Surface Areas and Volumes

Class 9 NCERT Maths

NCERT

1   A plastic box $1.5 m$ long, $1.25 m$ wide and $65 cm$ deep, is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:$\\$ (i) The area of the sheet required for making the box.$\\$ (ii) The cost of sheet for it, if a sheet measuring $1 m^2$ costs Rs. $20.$

Solution :

It is given that, length (l) of box $= 1.5 m$$\\$ Breadth (b) of box $ = 1.25 m$$\\$ Depth (h) of box $= 0.65 m$$\\$ (i) Box is to be open at top.$\\$ Area of sheet required$\\$ $= 2lh + 2bh + lb$$\\$ $= [2 * 1.5 * 0.65 + 2 * 1.25 * 0.65 + 1.5 * 1.25] m^2$$\\$ $= (1.95 + 1.625 + 1.875) m^2 = 5.45 m^2$$\\$ (ii) Cost of sheet per $m^2$ area = Rs. $20$$\\$ Cost of sheet of $5.45 m^2$ area = Rs. $(5.45 * 20)$ = Rs.$ 109$

2   The length, breadth and height of a room are $5 m, 4 m$ and $3 m$ respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. $7.50$ per $m^2 .$

Solution :

It is given that$\\$ Length (l) of room $= 5 m$$\\$ Breadth (b) of room $= 4 m$$\\$ Height (h) of room $= 3 m$$\\$ It can be observed that four walls and the ceiling of the room are to be whitewashed. The floor of the room is not to be white-washed.$\\$ Area to be white-washed = Area of walls + Area of ceiling of room$\\$ $= 2lh + 2bh + lb$$\\$ $= [2 * 5 * 3 + 2 * 4 * 3 + 5 * 4] m^2$$\\$ $= (30 + 24 + 20) m^2$$\\$ $= 74 m^2$$\\$ Cost of white-washing per $m^2$ area = Rs. $7.50$$\\$ Cost of white-washing $74 m^2$ area = Rs. $(74 × 7.50)$ = Rs.$ 555$

3   The floor of a rectangular hall has a perimeter $250 m$. If the cost of panting the four walls at the rate of Rs.$10$ per $m^2$ is Rs. $15000,$ find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]

Solution :

Let length, breadth, and height of the rectangular hall be $l$ m, $b$ m, and $h$ m respectively.$\\$ Area of four walls $= 2lh + 2bh = 2(l + b) h$$\\$ Perimeter of the floor of hall $= 2(l + b) = 250 m$$\\$ $\therefore $ Area of four walls $= 2(l + b) h = 250h m^2$$\\$ Cost of painting per $m^2$ area = Rs. $10$$\\$ Cost of painting $250h m^2$ area = Rs. $(250h × 10)$ = Rs. $2500h$$\\$ However, it is given that the cost of paining the walls is Rs. $15000.$$\\$ $\therefore 15000 = 2500h$$\\$ $h = 6$$\\$ Therefore, the height of the hall is $6 m.$$\\$

4   The paint in a certain container is sufficient to paint an area equal to $9.375 m^2$ . How many bricks of dimensions $22.5 cm *10 cm * 7.5 cm $ can be painted out of this container?

Solution :

Total surface area of one brick $= 2(lb + bh + lh)$$\\$ $= [2(22.5 *10 + 10 * 7.5 + 22.5 * 7.5)] cm^2$$\\$ $= 2(225 + 75 + 168.75) cm^2$$\\$ $= (2 * 468.75) cm^2$$\\$ $= 937.5 cm^2$$\\$ Let $n$ bricks can be painted out by the paint of the container.$\\$ Area of $n$ bricks = $(n *937.5) cm^2 = 937.5n cm^2$$\\$ Area that can be painted by the paint of the container$ = 9.375 m^2 = 93750 cm^2$$\\$ $\therefore 93750 = 937.5n$$\\$ $n = 100$$\\$ Therefore, $100$ bricks can be painted out by the paint of the container.

5   A cubical box has each edge $10 cm$ and another cuboidal box is $12.5 cm$ long, $10 cm$ wide and $8 cm$ high.$\\$ (i) Which box has the greater lateral surface area and by how much?$\\$ (ii) Which box has the smaller total surface area and by how much?

Solution :

(i) Edge of cube $= 10 cm$$\\$ Length (l) of box $= 12.5 cm$$\\$ Breadth (b) of box $= 10 cm$$\\$ Height (h) of box $= 8 cm$$\\$ Lateral surface area of cubical box $= 4(\text{edge}) 2$$\\$ $= 4(10 cm)^2$$\\$ $= 400 cm^2$$\\$ Lateral surface area of cuboidal box $= 2[lh + bh]$$\\$ $= [2(12.5 * 8 + 10 * 8)] cm^2$$\\$ $= (2 * 180) cm^2$$\\$ $= 360 cm^2$$\\$ Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.$\\$ Lateral surface area of cubical box - Lateral surface area of cuboidal box $\\$$= 400 cm^2 - 360 cm^2\\ = 40 cm^2 .$$\\$ Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by $40 cm^2 .$$\\$ (ii) Total surface area of cubical box = $6(\text{edge}) 2 = 6(10 cm)^ 2 = 600 cm^2$$\\$ Total surface area of cuboidal box$\\$ $= 2[lh + bh + lb]$$\\$ $= [2(12.5 * 8 + 10 * 8 + 12.5 * 100] cm^2$$\\$ $= 610 cm^2$$\\$ Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.$\\$ Total surface area of cuboidal box - Total surface area of cubical box $\\$ $= 610 cm^2 -600 cm^2 = 10 cm^2$$\\$ Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by $10 cm^2 .$

6   A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is $30 cm$ long, $25 cm$ wide and $25 cm$ high.$\\$ (i) What is the area of the glass?$\\$ (ii) How much of tape is needed for all the $ 12$ edges?

Solution :

(i) Length (l) of green house = $30 cm$ $\\$ Breadth (b) of green house = $25 cm$ $\\$ Height (h) of green house =$ 25 cm $ $\\$ Total surface area of green house $\\$ $= 2[lb + lh + bh]$ $\\$ $= [2(30 × 25 + 30 × 25 + 25 × 25)] cm^2$ $\\$ $= [2(750 + 750 + 625)] cm ^2$ $\\$ $= (2 × 2125) cm ^2$ $\\$ $= 4250 cm^2$ $\\$ Therefore, the area of glass is $4250 cm ^2$ .$\\$ (ii)It can be observed that tape is required along side $AB, BC, CD, DA, EF, FG, GH, HE, AH, BE, DG,$ and $CF.$ $\\$ Total length of tape = $4(l + b + h)$ $\\$ $= [4(30 + 25 + 25)] cm$ $\\$ $= 320 cm$ $\\$ Therefore, $320 cm$ tape is required for all the $12$ edges.

7   Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions $25 cm × 20 cm × 5 cm $ and the smaller of dimensions $15 cm × 12 cm × 5 cm.$ For all the overlaps, $5\%$ of the total surface area is required extra. If the cost of the cardboard is Rs. $4$ for $1000 cm ^2$ , find the cost of cardboard required for supplying $250$ boxes of each kind.

Solution :

Length $(l _1 )$ of bigger box $= 25 cm$ $\\$ Breadth $(b _1 )$ of bigger box =$ 20 cm$ $\\$ Height $(h _1 )$ of bigger box =$ 5 cm$ $\\$ Total surface area of bigger box =$ 2(lb + lh + bh)$ $\\$ $= [2(25 × 20 + 25 × 5 + 20 × 5)] cm^2$ $\\$ $= [2(500 + 125 + 100)] cm ^2$ $\\$ $= 1450 cm ^2$ $\\$ Extra area required for overlapping =$(\dfrac{1450*5}{100})cm^2$ $\\$ $=72.5 cm^2$ $\\$ While considering all overlaps, total surface area of $1$ bigger box$\\$ $= (1450 + 72.5) cm ^2 =1522.5 cm ^2$ $\\$ Area of cardboard sheet required for $250$ such bigger boxes$\\$ $= (1522.5 × 250) cm ^2 = 380625 cm ^2$ $\\$ Similarly, total surface area of smaller box = $[2(15 ×12 + 15 × 5 + 12 × 5] cm ^2$ $\\$ $= [2(180 + 75 + 60)] cm ^2$ $\\$ $= (2 × 315) cm^ 2$ $\\$ $= 630 cm^ 2$

Therefore, extra area required for overlapping =$(\dfrac{630*5}{100})cm^2 =31.5 cm^2 $ $\\$ Total surface area of 1 smaller box while considering all overlaps$\\$ $= (630 + 31.5) cm ^2 = 661.5 cm^ 2$ $\\$ Area of cardboard sheet required for $250$ smaller boxes = $(250 × 661.5) cm ^2$ $\\$ $= 165375 cm ^2$ $\\$ Total cardboard sheet required = $(380625 + 165375) cm ^2$ $\\$ $= 546000 cm ^2$ $\\$ Cost of $1000 cm^2$ cardboard sheet = Rs.$ 4$$\\$ Cost of $546000 cm^2$ cardboard sheet = Rs.$(\dfrac{546000*4}{1000})=Rs.2184$ $\\$Therefore, the cost of cardboard sheet required for $250$ such boxes of each kind will be Rs. $2184.$

8   Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height $2.5 m$, with base dimensions $4 m × 3 m$ ?

