NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume

Q1 ) A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of \( \pi \).

Answer :

Volume of the solid = Volume of the cone + Volume of the hemisphere

\( = \ \frac{1}{3} \pi r^2 h \ + \ \frac{2}{3} \pi R^3 \)

\( = \ \frac{1}{3} \pi r^2 \ × \ r \ + \ \frac{2}{3} \pi r^3 \) [\(\because \) h = r and R = r]

\( = \ \frac{ \pi}{3}r^3(1 \ + \ 2) \ = \ \pi r^3 \)

\( = \ \pi(1)^3 \ = \ \pi \) cm³ [\(\because \) r = 1 cm]

Q2 ) Rachel an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm,find the volume of air contained in the model the Rachel made. (Assume the outer and inner dimensions of the model be nearly the same).

Answer :

Volume of the air contained in the model = Volume of the cylindrical portion of the model + Volume of its two conical ends.

\( = \ \pi r^2 h_1 \ + \ 2 \ × \ \frac{1}{3} \pi r^2 h_2 \ \)
\( = \ \pi r^2(h_1 \ + \ \frac{2}{3} h_2) \)

where \( r \ = \ \frac{3}{2} \) cm , \(h_1 \ = \ 8 \) cm and \(h_2 \ = \ 2 \) cm

\( = \ \frac{22}{7} \ × \ ( \frac{3}{2})^2 \ × \ (8 \ + \ \frac{2}{3} \ × \ 2) \) \( = \ \frac{22}{7} \ × \ \frac{9}{4} \ × \ \frac{24 \ + \ 4}{3} \)

\(= \ \frac{22}{7} \ × \ \frac{9}{4} \ × \ \frac{28}{3} \) \( = \ 66 \) cm³

Q3 ) A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemsipherical ends with length 5 cm and diameter is 2.8 cm (see figure).

Answer :

Volume of the gulab jamun = Volume of the cylindrical portion + Volume of the hemispherical ends

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

where r = 1.4 cm, h = 2.2 cm

\( = \ \frac{22}{7} \ × \ (1.4)^2 \ × \ (2.2 \ + \ \frac{4}{3} \ × \ 1.4) \ \)
\( = \ \frac{22}{7} \ × \ 1.96 \ × \ ( \frac{6.6 \ + \ 5.6}{3} ) \)

\( = \ \frac{22}{7} \ × \ 1.96 \ × \ \frac{12.2}{3} \) cm³

Volume of 45 gulab jamuns

\( = \ 45 \ × \ \frac{22}{7} \ × \ 1.96 \ × \ \frac{12.2}{3} \) cm³

Quantity of syrup in gulab jamuns = 30% of their volume

\( = \ \frac{30}{100} \ × \ 45 \ × \ \frac{22}{7} \ × \ 1.96 \ × \ \frac{12.2}{3} \)

\(= \ \frac{9 \ × \ 11 \ × \ 1.96 \ × \ 12.2}{7} \ \)
\( = \ 338.184 \)

\(= \ 338 \) cm³

Q4 ) A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 m and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).

Answer :

Volume of wood in the entire stand = Volume of the cuboid - 4 × Volume of a depression (i.e., cone)

\( = \ l b h \ - \ 4 \ × \ \frac{1}{3} \pi r^2 h \)

\( = \ 15 \ × \ 10 \ × \ 3.5 \ - \ 4 \ × \ \frac{1}{3} \ × \ \frac{22}{7} \ × \ 0.5 \ × \ 0.5 \ × \ 1.4 \)

\( = \ 525 \ - \ 1.47 \ = \ 523.53 \) cm³

Q5 ) A vessel in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one- fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer :

Height of the conical vessel, h = 8 cm.

