WBBSE Solutions For Class 10 Maths Mensuration Chapter 3 Sphere

Mensuration Chapter 3 Sphere

Sphere 

If a semi-circle is rotated through an angle of 360° with its diameter as the axis, then the solid object thus produced is called a sphere.

 

For example, a football is a sphere. We see various spherical objects in our daily life, such as marbles ball, bullets, etc.

Surface of a sphere:

It is obvious that the surface of a sphere is a curved surface and the number of the curved surfaces is one.

So, we can say that the solid object surrounded by only one curved surface is known as a sphere. There is no plane surface in a sphere.

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Centre of a sphere:

The centre of the semi-circle, the diameter of which was taken as the axis to construct the sphere is known as the centre of the sphere.

 

 

O is the centre of the sphere.

The radius of a sphere:

The radius of the semi-circle, by the rotation of which the sphere is produced, is known as the radius of the sphere, i.e., the distance of any point on the circumference of the sphere from its centre is known as its radius.

OC is one of the radii of the sphere. Obviously, there can be infinite number of radii of a sphere, though the lengths of all of them are always equal.

The radius of the semi-circle and the radius of the sphere are the same.

AB is the diameter of the sphere.

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Height of a sphere:

The length of any diameter of the sphere is called the height of the sphere.

AB is a diameter of the semi-circle and so it is also the height of the sphere.

Wbbse Class 10 Maths Solutions

Formulae related to sphere

Let the radius of a sphere be r units, then

1. The curved surface or the total surface area of the sphere = 4nr² sq-units.
2. Height of the sphere = 2r units.
3. Volume of the sphere = \(\frac{4}{3}\) πr³ cubic-units.

Formulae related to hollow spheres:

Let the external radius of a hollow sphere be R units and its internal radius be r units. Then

1. the volume of the hollow sphere = \(\frac{4}{3}\)π(R³ -r³) cubic units.
2. the area of outer curved surface = 4πR² sq-units.
3. the area of inner curved surface = 4πr²sq-units.

Formulae related to solid hemispheres:

Let the radius of the solid hemisphere be r units. Then

1. the area of curved surface of the solid hemisphere = 2πr² sq-units.
2. the area of plane surface of the solid hemisphere = πr² sq-units.
3. the total area of the surfaces of the solid hemisphere = (2πr² + πr²) sq-units = 3πr² sq-units
4. the volume of the solid hemisphere = \(\frac{2}{3}\)πr³ cubic-units

Formulae related to a hollow hemisphere:

Let the radius of the hollow hemisphere be r units, and then

1. Whole surface area = curved surface area = 2πr² sq-units
2. Volume of the hollow hemisphere = \(\frac{2}{3}\) πr³ cubic-units.
3. Area of the whole surface of a hollow hemisphere = {2πR² + 27πr² + (πR² – πr²)} sq-units, where R and r are the radii of the outer and inner radius of the hollow hemisphere.

 

Mensuration Chapter 3 Sphere Multiple Choice Questions

Example 1. The volume of a solid sphere of radius 2r units is

1. \(\frac{32 \pi r^3}{3}\) cubic-units
2. \(\frac{16 \pi r^3}{3}\) cubic-units
3. \(\frac{8 \pi r^3}{3}\) cubic-units
4. \(\frac{64 \pi r^3}{3}\) cubic-units

Solution: Radius of the solid sphere = 2r units

∴ Volume of the solid sphere = \(\frac{4}{3}\)π x (2r)³ cubic-units

= \(\frac{2}{3}\) π x 8r³ cubic – units = \(\frac{32 \pi r^3}{3}\) cubic-units

∴ 1. \(\frac{32 \pi r^3}{3}\) cubic-units is correct

The volume of a solid sphere of radius 2r units is  \(\frac{32 \pi r^3}{3}\) cubic-units

Wbbse Class 10 Maths Solutions

Example 2. If the ratio of volumes of two solid spheres is 1: 8, the ratio of their curved surface areas is

1. 1: 2
2. 1: 4
3. 1: 8
4. 1: 16

Solution: Let the radii of the two spheres be r₁ and r₂ units.

So the volumes of two spheres are \(\frac{4}{3} \pi r_1^3\) cubic-units and \(\frac{4}{3} \pi r_2^3\) cubic-units.

As per question, \(\frac{4}{3} \pi r_1^3\) : \(\frac{4}{3} \pi r_2^3\) = 1:8

\(\frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3}=\frac{1}{8}\)

 

or, \(\frac{r_1^3}{r_2^3}=\frac{1}{8}\)

or, \(\left(\frac{r_1}{r_2}\right)^3=\left(\frac{1}{2}\right)^3\)

or, \(\frac{r_1}{r_2}=\frac{1}{2}\)

= \(\frac{4}{3} \pi r_1^3\) : \(\frac{4}{3} \pi r_2^3\)

= \(\frac{4 \pi r_1^2}{4 \pi r_2^2}=\left(\frac{r_1}{r_2}\right)^2=\left(\frac{1}{2}\right)^2\) = 1:4

Hence 2. 1: 4 is correct

The ratio of their curved surface areas is 1: 4

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Example 3. The whole surface area of a solid hemisphere with length of 7 cm radius is

1. 588π sq-cm
2. 392π sq-cm
3. 147π sq-cm
4. 98πsq-cm

Solution: The radius of the solid hemisphere = 7 cm.