Solution :

Length $(l)$ of shelter = $4 m$ $\\$ Breadth $(b)$ of shelter = $3 m$ $\\$ Height $(h)$ of shelter = $2.5 m$ $\\$ Tarpaulin will be required for the top and four wall sides of the shelter.$\\$ Area of Tarpaulin required = $2(lh + bh) + lb$ $\\$ $= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m^ 2$ $\\$ $= [2(10 + 7.5) + 12] m ^2$ $\\$ $= 47 m^ 2$ $\\$ Therefore, $47 m ^2$ tarpaulin will be required.

9   The curved surface area of a right circular cylinder of height $14 cm$ is $88 cm^ 2$ . Find the diameter of the base of the cylinder. [Assume $\pi =\dfrac{22}{7}]$

Solution :

Height $(h)$ of cylinder = $14 cm$ $\\$ Let the diameter of the cylinder be $d.$ $\\$ Curved surface area of cylinder = $88 cm ^2$ $\\$ $\Rightarrow 2\pi rh = 88 cm^ 2 (r$ is the radius of the base of the cylinder)$\\$ $\Rightarrow \pi dh = 88 cm ^2 (d = 2r)$ $\\$ $\Rightarrow [\dfrac{22}{7}*d*14]cm=88 cm^2$ $\\$ $\Rightarrow d=2 cm $ $\\$ Therefore, the diameter of the base of the cylinder is $ 2cm $

10   It is required to make a closed cylindrical tank of height $1 m$ and base diameter $140 cm$ from a metal sheet. How many square meters of the sheet are required for the same? [ Assume $\pi =\dfrac{22}{7}]$

Solution :

Height $(h)$ of cylindrical tank = $1 m$ $\\$ Base radius $(r)$ of cylindrical tank =$\dfrac{140}{2} cm = 70 cm = 0.7 m $ $\\$ Area of sheet required = Total surface area of tank = $2\pi r(r + h)$ $\\$ $=(2 * \dfrac{22}{7}*0.7 (0.7+1))m^2$ $\\$ $=(4.4*1.7)m^2$ $\\$ $=7.48 m^2$ $\\$ Therefore, it will require $7.48 m ^2$ area of sheet.

11   A metal pipe is $77 cm$ long. The inner diameter of a cross section is $4 cm$, the outer diameter being $4.4 cm.$ (i) Inner curved surface area,$\\$ (ii) Outer curved surface area,$\\$ (iii) Total surface area. [Assume $\pi = \dfrac{22}{7}]$

Solution :

Inner radius $(r1)$ of cylindrical pipe = $\dfrac{4}{2} cm = 2 cm$ $\\$ Outer radius $(r2)$ of cylindrical pipe =$\dfrac{ 4.4}{2} cm = 2.2 cm$ $\\$ Height $(h)$ of cylindrical pipe = Length of cylindrical pipe = $77 cm$$\\$ (i) $CSA$ of inner surface of pipe = $2\pi r1 h$$\\$ $=(2*\dfrac{22}{7}*2*77)cm^2$ $\\$ $=968 cm^2 $ $\\$ (ii) $CSA$ of outer surface of pipe = $2\pi r2 h$ $\\$ $=(2*\dfrac{22}{7}*2.2*77)cm^2$ $\\$ $=(22*22*2.2)cm^2$ $\\$ $=1064.8 cm^2$ $\\$ (iii) Total surface area of pipe = $CSA$ of inner surface + $CSA$ of outer surface + Area of both circular ends of pipe$\\$ $= 2\pi r_1h + 2\pi r_2h + 2\pi (r_2^2 - r_1^2) $ $\\$ $ = [968 + 1064.8 + 2\pi {(2.2)^2 - (2)^2}] cm^2 $ $\\$ $(2032.8+2*\dfrac{22}{7}*0.84) cm^2$ $\\$ $= (2032.8 + 5.28) cm^2$$\\$ $= 2038.08 cm^2$$\\$ Therefore, the total surface area of the cylindrical pipe is $2038.08 cm^2.$

12   The diameter of a roller is $84 cm $ and its length is $120 cm$. It takes $500$ complete revolutions to move once over to level a playground. Find the area of the playground in $m2$? [Assume $\pi=\dfrac{22}{7}]$

Solution :

It can be observed that a roller is cylindrical.$\\$ Height $(h) $ of cylindrical roller = Length of roller = $120 cm$$\\$ Radius $(r) $ of the circular end of roller =$ 84 cm = 42 cm ^2$ $\\$ $CSA$ of roller = $2\pi rh$ $\\$ $=(2*\dfrac{22}{7}*42*120)cm^2$ $\\$ $= 31680 cm^2$ $\\$ Area of field = $500 × CSA $ of roller = $(500 × 31680) cm^2$ $= 15840000 cm^2$$\\$ $= 1584 m^2$

13   A cylindrical pillar is $50 cm$ in diameter and $3.5 m $ in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.$12.50$ per $m^2.$ [Assume $\pi=\dfrac{22}{7}]$

Solution :

Height $(h)$ cylindrical pillar =$ 3.5 m$$\\$ Radius $(r)$ of the circular end of pillar =$\dfrac{50}{2} cm = 25 cm = 0.25 m$ $\\$ CSA of pillar = $2\pi rh $ $\\$ $=(2*\dfrac{22}{7}*0.25*3.5)m^2$ $\\$ $ = (44 * 0.125) m^2 = 5.5 m^2$ $\\$ Cost of painting $1 m^2$ area = Rs. $12.50$ $\\$ Cost of painting $5.5 m^2 $ area = Rs.$ (5.5 × 12.50) = Rs. 68.75$ $\\$ Therefore, the cost of painting the $CSA$ of the pillar is Rs.$ 68.75.$

14   Curved surface area of a right circular cylinder is $4.4 m^2$. If the radius of the base of the cylinder is $0.7 m$, find its height. [Assume $ \pi = \dfrac{22}{7}]$

Solution :

Let the height of the circular cylinder be $h.$ $\\$ Radius $(r)$ of the base of cylinder =$ 0.7 m CSA $of cylinder = $4.4 m^2$ $\\$ $2\pi rh = 4.4 m^2$ $\\$ $[2*\dfrac{22}{7}*0.7*h]=4.4 m^2$ $\\$ $h=1m$ $\\$ Therefore, the height of the cylinder is $1 m$

15   The inner diameter of a circular well is $3.5 m.$ It is $10 m$ deep. Find $\\$ (i) Its inner curved surface area,$\\$ (ii) The cost of plastering this curved surface at the rate of Rs. $40 per \ \ m^2.$ $\\$[Assume $\pi=\dfrac{ 22}{7}]$

Solution :

Inner radius $(r)$ of circular well = $(\dfrac{3.5}{2}) m = 1.75 m $ $\\$ Depth $(h)$ of circular well = $10 m$ $\\$ Inner curved surface area = $2\pi rh$ $\\$ $=(2*\dfrac{22}{7}*1.75*10)m^2$ $\\$ $ = (44 × 0.25 × 10) m^2 = 110 m^2$ $\\$ Therefore, the inner curved surface area of the circular well is $110 m^2$ .$\\$ Cost of plastering $1 m^2$ area = Rs. $40$$\\$ Cost of plastering $100 m^2 $ area = Rs.$ (110 × 40) $= Rs. $4400$ $\\$ Therefore, the cost of plastering the $CSA $ of this well is Rs.$ 4400.$

16   In a hot water heating system, there is a cylindrical pipe of length $28 m$ and diameter $5 cm.$ Find the total radiating surface in the system. $\\$ [Assume $\pi=\dfrac{ 22}{7}]$

Solution :

Height $(h)$ of cylindrical pipe = Length of cylindrical pipe = $28 m$$\\$ Radius $(r)$ of circular end of pipe = $52 = 2.5 cm = 0.025 m$$\\$ $CSA$ of cylindrical pipe = $2\pi rh $ $\\$ $=(2*\dfrac{22}{7}*0.025*28)m^2$ $\\$ $= 4.4 m^2$ $\\$ The area of the radiating surface of the system is $4.4 m^2.$

17   Find $\\$ (i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is $4.2 m$ in diameter and $4.5 m$ high.$\\$ (ii) How much steel was actually used, if $\dfrac{1}{12}$ of the steel actually used was wasted in making the tank. $\\$ [Assume $\pi =\dfrac{ 22}{7}]$

Solution :

Height $(h)$ of cylindrical tank = $4.5 m$ $\\$ Radius $(r)$ of the circular end of cylindrical tank = $(\dfrac{4.2}{2} m = 2.1 m$ $\\$ (i) Lateral or curved surface area of tank = $2\pi rh$ $=(2*\dfrac{22}{7}*2.1*4.5)m^2$ $\\$ $= (44 × 0.3 × 4.5) m^2$ $\\$ $= 59.4 m^2$ $\\$ Therefore, $CSA$ of tank is $59.4 m^2.$$\\$ (ii) Total surface area of tank = $2\pi r (r + h)$ $\\$ $=(2*\dfrac{22}{7}*2.18(2.1+4.5))m^2$ $\\$ $ = (44 × 0.3 × 6.6) m^2$ $\\$ $= 87.12 m^2$ $\\$ Let $A \ m^2$ steel sheet be actually used in making the tank.$\\$ $\therefore A(1-\dfrac{1}{12})=87.12 m^2$ $\\$ $\implies A=(\dfrac{12}{11}*87.12)m^2$ $\\$ $\implies A=95.04 m^2$ $\\$ Therefore , $95.04 m^2$ steel was used in actual while making such a tank.$\\$

18   In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of $20 cm$ and height of $30 cm.$ A margin of $2.5 cm$ is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. $ \\$ [Assume $\pi =\dfrac{22} {7}$ ]

Solution :

Height $(h)$ of the frame of lampshade = $(2.5 + 30 + 2.5) cm = 35 cm $$\\$ Radius $(r) $ of the circular end of the frame of lampshade =$\dfrac{ 20}{2} cm = 10 cm $ $\\$ Cloth required for covering the lampshade = $2\pi rh $ $\\$ $=(2*\dfrac{22}{7}*10*35)cm^2$ $\\$ $=2200 cm^2$ $\\$ Hence, for covering the lampshade, $2200 cm^2$ cloth will be required.

