**Question 1: A copper sphere of diameter 18 cm is drawn into wire of diameter 4 mm. Find the length of the wire.**

**Solution:**

We have, Radius of the sphere, R = \(\frac{18}{2}\)

Radius of the wire, r = \(\frac{4}{2}\)

Let the length of the wire be l.

Now,

Volume of the wire = Volume of the copper sphere

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, l = 24300 cm = 243 m

So, the length of the wire is 243 m.

**Question 2: The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm respectively. Find the slant height of the frustum.**

**Solution:**

We have,

Height of the frustum, h = 6 cm,

Radii of the circular ends, R = 14 cm and r = 6 cm

Let the slant height of the frustum be l.

Now,

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

= \(\sqrt{(14-6)^{2}+6^{2}}\)

= \(\sqrt{8^{2}+6^{2}}\)

= \(\sqrt{64+36}\)

= \(\sqrt{100}\)

= 10 cm

So, the slant height of the frustum is 10 cm.

**Question 3: Find the ratio of the volume of a cube to that of a sphere which will fit inside it.**

**Solution:**

Let the radius of the sphere be R and the edge of the cube be a.

As, the sphere is fit inside the cube.

So, diameter of the sphere = edge of the cube

\(\Rightarrow\)

Now,

The ratio of the volume of the cube to that of the sphere = \(\frac{Volume\, of\, the\, cube}{Volume\, of\, the\, sphere}\)

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

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

= \(\frac{3\times 8R^{3}}{4\Pi R^{3}}\)

= \(\frac{6}{\Pi }\)

= 6 : \(\Pi\)

So, the ratio of the volume of the cube to that of the sphere is 6 : \(\Pi\)

**Question 4: Find the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?**

**Solution:**

Let the radius of the sphere be r.

We have,

The radius of the cone = The radius of the cylinder = The radius of the sphere = r

and

The height of the cylinder = The height of the cone = The height of the sphere = 2r

Now,

Volume of the cylinder = \(\Pi r^{2}(2r)=2\Pi r^{3}\)

Volume of the cone = \(\frac{1}{3}\Pi r^{2}(2r)=\frac{2}{3}\Pi r^{3}\)

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

So,

The ratio of the volumes of the cylinder, the cone and the sphere = \(2\Pi r^{3}:\frac{2}{3}\Pi r^{3}:\frac{4}{3}\Pi r^{3}\)

= \(1:\frac{1}{3}:\frac{2}{3}\)

= 3 : 1 : 2

So, the ratio of the volumes of the cylinder, the cone and the sphere is 3 : 1 : 2.

**Question 5: Two cubes each of volume 125 cm ^{3} are joined end to end to form a solid. Find the surface area of the resulting cuboid.**

**Solution:**

Let the edge of the cube be a.

As, Volume of the cube = 125 cm^{3}

\(\Rightarrow\)^{3} = 125

\(\Rightarrow\)

\(\Rightarrow\)

So,

Length of the resulting cuboid, l = 2 x 5 = 10 cm,

Breadth of the resulting cuboid, b = 5 cm and

Height of the resulting cuboid, h = 5 cm

Now,

Surface area of the resulting cuboid = 2(lb + bh + hl)

= 2 x (10 x 5 + 5 x 5 + 5 x 10)

= 2 x (50 + 25 + 50)

= 2 x 125

= 250 cm^{2}

So, the surface area of the resulting cuboid is 250 cm^{2}.

**Question 6: Three metallic cubes whose edges are 3 cm, 4 cm and 5 cm, are melted and recast into a single large cube. Find the edge of the new cube formed.**

**Solution:**

We have,

Edges of the cubes : a_{1} = 3 cm, a_{2} = 4 cm and a_{3} = 5 cm

Let the edge of the new cube be a.

Now,

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

\(\Rightarrow\)^{3} = 3^{3} + 4^{3} + 5^{3}

\(\Rightarrow\)^{3} = 27 + 64 + 125

\(\Rightarrow\)^{3} = 216

\(\Rightarrow\)

Therefore, a = 6 cm

So, the edge of the new cube so formed is 6 cm.