Its radius r = 5 cm

Volume of cone = Volume of water in cone \(= \ \frac{1}{3} \pi r^2 h \)

\(= \ \frac{1}{3} \ × \ \frac{22}{7} \ × \ 5 \ × \ 5 \ × \ 8 \ = \ \frac{4400}{21} \) cm³

Volume of water flows out = Volume of lead shots

\( = \ \frac{1}{4} \) of the volume of water in the cone

\( = \ \frac{1}{4} \ × \ \frac{4400}{21} \ = \ \frac{1100}{21} \) cm³

Radius of the lead shot = 0.5 cm

Volume of one spherical lead shot \(= \ \frac{4}{3} \pi r^3 \ = \ \frac{4}{3} \ × \ \frac{22}{7} \ × \ ( \frac{5}{10})^3 \)

\(= \ \frac{11}{21} \) cm³

\(\therefore \) Number of lead shots dropped into the vessel

\( = \frac{ \ Volume \ of \ water \ flows \ out}{Volume \ of \ one \ lead \ shot} \ = \frac{ \frac{1100}{21}}{ \frac{11}{21}} \)

\( = \ \frac{1100}{21} \ × \ \frac{21}{11} \ = \ 100 \)

Q6 ) A solid iron pole consists of a cylindrical height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use \( \pi \ = \ 3.14 \) )

Answer :

Volume of the solid iron pole = Volume of the cylindrical portion + Volume of the other cylindrical portion

\( = \ \pi (r_1)^2 h_1 \ + \ \pi (r_2)^2 h_2 \)

\(= \ 3.14 \ × \ (12)^2 \ × \ 220 \ + \ 3.14 \ × \ (8)^2 \ × \ 60 \)

\(= \ 3.14 \ × \ 144 \ × \ 220 \ + \ 3.14 \ × \ 64 \ × \ 60 \)

\(= \ 99475.2 \ + \ 12057.6 \)

\(= \ 111532.8 \) cm³

\(\therefore \) The mass of the pole
\(= \ 111532.8 \ × \ 8 \) grams

\(= \ \frac{111532.8 \ × \ 8}{1000} \ = \ 892.26 \) kg

Q7 ) A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water circular such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Answer :

Volume of the cylinder \(= \ \pi r^2 h \ = \ \frac{22}{7} \ × \ (60)^2 \ × \ 180 \)

\(= \ \frac{22 \ × \ 3600 \ × \ 180}{7} \ = \ \frac{14256000}{7} \) cm³

Volume of the solid = Volume of cone + Volume of hemisphere

\(= \ \frac{1}{3} \ × \ \frac{22}{7} \ × \ 60^2 \ × \ 120 \ + \ \frac{2}{3} \ × \ \frac{22}{7} \ × \ 60^3 \)

\(= \ \frac{3168000}{7} \ + \ \frac{3168000}{7} \ = \ \frac{6336000}{7} \) cm³

Volume of water left in the cylinder = Volume of the cylinder - Volume of the solid

\(= \ \frac{14256000}{7} \ - \ \frac{6336000}{7} \ = \frac{7920000}{7} \)

\(= \ 1131428.57142 \ = \ 1.131 \) m³

Q8 ) A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter ; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds , a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements and \( \pi \ = \ 3.14 \)

Answer :

Volume of spherical vessel = Volume of sphere + Volume of cylinder

\(= \ \pi (1)^2 (8) \ + \ \frac{4}{3} \pi (4.25)^3 \)

\(= \ 3.14(8 \ + \ \frac{4}{3} \ × \ 76.765625) \ \)

\( = \ 3.14(8 \ + \ 102.353 ) \)

\( = \ 3.14 \ × \ 110.353 \ \)

\( = \ 346.51 \) cm³

\(\therefore \) She is incorrect. The correct answer is 346.51 cm³

Q1 ) A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Answer :

Volume of the sphere \( = \ \frac{4}{3} \pi r^3 \ = \ \frac{4}{3} \ × \ \pi \ × \ (4.2)^3 \) cm³

If h is the height of a cylinder of radius 6 cm. Then its volume \(= \ \pi (6)^2 h \ = \ 36\pi h \) cm³

\(\therefore \) The volume of metal in the form of sphere and cylinder remains the same , we have

\(\Rightarrow 36 \pi h \ = \ \frac{4}{3} \ × \ \pi \ × \ (4.2)^3 \)

\(\Rightarrow \ h \ = \ \frac{1}{36} \ × \ \frac{4}{3} \ × \ (4.2)^3 \)

\( \Rightarrow \ h \ = \ 2.744 \) cm

Q2 ) Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Answer :

Sum of the volumes of 3 gives spheres.