19   The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius $3 cm$ and height $10.5 cm.$ The Vidyalaya was to supply the competitors with cardboard. If there were $35$ competitors, how much cardboard was required to be bought for the competition? $\\$ [Assume $\pi=\dfrac{22}{7}]$

Solution :

Radius $(r)$ of the circular end of cylindrical penholder = $3 cm$$\\$ Height $(h)$ of penholder = $10.5 cm$$\\$ Surface area of $1$ penholder = $CSA$ of penholder + Area of base of penholder = $2\pi rh + \pi r^2$ $=(2*\dfrac{22}{7}*3*10.5*(3)^2) cm^2$ $\\$ $=(132*1.5+\dfrac{198}{7})cm^2$ $\\$ $=(198+\dfrac{198}{7})cm^2$ $\\$ $=\dfrac{1584}{7}cm^2$ $\\$ Area of cardboard sheet used by $1$ competitor $ =\dfrac{1584}{7}cm^2$ $\\$ Area of cardbord sheet used by $ 35 $ competitors =$ (\dfrac{1584}{7}* 35)cm^2$ $\\$ $=7920 cm^2$ $\\$ Therefore , $ 7920 cm62 $ cardboard sheet will be bought.

20   Diameter of the base of a cone is $10.5 cm $and its slant height is $10 cm.$ Find its curved surface area. $\\$ [Assume $\pi=\dfrac{ 22}{7}]$

Solution :

Radius $(r)$ of the base of cone = $10.5 cm = 5.25 cm ^2$$\\$ Slant height $(l)$ of cone = $10 cm$ $\\$ $ CSA$ of cone = $\pi rl$ $=(\dfrac{22}{7}*5.25*10)cm^2$$\\$ $= (22 *0.75 *10) cm^2$ $\\$ $= 165 cm^2$$\\$ Therefore, the curved surface area of the cone is $165 cm^2.$

21   Find the total surface area of a cone, if its slant height is $21 m$ and diameter of its base is $24 m.$ [Assume $\pi=\dfrac{ 22}{7}]$

Solution :

Radius $(r)$ of the base of cone= $24 m=12m ^2$ $\\$ Slant height $(l)$ of cone = $21 m$$\\$ Total surface area of cone = $\pi r(r + l)$ $\\$ $=(\dfrac{22}{7}*12*(12+21))m^2$ $\\$ $=(\dfrac{22}{7}*12*33)m^2$ $\\$ $= 1244.57 m^2$

22   Curved surface area of a cone is $308 cm^2$ and its slant height is $14 cm.$ Find$\\$ (i) radius of the base and $\\$ (ii) total surface area of the cone. $\\$ [Assume $\pi=\dfrac{22}{7}]$

Solution :

(i) Slant height $(l)$ of cone = $14 cm$ $\\$ Let the radius of the circular end of the cone be $r.$ We know, $CSA$ of cone $=\pi rl$ $\\$ $ (308)cm^2 = (\dfrac{22}{7}*r*14)cm $ $\\$ $r=\dfrac{308}{44}cm =7 cm$ $\\$ Therefore, the radius of the circular end of the cone is $7 cm.$$\\$ (ii) Total surface area of cone = $CSA$ of cone + Area of base = $\pi rl + \pi r_2$ $\\$ $=[308+\dfrac{22}{7}*(7)^2] cm^2$ $\\$ $= (308 + 154) cm^2$ $\\$ $= 462 cm^2$ $\\$ Therefore, the total surface area of the cone is $ 462 cm^2.$

23   A conical tent is $10 m$ high and the radius of its base is $24 m$. Find $\\$ (i) slant height of the tent $\\$ (ii) cost of the canvas required to make the tent, if the cost of $1 m^2$ canvas is Rs. $70.$ $\\$ [Assume $\pi =\dfrac{ 22}{7}]$

Solution :

(i) Let $ABC$ be a conical tent. $\\$ Height $(h)$ of conical tent = $10 m $ $\\$ Radius $(r)$ of conical tent = $24 m$$\\$ Let the slant height of the tent be 1. $\\$ In $ \Delta ABO,$ $\\$ $AB^2 = AO^2 + BO^2$ $\\$ $ l^2 =h^2 +r^2$ $\\$ $= (10 m)^2 + (24 m)^2$ $\\$ $= 676 m^2$ $\\$ $\therefore l = 26 m$ $\\$ Therefore, the slant height of the tent is $26 m.$ $\\$ (ii) $CSA$ of tent = $\pi rl $ $\\$ $=(\dfrac{22}{7}*24*26)m^2$ $\\$ $=(\dfrac{13728}{7}) m^2 $ $\\$ Cost of $1 m^2$ canvas = Rs.$ 70$ $\\$ Cost of $13728 m^2$ canvas = Rs. $(\dfrac{13728}{7}*70)$ $\\$ $ = Rs. 137280$ $\\$ Therefore, the cost of the canvas required to make such a tent is Rs. $137280.$

24   What length of tarpaulin $3 m$ wide will be required to make conical tent of height 8 m and base radius $6 m$? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $20 cm.$ [Use $\pi = 3.14$]

Solution :

Height $(h)$ of conical tent = $8 m$$\\$ Radius $(r)$ of base of tent = $6 m$ $\\$ Slant height $(l)$ of tent = $\sqrt{r^2 + h^2}$ $\\$ $=\sqrt{662+8^2} m = \sqrt{100} m=10m$ $\\$ $ CSA$ of conical tent = $\pi rl$ $\\$ $= (3.14 × 6 × 10) m^2$ $\\$ $= 188.4 m^2$ $\\$ Let the length of tarpaulin sheet required be $l.$ $\\$ As $20 cm$ will be wasted, therefore, the effective length will be $(l - 0.2 m)$. $\\$ Breadth of tarpaulin = $3 m$ $\\$ Area of sheet = $CSA$ of tent$\\$ $[(l - 0.2 m) × 3] m = 188.4 m^2$ $\\$ $l - 0.2 m = 62.8 m$ $\\$ $l = 63 m$ $\\$ Therefore, the length of the required tarpaulin sheet will be $63 m.$

25   The slant height and base diameter of a conical tomb are $25 m$ and $14 m$ respectively. Find the cost of white-washing its curved surface at the rate of Rs. $210$ per $100 m^2.$ $\\$[Assume $\pi =\dfrac{22}{7}]$ $\\$

Solution :

Slant height $(l)$ of conical tomb = $25 m$ $\\$ Base radius $(r)$ of tomb = $7 m$ $\\$ $CSA$ of conical tomb = $\pi rl$ $\\$ $=(\dfrac{22}{7}*7*25)cm^2$ $\\$ $=550 m^2$ $\\$ Cost of white-washing $100 m^2$ area = Rs. $210$ $\\$ Cost of white-washing 550 m2 area = Rs. $(\dfrac{ 210 * 550}{ 100 }$ $\\$ $= Rs. 1155$ $\\$ Therefore, it will cost Rs. $1155 $ while white-washing such a conical tomb.

26   A joker’s cap is in the form of right circular cone of base radius $7 cm $ and height $24 cm$. Find the area of the sheet required to make $10$ such caps. $\\$ [Assume $\pi =\dfrac{ 22}{7}]$

Solution :

Radius $(r)$ of conical cap = $7 cm$$\\$ Height $(h) $ of conical cap = $24 cm$ $\\$ Slant height $(l)$ of conical cap =$\sqrt{r^2 + h^2}$ $\\$ $ = \sqrt{(7)^2 +(24)^2} cm= \sqrt{625} cm=25cm$ $\\$ $CSA$ of $1$ conical cap = $\pi rl$ $\\$ $=(\dfrac{22}{7}*7*25)cm^2=550 cm^2$ $\\$ $CSA$ of $10$ such conical caps = $(10 × 550) cm^2 = 5500 cm^2$ $\\$ Therefore, $5500 cm^2 $ sheet will be required.

27   A bus stop is barricaded from the remaining part of the road, by using $50$ hollow cones made of recycled cardboard. Each cone has a base diameter of $40 cm$ and height $1 m.$ If the outer side of each of the cones is to be painted and the cost of painting is Rs. $12 $per $m^2$, what will be the cost of painting all these cones? (Use $\pi = 3.14$ and take $\sqrt{1.02}=1.02).$

Solution :

Radius $ (r) $ of cone= $\dfrac{40}{2}=20 cm=0.2m $ $\\$ Height $(h)$ of cone = $1 m$ $\\$ Slant height $(l) $ of cone = $\sqrt{r^2 + h^2}$ $\\$ $=\sqrt{ (1)^2 +(0.2)^2 }m=$ $\sqrt{1.04} m=1.02m$ $\\$ $CSA $ of each cone = $\pi rl$ $\\$ $= (3.14 × 0.2 × 1.02) m^2 = 0.64056 m^2$ $\\$ $CSA$ of $50$ such cones = $(50 × 0.64056) m^2$ $\\$ $= 32.028 m^2$ $\\$ Cost of painting $1 m^2$ area = Rs. $12$ $\\$ Cost of painting $32.028 m^2$ area = Rs. $(32.028 × 12)$ $\\$ $= Rs. 384.336$ $\\$ $= Rs. 384.34$ (approximately)$\\$ Therefore, it will cost Rs. $384.34 $ in painting $50$ such hollow cones.

