**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.