Rs Aggarwal 2018 for Class 10 Math Chapter 19 - Volume And Surface Area Of Solids

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Page No 874:

Question 1:

Two cubes each of volume 27 cm³ are joined end to end to form a solid. Find the surface area of the resulting cuboid. [CBSE 2011, 14]

Answer:

$As, volume of a cube = 27 {cm}^{3} \Rightarrow {(edge)}^{3} = 27 \Rightarrow edge = \sqrt[3]{27} \Rightarrow edge = 3 cm The length of the resulting cuboid, l = 3 + 3 = 6 cm, its breadth, b = 3 cm and its height, h = 3 cm Now, the surface area of the resulting cuboid = 2 (l b + b h + h l) = 2 (6 \times 3 + 3 \times 3 + 3 \times 6) = 2 (18 + 9 + 18) = 2 \times 45 = 90 {cm}^{2}$

Question 25:

A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of $\frac{3}{2}$ cm and its depth is $\frac{8}{9}$ cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape. [CBSE 2015]

Answer:

$We have, the base radius of the cylinder, R = 3 cm, the height of the cylinder, H = 5 cm, the base radius of the conical hole, r = \frac{3}{2} cm and the height of the conical hole, h = \frac{8}{9} cm Now, Volume of the cylinder, V = π R^{2} H = π \times 3^{2} \times 5 = 45 π {cm}^{3} Also, Volume of the cone removed from the cylinder, v = \frac{1}{3} π r^{2} h = \frac{π}{3} \times {(\frac{3}{2})}^{2} \times (\frac{8}{9}) = \frac{2 π}{3} {cm}^{3} So, the volume of metal left in the cylinder, V' = V - v = 45 π - \frac{2 π}{3} = \frac{133 π}{3} {cm}^{3} ∴ The required ratio = \frac{V'}{v} = \frac{(\frac{133 π}{3})}{(\frac{2 π}{3})} = \frac{133}{2} = 133 : 2$

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2.

Page No 876:

Question 26:

A spherical glass vessel has a cylindrical neck that is 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water the vessel can hold.

Answer:

Diameter of spherical part = 21 cm
Radius of the spherical part = 10.5 cm

Volume of spherical part of the vessel $= \frac{4}{3} {πr}^{3} = \frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 = 4851 {cm}^{3}$
Diameter of cylinder = 4 cm
Radius of cylinder = 2 cm
Height of cylinder = 7 cm

Volume of the cylindrical part of the vessel $= π r^{2} h = \frac{22}{7} \times 2 \times 2 \times 7 = 88 {cm}^{3}$

Total volume of the vessel $= 4851 + 88 = 4939 c m^{3}$

Page No 876:

Question 27:

The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other. Find the volume of the solid.

Answer:

Page No 877:

Question 31:

The inner diameter of a glass is 7 cm and it has a raised portion in the bottom in the shape of a hemisphere, as shown in the
figure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass.

Answer:

$We have, the height of the glass, h = 16 cm and the base radius of cylinder = the base radius of hemisphere, r = \frac{7}{2} cm Now, The apparent capacity of the glass = Volume of the cylinder = π r^{2} h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 16 = 616 {cm}^{3} Also, The actual capacity of the glass = Volume of cylinder - Volume of hemisphere = 616 - \frac{2}{3} π r^{3} = 616 - \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} = 616 - \frac{539}{6} = \frac{3157}{6} {cm}^{3} \approx 526.17 {cm}^{3}$

Page No 877:

Question 32:

A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure. The total height of the toy is 26 cm, while the height of the conical part is 6 cm. The diameter of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm. The conical part and the cylindrical part are respectively painted red and white. Find the area to be painted by each of these colours. [Take $π = \frac{22}{7}$ ]

Answer:

$We have, the base radius of the conical part, r = \frac{5}{2} = 2.5 cm, the base radius of the cylindrical part, R = \frac{4}{2} = 2 cm, the total height of the toy = 26 cm, the height of the conical part, h = 6 cm Also, the height of the cylindrical part, H = 26 - 6 = 20 cm And, the slant height of the conical part, l = \sqrt{r^{2} + h^{2}} = \sqrt{2 . 5^{2} + 6^{2}} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5 cm Now, The area to be painted by red colour = CSA of cone + Area of base of conical part - Area of base of cylindrical part = π r l + π r^{2} - π R^{2} = \frac{22}{7} \times 2.5 \times 6.5 + \frac{22}{7} \times 2.5 \times 2.5 - \frac{22}{7} \times 2 \times 2 = \frac{22}{7} \times 16.25 + \frac{22}{7} \times 6.25 - \frac{22}{7} \times 4 = \frac{22}{7} \times (16.25 + 6.25 - 4) = \frac{22}{7} \times 18.5 \approx 58.14 {cm}^{2} Also, The area to be painted by white colour = CSA of cylinder + Area of base of cylinder = 2 π R H + π R^{2} = π R (2 H + R) = \frac{22}{7} \times 2 \times (2 \times 20 + 2) = \frac{22}{7} \times 2 \times 42 = 264 {cm}^{2}$

Page No 912:

Question 20:

A fez, the cap used by the Turks, is shaped like the frustum of a cone. 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, then find the area of material used for making it. $[Use π = \frac{22}{7}]$

Answer:

$We have, Radius of open side, R = 10 cm, Radius of upper base, r = 4 cm and Slant height, l = 15 cm Now, The area of material used = π (R + r) l + π r^{2} = \frac{22}{7} \times (10 + 4) \times 15 + \frac{22}{7} \times 4 \times 4 = \frac{22}{7} \times 14 \times 15 + \frac{22}{7} \times 4 \times 4 = \frac{22}{7} \times (14 \times 15 + 4 \times 4) = \frac{22}{7} \times (210 + 16) = \frac{22}{7} \times 226 = \frac{4972}{7} {cm}^{2} \approx 710.28 {cm}^{2}$

So, the area of material used for making the fez is 710.28 cm².

Page No 912:

Question 21:

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, then find the area of the tin sheet required to make the funnel.

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 ₹25 per metre. [CBSE 2014]

Answer:

$We have, Width of the cloth, B = 5 m, Radius of the conical tent, r = \frac{14}{2} = 7 m and Height of the conical tent, h = 24 m Let the length of the cloth used for making the tent be L . Also, The slant height of the conical tent, l = \sqrt{r^{2} + h^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25 m Now, The curved surface of the conical tent = π r l = \frac{22}{7} \times 7 \times 25 \Rightarrow The area of the cloth used for making the tent = 550 m^{2} \Rightarrow L B = 550 \Rightarrow L = \frac{550}{B} \Rightarrow L = \frac{550}{5} \Rightarrow L = 110 m So, the cost of the cloth used = 25 \times 110 = ₹ 2750$

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

Page No 916:

Question 29:

A wooden toy was made by scooping out a hemisphere of 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, then find the volume of wood in the toy. [CBSE 2013]

Answer:

$We have, Radius of the cylinder = Radius of the hemispher = r = 3.5 cm and Height of the cylinder, h = 10 cm Now, Volume of the toy = Volume of the cylinder - Volume of the two hemispheres = π r^{2} h - 2 \times \frac{2}{3} π r^{3} = π r^{2} (h - \frac{4 r}{3}) = \frac{22}{7} \times 3.5 \times 3.5 \times (10 - \frac{4 \times 3.5}{3}) = 38.5 \times (10 - \frac{14}{3}) = 38.5 \times \frac{16}{3} = \frac{616}{3} {cm}^{3} \approx 205.33 {cm}^{3}$

So, the volume of wood in the toy is $\frac{616}{3}$ cm³ or 205.33 cm³.