28   Find the surface area of a sphere of radius:$\\$ (i) $10.5 cm$ $\\$ (ii) $5.6 cm$ $\\$ (iii)$ 14 cm $ $\\$ [Assume $\pi =\dfrac{ 22}{7}]$

Solution :

(i) Radius $(r)$ of sphere = $10.5 cm$ $\\$ Surface area of sphere = $4\pi r^2$ $\\$ $=[4*\dfrac{22}{7}*(10.5)^2]cm^2$ $\\$ $= (88 * 1.5 * 1.5) cm^2$ $\\$ $= 1386 cm^2$ $\\$ Therefore, the surface area of a sphere having radius $10.5 cm$ is $1386 cm^2$ $\\$ . (ii) Radius $(r)$ of sphere = $5.6 cm$ $\\$ Surface area of sphere = $4\pi r^2$ $\\$ $=[4*\dfrac{22}{7}*(5.6)^2]cm^2$ $\\$ $= (88 * 0.8 * 5.6) cm^2$ $\\$ $= 394.24 cm^2$ $\\$ Therefore, the surface area of a sphere having radius $5.6 cm$ is $394.24 cm^2$ $\\$. (iii) Radius $(r)$ of sphere = $14 cm$ $\\$ Surface area of sphere = $4\pi r^2$ $\\$ $=[4*\dfrac{22}{7}*(14)^2]cm^2$ $\\$ $= (4 * 44 * 14) cm^2$ $\\$ $= 2464 cm^2$ $\\$ Therefore, the surface area of a sphere having radius $14 cm $ is $2464 cm^2$ .

29   Find the surface area of a sphere of diameter:$\\$ (i) $14 cm$ $\\$ (ii)$ 21 cm $$\\$ (iii) $3.5 m$ $\\$ [Assume $\pi =\dfrac{22}{7}]$

Solution :

(i) Radius $(r)$ of sphere = $\dfrac{\text{Diameter}}{2} =\dfrac {14}{2} cm = 7 cm $ $\\$ Surface area of sphere = $4\pi r^2$ $\\$ $=[4*\dfrac{22}{7}*(7)^2] cm^2$ $\\$ $= (88 * 7) cm^2$ $\\$ $= 616 cm^2$ $\\$ Therefore, the surface area of a sphere having diameter $14 cm$ is $616 cm^2.$ $\\$ (ii) Radius $(r) $ of sphere =$\dfrac{\text{ Diameter}}{2} =\dfrac{ 21}{2} cm = 10.5 cm$ $\\$ Surface area of sphere = $4\pi r^2$ $\\$ $=[4*\dfrac{22}{7}*(10.5)^2]cm^2$ $\\$ $= 1386 cm^2$ $\\$ Therefore, the surface area of a sphere having diameter $21 cm$ is $1386 cm^2$

(iii) Radius $(r)$ of sphere = $\dfrac{\text{Diameter}}{2} =\dfrac{ 3.5}{2} m = 1.75 m$ $\\$ Surface area of sphere = $4\pi r^2$ $\\$ $=[4*\dfrac{22}{7}*(1.75)^2]m^2$ $\\$ $= 38.5 m^2$ $\\$ Therefore, the surface area of the sphere having diameter $3.5 m$ is $38.5 m^2$.

30   Find the total surface area of a hemisphere of radius $10 cm$. [Use $\pi = 3.14$]

Solution :

Radius $(r)$ of hemisphere = $10 cm$ $\\$ Total surface area of hemisphere = $CSA $of hemisphere + Area of circular end of hemisphere = $2\pi r^2 + \pi r^2$ $\\$ $= 3\pi r^2$ $\\$ = $[3 * 3.14 * (10)^2] cm^2$ $\\$ $= 942 cm^2$ $\\$ Therefore, the total surface area of such a hemisphere is $942 cm^2$.

31   The radius of a spherical balloon increases from $7 cm$ to $14 cm$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Solution :

Radius $(r_1)$ of spherical balloon = $7 cm$ $\\$ Radius $(r_2)$ of spherical balloon, when air is pumped into it = $14 cm$$\\$ Required Ratio =$\dfrac{\text{Initial surface area}}{\text{Surface area after pumping air into ballon}}$ $\\$ $=\dfrac{4\pi r^2_1}{4\pi r_2^2}=(\dfrac{r_1}{r_2})^2$ $\\$ $=(\dfrac{7}{14})^2=\dfrac{1}{4}$ $\\$ Therefore, the ratio between the surface areas in these two cases is $1:4.$

32   A hemispherical bowl made of brass has inner diameter $10.5 cm.$ Find the cost of tinplating it on the inside at the rate of Rs. $16$ per $100 cm^2.$$\\$ [Assume $\pi=\dfrac{ 22}{7}]$

Solution :

Inner radius $(r)$ of hemispherical bowl = $10.5 cm = 5.25 cm ^2$ $\\$ Surface area of hemispherical bowl = $2\pi r^2$ $\\$ $=[2*\dfrac{22}{7}*(5.25)^2] cm^2 $ $\\$ $= 173.25 cm^2$ $\\$ Cost of tin-plating $100 cm^2$ area = Rs. $16$ $\\$ Cost of tin-plating $173.25 cm^2$ area = Rs. $(\dfrac{16*173.25}{100})=Rs.27.72$ $\\$ Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs.$ 27.72.$

33   Find the radius of a sphere whose surface area is $154 cm^2.$ $\\$ [Assume $\pi =\dfrac{22}{7}]$

Solution :

Let the radius of the sphere be $r.$ $\\$ Surface area of sphere = $154$ $\\$ $\therefore 4\pi r^2=154 cm^2$ $\\$ $r^2=(\dfrac{154*7}{2*22})cm^2 =(\dfrac{7*7}{2*2})cm^2$ $\\$ $r=(\dfrac{7}{2})cm=3.5 cm$ $\\$ Therefore, the radius of the sphere whose surface area is $154 cm^2$ is $3.5 cm.$ $\\$

34   The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.

Solution :

Let the diameter of earth be $d.$$\\$ Therefore, the diameter of moon will be $\dfrac{d}{4} $ $\\$. Radius of earth = $\dfrac{d}{2}$ $\\$ Radius of moon = $\dfrac{1}{2}*\dfrac{d}{4}=\dfrac{d}{8}$ $\\$ Surface area of moon = $4 \pi (\dfrac{d}{8})^2$ $\\$ Surface area of earth = $4\pi (\dfrac{d}{2})^2$ $\\$ Required ratio = $\dfrac{4\pi (\dfrac{d}{8})^2}{4 \pi (\dfrac{d}{2})^2}$ $\\$ $=\dfrac{4}{64}=\dfrac{1}{16}$ $\\$ Therefore, the ratio between their surface areas will be $1:16.$

35   A hemispherical bowl is made of steel, $0.25 cm$ thick. The inner radius of the bowl is $5 cm.$ Find the outer curved surface area of the bowl. [Assume $\pi =\dfrac{22}{7}]$

Solution :

Inner radius of hemispherical bowl = $5 cm$ $\\$ Thickness of the bowl = $0.25 cm$ $\\$ $\therefore $ Outer radius $(r)$ of hemispherical bowl =$ (5 + 0.25) cm = 5.25 cm$ $\\$ Outer $CSA$ of hemispherical bowl = $2\pi r^2$$\\$ $=[2*\dfrac{22}{7}*(5.25)^2] cm^2=173.25 cm^2$ $\\$ Therefore, the outer curved surface area of the bowl is $173.25 cm^2.$

36   A right circular cylinder just encloses a sphere of radius $r$ (see figure). Find$\\$ (i) surface area of the sphere,$\\$ (ii) curved surface area of the cylinder,$\\$ (iii) ratio of the areas obtained in (i) and (ii).$\\$

Solution :

(i) Surface area of sphere = $4\pi r^2$$\\$ (ii) Height of cylinder = $r + r = 2r$$\\$ Radius of cylinder =$ r$ $\\$ $CSA$ of cylinder = $2\pi rh$ $ \\$ $= 2\pi r (2r)$ $\\$ $= 4\pi r^2$ $\\$ (iii) Required ratio = Surface area of sphere$\\$ $CSA$ of cylinder$\\$ $=\dfrac{4 \pi r^2}{4 \pi r^2}$ $\\$ $=\dfrac{1}{1}$ $\\$ Therefore, the ratio between these two surface areas is $1:1.$

37   A matchbox measures $4 cm × 2.5 cm × 1.5 cm$ . What will be the volume of a packet containing $12$ such boxes?

Solution :

Matchbox is a cuboid having its length $(l),$ $\\$ breadth $(b)$,$\\$ height $(h)$$\\$ as $4 cm, 2.5 cm,$ and $1.5 cm.$ Volume of 1 match box =$ l × b × h$$\\$ $=(4×2.5×1.5)cm^3 =15cm^3$$\\$ Volume of $12$ such matchboxes = $(15 × 12) cm^3$ $\\$ $= 180 cm^3$ $\\$ Therefore, the volume of $12$ match boxes is $180 cm^3.$

38   A cuboidal water tank is $6 m$ long, $5 m$ wide and $4.5 m $deep. How many litres of water can it hold? ($1 m^3 = 1000l$)

Solution :

The given cuboidal water tank has its length $(l) $ as $6 m, $$\\$ breadth $(b)$ as $5 m, $ and height$ (h) $as $4.5 m.$ $\\$ Volume of tank =$ l × b × h $=$(6×5×4.5)m^3 =135m^3$$\\$ Amount of water that $1 m^3$ volume can hold = $1000 $ litres$\\$ Amount of water that $135 m^3$ volume can hold = $(135 × 1000)$ litres =$ 135000$ litres $\\$ Therefore, such tank can hold up to $135000$ litres of water.

