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With the help of these solutions, you can not only crack the exam but can also obtain the best score in the exams.Â The easy to understand concepts, step by step solving of the questions help students in the preparation of the exam. TheÂ RS Aggarwal Class 10 Solutions Chapter 19 Volume And Surface Areas Of SolidsÂ help you in anticipating the type of questions which can be asked in the exam.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 19 – Volume And Surface Areas Of Solids Ex 19 F (19.6)

**Question 1: A river 1.5 m deep and 36 m wide is flowing at the rate of 3.5 km/hr. Find the amount of water (in cubic meters) that runs into the sea per minute.**

**Solution:**

We have,

Depth of the river, h = 1.5 m,

Width of the river, b = 36 m,

Speed of the flowing water, l = 3.5 km/hr = \(\frac{3.5\times 1000m}{60min}=\frac{175}{3}m/min\)

Now,

The amount of water that runs into the sea per minute = lbh

= \(\frac{175}{3}\times 36\times 1.5\)

= 3150 m^{3}/min

So, the amount of water that runs into the sea per minute is 3150 m^{3}.

**Question 2: The volume of a cube is 729 cm ^{3}. Find its surface area.**

**Solution:**

Let the edge of the cube be a.

As,

Volume of the cube = 729 cm^{3}

\(\Rightarrow\) a^{3} = 729

\(\Rightarrow\) a = \(\sqrt[3]{729}\)

\(\Rightarrow\) a = 9 cm

Now,

Surface area of the cube = 6a^{2}

= 6 x 9 x 9

= 486 cm^{2}

So, the surface area of the cube is 486 cm^{2}.

**Question 3: How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?**

**Solution:**

We have,

Edge of the cube, a = 10 cm and

Edge of the cubical box, l = 1 m = 100 cm

Now,

The number of cubes that can be put in the box = \(\frac{Volume\, of\, the\, cubical\, box}{Volume\, of\, the\, cube}\)

= \(\frac{l^{3}}{a^{3}}\)

= \(\frac{100^{3}}{10^{3}}\)

= 10^{3}

= 1000

So, the number of cubes that can be put in the cubical box is 1000.

**Question 4: Three cubes of iron whose edges are 6cm, 8 cm and 10 cm respectively are melted and formed into a single cube. Find the edge of the new cube formed.**

**Solution:**

We have,

Edges of the cubes: a_{1} = 6 cm, a_{2} = 8 cm and a_{3} = 10 cm

Let the edge of the new cube so formed be a.

As,

Volume of the new cube so formed = a_{1}^{3} + a_{2}^{3} + a_{3}^{3}

\(\Rightarrow\) a^{3} = 6^{3} + 8^{3} + 10^{3}

\(\Rightarrow\) a^{3} = 216 + 512 + 1000

\(\Rightarrow\) a^{3} = 1728

\(\Rightarrow\) a = \(\sqrt[3]{1728}\)

Therefore, a = 12 cm

So, the edge of the new cube so formed is 12 cm.

**Question 5: Five identical cubes, each of edge 5 cm, are placed adjacent to each other. Find the volume of the resulting cuboid.**

**Solution:**

We have,

Length of the resulting cuboid, l = 5 x 5 = 25 cm,

Breadth of the resulting cuboid, b = 5 cm and

Height of the resulting cuboid, h = 5 cm

Now,

Volume of the resulting cuboid = lbh

= 25 x 5 x 5

= 625 cm^{3}

So, the volume of the resulting cuboid is 625 cm^{3}

**Question 6: The volumes of two cubes are in the ratio 8 : 27. Find the ratio of their surface areas.**

**Solution:**

Let the edges of the cubes to a and b.

As,

\(\frac{Volume\, of\, the\, first\, cube}{Volume\, of\, the\, second\, cube}=\frac{8}{27}\)

\(\Rightarrow\) \(\frac{a^{3}}{b^{3}}=\frac{8}{27}\)

\(\Rightarrow\) \(\frac{a}{b}=\sqrt[3]{\frac{8}{27}}\)

\(\Rightarrow\) \(\frac{a}{b}=\frac{2}{3}\) – – – – – – – – (i)

Now,

The ratio of the surface areas of the cubes = \(\frac{Surface\, area\, of\, the\, first\, cube}{Surface\, area\, of\, the\, second\, cube}\)

= \(\frac{6a^{2}}{6b^{2}}\)

= \(\left ( \frac{a}{b} \right )^{2}\)

= \(\left ( \frac{2}{3} \right )^{2}\)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Using (i)]

= \(\frac{4}{9}\)

= 4 : 9

So, the ratio of the surface areas of the given cubes is 4 : 9.