Page No 916:

Question 34:

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. [CBSE 2014]

Answer:

$We have, Height of the cone, h = 84 cm and Base radius of the cone, r = 21 cm Let the radius of the solid sphere be R . Now, Volume of the solid sphere = Volume of the solid cone \Rightarrow \frac{4}{3} π R^{3} = \frac{1}{3} π r^{2} h \Rightarrow R^{3} = \frac{r^{2} h}{4} \Rightarrow R^{3} = \frac{21 \times 21 \times 84}{4} \Rightarrow R^{3} = 21 \times 21 \times 21 \Rightarrow R = 21 cm ∴ Diameter = 2 R = 2 \times 21 = 42 cm$

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

Page No 916:

Question 35:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total
surface area of the toy. [CBSE 2012]

Answer:

$We have, Radius of the hemisphere = Radius of the cone = r = 3.5 cm and Height of the cone = 15.5 - 3.5 = 12 cm Also, The slant height of the cone, l = \sqrt{h^{2} + r^{2}} = \sqrt{12^{2} + 3 . 5^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 cm Now, Total surface area of the toy = CSA of cone + CSA of hemisphere = π r l + 2 π r^{2} = π r (l + 2 r) = \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5 {cm}^{2}$

So, the total surface area of the toy is 214.5 cm².

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Page No 917:

Question 2:

A shuttlecock used for playing badminton is a combination of

(a) cylinder and a hemisphere
(b) frustum of a cone and a hemisphere
(c) a cone and a hemisphere
(d) a cylinder and a sphere

Answer:

(b) frustum of a cone and a hemisphere
A shuttlecock used for playing badminton is a combination of frustum of a cone and a hemisphere.

Page No 918:

Question 3:

A funnel is a combination of

(a) a cylinder and a cone
(b) a cylinder and a hemisphere
(c) a cylinder and frustum of a cone
(d) a cone and hemisphere

Answer:

Page No 918:

Question 9:

During conversion of a solid from one shape to another, the volume of the new shape will
(a) decrease
(b) increase
(c) remain unaltered
(d) be doubled

Answer:

(c) remain unaltered
During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

Page No 919:

Question 10:

In a right circular cone, the cross-section made by a plane parallel to the base is a
(a) sphere
(b) hemisphere
(c) circle
(d) a semicircle

Page No 925:

Question 74:

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 925:

Question 75:

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:

$We have, Height of the cylinder = Height of the cone = h = 15 cm and Radius of the cylinder = Radius of the cone = r = \frac{16}{2} = 8 cm Also, the slant height of the cone, l = \sqrt{h^{2} + r^{2}} = \sqrt{15^{2} + 8^{2}} = \sqrt{225 + 64} = \sqrt{289} = 17 cm Now, The total surface area of the remaining solid = CSA of the cone + CSA of the cylinder + Area of the base = π r l + 2 π r h + π r^{2} = π r (l + 2 h + r) = 3.14 \times 8 \times (17 + 2 \times 15 + 8) = 3.14 \times 8 \times 55 = 1381.6 {cm}^{2}$

So, the total surface area of the remaining solid is 1381.6 cm².

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Page No 938:

Question 18:

A solid rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness
5 cm. Find the length of the pipe.

Answer:

$We have, Length of the rectangular block, l = 4.4 m, Breadth of the rectangular block, b = 2.6 m, Height of the rectangular block, h = 1 m, Internal radius of the cylindrical pipe, r = 30 cm = 0.3 m and Thickness of the pipe = 5 cm = 0.05 m Also, the external radius of the pipe = 0.3 + 0.05 = 0.35 m Let the length of the pipe be H . Now, Volume of the pipe = Volume of the block \Rightarrow π R^{2} H - π r^{2} H = l b h \Rightarrow π (R^{2} - r^{2}) H = l b h \Rightarrow \frac{22}{7} \times (0 . 35^{2} - 0 . 3^{2}) H = 4.4 \times 2.6 \times 1 \Rightarrow \frac{22}{7} \times (0.1225 - 0.09) H = 4.4 \times 2.6 \Rightarrow \frac{22}{7} \times 0.0325 \times H = 4.4 \times 2.6 \Rightarrow H = \frac{4.4 \times 2.6 \times 7}{22 \times 0.0325} ∴ H = 112 m$

So, the length of the pipe is 112 m.