39   A cuboidal vessel is $10 m$ long and $8 m$ wide. How high must it be made to hold $380 $ cubic metres of a liquid?

Solution :

Let the height of the cuboidal vessel be $h.$ Length $(l)$ of vessel = $10 m$$\\$ Width $(b)$ of vessel = $8 m$ $\\$ Volume of vessel = $380 m^3$ $\\$ $\therefore l × b × h = 380$ $\\$ $[(10) (8) h] m^2= 380 m^3$ $\\$ $h = 4.75 m$ $\\$ Therefore, the height of the vessel should be $4.75 m.$

40   Find the cost of digging a cuboidal pit $8 m$ long, $6 m$ broad and $3 m$ deep at the rate of Rs $30$ per $m^3.$

Solution :

The given cuboidal pit has its length $(l) $ as $8 m,$ width $(b)$ as $6 m, $ and depth $(h)$ as $3 m.$$\\$ Volume of pit = $l × b × h$$\\$ $=(8×6×3)m^3 =144m^3$ $\\$ Cost of digging per $m^3 $ volume = Rs.$ 30$ $\\$ Cost of digging $144 m^3 $ volume = Rs. $(144 × 30)$ = Rs.$ 4320.$

41   The capacity of a cuboidal tank is $50000$ litres of water. Find the breadth of the tank, if its length and depth are respectively $2.5 m$ and $10 m.$

Solution :

Let the breadth of the tank be $b \ m.$ $\\$ Length $(l)$ and depth $(h)$ of tank is $2.5 m$ and $10 m$ respectively.$\\$ Volume of tank = $l × b × h$ $\\$ $= (2.5 × b × 10) m^3$ $\\$ $= 25b m^3$ $\\$ Capacity of tank = $25b m^3 = 25000 b$ litres$\\$ $\therefore 25000 b = 50000$ $\\$ $\implies b=2$ $\\$ Therefore, the breadth of the tank is $2 m.$

42   A village, having a population of $4000,$ requires $150$ litres of water per head per day. It has a tank measuring $20 m × 15 m × 6 m.$ For how many days will the water of this tank last?

Solution :

The given tank is cuboidal in shape having its length $(l)$ as $20 m,$$\\$ breadth $(b)$ as $15 m,$ and height $(h)$ as $6 m.$ $\\$ Capacity of tank = $l × b× h$ $\\$ $=(20×15×6)m^3 =1800m^3 =1800000$ litres$\\$ Water consumed by the people of the village in $1$ day =$ (4000 × 150)$ litres = $600000$ litres$\\$ Let water in this tank last for $n$ days.$\\$ Water consumed by all people of village in $n$ days = Capacity of tank $n × 600000 = 1800000$$\\$ $n=3$ Therefore, the water of this tank will last for $3$ days.

43   A godown measures $40 m × 25 m × 10 m.$ Find the maximum number of wooden crates each measuring $1.5 m × 1.25 m × 0.5 m$ that can be stored in the godown.

Solution :

The godown has its length $(l_1)$ as $40 m,$ breadth $(b_1)$ as $25 m,$ height $(h_1)$ as $10 m,$ while the wooden crate has its length $(l_2)$ as $1.5 m,$ breadth $(b_2)$ as $1.25 m,$ and height $(h_2)$ as $0.5 m.$ $\\$ Therefore, volume of godown =$l_1 × b_1 × h_1$ $\\$ $= (40 × 25 × 10) m^3$ $\\$ $= 10000 m^3$ $\\$ Volumeof1woodencrate=$l_2 ×b_2 ×h_2$ $\\$ $= (1.5 × 1.25 × 0.5) m^3$ $\\$ $= 0.9375 m^3$ $\\$ Let $n$ wooden crates can be stored in the godown. Therefore, volume of $n$ wooden crates = Volume of godown $0.9375 × n = 10000$ $\\$ $n= 10000 =10666.66 0.9375$ $\\$ Therefore, $10666 $wooden crates can be stored in the godown.

44   A solid cube of side $12 cm$ is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Solution :

Side $(a)$ of cube = $12 cm$$\\$ Volume of cube = $(a)^3 = (12 cm)^3 = 1728 cm^3$ $\\$ Let the side of the smaller cube be $a_1.$ $\\$ Volume of $1$ smaller cube =$ \dfrac{1728}{8} cm^3 = 216 cm^3 $ $\\$ $(a_1)^3= 216 cm^3$$\\$ $\implies a_1 =6cm$ $\\$ Therefore, the side of the smaller cubes will be $6 cm.$ $\\$ Ratiobetweensurfaceareasofcubes= $\dfrac{\text{Surface area of bigger cube}}{\text{ Surface area of smaller cube}}$ $\\$ $=\dfrac{6a^2}{6 a_1^2}=\dfrac{12^2}{6^2}$ $\\$ $=\dfrac{4}{1}$ $\\$ Therefore, the ratio between the surface areas of these cubes is $4:1.$

45   A river $3 m$ deep and $40 m$ wide is flowing at the rate of $2 km$ per hour. How much water will fall into the sea in a minute?

Solution :

Rate of water flow =$ 2 km$ per hour$\\$ $ =(\dfrac{2000}{60})m/min $ $\\$ $=(\dfrac{100}{3})m/min$ $\\$ Depth $(h)$ of river = $3 m$ $\\$ Width $(b)$ of river = $40 m$ $\\$ Volume of water flowed in $1min= (\dfrac{100}{3}*40*3)m^3=4000 m^3$ $\\$ Therefore, in $1$ minute, $4000 m^3$ water will fall in the sea.

46   The circumference of the base of cylindrical vessel is $132 cm$ and its height is $25 cm$. How many litres of water can it hold? ($1000 cm^3 = 1l$). [Assume $\pi=\dfrac{ 22}{7}]$

Solution :

Let the radius of the cylindrical vessel be $r.$ $\\$ Height $(h)$ of vessel = $25 cm$$\\$ Circumference of vessel = $132 cm$ $\\$ $2\pi r = 132 cm$ $\\$ $r=(\dfrac{132*7}{2*22})cm=21 cm$ $\\$ Volume of cylindrical vessel = $\pi r^2h$ $\\$ $=[\dfrac{22}{7}*(21)^2*25]cm^3$ $\\$ $=34650 cm^3$ $\\$ $=\dfrac{34650}{1000} $ litres $\qquad [\therefore 1$ litre $=1000 cm^3$]$\\$ $=34.65 $ litres $\\$ Therefore , such vessel can hold $ 34.65 $ litres of water.

47   The inner diameter of a cylindrical wooden pipe is $24 cm$ and its outer diameter is $28 cm.$ The length of the pipe is $35cm $.Find the mass of the pipe, if $1cm^3 $ of wood has a mass of $0.6g.$$\\$ [Assume $\pi =\dfrac{22}{7}]$

Solution :

Inner radius $(r_1)$ of cylindrical pipe = $\dfrac{24}{2} cm = 12 cm $ $\\$ Outer radius $(r_2)$ of cylindrical pipe = $\dfrac{28}{2} cm = 14 cm $ $\\$ Height $(h)$ of pipe = Length of pipe = $35 cm$ $\\$ Volume of pipe= $\pi (r^3_2 -r^3_1)h$ $\\$ $=[\dfrac{22}{7}*(14^2-12^2)*35] cm^3$ $\\$ $= 5720 cm^3$ $\\$ Mass of $1cm^3$ wood=$0.6g$ $\\$ Mass of $5720 cm^3$ wood = $(5720 × 0.6) g = 3432 g$$\\$ $= 3.432 kg$

48   A soft drink is available in two packs -$\\$(i) a tin can with a rectangular base of length $5 cm$ and width $4 cm$, having a height of $15 cm$ and $\\$(ii) a plastic cylinder with circular base of diameter $7 cm$ and height $10 cm$. Which container has greater capacity and by how much?$\\$ [Assume $ \pi=\dfrac{22}{7}]$

Solution :

The tin can will be cuboidal in shape while the plastic cylinder will be cylindrical in shape.

