(i) (b): We know that, volume of cylinder \(=\pi r^{2} h\)
\(\Rightarrow 1540=\frac{22}{7} \times r^{2} \times 10 \)
\(\Rightarrow \frac{154 \times 7}{22}=r^{2} \Rightarrow r^{2}=49 \Rightarrow r=7 \mathrm{~cm}\)
\(\therefore\) Diameter of the base of cylinder \(=2 r=2 \times 7=14 \mathrm{~cm}\)
(ii) (b): We know that, volume of sphere \(=\frac{4}{3} \pi r^{3}\)
\(\Rightarrow 38808=\frac{4}{3} \times \frac{22}{7} \times r^{3} \)
\(\Rightarrow r^{3}=\frac{38808 \times 3 \times 7}{4 \times 22}=441 \times 21=(21)^{3} \Rightarrow r=21 \mathrm{~cm}\)
\(\therefore\) Diameter of sphere = 42 cm.
(iii) (d): Total volume of shape formed = Volume of cylindrical shapes + Volume of sphere
= 11 x 1540 + 38808 = 16940 + 38808 = 55748 cm³
(iv) (c): Curved surface area of one cylindrical shape \(=2 \pi r h \)
\(=2 \times \frac{22}{7} \times 7 \times 10=440 \mathrm{~cm}^{2}\)
(v) (a): Area covered by cylindrical shapes on the surface of sphere
\(=11 \times \pi r^{2}=11 \times \frac{22}{7} \times 7 \times 7=1694 \mathrm{~cm}^{2}\)

(i) (b): Lateral surface area of Hermika which is cubical in shape = 4a² = 4 x (8)² = 256 m²
(ii) (a): Diameter of cylindrical base = 42 m
\(\therefore\) Radius of cylindrical base (r) = 21 m
Height of cylindrical base (h) = 12 m
\(\therefore\) Number of bricks used \(=\frac{\frac{22}{7} \times 21 \times 21 \times 12}{0.01}\)
= 1663200
(iii) (c) : Given, diameter of Anda which is
hemispherical in shape = 42 m
\(\Rightarrow\) Radius of Anda (r) = 21 m
\(\therefore \quad \text { Volume of } A n d a =\frac{2}{3} \pi r^{3}=\frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 \)
\(=44 \times 21 \times 21=19404 \mathrm{~m}^{3}\)
(iv) (d): Given, radius of Pradakshina Path (r) = 25 m
\(\therefore\) Perimeter of path = \(2\pi r\)
\(=\left(2 \times \frac{22}{7} \times 25\right) \mathrm{m}\)
· . Distance covered by priest \(=14 \times 2 \times \frac{22}{7} \times 25\)
= 2200 m
(v) (b):\(\because\) Radius of Anda (r) = 21 m
\(\therefore\) Curved surface area of Anda = \(2\pi r^{2}\)
\(=2 \times \frac{22}{7} \times 21 \times 21=2772 \mathrm{~m}^{2}\)

(i) (a)
(ii) (d): Slant height of conical cavity \(l=\sqrt{h^{2}+r^{2}}\)
\(=\sqrt{(24)^{2}+(7)^{2}}=\sqrt{576+49}=\sqrt{625}=25 \mathrm{~cm}\)
(iii) (b): Curved surface area of conical cavity = \(\pi r l\)
\(=\frac{22}{7} \times 7 \times 25=550 \mathrm{~cm}^{2}\)
(iv) (c) : External curved surface area of cylinder 
\(=2 \pi r h=2 \times \frac{22}{7} \times 7 \times 24=1056 \mathrm{~cm}^{2}\)
(v) (a): Volume of conical cavity \(=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24=1232 \mathrm{~cm}^{3}\)

(i) (b): Since, volume of sphere = volume of cylinder
\(\Rightarrow \frac{4}{3} \pi R^{3}=\pi r^{2} h\)
where R, r are the radii of sphere and cylinder respectively.
\(\Rightarrow R^{3}=\frac{6 \times 6 \times 8 \times 3}{4}=(6)^{3} \Rightarrow R=6 \mathrm{~cm}\)
\(\therefore\) Radius of sphere = 6 cm
(ii) (a): Volume of sphere \(=\frac{4}{3} \pi R^{3}\)
\(=\frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6=905.14 \mathrm{~cm}^{3}\)
(iii) (c): \(\because\) Volume of sphere = Volume of cylinder
\(\therefore\) Required ratio = 1 : 1
(iv) (a): Total surface area of the cylinder \(=2 \pi r(r+h)\)
\(=2 \times \frac{22}{7} \times 6(6+8)=2 \times \frac{22}{7} \times 6 \times 14=528 \mathrm{~cm}^{2}\)
(v) (c)

