RS Aggarwal Solutions Class 10 Ex 19B

All these RS Aggarwal class 10 solutions Chapter 19 Exercise 19.2 : Volume and Surface Areas of Solids are solved by Byju's top ranked professors as per CBSE guidelines.

QUESTION-1: The dimension of a metallic cuboid are 100cm x 80cm x 64cm. It is melted and recast into a cube. Find the surface area of the cube.

Solution:

Let the edge of the cube be a.

As, Volume of cube = volume of cuboid

\(\Rightarrow a^{3}=100\times 80\times 64\)
\(\Rightarrow a^{3}=512000\)
\(\Rightarrow a=\sqrt[3]{512000}\)

=> a = 80cm

Now, the surface area of the cube = \(6a^{2}\)

= 6 x 80 x 80

= 38400 \(cm^{2}\)

So, the surface area of the cube is 38400\(cm^{2}\).

 

QUESTION-2: A cone of height 20cm and radius of base 5cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.

Solution:

We have,

The radius of the cone, r = 5cm and the height of the cone, h = 20cm

Let the radius of the sphere be R.

As,

Volume of sphere = Volume of cone

\(\Rightarrow \frac{4}{3}\pi R^{3}=\frac{1}{3}\pi r^{2}h\)
\(\Rightarrow R^{3}=\frac{\pi r^{2}h\times 3}{3\times 4\pi }\)
\(\Rightarrow R^{3}=\frac{r^{2}h}{4}\)
\(\Rightarrow R^{3}=\frac{5\times 5\times 20}{4}\)
\(\Rightarrow R^{3}=125\)
\(\Rightarrow R=\sqrt[3]{125}=5cm\)

=> Diameter of the sphere = 2R = 2 x 5 = 10cm

So, the diameter of the sphere is 10cm

 

QUESTION-3: Metallic spheres of radii 6cm, 8cm and 10cm respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution:

We have,

The radii \(r_{1}=6cm,r_{2}=8cm\:and\:r_{3}=10cm\)

Let the radius of the resulting sphere be R.

As,

Volume of resulting sphere = Volume of three metallic spheres

\(\Rightarrow \frac{4}{3}R^{3}=\frac{4}{3}\pi r_{1}^{3}+\frac{4}{3}\pi r_{2}^{3}+\frac{4}{3}\pi r_{3}^{3}\)
\(\Rightarrow \frac{4}{3}R^{3}=\frac{4}{3}\pi \left ( r_{1} ^{3}+r_{2}^{3}+r_{3}^{3}\right )\)
\(\Rightarrow R^{3}=r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\)
\(\Rightarrow R^{3}=6^{3}+8^{3}+10^{3}\)
\(\Rightarrow R^{3}=216+512+1000\)
\(\Rightarrow R^{3}=1728\)
\(\Rightarrow R=\sqrt[3]{1728}\)

=> R = 12cm

So, the radius of the resulting sphere is 12cm.

 

QUESTION-4: A solid metal cone with the radius of base 12cm and height 24cm is melted to form solid spherical balls of diameter 6cm each. Find the number of balls thus formed.

Solution:

Radius of cone = 12cm

Height of the cone = 24cm

Volume = \(\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 12\times 12\times 24=48\times 24\times \pi cm^{3}\)

Radius of each ball = 3cm

Volume of each ball = \(\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi \times 3\times 3\times 3=36\pi cm^{3}\)

Total number of balls formed by melting the cone = \(\frac{Volume\:of\:cone}{Volume\:of\:a\:ball}=\frac{48\times 24\pi }{36\pi }=32\)

 

QUESTION-5: The radii of internal and external surfaces of a hollow spherical shell are 3cm and 5cm respectively. It is melted and recast into a solid cylinder of diameter 14cm. Find the height of the cylinder.

Solution:

We have,

The internal base radius of spherical shell, \(r_{1}\) = 3cm,

The external base radius of spherical shell, \(r_{2}\) = 5cm and

The base radius of solid cylinder, r = 14/2 = 7cm

Let the height of the cylinder be h.

