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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A hemispherical tank is made up of an iron sheet 1 cmthick. If the inner radius is 1 m,then find the volume of the iron used to make the tank |
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Answer» Inner radius `(r_(1))=1m=100 cm` Outer radius `(r_(2))=(100+1)cm " " (because` inner radius + thickness) = 101 cm Outer volume `=(2)/(3)pi(101)^(3) " " (because "volume" =(4)/(3)pi r^(3))` Inner volume `=(2)/(3)pi(100)^(3)` Hence, volume of the iron used to make the tank = outer volume - inner volume `= (2)/(3)pi[(101)^(3)-(100)^(3)]` `=(2)/(3)xx(22)/(7)(101-100)[(101)^(2)+101xx100+(100)^(2)]` `=(2)/(3)xx(22)/(7)(10201+10100+10000)` `=(44)/(21)(30301)=63487.80 cm^(3)` `=(63487.80)/(100xx100xx100)m^(3)=0.063 m^(2)` (approx.) |
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| 2. |
The diameters of three solid balls of iron are 6 cm, 8 cm and 10 cm. They are melted and recast a solid ball. Find the diameter of this ball. |
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Answer» Radius of first ball `r_(1)=(6)/(2)=3 cm` Radius of second ball `r_(2)=(8)/(2)=4 cm` Radius of third ball `r_(3) = (10)/(2)=5 cm` Let radius of new ball = r Now, volume of new ball = sum of the volumes of three balls `rArr (4)/(3)pi r^(3)=(4)/(3)pi r_(1)^(3)+(4)/(3)pi r_(2)^(3)+(4)/(3)pi r_(3)^(3)` `rArr r^(3)=r_(1)^(3)+r_(2)^(3)+r_(3)^(3)` `= 3^(3)+4^(3)+5^(3)` = 27 + 64 + 125 = 216 `rArr r = 6` `therefore` Diameter of new ball `= 2xx6 =12 cm` |
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| 3. |
The radius and height of a right circular cone are 4 cm and 9 cm respectively. Its volume will be : (i) `36pi cm^(3)` (ii) `48pi cm^(3)` (iii) `72pi cm^(3)` (iv) `100pi cm^(3)` |
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Answer» Here, radius r = 4 cm and heighth = 9 cm `therefore` Volume of cone `=(1)/(3)pi r^(2)hh=(1)/(3)pi^(2)h=(1)/(3)pi(4)^(2)xx9=48pi cm^(3)` |
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| 4. |
The length, breadth and height of a cuboid are in the ratio 6 : 5 : 3. If its total surface area is `504 cm^(2)`, find its dimensions. Also find the volume of the cuboid. |
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Answer» If length = 6x, breadth = 5x and height = 3x. `therefore` Total surface area `= 2(lb+bh+hl)` `therefore 504 = 2(6x xx5x + 5x xx 3x + 3x xx 6x)` `rArr 504 = 2(30x^(2)+15x^(2)+18x^(2))` `rArr 63x^(2)=252` `rArr x^(2)=4` or x = 2 Therefore, length `= 6xx2=12 cm` breadth `=5xx2=10 cm` and height `=3xx2=6 cm` Now, volume of cuboid `= l xx b xx h` `rArr` Volume `= 12xx10xx6` `rArr` Volume `=720 cm^(3)` |
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| 5. |
Find the volume and total surface area of a cuboid, whose length = 15 cm, breadth = 10 cm and height = 8 cm. |
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Answer» Given l = 15 cn, b = 10 cm and h = 8 cm Volume of cuboid `= l xx b xx h` Volume of cuboid `= 15xx10xx8=1200 cm^(3)` Total surface area of cuboid `=2(lb+bh+hl)` `=2(15xx10+10xx8+8xx15)` `=2(150+80+120)` `=2xx350` `=700 cm^(2)` |
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| 6. |
A semi-circular sheet of metal of diameter 28cm isbent into an open conical cup. Find the depth and capacity of cup. |
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Answer» Diameter of semi-circular sheet is 28 cm. It is bent to form an open conical cup. The radius of sheet becomes the slant height of the cup. The circumference of the sheetbecomes the circumference of the base of the cone. `therefore` l = slant height of conical cup = 14 cm. Let r cm be the radius and h cm be the height of the conical cup circumference of conical cup circmference of the semi-circular sheet. `therefore 2pi r=pi xx 14 rArr r = 7 cm` Now, `l^(2)=r^(2)+h^(2)` `rArr h=sqrt(l^(2)-r^(2))=sqrt((14)^(2)-(7)^(2))=sqrt(196-49)=sqrt(147)=12.12 cm` `therefore` Capacity of the cup `=(1)/(3)pi r^(2)h=(1)/(3)xx(22)/(7)xx7xx7xx12.12` `=622.16 cm^(3)` |