∴ the whole surface area of it = 3π x (7)² sq-cm = 3π x 49 sq-cm = 147π sq-cm.

∴ 3. 147π sq-cm is correct.

The whole surface area of it 147π sq-cm

Example 4. If the ratio of curved surface areas of two solid spheres is 16 : 9, the ratio of their volumes is

1. 64: 2 7
2. 4: 3
3. 27: 64
4. 3: 4

Solution: Let the radii of two solid spheres be units r₁ and r₂ units respectively.

∴ the curved surface areas of the spheres are \(4 \pi r_1^2\) sq-units and \(4 \pi r_2^2\) sq-units respectively.

As per question, \(4 \pi r_1^2\) : \(4 \pi r_2^2 s q\)

⇒ \(\frac{4 \pi r_1^2}{4 \pi r_2^2}=\frac{16}{9}\)

⇒ \(\left(\frac{r_1}{r_2}\right)^2=\left(\frac{4}{3}\right)^2\)

⇒ \(\frac{r_1}{r_2}=\frac{4}{3}\)

∴ The ratio of their volumes

= \(\frac{4}{3} \pi r_1^3:=\frac{4}{3} \pi r_2^3\)

= \(\frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3}\)

= \(\left(\frac{r_1}{r_2}\right)^3=\left(\frac{4}{3}\right)^3=\frac{64}{27}\)

= 64: 27

∴ 1. 64: 27 is correct

The ratio of their volumes is 64: 27

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Example 5. If the numerical value of curved surface area of a solid sphere is three times of its volume, the length of its radius is

1. 1 unit
2. 2 units
3. 3 units
4. 4 units

Solution: Let the radius of the sphere = r units

∴ curved surface area of the sphere = 4 πr² sq-units

∴ the volume of the sphere = \(\frac{4}{3}\)πr³ cubic-units

As per questions, 4πr² = 3x \(\frac{4}{3}\) πr³

or, 4πr² = 4πr³

⇒ 1 = r ⇒ r = 1

∴ radius = 1 unit

∴ 1. 1 unit is correct.

The length of its radius is 1 unit

Example 6. The radius of a copper sphere is twice the radius of an iron sphere. The numerical value of the whole surface area of the copper sphere is equal to the numerical value of the volume of the iron sphere. Then the radius of the copper sphere will be

1. 6 units
2. 12 units
3. 18 units
4. 24 units

Solution: Let the radius of the iron sphere be r units.

∴ the radius of the copper-sphere is 2r units.

∴ the whole surface area of the copper-sphere = 4π x (2r)² sq-units = 16πr² sq-units

Also, the volume of the iron sphere = \(\frac{4}{3}\)πr³ cubic-units

As per question, \(\frac{4}{3}\)πr³ = 16πr² ⇒ r = 12

∴ the radius of the copper-sphere = 2 x 12 units = 24 units.

∴ 4. 24 units is correct.

The radius of the copper sphere will be 24 units

Example 7. If the volume of a sphere be V cubic units, then the radius of the sphere will be

1. \(\left(\frac{3 V}{4 \pi}\right)^3\) units
2. \(\frac{3 \mathrm{~V}}{4 \pi}\) units
3. \(\sqrt{\frac{3 V}{4 \pi}}\) units
4. \(\sqrt[3]{\frac{3 V}{4 \pi}}\) units

Solution: Let the radius of the sphere be r units

∴ the volume of the sphere = \(\frac{4}{3}\)πr³cubic-units

As per question, \(\frac{4}{3}\)πr³ = V

or, r³ = \(\frac{3 \mathrm{~V}}{4 \pi}\) or r = \(\sqrt[3]{\frac{3 V}{4 \pi}}\)

∴ 4. \(\sqrt[3]{\frac{3 V}{4 \pi}}\) is correct.

The radius of the sphere will be \(\sqrt[3]{\frac{3 V}{4 \pi}}\)

Example 8. If the radius of a sphere is decreased by \(\frac{1}{2}\) times times, then the curved surface area of the sphere will be changed by

1. \(\frac{1}{2}\) times
2. \(\frac{1}{4}\) times
3. \(\frac{1}{8}\) times
4. \(\frac{1}{16}\) times

Solution: If the radius of a sphere be r units, then the curved surface area of it will be = S = 4πr² sq-units.

If the radius is decreased by \(\frac{1}{2}\) times times, then the curved surface area

= 4 x \(\left(\frac{r}{2}\right)^2\) sq – units = πr² sq-units = \(\frac{1}{4}\) x 4πr² sq-units

= \(\frac{1}{4}\) x S sq – units = \(\frac{S}{4}\) sq – units

∴ the new curved surface area of the sphere will be \(\frac{1}{4}\) times of the previous curved surface area.

∴ 2. \(\frac{1}{4}\) times is correct.

The curved surface area of the sphere will be changed by \(\frac{1}{4}\) times

Wbbse Class 10 Maths Solutions

Example 9. If the volume of a sphere is increased by 8 times, then its radius will be increased by

1. 2 times
2. 4 times
3. 6 times
4. 8 times

Solution: Let the radius of the sphere be r units and its volume be V cubic units.

∴ V = \(\frac{4}{3}\)πr³ (by formula)

Now, if the volume of the sphere be increased by 8 times, i.e., its volume be 8V cubic-units, then let its radius be r₁, units.