(i) (b): Volume of cuboidal part = 1 x b x h = (20 x 15 x 5) cm³ = 1500 cm³
(ii) (c): Radius of conical depression, r = 0.6 cm
Height of conical depression, h = 2.1 cm
\(\therefore\) Total volume of conical depressions \(=3 \times \frac{1}{3} \pi r^{2} h\)
\(=\frac{22}{7} \times 0.6 \times 0.6 \times 2.1=\frac{2.376}{1000}-2.376 \mathrm{~cm}^{3}\)
(iii) (d): Volume of wood used in the entire stand = Volume of cuboidal part - Total volume of conical depressions
= 1500 - 2.376 = 1497.624 cm³
(iv) (a)
(v) (d): Cost of wood per cm³ = Rs 5
\(\therefore\) Total cost of making the pen stand
= Rs (5 x 1497.624) = Rs 7488.12

We have, radius of each coin = 3.5 cm
\(=\frac{35}{10} \mathrm{~cm}=\frac{7}{2} \mathrm{~cm}\)
Thickness of each coin \(=0.5 \mathrm{~cm}=\frac{1}{2} \mathrm{~cm}\)
So,height of cylinder made by Meera (h1) \(=12 \times \frac{1}{2}=6 \mathrm{~cm}\)
and height of cylinder made by Dhara (h2)
\(=8 \times \frac{1}{2}=4 \mathrm{~cm}\)
(i) (b): Curved surface area of cylinder made by Meera \(=2 \times \frac{22}{7} \times \frac{7}{2} \times 6=132 \mathrm{~cm}^{2}\)
(ii) (b): Required ratio
\(=\frac{\text { Curved surface area of cylinder made by Meera }}{\text { Curved surface area of cylinder made by Dhara }}\)
\(=\frac{2 \pi r h_{1}}{2 \pi r h_{2}}=\frac{h_{1}}{h_{2}}=\frac{6}{4}=\frac{3}{2} \text { i.e., } 3: 2\)
(iii) (a): Volume of cylinder made by Dhara \(=\pi r^{2} h_{2}\)
\(=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4=154 \mathrm{~cm}^{3}\)
(iv) (c): Required ratio
\(=\frac{\text { Volume of cylinder made by Meera }}{\text { Volume of cylinder made by Dhara }} \)
\(=\frac{\pi r^{2} h_{1}}{\pi r^{2} h_{2}}=\frac{h_{1}}{h_{2}}=\frac{6}{4}=\frac{3}{2} \text { i.e., } 3: 2\)
(v) (a): When two coins are shifted from Meera's cylinder to Dhara's cylinder, then length of both cylinders become equal. So, volume of both cylinders become equal

(i) (d): Let r be the radius of the sphere. Then, diameter of sphere = 24 cm
\(\therefore \quad \text { Radius }(r)=\frac{24}{2}=12 \mathrm{~cm}\)
(ii) (a): Volume of ceramic block = 1 x b x h = 24 x 24 x 27 = 15552 cm³
(iii) (b): Volume of ceramic carved out \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \times \frac{22}{7} \times(12)^{3}=7241.14 \mathrm{~cm}^{3}\)
(iv) (c): Volume of cuboidal vase = Volume of ceramic block - Volume of sphere
= 15552 - 7241.14 = 8310.86 cm³
(v) (b): Surface area of the sphere carved out \(=4 \pi r^{2}\)
\(=4 \times \frac{22}{7} \times(12)^{2}=1810.28 \mathrm{~cm}^{2}\)

(i) (a): Type- I tank is cylindrical in shape with r = 1.5 m and h = 3.5 m.
\(\therefore \quad \text { Required volume } =\pi r^{2} h=\left(3.14 \times 1.5^{2} \times 3.5\right) \mathrm{m}^{3} =24.7275 \mathrm{~m}^{3}\)
Now, 1 m³ = 1000 litres
\(\therefore \quad Capacity of type-I tank =(24.7275 \times 1000) litres =24727.5 litres\)
(ii) (b): Capacity of type- II tank = 1 x b x h
= 5 x 4 x 3.5 m³ = 70 m³ = (70 x 1000) litres
= 70000 litres
(iii) (d): Volume of type-III tank
\(=\pi r^{2} h+\frac{2}{3} \pi r^{3}=3.14 \times(2.5)^{2}\left[(5.5-2.5)+\frac{2}{3}(2.5)\right]\)
= 91.58 m³ = 91.58 x 1000 litres = 91580 litres
\(\therefore\) Required difference = 91580 - 24727.5
= 66852.5 litres
(iv) (c): TSA of type-II tank = 2 (lb + bh + hI)
= 2 (5 x 4 + 4 x 3.5 + 3.5 x 5)
= 2(20 + 14 + 17.5) = 103 m²
\(\therefore\) Cost of cloth required = Rs (45 x 103) = Rs 4635
(v) (d): Required ratio \(=\frac{2 \pi r\left(r+h^{\prime}\right)}{2(l b+b h+h l)}\)
\(=\frac{2 \times 3.14 \times 1.5(1.5+3.5)}{103}=\frac{471}{1030} \text { i.e., } 471: 1030\)