As,

Volume of solid cylinder = Volume of spherical shell

\(\Rightarrow \pi r^{2}h=\frac{4}{3}\pi r_{2}^{3}-\frac{4}{3}\pi r_{1}^{3}\)
\(\Rightarrow \pi r^{2}h=\frac{4}{3}\pi \left ( r_{2}^{3}- r_{1}^{3} \right )\)
\(\Rightarrow r^{2}h=\frac{4}{3} \left ( r_{2}^{3}- r_{1}^{3} \right )\)
\(\Rightarrow 49\times h=\frac{4}{3}\left ( 125-27 \right )\)
\(\Rightarrow h=\frac{4}{3}\times \frac{98}{49}\)
\(Therefore, h=\frac{8}{3}\)

So, the height of the cylinder is \(\frac{8}{3}\)

 

QUESTION-6: The internal and external diameters of a hollow hemispherical shell are 6cm and 10cm respectively. It is melted and recast into a solid cone of base diameter 14cm. Find the height of the cone so formed.

Solution:

Internal radius = 3 cm and external radius = 5 cm

Volume of material in the shell = \( \frac{2}{3} \pi \times [5^{3} – 3^{3}] cm^{2} \)

= \( \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times h cm^{3} = \frac{154h}{3} cm^{3} \)

\( \Rightarrow \frac{154h}{3} = \frac{616}{3} \)
\( Rightarrow h = \frac{616}{154} = 4 cm \)

Hence, height of the cone = 4 cm.

 

QUESTION-7: A copper rod of diameter 2cm and length 10cm is drawn into a wire of uniform thickness and length 10m. Find the thickness of the wire.

Solution:

We have,

The radius of the copper rod, R = 2/2 = 1cm,

The height of the copper rod, H = 10cm and

The height of the wire , h = 10m = 1000cm

Let the radius of the wire be r.

As,

Volume of the wire = Volume of the rod

\(\Rightarrow \pi r^{2}h=\pi R^{2}H\)
\(\Rightarrow r^{2}h= R^{2}H\)
\(\Rightarrow r^{2}\times 1000=1\times 10\)
\(\Rightarrow r^{2}=\frac{10}{1000}\)
\(\Rightarrow r^{2}=\frac{1}{100}\)
\(\Rightarrow r=\sqrt{\frac{1}{100}}\)
\(\Rightarrow r=\frac{1}{10}=0.1cm\)

=> The diameter of the wire = 2r = 2 x 0.1 = 0.2cm

Therefore, The thickness of the wire = 0.2cm

So, the thickness of the wire is 0.2cm or 2mm.

 

QUESTION-8: A hemispherical bowl of internal diameter 30cm contains some liquid. This liquid is to be filled into cylindrical-shaped bottles each of diameter 5cm and height 6cm. Find the number of bottles necessary to empty the bowl.

Solution:

Inner diameter of the bowl = 30cm

Inner radius of the bowl = 15 cm

Volume of liquid in it = \( \frac{2}{3} \pi \times r^{3} = \frac{2}{3} \pi \times 15^{3} cm^{3} \)

Radius of each cylinder bottle = 2.5 cm and its height = 6 cm.

Volume of each cylindrical bottle = \( \pi r^{2} h = \pi \times (\frac{5}{2})^{2} \times 6 cm^{2} \)

\( = \frac{25}{4} \times 6 \pi = \frac{75 \pi }{2} cm^{3} \)

Required number of bottles = \( \frac{Volume \, of \, liquid}{Volume \, of \, each \, cylindrical \, bottle} \)

\( \frac{\frac{2}{3} \times \pi \times 15 \times 15 \times 15}{\frac{75}{2} \times \pi} = 60 \)

Hence, bottles required = 60.

 

QUESTION-9: A solid metallic sphere of diameter 21cm is melted and recast into a number of smaller cones, each of diameter 3.5cm and height 3cm. Find the number of cones so formed.

Solution:

Radius of the sphere = \( \frac{21}{2} cm \)

Volume of the sphere = \(\frac{4}{3} \pi \times r^{3} = \frac{4}{3} \times (\frac{21}{2})^{3} cm^{3} \)

Radius of cone = (7/4) cm and height 3 cm

Volume of cone = \( \frac{1}{3} \pi r^{2} h = \frac{1}{3} \times \pi \times (\frac{7}{4})^{2} \times 3 cm^{3} \)

Let the number of cones formed be n, then,

\( n \times \frac{1}{3} \pi \times (\frac{7}{4})^{2} \times 3 = \frac{4}{3} \pi (\frac{21}{2})^{3} \)
\( n = \frac{4}{3} \pi \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} \times \frac{3}{\pi} \times \frac{4}{7} \times \frac{4}{7} \times \frac{1}{3} \)

n = 504

Hence, number of cones formed = 504

 

QUESTION-10: A spherical cannon ball 28cm in diameter is melted and recast into a right circular conical mould, base of which is 35cm in diameter. Find the height of the cone.