∴ 8V = \(\frac{4}{3} \pi r_1^3\)

⇒ 8 x \(\frac{4}{3}\)πr³ = \(\frac{4}{3} \pi r_1^3\)

⇒ \(r_1^3=8 r^3\)

⇒ \(r_1^3=(2 r)^3\)

⇒ r₁= 2r, i.e., the radius will be increased by 2 times.

∴ Hence 1. 2 times is correct.

Radius will be increased by 1. 2 times

Example 10. If the external and internal radii of a hollow sphere be 6 cm and 3 cm respectively, then the volume (i.e.. the material) of the sphere will be

1. 629 cu-units
2. 792 cu-units
3. 829 cu-units
4. 929 cu-units

Solution: The required volume of the sphere = \(\left(\frac{4}{3} \pi \times 6^3-\frac{4}{3} \pi \times 3^3\right)\)

= \(\frac{4}{3} \times \frac{22}{7}\)(216-27) cu.units

= \(\frac{4}{3} \times \frac{22}{7}\) x 189 cubic. units

= 792 cubic.units.

∴ 2. 792 cu-units is correct.

The volume (i.e.. the material) of the sphere will be 792 cu-units

 

Mensuration Chapter 3 Sphere True Or False

Example 1. If the ratio of curved surface areas of two hemispheres is 4: 9, then the ratio of their lengths of radii is 2 : 3

Solution: Let the radii of the two spheres be r₁ and r₂ units.

As per question, \(2 \pi r_1^2: 2 \pi r_2^2=4: 9\)

⇒  \(\frac{2 \pi r_1^2}{2 \pi r_2^2}=\frac{4}{9}\)

⇒ \(\left(\frac{r_1}{r_2}\right)^2=\left(\frac{2}{3}\right)^2\)

⇒ \(\frac{r_1}{r_2}=\frac{2}{3}\)

⇒ r₁: r₂ = 2: 3

Hence the given statement is true.

Wbbse Class 10 Maths Solutions

Example 2. If the length of radius of a solid sphere be doubled, the volume of sphere will be doubled.

Solution: Let the radius of the sphere be r units, then the volume of the sphere will be \(\frac{4}{3}\)πr³ cu-units.

Now, if the radius be doubled, the volume of the sphere will be \(\frac{4}{3}\) π x(2r)³ cu-units= 8 x \(\frac{4}{3}\)πr³ cubic- units.

i,e, the volume of the sphere will be 8 times of its previous volume.

Hence th given statement is false.

Mensuration Chapter 3 Sphere Fill In The Blanks

Example 1. The name of solid which is composed of only one surface is ______

Solution: sphere

Example 2. The number of surfaces of a solid hemisphere is ______

Solution: 1

Example 3. If the length of radius of a solid hemisphere is 2r units, its whole surface area is _______ πr² sq-units.

Solution: 12

Mensuration Chapter 3 Sphere Short Answer Type Questions

Example 1. The numerical values of volume and whole surface area of a solid hemisphere are equal. Find the length of radius of the hemisphere.

Solution:

Given:

The numerical values of volume and whole surface area of a solid hemisphere are equal.

Let the radius of the solid hemisphere = r units.

∴ the volume of the solid hemisphere = \(\frac{2}{3}\)πr³ cu.units, and its whole surface area = 3πr² sq.units.

As per question, \(\frac{2}{3}\)πr³ = 3πr²

⇒ r = \(\frac{9}{2}\) = 4.5

Hence the radius of the solid hemisphere = 4.5 units.

Example 2. The curved surface area of a solid sphere is equal to the surface area of a solid right circular cylinder. The lengths of both height and diameter of cylinder are 12 cm. Find the length of radius of the sphere.

Solution:

Given:

The curved surface area of a solid sphere is equal to the surface area of a solid right circular cylinder. The lengths of both height and diameter of cylinder are 12 cm.

Let the radius of the sphere be r cm.

∴ curved surface area of the solid sphere = 4πr² sq-cm

The radius of the right circular cylinder = \(\frac{12}{2}\) cm= 6cm. and height of it = 12 cm.

∴ surface area of the right circular cylinder = 2π x 6 x 12 sq-cm.

As per question, 4πr² = 2π x 6 x 12 or, r² = 36 or, r = 6.

Hence the required radius of the sphere = 6cm.

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Example 3. Whole surface area of a solid hemisphere is equal to the curved surface area of a solid sphere. Find the ratio of lengths of radius of hemisphere and sphere.

Solution:

Given:

Whole surface area of a solid hemisphere is equal to the curved surface area of a solid sphere.

Let the radius of the solid hemisphere be r₁ unit and the radius of the solid sphere be r₂ unit.

∴ the whole surface area of the solid hemisphere = \(3 \pi r_1^2\) sq-unit and the curved surface area of the solid sphere = \(4 \pi r_2^2\) sq-unit

As per question, \(3 \pi r_1^2\) = \(4 \pi r_2^2\)

⇒ \(\frac{r_1^2}{r_2^2}=\frac{4}{3}\)

⇒ \(\left(\frac{r}{r_2}\right)^2=\left(\frac{2}{\sqrt{3}}\right)^2\)

⇒ \(\frac{r_1}{r_2}=\frac{2}{\sqrt{3}}\) = 2: 3

Hence the required ratio = 2 : √3.

Example 4. If curved surface area of a solid sphere is S and volume is V, then find the value of \(\frac{S^3}{V^2}\) [putting the value of π].

Solution:

Given:

If curved surface area of a solid sphere is S and volume is V

Let the radius of the solid sphere be r unit.