\( = \ \frac{4}{3} \pi ((r_1)^3 \ + \ (r_2)^3 \ + \ (r_3)^3) \)

\(= \ \frac{4}{3} \pi (6^3 \ + \ 8^3 \ + \ 10^3) \)

\(= \ \frac{4}{3} \pi (216 \ + \ 512 \ + \ 1000) \)

\(= \ \frac{4}{3} \pi (1728) \) cm³

Let R be the radius of the new spheres whose volume is the sum of the volumes of 3 given spheres.

\( \Rightarrow \ \frac{4}{3} \pi R^3 \ = \ \frac{4}{3} \pi (1728) \)

\(\Rightarrow \ R^3 \ = \ 1728 \ => \ R \ = \ 12 \) cm

\(\therefore \) The radius of the resulting sphere is 12 cm.

Q3 ) A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Answer :

Let \( h\) be the required height of the platform.

The shape of the platform will be like the shape of a cuboid of dimensions \(22 \ m \ × \ 14 \ m \ × \ h \) with a hole in the shape of cylinder of radius \(3.5 m\) and depth \(h\).

The volume of the platform will be equal to the volume of the earth dug out from the well.

Now, the volume of the earth = Volume of the cylindrical well

\(= \ \pi r^2 h \ = \ \frac{22}{7} \ × \ 12.25 \ × \ 20 \ = \ 770 \) m³

Also, the volume of the platform \( = \ 22 \ × \ 14 \ × \ h \) m³

But volume of the platform = Volume of the well

\(\Rightarrow \ 22 \ × \ 14 \ × \ h \ = \ 770 \)

\( \Rightarrow \ h \ = \ \frac{770}{22 \ × \ 14} \ = \ 2.5 \) m

\(\therefore \) Height of the platform = 2.5 m

Q4 ) A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Answer :

Let h be the required height of the embankment.



The shape of the embankment will be like the shape of a cylinder of internal radius 1.5 m and external radius \(4 \ + \ 1.5 \ = \ 5.5 \) m

The volume of the embankment will be equal to the volume of earth dug out from the well.

Now, the volume of the earth = Volume of the cylindrical well

\( = \ \pi \ × \ (1.5)^2 \ × \ 14 \ = \ 31.5\pi \) m³

Also, the volume of the embankment

\(= \ \pi (5.52 \ - \ 1.52)h \ = \ \pi \ × \ 7 \ × \ 4h \)

\(= \ 28 \pi h \) m³

\(\therefore 28 \pi h \ = \ 31.5 \pi \)

\( \Rightarrow \ h \ = \ \frac{31.5}{28} \ = \ 1.125 \) m

Hence, the required height of the embankment = 1.125 m

Q5 ) A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Answer :

Volume of the cylinder
\(= \ \pi r^2 h \ = \ \pi ( \frac{12}{2})^2 \ × \ 15 \ \)

\( = \ \pi \ × \ 6^2 \ × \ 15 \) cm³

Volume of a cone having hemispherical shape on the top

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

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

\( = \ \frac{1}{3} \pi ( \frac{6}{2})^2 (12 \ + \ 2 \ × \ \frac{6}{2} ) \ \)

\( = \ \frac{1}{3} \pi \ × \ 3^2 \ × \ 18 \) cm³

Let the number of cone that can be filled with ice cream be n.

Then, \( \frac{1}{3} \pi \ × \ 3^2 \ × \ 18 \ × \ n \ = \ \pi \ × \ 6^2 \ × \ 15 \)

\( \Rightarrow \ n \ = \ \frac{\pi \ × \ 6 \ × \ 6 \ × \ 15\ × \ 3 \ }{ \pi \ × \ 3 \ × \ 18 \ × \ 3 \ } = \ 10 \)

\(\therefore \) 10 cones can be filled with ice cream.