Solution:

Diameter of cannon ball = 28cm

So,

Radius of cannon ball = 14 cm

Volume of cannon ball = \( \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi \times 14^{3} \)

Radius of the cone = \( \frac{35}{2} cm \)

Let the height of cone be h cm.

Volume of cone = \( \frac{1}{3} \pi (\frac{35}{2})^{2} \times h \) cm3

Hence,

\( \frac{4}{3} \pi \times 14^{3} = \frac{1}{3} \pi \times (\frac{35}{2})^{2} \times h \)
\( h = \frac{4}{3} \pi \times 14 \times 14 \times 14 \times \frac{3}{\pi} \times \frac{2}{35} \times \frac{2}{35} \)

= 35.84 cm

Hence, the height of the cone = 35.84 cm.

 

QUESTION-11: A spherical ball of lead 3cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5cm and 2cm. Find the radius of the third ball.

Solution:

Let the radius of the third ball be r cm.

Then,

Volume of third ball = Volume of spherical ball – volume of 2 small balls

Volume of the third ball = \( \frac{4}{3} \pi \times 3^{3} – \frac{4}{3} \pi (\frac{3}{2})^{2} – \frac{4}{3} \pi ??\times 2^{3} \)

\( 36 \pi – \frac{9 \pi }{2} – \frac{32 \pi}{3} cm^{3} ??= \frac{125 \pi }{6} cm^{3} \)

i.e \( \frac{4}{3} \pi r^{3} = \frac{125 \pi }{6} \)

i.e \( r^{3} = \frac{125 \pi \times 3}{6 \times 4 \times \pi } = \frac{125}{8} \)

i.e \( r = \frac{5}{2} cm = 2.5 cm \)

Hence, the radius of the third ball is 2.5cm.

 

QUESTION-12: A spherical shell of lead, whose external and internal diameters are respectively 24cm and 18cm, is melted and recast into a right circular cylinder 37cm high. Find the diameter of the base of the cylinder.

Solution:

External diameter of shell = 24cm and internal diameter of shell = 18cm

So,

External radius of shell = 12 cm and internal radius = 9 cm

Volume of lead in the shell = \( \frac{4}{3} \pi [12^{3} – 9^{3}] cm^{3} \)

Let the radius of the cylinder be r cm

Its height = 37 cm

Volume of cylinder = \(\pi r^{2}h=\pi r^{2}\left ( 37 \right )\)

\( \frac{4}{3} \pi [12^{3} – 9^{3}] = \pi r^{2} \times 37 \)
\( \frac{4}{3} \pi \times 999 = \pi r^{2} \times 37 \)
\( r^{2} = \frac{4}{3} \times \pi \times 999 \times \frac{1}{37 \pi } = 36 cm^{2} \)
\( r = \sqrt{36} = 6 cm \)

Hence, diameter of the base of the cylinder = 12 cm

 

QUESTION-13: A hemisphere of lead of radius 9cm is cast into a right circular cone of height 72cm. Find the radius of the base of the cone.

Solution:

Volume of hemisphere of radius 9 cm

= \( \frac{2}{3} \times \pi \times 9 \times 9 \times 9 cm^{3} \)

Volume of circular cone (height = 72 cm)

\( \frac{1}{3} \times \pi \times r^{2} \times 72 cm \)

Volume of cone = volume of hemisphere

\( \frac{1}{3} \times \pi \times r^{2} \times 72 = \frac{2}{3} \pi \times 9 \times 9 \times 9 \)
\( r^{2} = \frac{2 \pi }{3} \times 9 \times 9 \times 9 \times \frac{1}{24 \pi } = 20.25 \)
\( r = \sqrt{20.25} = 4.5 cm \)

Hence, radius of the base of the cone = 4.5 cm

 

QUESTION-14: A spherical ball of diameter 21cm is melted and recast into cubes, each of side 1cm. Find the number of cubes so formed.