Page No 938:

Question 19:

An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres.

Answer:

$We have, Radius of the upper end of the frustum, R = \frac{45}{2} cm, Radius of the lower end of the frustum = Radius of the cylinder = r = \frac{25}{2} cm, Height of the cylinder, h = 6 cm and Total height of the bucket = 40 cm And, the height of the frustum, H = 40 - 6 = 34 cm Also, the slant height of frustum, l = \sqrt{{(R - r)}^{2} + H^{2}} = \sqrt{{(\frac{45}{2} - \frac{25}{2})}^{2} + 34^{2}} = \sqrt{10^{2} + 34^{2}} = \sqrt{100 + 1156} = \sqrt{1256} \approx 35.44 cm Now, The area of the metallic sheet used to make the bucket = CSA of the frustum + CSA of the cylinder + Area of the base of the cylinder = π (R + r) l + 2 πrh + {πr}^{2} = \frac{22}{7} \times (\frac{45}{2} + \frac{25}{2}) \times 35.44 + 2 \times \frac{22}{7} \times \frac{25}{2} \times 6 + \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2} = \frac{22}{7} \times 35 \times 35.44 + \frac{22}{7} \times 150 + \frac{22}{7} \times \frac{625}{4} = \frac{22}{7} \times (1240.4 + 150 + 156.25) = \frac{22}{7} \times 1546.65 = 4860.9 {cm}^{2} Also, The volume of the water that the bucket can hold = Volume of the frustum = \frac{1}{3} π H (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 34 \times [{(\frac{45}{2})}^{2} + {(\frac{25}{2})}^{2} + (\frac{45}{2}) (\frac{25}{2})] = \frac{1}{3} \times \frac{22}{7} \times 34 \times [{(\frac{45}{2})}^{2} + {(\frac{25}{2})}^{2} + (\frac{45}{2}) (\frac{25}{2})] = \frac{748}{21} \times (\frac{2025}{4} + \frac{625}{4} + \frac{1125}{4}) = \frac{748}{21} \times \frac{3775}{4} = 33615.48 {cm}^{3} = 33.61548 L (As, 1000 {cm}^{3} = 1 L) \approx 33.61 L$

Page No 938:

Question 20:

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km/hr, then in how much time will the tank be filled completely?

Answer:

$We have, Internal radius of the pipe, r = \frac{20}{2} = 10 cm = 0.1 m, Radius of the cylindrical tank, R = \frac{10}{2} = 5 m and Height of the cylindrical tank, H = 2 m Also, the speed of the water flow in the pipe, h = 4 km / hr = \frac{4 \times 1000 m}{1 hr} = 4000 m / hr Now, The volume of the water flowing out of the pipe in a hour = π r^{2} h = \frac{22}{7} \times 0.1 \times 0.1 \times 4000 = \frac{880}{7} m^{3} And, The volume of the cylindrical tank = π R^{2} H = \frac{22}{7} \times 5 \times 5 \times 2 = \frac{1100}{7} m^{3} So, The time taken to fill the tank = \frac{Volume of the cylindrical tank}{Volume of water flowing out of the pipe in a hour} = \frac{(\frac{1100}{7})}{(\frac{880}{7})} = \frac{1100}{880} = \frac{5}{4} hr = 1 \frac{1}{4} hr = 1 hr and \frac{1}{4} \times 60 \min = 1 hr 15 \min$

So, the tank will be completely filled in 1 hour 15 minutes.

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