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| 7. |
An open conical cup is formed from a thin metallic semi-circular sheet of diameter 28 cm. Find its volume. |
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Answer» Radius of semi-circle `R=(28)/(2)=14 cm` Length of arc of semi-circle `=pi R=(22)/(7)xx14=44 cm` Let radius of cone = r `therefore 2pi r=44cm` `rArr r = (44)/(2pi)cm` `rArr r=(44)/(2xx(22)/(7))=7cm` The radius of semi-circule will be equal to the slant height of cone `thereforel=R=14 cm` Now, `h^(2)=l^(2)-r^(2)=14^(2)-7^(2)` = 196 - 49 = 147 `rArr h = sqrt(147)=7sqrt(3)cm` `therefore` Volume of cup `= (1)/(3)pi r^(2)h` `=(1)/(3)xx(22)/(7)xx7xx7xx7sqrt(3)cm^(3)` `=(1078xx1.732)/(3)=622.36 cm^(3)` |
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| 8. |
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 completerevolutions to move once over to level a playground. Find the area of the playgroundin `m^2`. |
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Answer» We have, diameter of a roller = 84 cm r = radius of a roller = 42 cm h = 120 cm To cover 1 revolution = curved surface area of roller `= 2pi rh` `=2xx(22)/(7)xx42xx120 cm^(2)` `=44xx720 cm^(2)=31680 cm^(2)` `=(31680)/(100xx100)m^(2)=3.168 m^(2) " " (because 1m =100 cm)` Area of the playground = takes 500 complete revolutions `= 500xx3.168 m^(2)=1584 m^(2)` |
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| 9. |
How manyballs, each of radius 1 cm, can be made from a solid sphere of lead of radius8 cm? |
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Answer» Radius of metallic ball = 1 cm Volume of one ball `=(4)/(3)pi(1)^(3)=(4)/(3)pi cm^(3)` Radius of sphere = 8 cm `therefore` Volume of sphere `=(4)/(3)pi(8)^(3)cm^(3)` Now, the number of metallic balls `=("volume of sphere")/("volume of one ball")=((4)/(3)pi (8)^(3))/((4)/(3)pi)` `=(8)^(3)=512` |
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| 10. |
Tow solid spheres made of the same metal have weight 5920 g and 740 g, resopectively. Determine the radius of the radius of the larger sphere, if the diameter of the smaller one is 5 cm. |
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Answer» Teo solid sphere madeof the same metal have weight 5920 g and740 g, respectively. Mass per unit volume is called the density. `therefore (V_(1))/(V_(2))=((5920)/(D))/((740)/(D)) rArr ((4)/(3)pi r_(1)^(3))/((4)/(3)pi r_(2)^(3))=(5920)/(740)` `rArr (r_(1)^(3))/(r_(2)^(3))=(5920)/(740) rArr (r_(1)^(3))/(((5)/(2))^(3))=(5920)/(740) " " [because r_(2)=(1)/(2)xx5]` `rArr r_(1)^(3)=(5920)/(740)xx(125)/(8)=125 rArr r_(1)=(12.5)^(1//3)=5cm` Hence, the radiusof the larger sphere = 5 cm. |
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| 11. |
The height and radius of a right circular cone are increased by 20%and 25% respectively. Find the ratio of the volume of new cone and old cone. |
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Answer» Let for old cone, height = h and radius = r Volume `V_(1)=(1)/(3)pi r^(2)h` For new cone, Increase in height = 20% of `h=hxx(20)/(100)=(h)/(5)` `therefore` Height `H=h+(h)/(5)=(6h)/(5)` Increase in radius = 25% of `r=r xx (25)/(100)=(r )/(4)` `therefore` Radius `R = r+(r )/(4)=(5r)/(4)` Now, volume `V_(2)=(1)/(3)pi R^(2)H=(1)/(3)pi ((5pi)/(4))^(2).((6h)/(5))=(1)/(3)pi r^(2)h. (15)/(8)` The ratio of volume `=("Volume of new cone")/("Volume of old cone")=(V_(2))/(V_(1))=((1)/(3)pi r^(2)h.(15)/(8))/((1)/(3)pi r^(2)h)=(15)/(8)=15 : 8` |
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| 12. |