**Question 7: A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, find its width.**

**Solution:**

We have,

Radius of the metallic sphere, R = \(\frac{8}{2}\)

Height of the cylindrical wire, h = 12 m = 1200 cm

Let the radius of the base be r.

Now,

Volume of the cylindrical wire = Volume of the metallic sphere

\(\Rightarrow\)

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

\(\Rightarrow\)^{2} = \(\frac{4\times 4\times 4\times 4}{3\times 1200}\)

\(\Rightarrow\)^{2} = \(\frac{16}{225}\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, The width of the wire = 2r

= \(2\times \frac{4}{15}\)

= \(\frac{8}{15}cm\)

So, the width of the wire is \(\frac{8}{15}\)

**Question 8: A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used, at the rate of Rs. 25 per meter.**

**Solution:**

We have,

Width of the cloth, B = 5 m,

Radius of the conical tent, r = \(\frac{14}{2}\)

Height of the conical tent, h = 24 m

Let the length of the cloth used for making the tent be L.

Also,

The slant height of the conical tent, l = \(\sqrt{r^{2}+h^{2}}\)

= \(\sqrt{7^{2}+24^{2}}\)

= \(\sqrt{49+576}\)

= \(\sqrt{625}\)

= 25 m

Now,

The curved surface of the conical tent = \(\Pi rl\)

= \(\frac{22}{7}\times 7\times 25\)

\(\Rightarrow\)^{2}

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

So, the cost of the cloth used = 25 x 110 = 2750

So, the cost of the cloth used for making the tent is Rs. 2750.

**Question 9: A wooden toy was made by scooping out a hemisphere of the same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the volume of wood in the toy.**

**Solution:**

We have,

Radius of the cylinder = Radius of the hemisphere = r = 3.5 cm and

Height of the cylinder, h = 10 cm

Now,

Volume of the toy = Volume of the cylinder – Volume of the two hemispheres

= \(\Pi r^{2}h-2\times \frac{2}{3}\Pi r^{3}\)

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

= \(\frac{22}{7}\times 3.5\times 3.5\times \left ( 10-\frac{4\times 3.5}{3} \right )\)

= \(38.5\times \left ( 10-\frac{14}{3} \right )\)

= \(38.5\times \frac{16}{3}\)

= \(\frac{616}{3}cm^{3}\)

\(\approx 205.33cm^{3}\)

So, the volume of wood in the toy is \(\frac{616}{3}cm^{3}\)^{3}.

**Question 10: Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted and converted into a single cube whose diagonal is \(12\sqrt{3}\) cm. Find the edges of the three cubes.**

**Solution:**

Let the edge of the metal cubes be 3x, 4x and 5x.

Let the edge of the single cube be a.

As,

Diagonal of the single cube = \(12\sqrt{3}\)

\(\Rightarrow\)

\(\Rightarrow\)

Now,

Volume of the single cube = Sum of the volumes of the metallic cubes

\(\Rightarrow\)^{3} = (3x)^{3} + (4x)^{3} + (5x)^{3}

\(\Rightarrow\)^{3} = 27x^{3} + 64x^{3} + 125x^{3}

\(\Rightarrow\)^{3}

\(\Rightarrow\)^{3} = \(\frac{1728}{216}\)

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

\(\Rightarrow\)

\(\Rightarrow\)

So, the edges of the cubes are 3 x 2 = 6 cm, 4 x 2 = 8 cm and 5 x 2 = 10 cm.

Hence, the edges of the given three metallic cubes are 6 cm, 8 cm and 10 cm.

**Question 11: A hollow sphere of external and internal diameters 8 cm and 4 cm respectively is melted into a solid cone of base diameter 8 cm. Find the height of the cone.**

**Solution:**

We have,

External radius of the hollow sphere, R_{1} = \(\frac{8}{2}\)

Internal radius of the hollow sphere, R_{2} = \(\frac{4}{2}\)

Base radius of the cone, r = \(\frac{8}{2}\)

Let the height of the cone be h.

Now,

Volume of the cone = Volume of the hollow sphere

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, h = 14 cm

So, the height of the cone is 14 cm.