Q6 ) How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Answer :

The shape of the coin will be like the shape of a cylinder of radius
\( \frac{1.75}{2} \ = \ 0.875 \) cm and of height 2 mm = 0.2 cm

Its volume \(= \ \pi r^2 h \ \)

\( = \ \frac{22}{7} \ × \ 0.875 \ × \ 0.875 \ × \ 0.2 \)

\(= \ 0.48125 \) cm³

Volume of the cuboid \(= \ 5.5 \ × \ 10 \ × \ 3.5 \ = \ 192.5 \) cm³

Number of coins required to form the cuboid

\( = \ \frac{Volume \ of \ the \ cuboid}{Volume \ of \ the \ coin} \ \)

\( = \ \frac{192.5}{0.48125} \ = \ 400 \)

\(\therefore \) 400 coins must be melted to form a cuboid.

Q7 ) A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Answer :

Volume of the sand = Volume of the cylindrical bucket \(= \ \pi r^2 h \ = \ \pi \ × \ 18 \ × \ 18 \ × \ 32 \) cm3

Volume of the conical heap \(= \ \frac{1}{3} \pi r^2 h \) where, h = 24 cm

\(= \ \frac{1}{3} \pi r^2 \ × \ 24 \ = \ 8\pi r^2 \)

The volume of the conical heap will be equal to that of sand.

\( therefore 8\pi r^2 \ = \ \pi \ × \ 18 \ × \ 18 \ × \ 32 \)

\(\Rightarrow \ r^2 \ = \ 18 \ × \ 18 \ × \ 4 \ = \ 18^2 \ × \ 2^2 \)

\(\Rightarrow \ r \ = \ 18 \ × \ 2 \ = \ 36 \)

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

\(\Rightarrow \ l \ = \ \sqrt{36^2 \ + \ 24^2} \ \)
\( = \ 12 \sqrt{9 \ + \ 4} \ \)
\( = \ 12 \sqrt{13} \)

\(\therefore \) The radius of the conical heap is 36 cm and its slant height is \( 12 \sqrt{13} \) cm

Q8 ) Water in a canal 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Answer :

Width of the canal = 6 m

Depth of the canal = 1.5 m

Length of water column per hour = 10 km

Length of water column in 30 minutes or \(\frac{1}{2}\) hour \( = \ \frac{1}{2} \ × \ 10 \ = \ 5000 \) m

Volume of water flown in 30 minutes \(= \ 1.5 \ × \ 6 \ × \ 5000 \ = \ 45000 \) m³

\(\because \) \( 8 \) cm \( = \ \frac{8}{100} \) m

i.e., 0.08 m standing water is desired

\(\therefore \) Area irrigated in 30 minutes

\( = \ \frac{Volume}{Height} \ = \ \frac{45000}{0.08} \ = \ 562500 \) m² or \(56.25 \) hectares

Q9 ) A farmer connects a pipe of internal diameter 20 cm from a cannal into a cylindrical tank in his field, which is 10 m in diameter and 2m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Answer :

Diameter of the pipe = 20 cm

\( \therefore \) Radius of the pipe = 10 cm

Length of water column per hour
= 3 km = 3 × 1000 × 100 cm

Volume of water flown in one hour
\(= \ \pi \ × \ 100 \ × \ 300000 \) cm³

Tank to be filled = Volume of cylinder
(with r = 5m = 500 cm and h = 2m = 200 cm)

\(= \ \pi \ × \ 500 \ × \ 500 \ × \ 200 \) cm³

Time required to fill the tank
\(= \ \frac{Volume \ of \ tank}{Volume \ of \ water \ flown} \)

\(= \ \frac{ \pi \ × \ 500 \ × \ 500 \ × \ 200}{ \pi \ × \ 100 \ × \ 300000} \)

\( = \ \frac{5}{3} \ = \ 100 \) minutes.