Then the curved surface area of the sphere = 4πr²sq.units.

As per question, S = 4πr² ……..(1)

Also, the volume of the sphere = \(\frac{4}{3}\)πr³cu.units.

As per question, V = \(\frac{4}{3}\)πr³ ………..(2)

∴ \(\frac{\mathrm{S}^3}{\mathrm{~V}^2}=\frac{\left(4 \pi r^2\right)^3}{\left(\frac{4}{3} \pi r^3\right)^2}\)

[Divding cube of (1) by square of (2)]

= \(\frac{64 \pi^3 r^6}{\frac{16}{9} \pi^2 r^6}=\frac{64 \times 9}{16} \pi=36 \pi\)

Hence the required value of \(\frac{\mathrm{S}^3}{\mathrm{~V}^2}\) = 36π.

Wbbse Class 10 Maths Solutions

Example 5. If the length of radius of a sphere is increased by 50%, how much percent will be increased of its curved surface area?

Solution:

Given:

If the length of radius of a sphere is increased by 50%,

Let the radius of the sphere be r unit.

∴ the area of the curved surface of the sphere = 4πr²sq-unit.

If the radius of the sphere be increased by 50% then the new radius will be \(\left(r+r \times \frac{50}{100}\right)\) unit = \(\frac{3r}{2}\) unit.

Then the curved surface area of the sphere will be \(4 \pi\left(\frac{3 r}{2}\right)^2\) sq.units = \(4 \pi \cdot \frac{9 r^2}{4}\) sq.units = 9nr1 sq.units.

∴ increase of curved surface area = (9πr² – 4πr²) sq.unit = 5πr²sq.unit.

∴ the percent of increment of the curved surface area of the sphere = \(\frac{5 \pi r^2}{4 \pi r^2}\) x 100% = 125%

Hence the required percent = 125%

Example 6. The diameter of a sphere is double of the diameter of another sphere. Then how much times will be the volume of smaller sphere than the volume of the greater sphere?

Solution:

Given:

The diameter of a sphere is double of the diameter of another sphere.

Let the radius of the smaller sphere be r unit.

Then the radius of the greater sphere will be 2r unit, since the diameter of the greater sphere is double of the diameter of the smaller one.

Now, volume of the smaller sphere, V = \(\frac{4}{3}\)πr³cu.unit

and the volume of the greater sphere= \(\frac{4}{3}\)π x (2r)³cu. unit = \(\frac{4}{3}\)π x 8r³ cu.unit = 8 x \(\frac{4}{3}\) πr³cu.unit.

i.e., 8 times of the smaller sphere.

Hence the volume of the greater sphere is 8 times of the volume of the smaller sphere.

Example 7. What will be the ratio of diameter of a hemisphere and the length of its circumference?

Solution:

Let the radius of the hemisphere be r unit.

∴ its diameter = 2r unit

Also, the circumference of the hemisphere = (πr + 2r) unit.

As per question, 2r : (πr + 2r) = 2r : r (π + 2) = 2 : (π + 2)

Hence the ratio = 2 : (π + 2).

Example 8. If by melting a solid hemisphere be made into a sphere, then what will be the ratio of their radii?

Solution:

Let the radii of the hemisphere and the sphere be r₁ unit and r₂ unit respectively.

So, the volume of the hemisphere = \(\frac{2}{3} \pi r_1^3\) cubic.unit, and the volume of the original sphere = \(\frac{4}{3} \pi r_2^3\) cubic.unit

As per question, \(\frac{2}{3} \pi r_1^3=\frac{4}{3} \pi r_2^3\)

or, \(\frac{r_1^3}{r_2{ }^3}=\frac{4}{2}=2\)

or, \(\left(\frac{r_1}{r_2}\right)^3=2\)

or, \(\frac{r_1}{r_2}=\sqrt[3]{2}\)

∴ r₁: r₂ = \(\sqrt[3]{2}\) :1

Hence the required ratio = \(\sqrt[3]{2}\) : 1.

Example 9. A circular iron plate of thickness 2/3 cm is made by beating an iron sphere of diameter 4 cm. What will be the radius of the iron plate?

Solution:

Given:

A circular iron plate of thickness 2/3 cm is made by beating an iron sphere of diameter 4 cm.

The radius of the iron-sphere = \(\frac{4}{3}\) cm = 2cm.

∴ the volume of the iron-sphere = \(\frac{4}{3}\)π x 2³ cc. = \(\frac{32 \pi}{3}\) cc.

Let the radius of the iron plate be r cm.

∴ the area of the iron-plate = πr² sq-cm.

Since the iron-plate is of thickness \(\frac{2}{3}\) cm,its volume = \(\frac{2}{3}\)πr² cu.cm.

As per question, \(\frac{2}{3}\)πr² = \(\frac{32 \pi}{3}\)

⇒ r² = 16 ⇒ r = √16 = 4

Hence the radius of the circular iron plate = 4 cm.

Example 10. The external radius of a hollow hemisphere is 6 cm and it is of thickness 2 cm. What will be the whole surface area of the hemisphere?

Solution:

Given:

The external radius of the hemisphere is 6 cm. and the thickness of it is 2cm.

∴ the internal radius of the hemisphere = (6 – 2) cm = 4 cm.

∴ the whole surface area of the hemisphere

= (2π x 6² + 2π x 4² + π x 6² – π x 4²) sq.cm.