**Question 7: The volume of a right circular cylinder with its height equal to the radius is \(25\frac{1}{7}\) cm ^{3}. Find the height of the cylinder.**

**Solution:**

We have,

Height = Base radius i.e. h = r

As,

Volume of the cylinder = \(25\frac{1}{7}\) cm^{3}

\(\Rightarrow\) \(\Pi r^{2}h=\frac{176}{7}\)

\(\Rightarrow\) \(\frac{22}{7}\times h^{2}\times h=\frac{176}{7}\)

\(\Rightarrow\) h^{3} = \(\frac{176\times 7}{7\times 22}\)

\(\Rightarrow\) h^{3} = 8

\(\Rightarrow\) h = \(\sqrt[3]{8}\)

Therefore, h = 2 cm

So, the height of the cylinder is 2 cm.

**Question 8: The ratio between the radius of the base and the height of a cylinder is 2 : 3. If the volume of the cylinder is 12936 cm ^{3}, find the radius of the base of the cylinder.**

**Solution:**

Let the radius of the base and the height of the cylinder be r and h, respectively.

We have,

r : h = 2 : 3 i.e. \(\frac{r}{h}=\frac{2}{3}\)

or h = \(\frac{3r}{2}\) – – – – – – – – – (i)

As,

Volume of the cylinder = 12936 cm^{3}

\(\Rightarrow\) \(\Pi r^{2}h=12936\)

\(\Rightarrow\) \(\frac{22}{7}\times r^{2}\times \frac{3r}{2}=12936\)Â Â Â Â Â Â Â Â Â [Using (i)]

\(\Rightarrow\) \(\frac{33}{7}\times r^{3}=12936\)

\(\Rightarrow\) r^{3} = \(12936\times \frac{7}{33}\)

\(\Rightarrow\) r^{3} = 2744

\(\Rightarrow\) r = \(\sqrt[3]{2744}\)

Therefore, r = 14 cm

So, the radius of the base of the cylinder is 14 cm.

**Question 9: The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. Find the ratio of their volumes.**

**Solution:**

Let the radii of the cylinders be r_{1} and r_{2}; and their heights be h_{1} and h_{2}.

We have,

r_{1} : r_{2} = 2 : 3 or \(\frac{r_{1}}{r_{2}}=\frac{2}{3}\) – – – – – – – (i)

and h_{1} : h_{2} = 5 : 3 or \(\frac{h_{1}}{h_{2}}=\frac{5}{3}\) – – – – – – – – – – (ii)

Now,

The ratio of the volumes of the cylinders = \(\frac{Volume\, of\, first\, cylinder}{Volume\, of\, second\, cylinder}\)

= \(\frac{\Pi r_{1}^{2}h_{1}}{\Pi r_{2}^{2}h_{2}}\)

= \(\left ( \frac{r_{1}}{r_{2}} \right )^{2}\times \frac{h_{1}}{h_{2}}\)

= \(\left ( \frac{2}{3} \right )^{2}\times \frac{5}{3}\)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Using (i) and (ii)]

= \(\frac{20}{27}\)

= 20 : 27

So, the ratio of the volumes of the given cylinders is 20 : 27.

**Question 10: 66 cubic cm of silver is drawn into a wire 1 mm in diameter. Calculate the length of the wire in metres.**

**Solution:**

We have,

Radius of wire, r = \(\frac{1}{2}\) = 0.5 mm = 0.05 cm

Let the length of the wire be l.

As,

Volume of the wire = 66 cm^{3}

\(\Rightarrow\) \(\Pi r^{2}l=66\)

\(\Rightarrow\) \(\frac{22}{7}\times 0.05\times 0.05\times l=66\)

\(\Rightarrow\) l = \(66\times \frac{7}{22\times 0.05\times 0.05}\)

Therefore, l = 8400 cm = 84 m

So, the length of the wire is 84 m.

**Question 11: If the area of the base of a right circular cone is 3850 cm ^{2} and its height is 84 cm, find the slant height of the cone.**

**Solution:**

We have,

Height = 84 cm

Let the radius and the slant height of the cone be r and l, respectively.