Q.9 A farmer connects a pipe of internal diameter 20 cm from a cannal into a cylindrical tank in his field, which is 10 m in diameter and 2m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Answer :

Diameter of the pipe = 20 cm

=> Radius of the pipe = 10 cm

Length of water column per hour = 3 km = 3 × 1000 × 100 cm

Volume of water flown in one hour \(= \ \pi \ × \ 100 \ × \ 300000 \) cm³

Tank to be filled = Volume of cylinder (with r = 5m = 500 cm and h = 2m = 200 cm) \(= \ \pi \ × \ 500 \ × \ 500 \ × \ 200 \) cm³

Time required to fill the tank \(= \ \frac{Volume \ of \ tank}{Volume \ of \ water \ flown} \)

\(= \ \frac{ \pi \ × \ 500 \ × \ 500 \ × \ 200}{ \pi \ × \ 100 \ × \ 300000} \)

\( = \ \frac{5}{3} \ = \ 100 \) minutes.

Q1 ) A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Answer :

Capacity of the glass, \(V \ = \ \frac{1}{3} \ × \pi \ × \ h \ (R^2 \ + \ r^2 \ + \ Rr) \)

Here, R = 2 cm , r = 1 cm , h = 14 cm

\( \therefore V \ = \ \frac{1}{3} \ × \frac{22}{7} \ × \ 14 \ (2^2 \ + \ 1^2 \ + \ 2 \ × \ 1) \ \)
\( = \ \frac{44}{3} \ × \ (4 \ + \ 1 \ + \ 2) \)

\(= \ \frac{44}{3} \ × \ 7 \ = \ \frac{308}{3} \) cm³

Q2 ) The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Answer :

Slant height, l = 4 cm (Given)

\(2 \pi r_1 \ = \ 6 \ \)

\( \Rightarrow \ \pi r_1 \ = \ 3 \)

and \(2 \pi r_2 \ = \ 18 \ \)

\( \Rightarrow \ \pi r_2 \ = \ 9 \)

Curved surface of the frustum \(= \ ( \pi r_1 \ + \ \pi r_2)l \)

\(= \ (3\ + \ 9) \ × \ 4 \ = \ 48 \) cm²

Q3 ) A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Answer :

Here R = 10 cm, r = 4 cm and l = 15 cm



Area of the material used for making the fez = Surface area of frustum + The surface of top circular section

\( = \ \pi (R \ + \ r)l \ + \ \pi r^2 \)

\(= \ \frac{22}{7} (10 \ + \ 4) 15 \ + \ \frac{22}{7} \ × \ 4 \ × \ 4 \)

\(= \ \frac{22}{7} \ × \ 14 \ × \ 15 \ + \ \frac{352}{7} \)

\(= \ \frac{4620 \ + \ 352}{7} \ = \frac{4972}{7} \) cm²

Q4 ) A container opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm². (Take \( \pi \ = \ 3.14 \) ).

Answer :

Here R = 20 cm, r = 8 cm and h = 16 cm



Capacity of the container = Volume of the frustum \(= \ \frac{1}{3} \pi h(R^2 \ + \ r^2 \ + \ Rr) \)

\( = \ \frac{1}{3} \ × \ 3.14 \ × \ 16(20^2 \ + \ 8^2 \ + \ 20 \ × \ 8) \ \)

\( = \ \frac{50.24}{3} \ × \ (400 \ + \ 64 \ + \ 160) \)

\(= \ \frac{50.24}{3} \ × \ 624 \ \)

\( = \ 50.24 \ × \ 208 \)

\(= \ \frac{10449.92}{1000} \) litres

Cost of milk @ Rs 20 per litre \( = \ 20 \ × \ \frac{10449.92}{1000} \)

\( = \ 208.99 \) \( \approx \) Rs \(209 \)

To find the slant height



\( l \ = \ \sqrt{16^2 \ + \ 12^2} \ \)
\( = \ \sqrt{256 \ + \ 144} \)
\( = \ \sqrt{400} \ \)
\( = \ 20 \) cm