In a right angle, the sides conrtaining right angles are 5 cm and 12 cm. It is rotated aboutits hypotenuse taking it as axis. Find the total surface area and volume of formed figure. |
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Answer» Let `Delta ABC` is given in which AB = 5cm BC = 12cm and `angleABC= 90^(@)` Now, from Phythagoarstheorem `AC^(2)=AB^(2)+BC^(2)` `=5^(2)+12^(2)` = 25 + 144 = 169 `rArr AC=13cm` Let BO = r Now, area of `Delta ABC =(1)/(2)xxABxxBX=(1)/(2)xx5xx12` Again, area of `Delta ABC = (1)/(2)xxACxxBO=(1)/(2)xx13xx r` `therefore (1)/(2)xx13xx r=(1)/(2)xx5xx12` `rArr r = (60)/(13) cm` Now, the volume of solid = volume of cone ABD + volume of cone CBD `=(1)/(3)pi r^(2)xxAO+(1)/(3)pi r^(2)xxOC=(1)/(3)pi r^(2)(AO+OC)` `=(1)/(3)pi r^(2)xxAC=(1)/(3)xx pi xx(60)/(13)xx(60)/(13)xx13=(1200pi)/(13)cm^(3)` and total surface area = curved surface of cone ABD + curved surface of cone CBD `= pi r(AB)+ pi r(BC)` `= pi r(AB+BC)` `= pi xx(60)/(13)(5+12)` `=(1020pi)/(13)cm^(2)` |
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| 13. |
If `h , c ,V`are respectively the height, the curved surfaceand the volume of a cone, prove that `3piV h^3-C^2h^2+9V^2=0.` |
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Answer» We know that : `C = pi rl, V=(1)/(3)pi r^(2)h, l^(2)=h^(2)+r^(2)` `L.H.S. =3pi Vh^(3)-C^(2)h^(2)+9V^(2)` `=3pi((1)/(3)pir^(2)h)h^(3)-(pi rl)^(2)h^(2)+9 ((1)/(3)pi r^(2)h)^(2)` `pi^(2)r^(2)h^(4)-pi^(2)r^(2)h^(2)(h^(2)+r^(2))+9xx(1)/(9)pi^(2)r^(4)h^(2) " " (because l^(2)=h^(2)+r^(2))` `= pi^(2)r^(2)h^(4)-pi^(2)r^(2)h^(4)-pi^(2)r^(4)h^(2)+pi^(2)r^(4)h^(2)` = 0 = R.H.S. |
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| 14. |
A cylinder is formed by folding a square of side 5 cm. Find its curved surface area and the volume. |
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Answer» By folding the square in the shape of cylinder, height of cylinder h = 5 cm and `2pi r=5` `rArr r=(5)/(2pi)=(5xx7)/(2xx22)=(35)/(44)cm` Now, curved surface area of cylinder = area of paper `=5xx5=25 cm^(2)` and volume `= pi r^(2)h` `=(22)/(7)xx(35)/(44)xx(35)/(44)xx5` `=(875)/(88) cm^(3)` |
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| 15. |
The radius of base of a cylinder is 7 cm and height is 10 cm. Find its curved surface area and volume. |
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Answer» Radius of the base of cylinder r = 7 cm, height h = 10 cm `therefore` Curved surface area of cylinder `=2pi rh` `=2xx(22)/(7)xx7xx10` `=440 cm^(2)` and volume `=pi r^(2)h` `=(22)/(7)xx7xx7xx10` `=1540 cm^(3)` |
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| 16. |
Find the curved surface area of the cylinder whose height is 20 cm and the radius of base is 7 cm. |
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Answer» Here, height h = 20 cm, radius r = 7 cm `therefore` Curved surface area `= 2pi rh=2xx(22)/(7)xx7xx20=880` sq cm |
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| 17. |
The area of the base of a right circular cylinder is `42pi cm^(2)` and height is 3.5 cm. Find its volume. |
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Answer» Volume of cylinder = Area of base `xx` Height `=42pi xx3.5 cm^(3)` `=42xx(22)/(7)xx3.5 cm^(3)` `=462 cm^(3)` |
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| 18. |
The diameter of a right circular cylinder is increased by 20%. Find the percentage decrease in its height if its volume remains unchanged. |