= π(2 x 36 + 2 x 16 + 36 – 16) sq.cm

= \(\frac{22}{7}\) x(72 + 32 + 20) sq.cm = \(\frac{22}{7}\) x 124 sq.cm = \(\frac{1728}{7}\) sq.cm = 389 \(\frac{5}{7}\) sq.cm.

Hence the whole surface area of the hemisphere =389 \(\frac{5}{7}\)sq.cm.

Class 10 Maths Solutions Wbbse

Example 11. A large new sphere is made by melting two metal spheres of radii rj unit and r2 unit respectively. Find the radius of the large sphere.

Solution:

Given:

A large new sphere is made by melting two metal spheres of radii rj unit and r2 unit respectively.

The total volume of the Mo spheres = \(\frac{4}{3}\)π \(\left(r_1^3+r_2^3\right)\) cubic. unit.

Now, let the radius of the large sphere be r unit.

∴ the volume of the large sphere = \(\frac{4}{3}\)πr³cu.umt.

As per condition,\(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)πr³ \(\left(r_1^3+r_2^3\right)\)

⇒ r = \(\sqrt[3]{r_1^3+r_2^3}\)

Hence the required radius of the large sphere = \(\sqrt[3]{r_1^3+r_2^3}\) unit.

 

Mensuration Chapter 3 Sphere Long Answer Type Questions

Example 1. If the cost of making a leather ball is ₹431.20 at the rate of ₹17.50 per square cm. Calculate the length of diameter of the ball.

Solution:

Given:

If the cost of making a leather ball is ₹431.20 at the rate of ₹17.50 per square cm.

Let the radius of the ball = r cm.

∴ the whole surface area of the ball = 4πr² sq-cm.

₹17.50 is required to make a ball of leather of 1 sq.cm

∴ ₹431.20 is required to make a ball of leather of \(\frac{431 \cdot 20}{17 \cdot 50}\) sq.cm

∴ 4πr² = \(\frac{431 \cdot 20}{17 \cdot 50}\)

or, 4 x \(\frac{22}{7}\) x r² = \(\frac{4312}{175}\)

or, \(r^2=\frac{4312 \times 7}{175 \times 4 \times 22}=\frac{49}{25}\)

or, \(r=\sqrt{\frac{49}{25}}=\frac{7}{5}\) =1.4

∴ the diameter of the ball = 2 x 1.4 cm = 2.8 cm.

Hence the diameter of the leather ball = 2.8 cm.

Example 2. If the length of diameter of a solid sphere is 28 cm and it is completely immersed into the water then calculate the volume of water displaced by the sphere.

Solution:

Given:

If the length of diameter of a solid sphere is 28 cm and it is completely immersed into the water

The diameter of the sphere = 28 cm

∴ radius of the sphere = \(\frac{28}{2}\) cm = 14 cm

∴ the volume of the sphere = \(\frac{4}{3}\) x \(\frac{22}{7}\) x(14)³cc = 11498 \(\frac{2}{3}\) cc

Hence the required volume of water = 11498 \(\frac{2}{3}\) cc

Example 3. The length of radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it, then find the ratio of surface areas of the balloon in two cases.

Solution:

Given:

The length of radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it

The previous radius of the balloon = 7 cm.

∴ the previous surface area of the balloon = 4 x \(\frac{22}{7}\) x 7² sq.cm

∴ the ratio of the whole surface areas of the balloon in two cases

= 4 x \(\frac{22}{7}\) x 7²: 4 x \(\frac{22}{7}\) x 21² = \(\frac{7 \times 7}{21 \times 21}\) = \(\frac{1}{9}\)= l: 9.

Hence the required ratio = 1:9.

Example 4. 127 \(\frac{2}{7}\)sq.cm of sheet is required to make a hemispherical bowl. Calculate the length of diameter of the forepart of the bowl.

Solution:

Given:

127 \(\frac{2}{7}\)sq.cm of sheet is required to make a hemispherical bowl.

Let the radius of the forepart of the bowl be r cm.

∴ the quantity of sheet required to make the bowl = 2πr² sq.cm.

As per question, 2πr²=127 \(\frac{2}{7}\)

⇒ 2 x \(\frac{22}{7}\) x r² = \(\frac{891}{7}\)

⇒ r² = \(\frac{891 \times 7}{7 \times 2 \times 22}\)

⇒ r² = \(\frac{81}{4}\)

⇒ r = \(\frac{9}{2}\)

So, the diameter of the bowl = 2 x \(\frac{9}{2}\) cm = 9cm.

Hence the required diameter of the forepart of the bowl = 9 cm.

Class 10 Maths Solutions Wbbse

Example 5. The length of diameter of a solid sphere of lead is 14 cm. If the sphere is melted, then calculate how many spheres with length of 3.5 cm radius can be made?

Solution:

Given:

The length of diameter of a solid sphere of lead is 14 cm. If the sphere is melted

The radius of the large solid sphere = \(\frac{14}{2}\) cm= 7cm

∴ volume of it = \(\frac{4}{3}\) x 7³ cc.

The radius of each smaller sphere = 3.5 cm

∴ volume of each of it = \(\frac{4}{3}\)π x (3.5)³ cc

Let the number of smaller spheres which can be made is x.

