{3}πR^3 \\[1em] \Rightarrow n = \dfrac{\dfrac{4}{3}πR^3}{\dfrac{1}{3}πr^2h} \\[1em] \Rightarrow n = \dfrac{4πR^3 \times 3}{πr^2h \times 3} \\[1em] \Rightarrow n = \dfrac{4R^3}{r^2h} \\[1em] \Rightarrow n = \dfrac{4 \times 6^3}{2^2 \times 3} \\[1em] \Rightarrow n = \dfrac{864}{12} \\[1em] \Rightarrow n = 72.$

Hence, 72 cones need to be melted to recast into a solid sphere.

A conical tent is to accommodate 77 persons. Each person must have 16 m³ of air to breathe. Given the radius of the tent as 7 m, find the height of the tent and also its curved surface area.

Answer

Given,

Each person must have 16 m³ of air to breathe.

∴ 77 persons need 77 × 16 m³ = 1232 m³.

Radius of tent (r) = 7 m.

Let height of conical tent be h meters.

Since, conical tent needs to accommodate 77 persons, so its volume will be equal to volume of air required for 77 persons.

$\Rightarrow \dfrac{1}{3}πr^2h = 1232 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times h = 1232 \\[1em] \Rightarrow \dfrac{1}{3} \times 22 \times 7 \times h = 1232 \\[1em] \Rightarrow h = \dfrac{1232 \times 3}{22 \times 7} \\[1em] \Rightarrow h = \dfrac{3696}{154} \\[1em] \Rightarrow h = 24 \text{ m}.$

By formula,

⇒ l² = r² + h²

⇒ l² = 7² + 24²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$ = 25 m.

Curved surface area of tent = πrl

= $\dfrac{22}{7} \times 7 \times 25$

= 22 × 25

= 550 m².

Hence, height of tent = 24 m and curved surface area of tent = 550 m².

A metallic solid cylinder has one end surmounted by a cone of the same radius and a hemisphere, which is also of same radius as that of the cone. The conical portion, the cylindrical portion and the hemispherical portion are separated as shown below. If the radius of the given solid is r cm and the height of the cone = the height of the cylinder = h cm, find expression for the total surface area of the three parts obtained.

Answer

When the three portions are separated, each becomes a complete solid. So, we add the entire surface area of the cone, the cylinder and the hemisphere.

For the cone :

Radius = r cm and height = h cm.

Slant height (l) = $\sqrt{h^2 + r^2}$ cm.

Total surface area of cone = πrl + πr² = $πr\sqrt{h^2 + r^2}$ + πr².

For the cylinder :

Radius = r cm and height = h cm.

Total surface area of cylinder = 2πrh + 2πr².

For the hemisphere :

Radius = r cm.

Total surface area of hemisphere = 2πr² + πr² = 3πr².

∴ Total surface area of the three parts

$= \left(πr\sqrt{h^2 + r^2} + πr^2\right) + \left(2πrh + 2πr^2\right) + 3πr^2 \\[1em] = πr\sqrt{h^2 + r^2} + 2πrh + 6πr^2 \\[1em] = πr\left(\sqrt{h^2 + r^2} + 2h + 6r\right).$

Hence, total surface area of the three parts = πr$\left(\sqrt{h^2 + r^2} + 2h + 6r\right)$ cm².

A solid cone of radius 20 cm is cut from its middle, parallel to its base, into two parts as shown. Find :

The ratio between the volumes of the two parts obtained.

Answer

Let the radius of the base of the cone be R = 20 cm and its height be H cm.

The cone is cut at the middle of its height by a plane parallel to the base. So the cut is at a height $\dfrac{H}{2}$ from the apex.

The upper part is a small cone similar to the whole cone. Since the height is halved, by similar triangles its radius is also halved.

∴ Radius of small cone (r) = $\dfrac{R}{2}$ = 10 cm and height = $\dfrac{H}{2}$.

Volume of small cone (upper part) :

$V_1 = \dfrac{1}{3}πr^2\left(\dfrac{H}{2}\right) = \dfrac{1}{3}π(10)^2 \times \dfrac{H}{2} = \dfrac{1}{3}π \times 50H.$

Volume of the whole cone :

$V = \dfrac{1}{3}πR^2H = \dfrac{1}{3}π(20)^2H = \dfrac{1}{3}π \times 400H.$

Volume of lower part (frustum) = Volume of whole conce - Volume of small cone

$V_2 = V - V_1 = \dfrac{1}{3}π \times 400H - \dfrac{1}{3}π \times 50H = \dfrac{1}{3}π \times 350H.$

∴ Ratio of the two parts :

$V_1 : V_2 = \dfrac{1}{3}π \times 50H : \dfrac{1}{3}π \times 350H = 50 : 350 = 1 : 7.$

Hence, the ratio between the volumes of the two parts (smaller cone : frustum) is 1 : 7.