As,

Area of the base of the cone = 3850 cm^{2}

\(\Rightarrow\) \(\Pi r^{2}=3850\)

\(\Rightarrow\) \(\frac{22}{7}\times r^{2}=3850\)

\(\Rightarrow\) r^{2} = \(3850\times \frac{7}{22}\)

\(\Rightarrow\) r^{2} = 1225

\(\Rightarrow\) r = \(\sqrt{1225}\)

Therefore, r = 35 cm

Now,

l = \(\sqrt{h^{2}+r^{2}}\)

= \(\sqrt{84^{2}+35^{2}}\)

= \(\sqrt{7056+1225}\)

= \(\sqrt{8281}\)

= 91 cm

So, the slant height of the given cone is 91 cm.

**Question 12: A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm. Calculate the radius of the base of the cone.**

**Solution:**

We have,

Base radius of the cylinder, r = 8 cm,

Height of the cylinder, h = 2 cm and

Height of the cone, H = 6 cm

Let the base radius of the cone be R.

Now,

Volume of the cone = Volume of the cylinder

\(\Rightarrow\) \(\frac{1}{3}\Pi R^{2}H=\Pi r^{2}h\)

\(\Rightarrow\) \(R^{2}=\frac{3r^{2}h}{H}\)

\(\Rightarrow\) \(R^{2}=\frac{3\times 8\times 8\times 2}{6}\)

\(\Rightarrow\) R^{2} = 64

\(\Rightarrow\) R = \(\sqrt{64}\)

Therefore, R = 8 cm

So, the radius of the base of the cone is 8 cm.

**Question 13: A right cylindrical vessel is full of water. How many right cones having the same radius and height as those of the right cylinder will be needed to store that water?**

**Solution:**

Let the radius and height of the cone be r and h, respectively. Then,

Radius of the cylindrical vessel = r and

Height of the cylindrical vessel = h

Now,

The number of cones = \(\frac{Volume\, of\, the\, cylindrical\, vessel}{Volume\, of\, a\, cone}\)

= \(\frac{\Pi r^{2}h}{\left ( \frac{1}{3}\Pi r^{2}h \right )}\)

= 3

So, the number of cones that will be needed to store the water is 3.

**Question 14: The volume of a sphere is 4851 cm ^{3}. Find its curved surface area.**

**Solution:**

Let the radius of the sphere be r.

As,

Volume of the sphere = 4851 cm^{3}

\(\Rightarrow\) \(\frac{4}{3}\Pi r^{3}=4851\)

\(\Rightarrow\) \(\frac{4}{3}\times \frac{22}{7}\times r^{3}=4851\)

\(\Rightarrow\) r^{3} = \(4851\times \frac{3\times 7}{4\times 22}\)

\(\Rightarrow\) r^{3} = \(\frac{9261}{8}\)

\(\Rightarrow\) r = \(\sqrt[3]{\frac{9261}{8}}\)

\(\Rightarrow\) r = \(\frac{21}{2}cm\)

Now,

Curved surface area of the sphere = \(4\Pi r^{2}\)

= \(4\times \frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}\)

= 1386 cm^{2}

So, the curved surface area of the sphere is 1386 cm^{2}

**Question 15: The curved surface area of a sphere is 5544 cm ^{3}. Find its volume.**

**Solution:**

Let the radius of the sphere be r.

As,

Curved surface area of the sphere = 5544 cm^{2}

\(\Rightarrow\) \(4\Pi r^{2}=5544\)

\(\Rightarrow\) \(4\times \frac{22}{7}\times r^{2}=5544\)

\(\Rightarrow\) r^{2} = \(5544\times \frac{7}{4\times 22}\)

\(\Rightarrow\) r^{2} = 441

\(\Rightarrow\) r = \(\sqrt{441}\)

\(\Rightarrow\) r = 21 cm

Now,

Volume of the sphere = \(\frac{4}{3}\Pi r^{3}\)

= \(\frac{4}{3}\times \frac{22}{7}\times 21\times 21\times 21\)

= 38808 cm^{3}

So, the volume of the sphere is 38808 cm^{3}.

**Question 16: The surface areas of two spheres are in the ratio of 4 : 25. Find the ratio of their volumes.**

**Solution:**

Let the radii of the two spheres be r and R.