∴ the volume of x smaller sphere = x x \(\frac{4}{3}\)π x (3.5)³ cc

As per condition, x x \(\frac{4}{3}\)π x (3.5)³ = \(\frac{4}{3}\)π x 7³

⇒ x = \(\frac{7 \times 7 \times 7}{3 \cdot 5 \times 3 \cdot 5 \times 3 \cdot 5}\)

∴ x = 8

Hence the required number of smaller spheres = 8.

Example 6. Three spheres made of copper having the lengths of 3 cm, 4 cm and 5 cm radii are melted and a large sphere is made. Calculate the length of radius of the large sphere.

Solution:

Given:

Three spheres made of copper having the lengths of 3 cm, 4 cm and 5 cm radii are melted and a large sphere is made.

The radii of three smaller spheres are 3 cm, 4 cm and 5 cm.

∴ the total volume of the three smaller spheres

= \(\frac{4}{3}\)π(33 + 43 +53)cc = \(\frac{4}{3}\)π(27 + 64 + 125)cc = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 216 cc.

Let the radius of the large sphere be R cm.

∴ volume ot the large sphere = \(\frac{4}{3}\) x \(\frac{22}{7}\) x R³ cc.

As per question, \(\frac{4}{3}\) x \(\frac{22}{7}\) x R³ = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 216

⇒ R³ = 6³

⇒  R = 6.

Hence the radius of the large sphere = 6 cm.

Example 7. The length of diameter of base of a hemispherical tomb is 42 dcm. Calculate the cost of colouring the upper surface of the tomb at the rate of 35 per square metre.

Class 10 Maths Solutions Wbbse

Solution:

Given:

The length of diameter of base of a hemispherical tomb is 42 dcm.

The diameter of base of the hemispherical tomb = 42 dcm

∴ its radius = \(\frac{42}{2}\) dcm = 21 dcm

∴ the surface area of the upper surface of the tomb

= 2 x \(\frac{22}{7}\) x 21² sq.dcm = 2772 sq.dcm = 27-72 sq. metres

So the cost of colouring the tomb = ₹27.72 x 35 = ₹970.2

Hence the required cost = ₹970.2.

Class 10 Maths Solutions Wbbse

Example 8. Two hollow spheres with the lengths of diameter 21 cm and 17.5 cm respectively are made from the sheets of the same metal, Calculate the volumes of sheets of metal required to make the two spheres.

Solution:

Given:

Two hollow spheres with the lengths of diameter 21 cm and 17.5 cm respectively are made from the sheets of the same metal

The lengths of diameters of the two hollow spheres are 21 cm and 17.5 cm.

∴ their radii are \(\frac{21}{2}\) cm and \(\frac{17.5}{2}\) cm

So, the ratio of the curved surface areas of the two hollow spheres

= 4π x \(\left(\frac{21}{2}\right)^2\) : \(4 \pi \times\left(\frac{17 \cdot 5}{2}\right)^2\)

= \(\frac{21^2}{4}: \frac{(17 \cdot 5)^2}{4}\)

= 21 x 21: 17.5 x 17.5 = \(\frac{21 \times 21 \times 100}{175 \times 175}=\frac{36}{25}\) = 36: 25

Hence the required ratio = 36: 25

Example 9. The curved surface of a solid metalic sphere is cut in such a way that the curved surface area of the new sphere is half of that previous one. Calculate the ratio of the volumes of the portion cut off and remaining portion of the sphere.

Solution:

Given:

The curved surface of a solid metalic sphere is cut in such a way that the curved surface area of the new sphere is half of that previous one.

Let the radius of the metalic sphere be R unit and that of the new sphere produced be r unit.

As per question, 4πr² = \(\frac{1}{2}\) x 4πR²

⇒ R² = 2r²

⇒ R = √2r.

∴ the volume of the new sphere = \(\frac{4}{3}\)πr³ cubic.units.

Also, the volume of the portion cut off = \(\left(\frac{4}{3} \pi R^3-\frac{4}{3} \pi r^2\right)\) cubic.units.

= \(\frac{4}{3}\)πr³(R³-r³) cubic.units.

= \(\frac{4}{3}\) x \(\frac{22}{7}\)\(\left\{(\sqrt{2} r)^3-r^3\right\}\) cubic.units

= \(\frac{4}{3}\) x \(\frac{22}{7}\)(2√2r³ -r³) cubic.units

= \(\frac{4}{3}\) x \(\frac{22}{7}\)(2√2-1)r³ cubic, units.

∴ ratio of the volumes of the cut off the large sphere and the volume of the remaining part.

= \(\frac{4}{3}\) x \(\frac{22}{7}\) x (2√2-1)r³ : \(\frac{4}{3}\)πr³ = (2√2-1):1

Hence the required ratio =(2√2-1) : 1.

Example 10. On the curved surface of the axis of a globe with the length of 14 cm radius, two circular holes are made each of which has the length of radius 0.7 cm. Calculate the area of metal sheet surrounding its curved surface.

Solution:

Given:

On the curved surface of the axis of a globe with the length of 14 cm radius, two circular holes are made each of which has the length of radius 0.7 cm.

The whole surface area of the globe = 4π x (14)² sq-cm.

The areas of the holes = 2π x (0.7)² sq-cm.

So the required surface area of the two circular holes

= {4π x (14)² – 2π x (0.7)²} sq-cm = 2π [2 x (14)² – (0.7)²] sq-cm

= 2 x \(\frac{22}{7}\) x (392 – 0.49) sq.cm = 2 x \(\frac{22}{7}\) x 391.51 sq.cm = 2460.92 sq.cm.