**Question 12: A bucket of height 24 cm is in the form of the frustum of a cone whose circular ends are of diameter 28 cm and 42 cm. Find the cost of milk at the rate of Rs. 30 per liter, which the bucket can hold.**

**Solution:**

We have,

Height of the frustum, h = 24 cm,

Radius of the open end, R = \(\frac{42}{2}\)

Radius of the close end, r = \(\frac{28}{2}\)

Now,

Volume of the bucket = \(\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)\)

= \(\frac{1}{3}\times \frac{22}{7}\times 24\times (21^{2}+12^{2}+21\times 14)\)

= \(\frac{176}{7}\times (441+196+294)\)

= \(\frac{176}{7}\times 931\)

= 23408 cm^{3}

= 23.408 L (As, 1000 cm^{3} = 1 L)

Therefore, The cost of the milk = 30 x 23.408 = 702.24

So, the cost of the milk which the bucket can hold is Rs. 702.24.

**Question 13: The interior of a building is in the form of a right circular cylinder of diameter 4.2 m and height 4 m surmounted by a cone of the same diameter. The height of the cone is 2.8 m. Find the outer surface area of the building.**

**Solution:**

We have,

Radius of the cylinder = Radius of the cone = r = \(\frac{4.2}{2}\)

Height of the cylinder, H = 4 m and

Height of the cone, h = 2.8 m

Also,

The slant height of the cone, l = \(\sqrt{r^{2}+h^{2}}\)

= \(\sqrt{2.1^{2}+2.8^{2}}\)

= \(\sqrt{4.41+7.84}\)

= \(\sqrt{12.25}\)

= 3.5 m

Now,

The outer surface area of the building = CSA of the cylinder + CSA of the cone

= \(2\Pi rH+\Pi rl\)

= \(\Pi r(2H+l)\)

= \(\frac{22}{7}\times 2.1\times (2\times 4+3.5)\)

= 6.6 x 11.5

= 75.9 m^{2}

So, the outer surface area of the building is 75.9 m^{2}.

**Question 14: A metallic solid right circular cone is of height 84 cm and the radius of its base is 21 cm. It is melted and recast into a solid sphere. Find the diameter of the sphere.**

**Solution:**

We have,

Height of the cone, h = 84 cm and

Base radius of the cone, r = 21 cm

Let the radius of the solid sphere be R.

Now,

Volume of the solid sphere = Volume of the solid cone

\(\Rightarrow\)

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

\(\Rightarrow\)^{3} = \(\frac{21\times 21\times 84}{4}\)

\(\Rightarrow\)^{3} = 21 x 21 x 21

\(\Rightarrow\)

Therefore, Diameter = 2R = 2 x 21 = 42 cm

So, the diameter of the solid sphere is 42 cm.

**Question 15: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.**

**Solution:**

We have,

Radius of the hemisphere = Radius of the cone = r = 3.5 cm and

Height of the cone = 15.5 – 3.5 = 12 cm

Also,

The slant height of the cone, l = \(\sqrt{h^{2}+r^{2}}\)

= \(\sqrt{12^{2}+3.5^{2}}\)

= \(\sqrt{144+12.25}\)

= \(\sqrt{156.25}\)

= 12.5 cm

Now,

Total surface area of the toy = CSA of cone + CSA of hemisphere

= \(\Pi rl+2\Pi r^{2}\)

= \(\Pi r(l+2r)\)

= \(\frac{22}{7}\times 3.5\times (12.5+2\times 3.5)\)

= 11 x (12.5 + 7)

= 11 x 19.5

= 214.5 cm^{2}

So, the total surface area of the toy is 214.5 cm^{2}.