Curved Surface Area of the container = \( \pi (R \ + \ r)l \)

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

\( = \ \frac{22}{7} \ × \ 28 \ × \ 20 \ = \ 1758.4 \) cm²

Area of the bottom of the container = \( \pi r^2 \)

\( = \ \frac{22}{7} \ × \ (8)^2 \ = \ \frac{22}{7} \ × \ 64 \ = \ 200.96 \) cm²

\(\therefore \) Total area of metal required \(= 1758.4 \ + \ 200.96 \ = \ 1959.36 \) cm²

Cost of metal sheet used to manufacture the container @ Rs 8 per 100 cm² \(= \ \frac{8}{100} \ × \ 1959.36 \ = \) Rs \(156.75 \)

Q5 ) A metallic right circular cone 20 cm high and whose vertical angle is 60º is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into wire of diameter \( \frac{1}{16} \) cm, find the length of the wire.

Answer :

Let ABC be the metallic cone, DECB is the required frustum.



Let the two radii of the frustum be \( DO' \ = \ r_2 \) and \( BO \ = \ r_1 \)

From the ∆ ADO' and ABO,

\( r_2 \ = \ h_1 \) tan 30º \( = \ 10 \ × \ \frac{1}{ \sqrt{3}} \)

\( r_1 \ = \ (h_1 \ + \ h_2 \) tan30º \( = \ 20 \ × \ \frac{1}{ \sqrt{3}} \)

Volume of the frustum DBCE \(= \ \frac{1}{3} \pi h_2 \ × [ (r_1)^2 \ + \ (r_1)(r_2) \ + \ (r_2)^2 ] \)

\(= \ \frac{1}{3} \pi h_2 \ × \ [ \frac{400}{3} \ + \ \frac{200}{3} \ + \ \frac{100}{3}] \ \)
\( = \frac{ \pi \ × \ 10}{3} \ × \ \frac{700}{3} \)

Volume of the wire of length l and diameter D

\(= \ \pi ( \frac{D}{2})^2 \ × \ l \ \)
\( = \ \frac{ \pi D^2}{4} \ × \ l \)
[ \( V \ = \ \pi r2h \) ]

\(\therefore \) Volume of the frustum = Volume of the wire drawn from it

\(\Rightarrow \ \frac{ \pi \ × \ 10}{3} \ × \ \frac{700}{3} \ \)
\( = \ \frac{ \pi D^2}{4} \ × \ l \)
[ \(\because \) D=116]

\(l \ = \ \frac{10 \ × \ 700 \ × \ 4}{3 \ × \ 3 \ × \ 16 \ × \ 16} \ \)
\( = \ 7964.44 \) m

Q1 ) A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³ .

Answer :

Given that,

Diameter of cylinder = 10 cm

So, radius of the cylinder (r) \( \frac{10}{2} \) cm
= 5 cm

Length of wire in completely one round
\( = \ 2 \pi r \ = \ 2 × \ 3.14 \ × \ 5 \ = \ 31.4 \) cm

It is given that diameter of wire = 3 mm \( = \ \frac{3}{10} \) cm

The thickness of cylinder covered in one round = \( \frac{3}{10} \) m

Hence, the number of turns(rounds) of the wire to cover 12 cm will be \( = \ \frac{12}{ \frac{3}{10}} \ = \ 12 \ × \ \frac{10}{3} = 40 \)

Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds

\(\Rightarrow 40 \ × \ 31.4 \ = \ 1256 \) cm

Radius of the wire \( = \ \frac{0.3}{2} \ = \ 0.15 \) cm

Volume of wire = Area of cross-section of wire × Length of wire

\( = \ \pi(0.15)^2 \ × \ 1257.14 \ = \ 88.898 \) cm³

We know,

Mass = Volume × Density

\( = \ 88.898 \ × \ 8.88 \ = \ 789.41 \) gm

Q2 ) A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of \( \pi \) as found appropriate.)