Hence the required surface area of the circular metalic sheet = 2460.92 sq.cm.

Example 11. Calculate how many marbles with lengths of 1 cm radius may be formed by melting a solid sphere of iron having 8 cm length of radius.

Solution:

The radius of solid iron-sphere = 8 cm

∴ the volume of the iron-sphere = \(\frac{4}{3}\) x π x 8³cc

The radius of each marble = 1 cm

∴ volume of each marble = \(\frac{4}{3}\)π x 1³cc.

Let x marbles can be made.

∴ as per condition, x x \(\frac{4}{3}\)π x 1³ = \(\frac{4}{3}\)π x 8³

⇒ x = \(\frac{8^3}{1^3}\) = 512

Hence 512 marbles can be made.

Example 12. The external and internal radii of a hollow copper-sphere are 20 cm and 16 cm respectively. By melting this sphere, how many solid bullets of diameter 40 cm each can be made?

Solution:

Given:

The external and internal radii of a hollow copper-sphere are 20 cm and 16 cm respectively.

The external radius of the hollow sphere = \(\frac{20}{2}\) cm = 10 cm.

and the internal radius of the hollow-sphere = \(\frac{16}{2}\) cm = 8 cm

∴ the volume of materials of the hollow sphere

= \(\left[\frac{4}{3} \pi \times 10^3-\frac{4}{3} \pi \times 8^3\right]\) cc. = \(\frac{4}{3}\)π(1000- 512)cc = \(\frac{4}{3}\)π x 488cc

The radius of solid bullets = \(\frac{4}{2}\) = 2cm

∴ volume of solid bullets = \(\frac{4}{3}\)π x 2³ cc.

Let the number of solid bullets that can be made be x.

∴ x x \(\frac{4}{3}\)³ x 2³ = \(\frac{4}{3}\)π x 488

⇒ x = \(\frac{488}{8}\) = 61.

Hence 61 solid bullets can be made.

Class 10 Maths Solutions Wbbse

Example 13. If the radius of a sphere be increased by 3 cm, the volume of the sphere is increased by 264 cc. Find the radius of the sphere.

Solution:

Given:

If the radius of a sphere be increased by 3 cm, the volume of the sphere is increased by 264 cc.

Let the radius of the sphere be r cm.

∴ the volume of the sphere = \(\frac{4}{3}\)πr³ cc.

If the radius of the sphere be increased by 3 cm, then its volume will be \(\frac{4}{3}\)π(r + 3)³ cc.

As per question, \(\frac{4}{3}\)π(r + 3)³ – \(\frac{4}{3}\)πr³ = 264

or, \(\frac{4}{3}\)π[(r +3)³-r³] = 264 or, \(\frac{4}{3}\) x \(\frac{22}{7}\)(r³ +9r² +27r + 27-r³) = 264

or, \(\frac{4}{3}\) x \(\frac{22}{7}\)(9r²+ 27r + 27) = 264

or, 9r² +27r + 27 = \(\frac{264 \times 3 \times 7}{4 \times 22}\)

or, 9(r² + 3r + 3) = 63 or, r² + 3r + 3 = 7 or, r² + 3r + 3 – 7 = 0 or, r² + 3r- 4 = 0

or, r² + 4r-r-4=0 or, r (r + 4) – 1 (r + 4) = 0 or, (r + 4)(r – 1) = 0

∴ either r + 4 = 0, or, r- 1 = 0 ⇒ r = – 4 or, r = 1

But the value of r cannot be negative, ∴ r = 1.

Hence the required radius of the sphere was 1 cm.

Example 14. If the radius of a solid sphere be decreased by 1 cm, then the curved surface area of it is decreased by 88 sq-cm. What was the radius of the square?

Solution:

Given:

If the radius of a solid sphere be decreased by 1 cm, then the curved surface area of it is decreased by 88 sq-cm

Let the radius of the solid sphere be r cm.

∴ the curved surface area of the sphere = 4πr² sq-cm.

Now, if the radius of the sphere is decreased by 1 cm, then the curved surface of it will be 4 (r-1)² sq-cm

As per question, 4πr² – 4π(r- 1)² = 88

or, 4π[r² – (r – 1)²] = 88

or, 4 x \(\frac{22}{7}\)[(r+r-1)(r-r + 1)] = 88 or, 4 x \(\frac{22}{7}\)[(2r- 1)] = 88

or, 2r-1 = \(\frac{88 \times 7}{4 \times 22}\) or, 2r- 1 = 7 or, 2r = 8 or, r = 4.

Hence the radius of the solid sphere was 4 cm.

Example 15. The external diameter of a hollow gold-sphere is 12 cm and it is made of a plate of thickness 1 cm. If the mass of 1 cc gold be 19.5 gm and the price of 1 gm gold be X 1280, then what is the price of the whole solid sphere?

Solution:

Given:

The external diameter of a hollow gold-sphere is 12 cm and it is made of a plate of thickness 1 cm. If the mass of 1 cc gold be 19.5 gm and the price of 1 gm gold be X 1280

The external diameter of the gold-sphere is 12 cm and the thickness of it is 1 cm.

∴ the external radius of the gold-radius is \(\frac{12}{2}\)cm = 6 cm and its internal radius = (6- 1) cm = 5 cm.