**Question 16: If the radii of the circular ends of a bucket 28 cm high, are 28 cm and 7 cm, find its capacity and total surface area.**

**Solution:**

We have,

Height, h = 28 cm,

Radius of the upper end, R = 28 cm and

Radius of the lower end, r = 7 cm

Also,

The slant height, l = \(\sqrt{(R-r)^{2}+h^{2}}\)

= \(\sqrt{(28-7)^{2}+28^{2}}\)

= \(\sqrt{21^{2}+28^{2}}\)

= \(\sqrt{441+784}\)

= \(\sqrt{1225}\)

= 35 cm

Now,

Capacity of the bucket = \(\frac{1}{3}\Pi h(R^{2}+r^{2}+Rr)\)

= \(\frac{1}{3}\times \frac{22}{7}\times 28\times (28^{2}+7^{2}+28\times 7)\)

= \(\frac{88}{3}\times (784+49+196)\)

= \(\frac{88}{3}\times 1029\)

= 30184 cm^{3}

Also,

Total surface area of the bucket = \(\Pi l(R+r)+\Pi r^{2}\)

= \(\frac{22}{7}\times 35\times (28+7)+\frac{22}{7}\times 7\times 7\)

= 110 x (35) + 154

= 4004 cm^{2}

**Question 17: A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm ^{3} of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket. (Use \(\Pi =3.14\))**

**Solution:**

We have,

Radius of the upper end, R = 20 cm and

Radius of the lower end, r = 12 cm

Let the height of the bucket be h.

As,

Volume of the bucket = 12308.8 cm^{3}

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, h = 15 cm

So, the height of the bucket is 15 cm.

**Question 18: A milk container is made of metal sheet in the shape of the frustum of a cone whose volume is \(10459\frac{3}{7}\) cm ^{3}. The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 cm^{2}.**

**Solution:**

We have,

Radius of the upper end, R = 20 cm and

Radius of the lower end, r = 8 cm

Let the height of the container be h.

As,

Volume of the container = \(10459\frac{3}{7}\)^{3}

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Also,

The slant height of the container, l = \(\sqrt{(R-r)^{2}+h^{2}}\)

= \(\sqrt{(20-8)^{2}+16^{2}}\)

= \(\sqrt{12^{2}+16^{2}}\)

= \(\sqrt{144+256}\)

= \(\sqrt{400}\)

= 20 cm

Now,

Total surface area of the container = \(\Pi l(R+r)+\Pi r^{2}\)

= \(\frac{22}{7}\times 20\times (20+8)+\frac{22}{7}\times 8\times 8\)

= \(\frac{22}{7}\times 20\times 28+\frac{22}{7}\times 64\)

= \(\frac{22}{7}\times (560+64)\)

= \(\frac{22}{7}\times 624\)

= \(\frac{13728}{7}\)^{2}

So, the cost of metal sheet = \(1.4\times \frac{13728}{7}=2745.60\)

Hence, the cost of the metal sheet used for making the milk container is Rs. 2745.60.

**Question 19: ****A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter \(4\frac{2}{3}\) and height 3 cm. Find the number of cones so formed.**

**Solution:**

We have,

Radius of the metallic sphere, R = \(\frac{28}{2}\)

Radius of the smaller cone, r = \(\frac{1}{2}\times \left ( 4\frac{2}{3} \right )=\frac{1}{2}\times \frac{14}{3}=\frac{7}{3}\)

Height of the smaller cone, h = 3 cm

Now,

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

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

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

= \(\frac{4\times 14\times 14\times 14}{\left ( \frac{7}{3}\times \frac{7}{3}\times 3 \right )}\)

So, the number of cones so formed is 672.

**Question 20: A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of water **

**(i) displaced out of the cylinder**

**(ii) left in the cylinder.**

**Solution:**

We have,

Internal radius of the cylindrical vessel, R = \(\frac{10}{2}\)

Height of the cylindrical vessel, H = 10.5 cm,

Radius of the solid cone, r = \(\frac{7}{2}\)

Height of the solid cone, h = 6 cm

**(i)** Volume of water displaced out of the cylinder = Volume of the solid cone

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

= \(\frac{1}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 6\)^{3}

**(ii)** As, Volume of the cylindrical vessel = \(\Pi R^{2}H\)

= \(\frac{22}{7}\times 5\times 5\times 10.5\)

= 825 cm^{3}

So, the volume of water left in the cylindrical vessel = Volume of the cylindrical vessel – Volume of the solid cone

= 825 – 77 = 748 cm^{3}