Length $(l)$ of tin can = $5 cm$ $\\$ Breadth $(b)$ of tin can = $4 cm$ $\\$ Height (h) of tin can = $15 cm $ $\\$ Capacity of tin can =$ l × b × h = (5 × 4 × 15) cm^3$ $\\$ $= 300 cm^3$ $\\$ Radius $(r)$ of circular end of plastic cylinder = $72 cm = 3.5 cm $ $\\$ Height $(H)$ of plastic cylinder = $10 cm$ $\\$ Capacity of plastic cylinder = $\pi r^2h$ $\\$ $=[\dfrac{22}{7}*(3.5)^2*10]cm^3$ $\\$ $=11*35 cm^3 $ $\\$ $=385 cm^3 $ $\\$ Therefore, plastic cylinder has the greater capacity. $\\$ Difference in capacity = $(385 - 300) cm^3 = 85 cm^3$

49   If the lateral surface of a cylinder is $94.2 cm^2$ and its height is $5 cm,$ then find$\\$ (i) radius of its base$\\$ (ii) its volume. [Use $\pi = 3.14]$

Solution :

(i) Height $(h)$ of cylinder = $5 cm $ $\\$ Let radius of cylinder be $r.$ $\\$ $CSA $ of cylinder = $94.2 cm^2$ $\\$ $ 2\pi rh = 94.2 cm^2$ $\\$ $(2 × 3.14 × r × 5) cm = 94.2 cm^2$ $\\$ $ r = 3 cm$ $\\$ (ii) Volume of cylinder = $\pi r^2h = (3.14 × (3)^2 × 5) cm^3$ $\\$ $= 141.3 cm^3$

50   It costs Rs.$ 2200$ to paint the inner curved surface of a cylindrical vessel $10 m $deep. If the cost of painting is at the rate of Rs. $20$ per $m^2,$ find$\\$ (i) Inner curved surface area of the vessel$\\$ (ii) Radius of the base$\\$ (iii) Capacity of the vessel $\\$ [Assume $\pi =\dfrac{ 22}{7}]$

Solution :

(i) Rs. $20$ is the cost of painting $1 m^2$ area.$\\$ Rs. $2200$ is the cost of painting = $( \dfrac{1 }{20}* 2200) m^2$ area$\\$ $= 110 m^2$ area$ \\$ Therefore, the inner surface area of the vessel is $110 m^2$.$\\$ (ii) Let the radius of the base of the vessel be $r.$ $\\$ Height $(h)$ of vessel = $10 m$ $\\$ Surface area =$ 2\pi rh = 110 m^2$ $\\$ $\implies [2*\dfrac{22}{7}*r*10]m=110 m^2$ $\\$ $\implies r=\dfrac{7}{4}m =1.75 m$ $\\$ (iii) Volume of vessel = $\pi r^2h$ $\\$ $=[\dfrac{22}{7}*(1.75)^*10]m^3 $ $\\$ $= 96.25 m^3$ $\\$ Therefore, the capacity of the vessel is $96.25 m^3$ or $96250 $ litres.

51   The capacity of a closed cylindrical vessel of height $1 m$ is $15.4$ litres. How many square metres of metal sheet would be needed to make it? [Assume $\pi =\dfrac{ 22}{7}]$

Solution :

Let the radius of the circular end be $r.$ $\\$ Height $(h)$ of cylindrical vessel = $1 m$ $\\$ Volume of cylindrical vessel = $15.4$ litres = $0.0154 m^3 \pi r^2h = 0.0154 m^3$ $\\$ $(\dfrac{22}{7}*r^2*1)m=0.0154 m^3 $ $\\$ $\implies r = 0.07 m$ $\\$ Total surface area of vessel = $2\pi r(r + h)$ $\\$ $=[2*\dfrac{22}{7}*0.07(0.07+1)]m^2 $ $\\$ $= 0.44 * 1.07 m^2$ $\\$ $= 0.4708 m^2$ Therefore,$ 0.4708 m^2$ of the metal sheet would be required to make the cylindrical vessel.

52   A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is $7 mm $and the diameter of the graphite is $1 mm$. If the length of the pencil is $14 cm, $find the volume of the wood and that of the graphite. [Assume $\pi =\dfrac{ 22}{7}]$

Solution :

Radius $(r_1)$ of pencil= $\dfrac{7}{2} mm= \dfrac{0.7}{2} cm =0.35 cm $ $\\$ Radius $(r_2)$ of graphite = $\dfrac{1}{2} mm = \dfrac{0.1}{2} cm = 0.05 cm $ $\\$ Height $(h)$ of pencil = $14 cm$ $\\$ Volume of wood in pencil = $\pi(r_2^3 -r^3_1)h$ $\\$ $=[\dfrac{22}{7}*{(0.35)^2-(0.05)^2}*14] cm^3$ $\\$ $=[\dfrac{22}{7}*{0.1225-0.0025}*14]cm^3 $ $\\$ $=(44*0.12)cm^3$ $\\$ $=5.28 cm^3$ $\\$ Volume of graphite =$\pi(r_2)^2h$ $\\$ $=[\dfrac{22}{7}*(0.05)^2*14]cm^3 $ $\\$ $=(44* 0.0025)cm^3$ $\\$ $=0.11 cm^3 $

53   A patient in a hospital is given soup daily in a cylindrical bowl of diameter $7 cm.$ If the bowl is filled with soup to a height of $4 cm, $how much soup the hospital has to prepare daily to serve $250$ patients? [Assume $\pi =\dfrac{22}{7}]$

Solution :

Radius $(r)$ of cylindrical bowl = $\dfrac{7}{2} cm = {3.5} cm$$\\$ Height $(h)$ of bowl, up to which bowl is filled with soup = $4 cm$ $\\$ Volume of soup in $1$ bowl = $\pi r^2h$ $\\$ $=[\dfrac{22}{7}*(3.5)^2*4] cm^3 $ $\\$ $= (11 × 3.5 × 4) cm^3$ $\\$ $= 154 cm^3$ $\\$ Volume of soup given to $250$ patients = $(250 × 154) cm^3 = 38500 cm^3$$\\$ =$ 38.5$ litres.

54   Find the volume of the right circular cone with$\\$ (i) radius $6 cm$, height $7 cm$ $\\$ (ii) radius $3.5 cm,$ height $12 cm$ [Assume $\pi =\dfrac{ 22}{7}]$

Solution :

(i) Radius $(r)$ of cone = $6 cm$ $\\$ Height $(h)$ of cone = $7 cm$ $\\$ Volume of cone = $\dfrac{1}{3}r^2h $ $\\$ $=[\dfrac{1}{3}*\dfrac{22}{7}*(6)^2*7]cm^3 $ $\\$ $= (12 * 22) cm^3 = 264 cm^3$ $\\$ Therefore, the volume of the cone is $264 cm^3.$ $\\$ (ii) Radius $(r)$ of cone = $3.5 cm$ $\\$ Height $(h)$ of cone = $12 cm$$\\$ Volume of cone = $\dfrac{1}{2} \pi r^2 h$ $\\$ $= 154 cm^3$ $\\$ Therefore, the volume of the cone is$154 cm^3.$ $\\$

55   Find the capacity in litres of a conical vessel with $\\$ (i) radius $7 cm,$ slant height $25 cm$ $\\$ (ii) height $12 cm,$ slant height $12 cm $ [Assume $\pi =\dfrac{22}{7}]$

Solution :

(i) Radius $(r)$ of cone = $7 cm$ $\\$ Slant height $(l)$ of cone = $25 cm$ $\\$ Height $(h)$ of cone= $\sqrt{l^2-r^2} =\sqrt{ 25^2-7^2} cm$$\\$ $=24 cm$ $\\$ Volume of cone =$\dfrac{1}{3}\pi r^2 h$ $\\$ $=[\dfrac{1}{3}*\dfrac{22}{7}*(7^2)*24] cm^3$ $\\$ $=(154 * 8)cm^3 $ $\\$ $=1232 cm^3 $ $\\$ Therefore , capacity of the conical vessel $ \\$ $=(\dfrac{1232}{1000}) $litres $ \qquad (1$ litre=$ 1000 cm^3)$ $\\$ $=1.232 $ litres $ \\$ (ii) Height $(h)$ of cone = $12 cm$ $\\$ Slant height $(l)$ of cone = $13 cm$ $\\$ Radius $(r)$ of cone= $\sqrt{l^2 -h^2} $ $\\$ $=\sqrt{ 13^2 -12^2} cm $ $\\$ $= 5 cm$ $\\$ Volume of cone = $\dfrac{1}{3}\pi r^2h$ $\\$ $=[\dfrac{1}{3}*\dfrac{22}{7}*(5)^2*12]cm^3 $ $\\$ $=[4*\dfrac{22}{7}*25] cm^3 $ $\\$ =$\dfrac{ 2200}{7} cm^3$ $\\$ Therefore, capacity of the conical vessel $\\$ $(\dfrac{2200}{7000})litres \qquad (1 litre = 1000 cm^3)$ $\\$ $=\dfrac{11}{35} $ litres

56   The height of a cone is $15 cm$. If its volume is $1570 cm^3,$ find the diameter of its base. [Use $\pi= 3.14$]

Solution :

Height $(h)$ of cone = $15 cm$$\\$ Let the radius of the cone be $r.$ $\\$ Volume of cone =$ 1570 cm^3$$\\$ $\dfrac{1}{3}\pi r^2h=1570 cm^3 $ $\\$ $r^2 =100cm^2$ $\\$ $ r = 10 cm$ $\\$ Therefore, the radius of the base of cone is $10 cm.$

57   If the volume of a right circular cone of height $9 cm$ is $48\pi cm^3$, find the diameter of its base.

Solution :

Height $(h)$ of cone = $9 cm$$\\$ Let the radius of the cone be $r.$ $\\$ Volume of cone = $48\pi cm^3$ $\\$ $\dfrac{1}{3}\pi r^2h=48 \pi cm^3 $ $\\$ $r^2 =16cm^2$ $\\$ $ r = 4 cm$ $\\$ Diameter of base = $2r = 8 cm$

58   A conical pit of top diameter $3.5 m$ is $12 m$ deep. What is its capacity in kilolitres? [Assume $\pi=\dfrac{ 22}{7}$]

Solution :

Radius $(r)$ of pit= $\dfrac{3.5}{2} m=1.75m $ $\\$ Height $(h)$ of pit = Depth of pit = $12 m$ $\\$ Volume of pit= $\dfrac{1}{3}\pi r^2h$ $\\$ $=[\dfrac{1}{3}*\dfrac{22}{7}*(1.75)^2*12]cm^3$ $\\$ $= 38.5 m^3$ $\\$ Thus, capacity of the pit = $(38.5 × 1)$ kilolitres = $38.5$ kilolitres

59   The volume of a right circular cone is $9856 cm^3$. If the diameter of the base is $28 cm,$ find $\\$ (i) height of the cone$\\$ (ii) slant height of the cone$\\$ (iii) curved surface area of the cone$\\$ [Assume $ \pi =\dfrac{22}{7}]$

Solution :

(i)Radius of cone= $\dfrac{28}{2} cm=14cm $ $\\$ Let the height of the cone be $h.$ $\\$ Volume of cone = $9856 cm^3$ $\\$ $\dfrac{1}{3}\pi r^2 h=9856 cm^3 $ $\\$ $\implies [\dfrac{1}{3}*\dfrac{22}{7}*(14)^2*h]cm^2=48 \pi cm^3$ $\\$ $h = 48 cm$ $\\$ Therefore, the height of the cone is $48 cm.$ $\\$ (ii) Slant height $(l)$ of cone = $\sqrt{r_2 - h_2} $ $\\$ $=\sqrt{14^2+48^2} cm$ $\\$ $\sqrt{196+2304}cm$ $\\$ $=50 cm$ $\\$ Therefore, the slant height of the cone is $50 cm. $ $\\$ (iii) $CSA$ of cone = $\pi rl$ $\\$ $=(\dfrac{22}{7}*14*50) cm^2$ $\\$ $= 2200 cm^2$ $\\$ Therefore, the curved surface area of the cone is $2200 cm^2$.