(i) (d): Volume of cone \(=\frac{1}{3} \pi r^{2} h\)
\(= \frac{1}{3} \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} \times \frac{3.5}{2}\)
\(\left[\because r=\frac{3.5}{2} \text { and } h=3.5-\frac{3.5}{2}=\frac{3.5}{2}\right] \)
\(= 5.614 \mathrm{~m}^{3}\)
(ii) (a): Volume of hemisphere \(=\frac{2}{3} \pi r^{3}\)
\(=\frac{2}{3} \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} \times \frac{3.5}{2}=11.23 \mathrm{~m}^{3}\)
(iii) (d): Volume of cylinder that circumscribe the cone and hemisp here \(=\frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} \times 3.5\)
= 33.69 m³
(iv) (d): Additional space enclosed by cylinder
= Volume of cylinder - (volume of cone + volume of hemisphere)
= ;33.69- (11.23 + 5614) = 16.846 m³
(v) (a): Required ratio
\(=\frac{\text { Curved surface area of cone }}{\text { Curved surface area of hemisphere }}=\frac{\pi r \sqrt{r^{2}+h^{2}}}{2 \pi r^{2}} \\ \)
\(=\frac{\sqrt{2 r^{2}}}{2 r}=\frac{\sqrt{2} r}{2 r}=\frac{1}{\sqrt{2}} \text { i.e., } 1: \sqrt{2}\)

Diameter of each glass = 4.6 cm
\(\therefore\) Radius of each glass = 2.3 cm
Height of each glass = 11 cm
(i) (c): Volume of type (A) glass \(=\pi r^{2} h\)
\(=\frac{22}{7} \times 2.3 \times 2.3 \times 11=182.88 \mathrm{~cm}^{3}\)
(ii) (d): Volume of type (B) glass
= Volume of type (A) glass - Volume of hemisphere
\(=182.88-\frac{2}{3} \pi r^{3}=182.88-\frac{2}{3} \times \frac{22}{7} \times 2.3 \times 2.3 \times 2.3 \)
\(=182.88-25.49=157.39 \mathrm{~cm}^{3}\)
(iii) (d) : Volume of type (C) glass = Volume of type (A) glass - Volume of cone
\(=182.88-\frac{1}{3} \pi r^{2} h=182.88-\frac{1}{3} \times \frac{22}{7} \times 2.3 \times 2.3 \times 1.6 \)
\(=182.88-8.86=174.02 \mathrm{~cm}^{3}\)
\(\therefore\)  Required difference = 182.88 - 174.02
= 8.86 cm³ = 8.86 mL
(iv) (b) : Glass of type B has minimum capacity.
(v) (c)

(i) (b): Area to be painted = Area of 14 square faces = 14 x (20)² = 5600 cm²
(ii) (a): Height of pot = 20 cm
Length of pot = 20 x 4 = 80 cm
Breadth of pot = 20 cm
\(\therefore\) Volume of pot = 20 x 80 x 20 = 32000 cm³
(iii) (d) : Required area = 2(l + b) x h
= 2(80 + 20) x 20 = 4000 cm²
Side of coloured square paper = 10 cm
\(\therefore\) Number of pieces of paper required \(=\frac{4000}{10 \times 10}=40\)
(iv) (a) : We have,
\(\text { Radius }(r)=\frac{14.2}{2} \mathrm{~cm}=7.1 \mathrm{~cm}\)
\(\text { Height }(h)=11 \mathrm{~cm} \)
\(\therefore \quad \text { Volume of each sapling }=\pi r^{2} h=\frac{22}{7} \times(7.1)^{2} \times 11 \)
\(=1742.75 \mathrm{~cm}^{3}\)
(v) (b): Total volume of pot = 32000 cm³
Volume of 4 saplings = 1742.75 x 4 = 6971 cm³
So, volume of compost and soil = 32000 - 6971 = 25029 cm³