Solution:

Diameter of sphere = 21 cm

Hence, radius of the sphere = \( \frac{21}{2} \)

Volume of sphere = \( \frac{4}{3} \pi \times r^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} \)

Volume of cube = a3 = 13

Let number of cubes formed be n

Volume of sphere = n(volume of cube)

\( \frac{4}{3} \times \frac{22}{7} \times \frac{441}{4} \times \frac{21}{2} = n \times 1 \)

= \( 441 \times 11 = n \)

4851 = n

Hence, number of cubes is 4851.

 

QUESTION-15: How many lead balls, each of radius 1cm, can be made from a sphere of radius 8cm?

Solution:

Volume of sphere (when r = 1 cm) = \( \frac{4}{3} \pi r^{3} = \frac{4}{3} \times 1 \times 1 \times 1 \times \pi cm^{3} \)

Volume of sphere (when r = 8 cm) = \( \frac{4}{3} \pi r^{3} = \frac{4}{3} \times 8 \times 8 \times 8 \times \pi cm^{3} \)

Let the number of balls = n

\( n \times \frac{4}{3} \times 1 \times 1 \times 1 \times \pi = \frac{4}{3} \times 8 \times 8 \times 8 \times \pi \)
\( n = \frac{4 \times 8 \times 8 \times 8 \times 3}{3 \times 4} = 512 \)

Hence, the number of lead balls can be made is 512.

 

QUESTION-16: A solid sphere of radius 3cm is melted and then cast into small spherical balls, each of diameter 0.6cm. Find the number of small balls so obtained.

Solutions:

Radius of sphere = 3 cm

Volume of sphere = \( \frac{4}{3} \pi r^{3} = \frac{4}{3} \times \pi \times 3 \times 3 \times 3 cm^{3} = 36 \pi cm^{3} \)

Radius of small sphere = \( \frac{0.6}{2} cm \) = 0.3 cm

Volume of small sphere = \( \frac{4}{3} \times \pi \times 0.3 \times 0.3 \times 0.3 cm^{3} cm^{3} \)

= \( \frac{4}{3} \pi \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10} cm^{3}\)

Let the number of small balls be n

\( n \times \frac{4}{3} \pi \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10} = \frac{4}{3} \times \pi \times 3 \times 3 \times 3 \)

n = 1000

Hence, the number of small balls = 1000.

 

QUESTION-17: The diameter of a sphere is 42cm. It is melted and drawn into a cylindrical wire of diameter 2.8cm. Find the length of the wire.

Solution:

Diameter of sphere = 42 cm

Radius of sphere = 21 cm

Volume of sphere = \( \frac{4}{3} \pi r^{3} = \frac{4}{3} \times \pi \times 21 \times 21 \times 21 cm^{3} \)

Diameter of cylindrical wire = 2.8 cm

Radius of cylindrical wire = 1.4 cm

Volume of cylindrical wire = \( \pi r^{2} h = \pi \times 1.4 \times 1.4 \times h cm^{3} = 1.96 \pi h cm^{3} \)

Volume of cylindrical wire = volume of sphere

\( 1.96 \pi h = \frac{4}{3} \times \pi \times 21 \times 21 \times 21 \)

\( h = \frac{4}{3} \times \pi \times 21 \times 21 \times 21 \times \frac{1}{1.96} \times \frac{1}{\pi} \) cm

h = 6300

\( h (\frac{6300}{100}) m = 63 m \)

Hence, length of the wire = 63 m.

 

QUESTION-18: The diameter of a sphere is 18cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108m, find its diameter.

Solution:

Diameter of sphere = 18 cm

Radius of copper sphere = \( \frac{3600}{100} m = 36 m \)

Volume of sphere =??\( \frac{4}{3} \pi r^{3} = \frac{4}{3} \times \pi \times 9 \times 9 \times 9 cm^{3} = 972 \pi cm^{3} \)

Length of wire = 108 m = 10800 cm

Let the radius of wire be r cm

= \( \pi r^{2} l = \pi r^{2} \times 10800 \)

But the volume of wire = volume of sphere

\( \Rightarrow \pi r^{2} \times 10800 = 972 \pi ??\)
\( r^{2} = \frac{972 \pi }{10800 \pi } = 0.09 cm^{2} \)
\( r = \sqrt{0.09} cm = 0.3 \)

Hence, the diameter = 2r = 0.6 cm.

 


Practise This Question

Any positive odd integer can be written in the form 4q+1 or 4q+3, where q is any integer.