As,

\(\frac{Surface\, area\, of\, the\, first\, sphere}{Surface\, area\, of\, the\, second\, sphere}=\frac{4}{25}\)

\(\Rightarrow\) \(\frac{4\Pi r^{2}}{4\Pi R^{2}}=\frac{4}{25}\)

\(\Rightarrow\) \(\left ( \frac{r}{R} \right )^{2}=\frac{4}{25}\)

\(\Rightarrow\) \(\frac{r}{R}=\sqrt{\frac{4}{25}}\)

\(\Rightarrow\) \(\frac{r}{R}=\frac{2}{5}\) – – – – – – – – – – – – (i)

Now,

The ratio of the volumes of the two sphere = \(\frac{Volume\, of\, the\, first\, sphere}{Volume\, of\, the\, second\, sphere}\)

= \(\frac{\left ( \frac{4}{3}\Pi r^{3} \right )}{\left ( \frac{4}{3}\Pi R^{3} \right )}\)

= \(\left ( \frac{r}{R} \right )^{3}\)

= \(\left ( \frac{2}{5} \right )^{3}\)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Using (i)]

= \(\frac{8}{125}\)

= 8 : 125

So, the ratio of the volumes of the given spheres is 8 : 125.

**Question 17: A solid metallic sphere of radius 8 cm is melted and recast into spherical balls each of radius 2 cm. Find the number of spherical balls obtained.**

**Solution:**

We have,

Radius of the solid metallic sphere, R = 8 cm and

Radius of the spherical ball, r = 2 cm

Now,

The number spherical balls obtained = \(\frac{Volume\, of\, the\, solid\, metallic\, sphere}{Volume\, of\, a\, spherical\, ball}\)

= \(\frac{\left ( \frac{4}{3}\Pi R^{3} \right )}{\left ( \frac{4}{3}\Pi r^{3} \right )}\)

= \(\left ( \frac{R}{r} \right )^{3}\)

= \(\left ( \frac{8}{2} \right )^{3}\)

= 4^{3}

= 64

So, the number of spherical balls obtained is 64.

**Question 18: How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9cm x 11 cm x 12 cm?**

**Solution:**

We have,

Radius of a lead shot, r = \(\frac{3}{2}\) = 1.5 mm = 0.15 cm and

Dimensions of the cuboid are 9 cm x 11 cm x 12 cm

Now,

The number of the lead shots = \(\frac{Volume\, of\, the\, cuboid}{Volume\, of\, a\, shot}\)

= \(\frac{9\times 11\times 12}{\left ( \frac{4}{3}\Pi r^{3} \right )}\)

= \(\frac{9\times 11\times 12}{\left ( \frac{4}{3}\times \frac{22}{7}\times 0.15\times 0.15\times 0.15 \right )}\)

= 84000

So, the number of lead shots that can be made from the cuboid is 84000.

**Question 19: A metallic cone of radius 12 cm and height 24 cm is melted and made into spheres of radius 2 cm each. How many spheres are formed?**

**Solution:**

We have,

Radius of the metallic cone, r = 12 cm,

Height of the metallic cone, h = 24 cm and

Radius of the sphere, R = 2 cm

Now,

The number of spheres so formed = \(\frac{Volume\, of\, the\, metallic\, cone}{Volume\, of\, a\, sphere}\)

= \(\frac{\left ( \frac{1}{3}\Pi r^{2}h \right )}{\left ( \frac{4}{3}\Pi R^{3} \right )}\)

= \(\frac{r^{2}h}{4R^{3}}\)

= \(\frac{12\times 12\times 24}{4\times 2\times 2\times 2}\)

= 108

So, the number of spheres so formed is 108.

**Question 20: A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. Find the radius of the base of the cone.**

**Solution:**

We have,

Radius of the hemisphere, R = 6 cm and

Height of the cone, h = 75 cm

Let the radius of the base of the cone be r.

Now,

Volume of the cone = Volume of the hemisphere

\(\Rightarrow\) \(\frac{1}{3}\Pi r^{2}h=\frac{2}{3}\Pi R^{3}\)

\(\Rightarrow\) r^{2} = \(\frac{2R^{3}}{h}\)

\(\Rightarrow\) r^{2} = \(\frac{2\times 6\times 6\times 6}{75}\)

\(\Rightarrow\) r^{2} = 5.76

\(\Rightarrow\) r = \(\sqrt{5.76}\)

Therefore, r = 2.4 cm

So, the radius of the base of the cone is 2.4 cm.

### Key Features of RS Aggarwal Class 10 Solutions ChapterÂ 19 – Volume And Surface Areas Of Solids Ex 19 F (19.6)

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