Diameter of football = Length of base of the box =23cm
\(\therefore\) Radius of football \(=\left(\frac{23}{2}\right) \mathrm{cm}\)
(i) (b): Volume of the football \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \times \frac{22}{7} \times \frac{23}{2} \times \frac{23}{2} \times \frac{23}{2}=6373.19 \mathrm{~cm}^{3}\)
(ii) (a): Area of wrapping sheet = Total surface area of the cuboidal box
= 2(lb + bh + hl) = 2(23 x 23 + 23 x 28 + 28 x 23)
= 2(529 + 644 + 644) = 3634 cm²
(iii) (c) : Volume of the box = I x b x h = 23 x 23 x 28 = 14812 cm³
(iv) (b) : Volume of thermocal balls used
\(=\frac{1}{2} \text { (Volume of box - Volume of football) } \)
\(=\frac{1}{2}(14812-6373.19)=\frac{1}{2} \times 8438.81=4219.405 \mathrm{~cm}^{3}\)
(v) (d): Surface area of the football = \(4 \pi r^{2}\)
\(=4 \times \frac{22}{7} \times \frac{23}{2} \times \frac{23}{2}=1662.57 \mathrm{~cm}^{2}\)

(i) (c) : Required area of canvas = Curved surface area of cone + Curved surface area of cylinder

\(=\pi r l+2 \pi r h=\pi r(l+2 h) \)
\(=\frac{22}{7} \times 21(29+44)=4818 \mathrm{~m}^{2} \)
\(\because l=\sqrt{r^{2}+h_{1}^{2}} =\sqrt{(21)^{2}+(20)^{2}} \)
\(=\sqrt{841}=29 \mathrm{~m} \)
(ii) (b): Area of floor \(=\pi r^{2}\)
\(=\frac{22}{7} \times 21 \times 21=1386 \mathrm{~m}^{2}\)
Number of persons that can be accommodated in the tent \(=\frac{1386}{126}=11\)
(iii) (d) : Since, cost of 100 m2 of canvas = Rs 425
\(\therefore\) Cost of 1 m 2 of canvas = Rs 4.25
Thus, cost of 4818 m2 of canvas = Rs 20476.50
(iv) (c) : Volume of tent = Volume of cone + Volume of cylinder
\(=\frac{1}{3} \pi r^{2} h_{1}+\pi r^{2} h=\pi r^{2}\left(\frac{1}{3} h_{1}+h\right)\)
\(=\frac{22}{7} \times(21)^{2}\left[\frac{20}{3}+22\right]=\frac{9702}{7} \times \frac{86}{3}=39732 \mathrm{~m}^{3}\)
(v) (a): Required rrumber of persons
\(=\frac{\text { Volume of tent }}{\text { Space required by one person }}=\frac{39732}{1892}=21\)

(i) (b): Quantity of ice-cream in the brick
= volume of the brick = (30 x 25 x 10) cm³= 7500 cm³
\(=\frac{7500}{1000} l \quad\left[\because 1 l=1000 \mathrm{~cm}^{3}\right] \)
\(=7.5 \mathrm{l}\)
(ii) (c): Volume of hemispherical scoop \(=\frac{2}{3} \pi r^{3}\)
\(=\frac{2}{3} \times \frac{22}{7} \times(3.5)^{3}=\frac{1886.5}{21}=89.83 \mathrm{~cm}^{3}\)
(iii) (d) : Volume of cone \(=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 15=\frac{4042.5}{21}=192.5 \mathrm{~cm}^{3}\)
(iv) (a) : Number of scoops required to fill one cone
\(=\frac{\text { Volume of a cone }}{\text { Volume of a scoop }}=\frac{192.5}{89.83}=2.14 \approx 2\)
(v) (b): Number of cones that can be filled using the whole brick \(=\frac{\text { Volume of brick }}{\text { Volume of } 1 \text { cone }}\)
\(=\frac{7500}{192.5}=38.96 \approx 39\)

(i) (b):Curved surface area of two identical cylindrical parts \(=2 \times 2 \pi r h=2 \times 2 \times \frac{22}{7} \times \frac{2.5}{2} \times 5\)
= 78.57 cm²
(ii) (a): Volume of big cylindrical part \(=\pi r^{2} h\)
\(=\frac{22}{7} \times \frac{4.5}{2} \times \frac{4.5}{2} \times 12=190.93 \mathrm{~cm}^{3}\)
(iii) (b) :Volume of two hemispherical ends \(=2 \times \frac{2}{3} \pi r^{3}\)
\(=\frac{2 \times 2}{3} \times \frac{22}{7} \times\left(\frac{2.5}{2}\right)^{3}=8.18 \mathrm{~cm}^{3}\)
(iv) (c) : Curved surface area of two hemispherical ends \(=2 \times 2 \pi r^{2}=2 \times 2 \times \frac{22}{7} \times \frac{2.5}{2} \times \frac{2.5}{2}=19.64 \mathrm{~cm}^{2}\)
(v) (b): Difference of volume of bigger cylinder to two small hemispherical ends = 190.93 - 8.18 = 182.75 cm³