Answer :

Let us consider the ∆ ABC

Here,

AB = 3 cm, AC = 4 cm

So, Hypotenuse BC = 5 cm

We have got 2 cones on the same base AA’ where the radius = DA or DA’

Now, \( \frac{AD}{CA} = \frac{AB}{CB} \)

By putting the value of CA, AB and CB we get,

AD = \( \frac{12}{5} \) cm

We also know,

\( \frac{DB}{AB} = \frac{AB}{CB} \)

So, DB = \( \frac{9}{5} \) cm

As, CD = BC - DB,

CD = \( \frac{16}{5} \) cm

Now, volume of double cone will be

\( [ \frac{1}{3} \pi \ × \ ( \frac{12}{5})^2 \frac{9}{5} \ + \ \frac{1}{3} \pi \ × \ ( \frac{12}{5})^2 \ × \ \frac{16}{5} ] \) cm³

Solving this we get,

V = 30.14 cm³

The surface area of the double cone wil be

\( ( \pi \ × \ \frac{12}{5} \ × \ 3) \ + \ ( \pi \ × \ \frac{12}{5} \ × \ 4) \) \( = \ \pi \ × \ \frac{12}{5}[3 \ + \ 4] \)

\( = \ 52.75 \) cm³

Q3 ) A cistern, internally measuring 150 cm × 120 cm × 100 cm, has 129600 \(cm^3\) of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without ocerflowing the water, each being 22.5 cm × 7.5 cm × 6.5 cm?

Answer :

Given that the dimension of the cistern = 150 × 120 × 110

So, volume = 1980000 cm³

Volume to be filled in cistern = 1980000 - 129600

= 1850400 cm³

Now, let the number of bricks placed be “n”

So, volume of n bricks will be = \( n × 22.5 × 7.5 × 6.5\)

Now as each brick absorbs one-seventeenth of its volume, the volume will be

\(= \ \frac{n}{17} \ × \ 22.5 \ × \ 7.5 \ × \ 6.5 \)

For the condition given in the question,

The volume of n bricks has to be equal to volume absorbed by n bricks + Volume to be filled in cistern

\(\Rightarrow n × 22.5 × 7.5 × 6.5 = 1850400 + \frac{n}{17} × (22.5 × 7.5 × 6.5) \)

Solving this we get,

n = 1792.41

Therefore, 1792 bricks were placed in the cistern.

Q4 ) In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 \(km^2\), show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer :

From the question, it is clear that

Total volume of 3 rivers
= 3 × [(Surface area of a river) × Depth]

Given,

Surface area of a river
= [\( 1072 × \frac{75}{1000} \)] km

And,

Depth = \( \frac{3}{1000} \) km

Now, volume of 3 rivers
= 3 × [1072 × \( \frac{75}{1000} \)] × \( \frac{3}{1000} \)

= 0.72 km³

Now, volume of rainfall in one fortnight (14 days)
= total surface area × total height of rain

\( = \ 9780 \ × \ \frac{10}{100 \ × \ 1000} \)

= 9.7 km³

Volume of rainfall in 1 day
= \( \frac{9.7}{14} = 0.694 \ km^3 \)

For the total rainfall to be approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.

Since 0.694 km³ is nearly equivalent to 0.72 km³,

We can say that the total rainfall was approximately equivalent to the addition to the normal water of the three rivers.

Q5 ) An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig.).

Answer :

Given,

Diameter of upper circular end of frustum part = 18 cm

So, radius (r₁) = 9 cm

Now, the radius of the lower circular end of frustum (r₂) will be equal to the radius of the circular end of the cylinder

So, r₂ = \(\frac{8}{2} \) = 4 cm

Now, height (h₁) of the frustum section = 22 - 10 = 12 cm

And,

Height (h₂) of cylindrical section = 10 cm (given)

Now, the slant height will be, \( l \ = \ \sqrt{(r_1 \ - \ r_2)^2 \ + \ (h_1)^2} \)

Or, l = 13 cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

\( = \ (r_1 \ + \ r_2)l \ + \ 2 \pi r_2 h_2 \)

Solving this we get,

Area of tin sheet required = 782 × \( \frac{4}{7} \) cm²

Q6 ) Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Answer :

Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r₁ and r₂ are the radii of the frustum ends of the cone and h be the frustum height.