60   A right triangle $ABC$ with sides $5 cm, 12 cm$ and $13 cm$ is revolved about the side $12 cm$. Find the volume of the solid so obtained.

Solution :

When right-angled $\Delta ABC$ is revolved about its side $12 cm$, a cone with height $(h)$ as $12 cm, $radius $(r)$ as $5 cm,$ and slant height $(l) 13 cm$ will be formed.$\\$ Volume of cone = $\dfrac{2}{3}\pi r^2 h$ $\\$ $[\dfrac{1}{3}*\pi *(5)^2*12] cm^3 $ $\\$ $= 100\pi cm^3$ $\\$ Therefore, the volume of the cone so formed is $100\pi cm^3$

61   If the triangle $ABC$ in the Question 7 above is revolved about the side $5 cm,$ then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Solution :

When right-angled $\Delta ABC$ is revolved about its side $5 cm$, a cone will be formed having radius $(r)$ as $12 cm$, height $(h)$ as $5 cm$, and slant height $(l)$ as $13 cm.$ $\\$ Volume of cone = $\dfrac{1}{3}\pi r^2 h$ $\\$ $=[\dfrac{1}{3}*\pi *(12)^2*5] cm^3 $ $\\$ $= 240\pi cm^3$ $\\$ Therefore, the volume of the cone so formed is $240\pi cm^3$. Required ratio = $\dfrac{100 \pi}{240 \pi}$ $\\$ $=\dfrac{5}{10}=5:12$

62   A heap of wheat is in the form of a cone whose diameter is $10.5 m$ and height is $3 m.$ Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. [Assume $\pi =\dfrac{ 22}{7}$]

Solution :

Radius $(r)$ of heap = $10.5 m = 5.25 m^2$ $\\$ Height $(h)$ of heap = $3 m$$\\$ Volume of heap = $\dfrac{1}{3}\pi r^2h$ $\\$ $=[\dfrac{1}{3}*\dfrac{22}{7}*(5.25)^2*3]m^3$ $\\$ $= 86.625 m^3$ $\\$ Therefore, the volume of the heap of wheat is $86.625 m^3.$ $\\$ Area of canvas required = $CSA$ of cone $\\$ $=\pi r l=\pi r\sqrt{r^2+h^2}$ $\\$ $=\dfrac{22}{7}*5.25*\sqrt{(5.25)^2+(3)^2}m62$ $\\$ $=\dfrac{22}{7}*5.25* 6.05 m^2$ $\\$ $= 99.825 m^2$ $\\$ Therefore, $99.825 m^2$ canvas will be required to protect the heap from rain.

63   Find the volume of a sphere whose radius is $\\$ (i)$ 7 cm$ $\\$ (ii)$ 0.63 m $ $\\$[Assume $\pi=\dfrac{ 22}{7}$]

Solution :

(i) Radius of sphere = $7 cm $ $\\$ Volume of sphere = $\dfrac{4}{3}\pi r^3$ $\\$ $=\dfrac{4}{3}*\dfrac{22}{7}*(7)^3 cm^3$ $\\$ $=\dfrac{4312}{3} cm^3 $ $\\$ $=1437\dfrac{1}{3} cm^3 $$\\$ Therefore, the volume of the sphere is $1437 \dfrac{1 }{3}cm^3. $ $\\$ (ii) Radius of sphere = $0.63 m$ $\\$ Volume of sphere = $\dfrac{4}{3}\pi r^3$ $\\$ $=\dfrac{4}{3}*\dfrac{22}{7}*(0.63)^3m^3$ $\\$ $= 1.0478 m^3$ $\\$ Therefore, the volume of the sphere is $1.05 m^3$ (approximately).

64   Find the amount of water displaced by a solid spherical ball of diameter$\\$ (i) $28 cm$ $\\$ (ii) $0.21 m$ $\\$ [Assume $\pi =\dfrac{ 22}{7}$]

Solution :

(i)Radius $(r)$ of ball $=\dfrac{28}{2} cm=14cm $ $\\$ Volume of ball = $\dfrac{4}{3}\pi r^3 $ $\\$ $=\dfrac{4}{3}*\dfrac{22}{7}*(14)^3 cm^3 $ $\\$ $=11498\dfrac{2}{3}cm^3 $ $\\$ Therefore, the volume of the sphere is $11498 \dfrac{2}{8} cm^3. $ $\\$ (ii)Radius $(r)$ of ball $=\dfrac{ 0.21}{2} m = 0.105 m $ $\\$ Volume of ball = $\dfrac{4}{3}\pi r^3$ $\\$ $=\dfrac{4}{3}*\dfrac{22}{7}*(0.105)^3 m^3 $ $\\$ $= 0.004851 m^3$ Therefore, the volume of the sphere is $0.004851 m^3.$

65   The diameter of a metallic ball is $4.2 cm.$ What is the mass of the ball, if the density of the metal is $8.9 g$ per $cm^3$? [Assume $\pi =\dfrac{22}{7}$]

Solution :

Radius $(r)$ of metallic ball $=\dfrac{4.2}{2} cm = 2.1 cm $ $\\$ Volume of metallic ball = $\dfrac{4}{3}\pi r^3$ $\\$ $=\dfrac{4}{3}*\dfrac{22}{7}*(2.1)^3 cm^3 $ $\\$ $= 38.808 cm^3$ $\\$ Density $=\dfrac{\text{Mass}}{\text{ Volume}}$ $\\$ Mass = Density × Volume $\\$ $= (8.9 × 38.808) g$ $\\$ $= 345.3912 g$ $\\$ Hence, the mass of the ball is $345.39 g$ (approximately).

66   The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Solution :

Let the diameter of earth be $d.$ $\\$ Therefore, the radius of earth will be $\dfrac{d}{2}$ .$\\$ Diameter of moon will be $\dfrac{d}{4 }$ and the radius of moon will be $\dfrac{d}{8} .$ $\\$ Volume of moon= $\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi (\dfrac{d}{8})^3=\dfrac{1}{512}*\dfrac{4}{3}\pi d^3 $ $\\$ Volume of earth= $\dfrac{4}{3}\pi r^3 =\dfrac{4}{3}\pi(\dfrac{d}{2})^3=\dfrac{1}{8}*\dfrac{4}{3}\pi d^3$ $\\$ $\dfrac{\text{Volume of moon}}{\text{Volume of earth}}=\dfrac{\dfrac{1}{512}*\dfrac{4}{3}\pi d^3}{\dfrac{1}{8}*\dfrac{4}{3}\pi d^3}=\dfrac{1}{64}$ $\\$ Volume of moon =$\dfrac {1}{64}$ Volume of earth $ \\$ Therefore, the volume of moon is $\dfrac{1}{64}$ of the volume of earth.

67   How many litres of milk can a hemispherical bowl of diameter $10.5 cm$ hold? [Assume $\pi =\dfrac{22}{7}$]

Solution :

Radius $(r)$ of hemispherical bowl $=\dfrac{10.5}{2} cm = 5.25 cm $ $\\$ Volume of hemispherical bowl = $\dfrac{2}{3}\pi r^3$ $\\$ $=\dfrac{2}{3}*\dfrac{22}{7}*(5.25)^3 cm^3$ $\\$ $= 303.1875 cm^3$ $\\$ Capacity of the bowl = $\dfrac{303.1875 }{1000}$ litre $\\$ $= 0.3031875 $ litre $\\$ $= 0.303$ litre (approximately)$\\$ Therefore, the volume of the hemispherical bowl is $0.303$ litre.$\\$

68   A hemispherical tank is made up of an iron sheet $1 cm $ thick. If the inner radius is $1 m,$ then find the volume of the iron used to make the tank. [Assume $\pi =\dfrac{ 22}{7}$]

Solution :

Inner radius $(r_1)$ of hemispherical tank = $1 m$ $\\$ Thickness of hemispherical tank = $1 cm = 0.01 m$ $\\$ Outer radius $(r_2)$ of hemispherical tank = $(1 + 0.01) m = 1.01 m$ $\\$ Volumeofironusedtomakesuchatank= $\dfrac{2}{3}\pi(r_2^3-r_1^3)$ $\\$ $=[\dfrac{2}{3}*\dfrac{22}{7}*{(1.01)^3-(1)^3}] m^3$ $\\$ $=[\dfrac{44}{21}*{1.030301-1}] m^3$ $\\$ $=0.06348 m^3$ (approximately).