So the volume of the sphere = \(\left(\frac{4}{3} \pi \times 6^3-\frac{4}{3} \pi \times 5^3\right)\) cc = \(\frac{4}{3}\)π(216-125) cc

= \(\frac{4}{3}\) x \(\frac{22}{7}\) x 91 cc = \(\frac{88 \times 13}{3}\) cc

The mass of 1 cc gold = 19.5 gm.

∴ the mass of \(\frac{88 \times 13}{3}\) cc gold = \(\frac{88 \times 13}{3}\) x 19.5 gm

∴ the price of the gold-sphere = ₹\(\frac{88 \times 13}{3}\) x 19.5 x 1280 = ₹9518080

Hence the required cost of the hollow gold-sphere = ₹9518080.

Example 16. Two solid spheres of radii 3 cm and 4 cm are melted to make a hollow sphere of external radius of 6 cm, then what will be the thickness of the new hollow sphere?

Solution:

Given:

Two solid spheres of radii 3 cm and 4 cm are melted to make a hollow sphere of external radius of 6 cm

The radii of the two solid spheres are 3 cm and 4 cm.

So, the tqtal volume of the two spheres

= \(\left(\frac{4}{3} \pi \times 3^3+\frac{4}{3} \pi \times 4^3\right)\)cc = \(\frac{4}{3}\) (27 + 64)cc = \(\frac{4}{3}\)π x 91 cc.

The external radius of the hollow sphere = 6 cm

Let the internal radius of it = r cm.

∴ the volume of material of the hollow sphere = \(\frac{4}{3}\)(63 -r³)cc = \(\frac{4}{3}\)(216-r³)cc

= \(\frac{4}{3}\)π(216-r³) = \(\frac{4}{3}\)π x 91 or, 216 – r³ = 91 or, r³ = 125 or, r³ = 5³ ⇒ r = 5.

∴ thickness of the hollow sphere = (6 – 5) cm = 1 cm

Hence the thickness of the hollow sphere = 1 cm.

Class 10 Maths Solutions Wbbse

Example 17. The ratio of the volumes of two spheres is 216: 125. If the sum of the radii of the two spheres be 22 cm; then find the radii of the spheres.

Solution:

Given:

The ratio of the volumes of two spheres is 216: 125. If the sum of the radii of the two spheres be 22 cm

Let the radii of the two spheres be r₁cm and r₂ cm respectively.

As per question, \(\frac{4}{3} \pi r_1^3: \frac{4}{3} \pi r_2^3\) = 216: 125

⇒ \(\left(\frac{r_1}{r_2}\right)^3=\frac{216}{125}=\left(\frac{6}{5}\right)^3\)

⇒ \(\frac{r_1}{r_2}=\frac{6}{5}\)

⇒ \(r_1=\frac{6 r_2}{5}\)……(1)

Again, by question, r₁ + r₂ = 22

⇒ \(\frac{6 r_2}{5}+r_2\) = 22 [by (1)]

⇒ 11r₂= 22 x 5

⇒ r₂ = \(\frac{22 \times 5}{11}\) = 10

∴ from (1) we get, r₁ = \(\frac{6 \times 10}{5}\) = 12

Hence the radii of the two spheres were 10 cm and 12 cm respectively.

Example 18. A sphere of greatest volume is cut off from a metal hemisphere of radius 6 cm. If the cost of 1 cc metal be ₹42, then what will be the cost of the remaining hemisphere?

Solution:

Given:

A sphere of greatest volume is cut off from a metal hemisphere of radius 6 cm. If the cost of 1 cc metal be ₹42

The radius of the metal hemisphere = 6 cm.

So, the volume of the hemisphere = \(\frac{2}{3}\)π x 6³ cc.

Again, the diameter of the greatest sphere that can be cut off from the hemisphere will be 6 cm.

So, its radius will be \(\frac{6}{2}\)cm = 3cm.

∴ the volume of the greatest sphere cut off from the hemisphere = \(\frac{4}{3}\)π x 3³ cc.

∴ The volume of the remaining part of the hemisphere

= \(\left(\frac{2}{3} \pi \times 6^3-\frac{4}{3} \pi \times 3^3\right)\)cc = \(\frac{2}{3}\)π(216-54)cc = \(\frac{2}{3}\)π x 162 cc = 108π cc.

So, the cost of the remaining part of the metal hemisphere = ₹108 x \(\frac{22}{7}\) x 42 = ₹14256.

Hence the required cost = ₹14256.

Example 19. The external diameter of the forepart of a bowl made of a copper plate of thickness I cm is 12 cm. If the weight of 1 cc of copper be 12 gm, then what will be the weight of the bowl?

Solution:

Given:

The external diameter of the forepart of a bowl made of a copper plate of thickness I cm is 12 cm. If the weight of 1 cc of copper be 12 gm

The external radius of the semicircular bowl = \(\frac{12}{2}\) cm = 6 cm.

The bowl is of thickness 1 cm.

So, the internal radius of the bowl = (6 – 1) cm = 5 cm.

∴ the volume of the material of the bowl

= \(\left(\frac{2}{3} \pi \times 6^3-\frac{2}{3} \pi \times 5^3\right)\)cc = \(\frac{2}{3}\)π(216-125) cc

= \(\frac{2}{3}\) x \(\frac{22}{7}\) x 91 cc

∴ the weight of the bowl = \(\frac{2}{3}\) x \(\frac{22}{7}\) x 91 x 21 gm = 2288 gm = 2.288 kg

Hence the weight of the semicircle bowl = 2.288 kg