Now, consider the \(\triangle \) ABG \(\triangle \) ADF,

Here, DF||BG

So, \( \triangle \ ABG \ \sim \ \triangle \ ADF \)

\( \Rightarrow \frac{DF}{BG} \ = \ \frac{AF}{AG} \ = \ \frac{AD}{AB} \)

\( \Rightarrow \frac{r_2}{r_1} \ = \ \frac{h_1 \ - \ h}{h_1} \ = \ \frac{l_1 \ - \ l}{l_1} \)

\(\Rightarrow \ l_1 \ = \ \frac{r_1 l}{r_1 \ - \ r_2} \)

The total surface area of frustum will be equal to the total CSA ot frustum + the area ol upper circular end + area of the lower circular end

\( = \pi (r_1 \ + \ r_2)l \ + \ \pi (r_2)^2 \ + \ \pi (r_1)^2 \)

\(\therefore \) Surface area of frustum \(= \ \pi [r_1 \ + \ r_2)l \ + \ (r_1)^2 \ + \ (r_2)^2] \)

Q7 ) Derive the formula for the volume of the frustum of a cone.

Answer :

Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r₁ and r₂ are the radii of the frustum ends of the cone and h be the frustum height.

Now, consider the ∆ ABG ∆ ADF,

Here, DF||BG

So, \( ∆ \ ABG \ \sim \ ∆ \ ADF \)

\( \Rightarrow \frac{DF}{BG} \ = \ \frac{AF}{AG} \ = \ \frac{AD}{AB} \)

\( \Rightarrow \frac{r_2}{r_1} \ = \ \frac{h_1 \ - \ h}{h_1} \ = \ \frac{l_1 \ - \ l}{l_1} \)

\( \Rightarrow \frac{r_2}{r_1} \ = \ 1 \ - \ \frac{h}{h_1} \ = \ 1 \ - \ \frac{l}{l_1} \)

\( \Rightarrow 1 \ - \ \frac{h}{h_1} \ = \ \frac{r_2}{r_1} \)

\( \Rightarrow \ \frac{h}{h_1} \ = \ \frac{r_1 \ - \ r}{r_1} \)

\( \Rightarrow h_1 \ = \ \frac{r_1 h}{r_1 \ - \ r_2} \)

The total volume of frustum of the cone will be = Volume of cone ABC – Volume of cone ADE

\( = \ \frac{1}{3} \pi (r_1)^2 h_1 \ – \ \frac{1}{3} \pi (r_2)^2 (h_1 \ – \ h) \)

\(= \ ( \frac{ \pi}{3})[(r_1)^2 h_1 \ - \ (r_2)^2 (h_1 \ - \ h)] \)

\(= \ ( \frac{ \pi}{3})[ ( \frac{(r_1)^3 h}{r_1 \ - \ r_2} ) \ - \ (r_2)^2 ( \frac{r_1 h}{r_1 \ - \ r_2} \ - \ h)] \)

\(= \ ( \frac{ \pi}{3})[ ( \frac{(r_1)^3 h}{r_1 \ - \ r_2} ) \ - \ (r_2)^2 \ ( \frac{hr_1 \ - \ hr_1 \ + \ hr_2}{r_1 \ - \ r_2} ) ] \)

\( = \ \frac{ \pi }{3} h [ \frac{(r_1)^3 \ - \ (r_2)^3}{r_1 \ - \ r_2} ] \)

\( = \ \frac{ \pi }{3} h [ \frac{(r_1 \ - \ r_2)(r_1 ^2 \ + \ r_2^2 \ + \ r_1 r_2}{r_1 \ - \ r_2} ] \)

\(\therefore \) Volume of frustum of the cone \( = \ \frac{1}{3} \pi h \ [r_1^2 \ + \ r_2^2 \ + \ r_1 r_2 \ ] \)