69   Find the volume of a sphere whose surface area is $154 cm^2.$ [Assume $\pi=\dfrac{22}{7}]$

Solution :

Let radius of sphere be $r.$ $\\$ Surface area of sphere = $154 cm^2$ $\\$ $\implies 4\pi r^2 = 154 cm^2$ $\\$ $\implies r^2=(\dfrac{154*7}{4*22}) cm^2$ $\\$ $\implies r=(\dfrac{7}{2}) cm=3.5 cm$ $\\$ Volume of aphere =$\dfrac{4}{3}\pi r^3$ $\\$ $=\dfrac{4}{3}*\dfrac{22}{7}*(3.5)^3 cm^3 $ $\\$ $=179 \dfrac{2}{3} cm^3 $ $\\$ Therefore , the volume of the sphere is $ 179\dfrac{2}{3} cm^3 $

70   A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of Rs. $498.96.$ If the cost of white-washing is Rs.$ 2.00$ per square meter, find the $\\$ (i) inside surface area of the dome,$\\$ (ii) volume of the air inside the dome. [Assume $\pi =\dfrac{22}{7}$]

Solution :

(i) Cost of white-washing the dome from inside = Rs. $498.96$$\\$ Cost of white-washing $1 m^2$ area = Rs.$ 2$ $\\$ Therefore, $CSA$ of the inner side of dome =$ (\dfrac{498.96} {2}) m^2$ $\\$ $= 249.48 m^2$ $\\$ (ii) Let the inner radius of the hemispherical dome be $r.$ $\\$ $ CSA$ of inner side of dome =$ 249.48 m^2$ $\\$ $2\pi r^2 = 249.48 m^2$ $\\$ $\implies 2*\dfrac{22}{7}*r^2=249.48 m^2 $ $\\$ $\implies r^2=(\dfrac{249.48*7}{2*22}) m^2 =39.69 m^2 $ $\\$ $\implies r=6.3 m$ $\\$ Volume of air inside the dome=Volume of hemispherical dome $ \\$ $=\dfrac{2}{3}\pi r^3$ $\\$ $=[\dfrac{2}{3}*\dfrac{22}{7}*(6.3)^3]m^3 $ $\\$ $=523.908 m^3 $ $\\$ $=523.9 m^3 $ (approximately)$\\$ Therefore, the volume of air inside the dome is $ 523.9 m^3$

71   Twenty-seven solid iron spheres, each of radius $r$ and surface area $S$ are melted to form a sphere with surface area $S'$. Find the$\\$ (i) radius $r'$ of the new sphere,$\\$ (ii) ratio of $S$ and $S'.$

Solution :

(i)Radius of $1$ solid iron sphere =$ r$ $\\$ Volume of $1$ solid iron sphere = $\dfrac{4}{3}\pi r^3$ $\\$ Volume of $27$ solid iron spheres = $27 * \dfrac{4}{3}\pi r^3 $ $\\$ $27$ solid iron spheres are melted to form $1$ iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of $27$ solid iron spheres. Let the radius of this new sphere be $r'.$ $\\$ Volume of new solid iron sphere = $\dfrac{4}{3}\pi r^{'3}$ $\\$ $\dfrac{4}{3}\pi r^{'3}=27*\dfrac{4}{3} \pi r^3 $ $\\$ $r^{'3}=27 r^3 $ $\\$ $r^{'}=3r $ $\\$ (ii) Surface area of $1$ solid iron sphere of radius $r = 4\pi r^2$ $\\$ Surface area of iron sphere of radius $r' = 4\pi (r^{'})^2$ $\\$ $= 4\pi (3r)^2 = 36 \pi r^2$ $\\$ $\dfrac{S}{S^{'}}=\dfrac{4 \pi r^2}{36 \pi r^2}=\dfrac{1}{9}=1:9 $

72   A capsule of medicine is in the shape of a sphere of diameter $3.5 mm.$ How much medicine (in $mm^3$) is needed to fill this capsule? [Assume $\pi =\dfrac{ 22}{7}$]

Solution :

Radius $(r)$ of capsule $=(\dfrac{3.5}{2})mm=1.75 mm $ $\\$ Volume of spherical capsule =$\dfrac{4}{3} \pi r^3 $ $\\$ $[\dfrac{4}{3}*\dfrac{22}{7}*(1.75)^3] mm^3 $ $\\$ $=22.458 mm^3 $ $\\$ $=22.46 mm^3 $ (approximately)$\\$ Therefore , the volume of the spherical capsule is $ 22.46 mm^3 $

73   A wooden bookshelf has external dimensions as follows: Height = $110 cm,$ Depth = $25 cm,$ Breadth = $85 cm$ (see the given figure). The thickness of the plank is $5 cm$ everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is $20$ paise per $cm^2$ and the rate of painting is $10$ paise per $cm^2$, find the total expenses required for polishing and painting the surface of the bookshelf

Solution :

External height $(l)$ of book self = $85 cm$ $\\$ External breadth $(b)$ of book self = $25 cm$ $\\$ External height $(h)$ of book self = $110 cm$ $\\$ External surface area of shelf while leaving out the front face of the shelf =$ lh + 2(lb + bh) = [85 × 110 + 2 (85 × 25 + 25 × 110)] cm^2$ $\\$ $= (9350 + 9750) cm^2$ $\\$ $= 19100 cm^2$ $\\$ Area of front face = $[85 × 110 - 75 × 100 + 2 (75 × 5)] cm^2 = 1850 + 750 cm^2$ $\\$ $= 2600 cm^2$ $\\$ Area to be polished = $(19100 + 2600) cm^2 = 21700 cm^2$ $\\$ Cost of polishing $1 cm^2 $ area = Rs.$ 0.20$ $\\$ Cost of polishing $21700 cm^2$ area Rs. $(21700 × 0.20)$ = Rs.$ 4340$

It can be observed that length $(l),$ breadth $(b)$, and height $(h)$ of each row of the book shelf is $75 cm, 20 cm,$ and $30 cm$ respectively.$\\$ Area to be painted in $1$ row =$ 2(l + h)b + lh$ $\\$ $= [2 (75 + 30) × 20 + 75 × 30] cm^2$ $\\$ $= (4200 + 2250) cm^2 = 6450 cm^2$ $\\$ Area to be painted in $3$ rows =$ (3 × 6450) cm^2 = 19350 cm^2 $ $\\$ Cost of painting $1 cm^2$ area = Rs.$ 0.10$ $\\$ Cost of painting $19350 cm^2$ area = Rs. $(19350 × 0.1)$ $\\$ = Rs. $1935$ $\\$ Total expense required for polishing and painting = Rs. $(4340 + 1935) $= Rs. $6275$ $\\$ Therefore, it will cost Rs.$6275$ for polishing and painting the surface of the bookshelf.

74   The front compound wall of a house is decorated by wooden spheres of diameter $21 cm$, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius $1.5 cm$ and height $7 cm$ and is to be painted black. Find the cost of paint required if silver paint costs $25$ paise per $cm^2 $ and black paint costs $5$ paise per $cm^2.$

Solution :

Radius $(r)$ of wooden sphere = $(\dfrac{21}{2} cm = 10.5 cm $ $\\$ Surface area of wooden sphere = $4\pi r^2$ $\\$ $=[4*\dfrac{22}{7}*(10.5)^2]cm^2=1386 cm^2 $ $\\$ Radius $(r_1)$ of the circular end of cylindrical support = $1.5 cm$ $\\$ Height $(h)$ of cylindrical support = $7 cm $ $\\$ $CSA$ of cylindrical support =$ 2\pi rh$ $\\$ $=[2*\dfrac{22}{7}*(1.5)*7]cm^2 =66 cm^2 $ $\\$ Area of the circular end of cylindrical support = $\pi r^2$$\\$ $=[\dfrac{22}{7}*(1.5)^2]cm^2 $ $\\$ $= 7.07 cm^2$ $\\$ Area to be painted silver =$ [8 × (1386 - 7.07)] cm^2$ $\\$ $ = (8 × 1378.93) cm^2 = 11031.44 cm^2$ $\\$ Cost for painting with silver color = Rs. $(11031.44 × 0.25) = Rs. 2757.86 $ $\\$ Area to be painted black =$ (8 × 66) cm^2 = 528 cm^2$ $\\$ Cost for painting with black color = Rs. $(528 × 0.05) = Rs. 26.40$ $\\$ Total cost in painting = Rs. $(2757.86 + 26.40) = Rs. 2784.26$ $\\$ Therefore, it will cost Rs. $2784.26$ in painting in such a way.

75   The diameter of a sphere is decreased by $25\%$. By what per cent does its curved surface area decrease?

Solution :

Let the diameter of the sphere be $d.$ $\\$ Radius $(r_1) $ of sphere = $\dfrac{d}{2}$ $\\$ New radius $(r_2)$ of sphere $=\dfrac{d}{2}(1-\dfrac{25}{100})=\dfrac{3}{8}d $ $\\$ $CSA (S_1)$ of sphere =$ 4 \pi r_1^2 $ $\\$ $=4\pi (\dfrac{d}{2})^2=\pi d^2 $ $\\$ Decrease in surface area of sphere = $S_1 -S_2$$\\$ $=\pi d^2-\dfrac{9}{16}\pi d^2 $ $\\$ Percentage decrease in surface area of sphere = $\dfrac{S_1-S_2}{S_1}*100$ $\\$ $=\dfrac{7 \pi d^2}{16 \pi d^2}*100=\dfrac{700}{16}=43.75 \%$