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Answer» Let height of cylinder = 100 units and diameter `= D rArr` radius `=(D)/(2)` Volume of cylinder `V_(1)=pi((D)/(2))^(2)xx100=25pi D^(2)` Now, increase in diameter = 20% of D `=Dxx(20)/(100)=(D)/(5)` and new diameter `=D+(D)/(5)=(6D)/(5)` `rArr` new radius `=(6D)/(5xx2)=(3D)/(5)` Let new height = h `therefore` Volume `=pi ((3D)/(5))^(2) h = (9)/(25)pi D^(2)h` Given that, `(9)/(25)pi D^(2)h=25 pi D^(2)` `rArr h = (625)/(9)` `therefore` decrease in height `=100-(625)/(9)=(900-625)/(9)=(275)/(9)` and % decrease in height`=(275)/(9xx100)xx100%=30(5)/(9)%` Alternative Method (Short Trick) : `V=pi r^(2)h i.e., r,r,h` (`pi` is constant) % change in volume `-a+b+c+(ab+bc+ca)/(100)+(abc)/((100)^(2))` Here `0=20+20-h+(20(20)+20(-h)+(-h)(20))/(100)+(20(20)(-h))/(100xx100)` `rArr 0=40-h+(400-40h)/(100)-(4h)/(100)` `rArr h=40=(400-44h)/(100)` `rArr 100h-4000=400-44h` `rArr 144h=4400 rArr h=(4400)/(144)=30(5)/(9)%` |
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| 19. |
The diameter and height of a right circular cylinder are equal to the diameter of a sphere. Find the ratio of volumes of cylinder and sphere. |
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Answer» Let radius of sphere = r `therefore` Volume of sphere `V_(1)=(4)/(3)pi r^(3)` Now, the diameter of cylinder = diameter of sphere = 2r `therefore` radius of cylinder = r and height of cylinder = 2r Now, volume of cylinder `= pi r^(2)(2r)=2pi r^(3)` `therefore ("Volume of cylinder")/("Volume of sphere")=(2pi r^(3))/((4)/(3)pi r^(3))=(3)/(2)` `rArr` Required ratio = 3 : 2 |
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| 20. |
A cone, a hemisphere and a cylinder stand on equalbases and have the same height. Show that their volumes are in the ratio1:2:3. |
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Answer» A cone, a hemisphere and a cylinder stand on equal bases and same height. Let radius of each of a cone, a hemisphere and cylinder be r (equal bases) Heightof hemisphere = r, So, the height of the cone = r, and height of cylinder = r. Now, volume of cone, `V_(1)=(1)/(3)pi r^(2)h=(1)/(3)pi r^(2)xx r=(1)/(3)pi r^(3)` Volume of hemisphere, `V_(2)=(2)/(3)pi r^(3)` and volume of cylinde, `V_(3)=pi r^(2)h=pi r^(2)xx r=(3)/(3)pi r^(3)` `therefore V_(1):V_(2):V_(3)=(1)/(3)pi r^(3):(2)/(3)pi r^(3) : (3)/(3)pi r^(3)=1 : 2 : 3` i.e.,the ratio of their volumes is 1 : 2 : 3. |
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| 21. |
Shanti Sweet Stall was placing an order for making carboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions `25cm xx20cm xx5cm` and the smaller of dimensions `15cm xx 12cm xx5cm`. For all the overlaps, 5% of the total surface area is required extra. If the cost of the carboard is Rs. 4 for `1000 cm^(2)`, find the cost of carboard required for supplying 250 boxes of each kind. |
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Answer» Dimension for bigger box, l = 25 cm, b = 20 cm and h = 5 cm Total surface area of the bigger size bos `=2(l xx b+b xx h+h xxl)` `=2(25xx20+20xx5+5xx25)cm^(2)` `=2(500+100+125)cm^(2)` `=2(725)cm^(2)=1450 cm^(2)` Dimension for smaller box, l = 15 cm, b = 12 cm and h = 5 cm Total surface area of the smaller size box `=2(15xx12+12xx5+5xx15)` `=2(180+60+75)cm^(2)=2(315)cm^(2)=630 cm^(2)` Area for all the overlaps `=5%xx(1450+630)cm^(2)=(5)/(100)xx2080 cm^(2)=104 cm^(2)` Total surface area of both boxes and area of overlaps `=(2080+104)cm^(2)=2184 cm^(2)` Total surface area for 250 boxes `= 2184xx250 cm^(2)` Cost of the cardboard for `1 cm^(2)= Rs. (4)/(1000)` Cost of the cardboard for `2184xx250 cm^(2)= Rs. (4xx2184xx250)/(1000)=Rs. 2184` |
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| 22. |
Find the volume and curved surface of a sphere whose diameter is 6 cm. |
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Answer» Here, 2r = 6 `rArr r - 3 cm` `therefore` Volume of shpere `=(4)/(3)pi r^(3)=(4)/(3)pi(3)^(3)` `= 36pi cm^(3)` and curved surface `= 4pi r^(2)` `= 4 pi (3)^(2)` `= 36 pi cm^(2)` |
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