Rs Aggarwal 2015 for Class 10 Math Chapter 19 - Volume And Surface Area Of Solids

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Page No 755:

Question 1:

A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre.

Answer:

Radius of the cylinder = 14 m
Radius of the base of the cone = 14 m
Height of the cylinder (h) = 3 m
Total height of the tent = 13.5 m
Surface area of the cylinder $= 2 π r h = (2 \times \frac{22}{7} \times 14 \times 3) m^{2} = 264 m^{2}$

Height of the cone $= Total height - Height of cone = (13.5 - 3) m = 10.5 m$

Surface area of the cone = $π r \sqrt{r^{2} + h^{2}}$

$= π r \sqrt{r^{2} + h^{2}} = (\frac{22}{7} \times 14 \times \sqrt{14^{2} + 10 . 5^{2}}) m^{2} = (44 \times \sqrt{196 + 110.25}) m^{2} = (44 \times \sqrt{306.25}) m^{2} = (44 \times 17.5) m^{2} = 770 m^{2}$

Total surface area $= (264 + 770) m^{2} = 1034 m^{2}$

∴ Cost of cloth $= Rs 1034 \times 80 = Rs 82720$

Match the following columns:

Column I	Column II
(a) A solid metallic sphere of radius 8 cm is melted and the material is used to make solid right cones with height 4 cm and base radius of 8 cm. How many cones are formed?	(p) 18
(b) A 20-m-deep well with diameter 14 m is dug up and the earth from digging is evenly spread out to form a platform 44 m by 14 m. The height of the platform is ...........m.	(q) 8
(c) A sphere of radius 6 cm is melted and recast in the shape of a cylinder of radius 4 cm. Then, the height of the cylinder is ......... cm.	(r) 16 : 9
(d) The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is ....... .	(s) 5

Answer:

(a)
Volume of the sphere $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3} π \times {(8)}^{3}) {cm}^{3}$

Volume of each cone $= \frac{1}{3} π r^{2} h$
$= \frac{1}{3} π \times {(8)}^{2} \times 4 {cm}^{3}$

Number of cones formed $= \frac{Volume of the sphere}{Volume of each cone}$

$= \frac{4 π \times 8 \times 8 \times 8 \times 3}{3 \times π \times 8 \times 8 \times 4} = 8$
Hence, $(a) \Rightarrow (q)$

(b)
Volume of the earth dug out = Volume of the cylinder
$= π r^{2} h = \frac{22}{7} \times 7 \times 7 \times 20$
Let the height of the platform be h.
Then, volume of the platform = volume of the cuboid
$= (44 \times 14 \times h) m^{3}$
Therefore,

$\frac{22}{7} \times 7 \times 7 \times 20 = 44 \times 14 \times h \Rightarrow 3080 = 616 \times h \Rightarrow h = \frac{3080}{616} \Rightarrow h = 5 m$

Hence, $(b) \Rightarrow (s)$

(c)
Volume of the sphere
$= \frac{4}{3} π r^{3}$

$= \frac{4}{3} π \times 6 \times 6 \times 6$

Let h be the height of the cylinder.
Then, volume of the cylinder $= {πr}^{2} h$
$= π \times 4 \times 4 \times h$
Therefore,

$\frac{4}{3} π \times 6 \times 6 \times 6 = π \times 4 \times 4 \times h \Rightarrow \frac{4}{3} \times 6 \times 6 \times 6 = 4 \times 4 \times h \Rightarrow 228 = 16 \times h \Rightarrow h = \frac{228}{16} \Rightarrow h = 18 cm$

Hence, $(c) \Rightarrow (p)$

(d)
Let the radii of the spheres be R and r respectively.
Then, ratio of their volumes $= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}}$
Therefore,
$\frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}} = \frac{64}{27} \Rightarrow \frac{R^{3}}{r^{3}} = \frac{64}{27} \Rightarrow {(\frac{R}{r})}^{3} = {(\frac{4}{3})}^{3} \Rightarrow \frac{R}{r} = \frac{4}{3}$
Hence, the ratio of their surface areas $= \frac{4 π R^{2}}{4 π r^{2}}$
$= \frac{R^{2}}{r^{2}} = {(\frac{R}{r})}^{2} = {(\frac{4}{3})}^{2} = \frac{16}{9} = 16 : 9$
Hence, $(d) \Rightarrow (r)$

Page No 789:

Question 62:

Match the following columns:

Column I	Column II
(a) The radii of the circular ends of a bucket, in the form of the frustum of a cone of height 30 cm, are 20 cm and 10 cm respectively. The capacity of the bucket is ........cm³.	(p) 2418π
(b) The radii of the circular ends of a conical bucket of height 15 cm are 20 and 12 cm respectively. The slant height of the bucket is ........ cm.	(q) 22000
(c) The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. The total surface area of the bucket is .........cm².	(r) 12
(d) Three solid metallic spheres of radii 3 cm, 4 cm and 5 cm are melted to form a single solid sphere. The diameter of the resulting sphere is ........ cm.	(s) 17

Answer:

Radius of cylinder = $\frac{12}{2} = 6 cm$
Height of cylinder = 15 cm
Volume of cylinder $= π r^{2} h$
   $= (\frac{22}{7} \times 6 \times 6 \times 15) {cm}^{3} [d = 12 cm, r \Rightarrow \frac{12}{2} cm = 6 cm] = \frac{11880}{7} {cm}^{3}$

Radius of the hemisphere = 3 cm

Height of the cone = Total height of the toy - Radius of the hemisphere

                              $= (12 - 3) cm = 9 cm$

Volume of each toy = (Volume of the hemisphere) + (Volume of the cone)
$= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} (2 r + h) = \frac{1}{3} \times \frac{22}{7} \times {(3)}^{2} \times [(2 \times 3) + 9] {cm}^{3} = (\frac{1}{3} \times \frac{22}{7} \times 9 \times 15) {cm}^{3} = \frac{990}{7} {cm}^{3}$

Number of toys $= \frac{Volume of the cylinder}{Volume of each toy}$

   $= (\frac{11880 \times 7}{7 \times 990}) = 12$

Page No 801:

Question 20:

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

Answer:

Let the radii of the two cylinders be 2r and 3r and their heights be 5h and 3h respectively.

Then, ratio of their volumes $= \frac{π \times 4 r^{2} \times 5 h}{π \times 9 r^{2} \times 3 h}$

$= \frac{20}{27} = 20 : 27$

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