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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. |
The radius of the largest sphere that can be cut off from a solid cube of side x unit each will beA. x unitB. 2x unitC. `x/2` unitD. 4x unit |
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Answer» Correct Answer - C The length of each side of the cube = x unit ` :. ` the diameter of the largest sphere ` :.` the diameter of the largest sphere = ` x/2` unit |
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| 2. |
The height of the water - level when the water of a right circular cone - shaped bottle of radius r unit and of height h unit is poured into a right circular cylinder shaped pot of radius mr unit will beA. `h/2m^(2)` unitB. `(2h)/m` unitC. `h/(3m^(2))` unitD. `m/(2h)` unit |
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Answer» Correct Answer - C The radius of the conical bottle = r unit and height = h unit ` :. ` the volume of the bottle ` = 1/3 pi^(2) h ` cubic - unit Again , the radius of the cylindrical pot = mr unit ` :. ` the volume of the water poured into the pot ` = pi xx (mr)^(2) xx x` cubic - unit ` :. pi (mr)^(2) x = 1/3 r^(2) h ` ` rArr m^(2) r^(2) x = (r^(2)h)/3` ` rArr x = h/(3m^(2))` ` :. ` the water - level of the pot will rise `h/(3m^(2))` unit . |
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| 3. |
A hemisphere of internal diameter 36 cm is filled with water . If this water is poured into right circular bottles of radius 3 cm and of height 6 cm , then how many bottles will be needed ? |
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Answer» The radius of the hemisphere = `36/2 ` cm = 18 cm ` :. ` the volume of the hemisphere ` = 2/3 pi xx 18^(3) ` cc The radius of each right circular bottle = 3 cm and its height = 6 cm ` :. ` The volume of each bottle ` = pi xx 3^(2) 6 ` cc Let the number of bottles to be inded be x ` :. x xx pi xx 3^(2) xx 6 = 2/3 xx pi xx 18^(3) ` ` rArr x = ( 2 xx 18 xx 18 x 18)/(3 xx 3 xx 3 xx 6 ) rArr = 72 ` Hence the required number of bottles = 72 |
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| 4. |
There is some water in a right circular cylindrical pot of diameter 24 cm . How much the height of water - level will be increased if 60 solid conical iron piece of diameter 6 cm and of height 4 cm each are completely immersed into that water ? |
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Answer» The diameter of the conical iron - piece = 6 cm ` :. ` the radius of the conical iron piece = `6/2 ` cm = 3 cm Also , the volume of each iron - piece `= 1/2 xx pi xx (3)^(2) xx 4 "cc"` ` 12 pi "cc "` Again , the diameter of the conical pot = 24 cm ` :.` the radius of the coniacal pot = 24 cm Let water - level will rise h cm ` :. ` the volume of the raised water = ` pi xx (12)^(2) xx h "cc"` = 144 `pih " cc"` As per question , `144 ph = 60 xx 12 pi` ` rArr h = (60 xx 12pi)/(144 pi)` ` rArr h = 5 ` Hence the water - level of the pot will rise 5 cm |
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| 5. |
A conical flask of height 24 cm is full of water . If this water is poured into another cylindrical flask of radius half of the conical flask , then how much water - level of the cylindrical flask will be raised ? |
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Answer» Let the radiuse of the base of the conical flask = r cm ` :. ` the radius of base of the cylindrical flask = `r/2 ` cm ` :.` the volume of the conical flask ` = 1/3 pi r^(2) h ` cc ` = 1/3 pir^(2) xx 24 ` cc ` = 8pir^(2) ` cc Let water - level into the cylindrical flask will rise h cm As per condition ` pi xx (r/2)^(2) xx = 8 pir^(2)` ` rArr 1/4 pir^(2) h = 8pir^(2) rArr h = 32 ` Hence the water - level into cylindrical flask will rise 32 cm |
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| 6. |
The ratio of the radii of a right circular cylinder cone is 3: 4 and the ratio of their heights is 2:3 Then find the ratio of the volumes of them |
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Answer» Let the radii of the cylinder and the cone be 3r unit and 4r unit respectively and their heights be 2h unit and 3h unit respectively . ` :. ` Volume of the cylinder ` = pi xx (3r)^(2) xx 2h` cubic - unit = `18pir^(2)h` cubic - unit and the volume of the cone ` = 1/3 pi xx (4r)^(2) xx 3h ` cubic - unit = ` 16pir^(2) h ` cubic - unit Hence the ratio of the volumes of cylinder and the cone is ` 18pir^(2) h : 16pir^(2) h` ` = (18pir^(2)h)/(16pir^(2)h) = 9/8 = 9 :8 ` Hence the required ratio = 9 : 8 |
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| 7. |
The lower part of a tent is of shape of a right circular cylinder , the height of which is 3.5 metres . The upper part of the the tent is of a shape of a right circular cone . If the total height of the tent be 9.5 metre and it its diameter of base be 5 metre , then how much quantity of tarpailin will be required to make 14 such tents ? |
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Answer» The diameter of the base of the tent = 5 cm ` :. ` the radius of the base of the tent = `5/2 ` m = 2.5 m height of the cylindrical part = 3.5 m ` :.` the curve surface area of this ppart = ` 2pirh = 2 xx 22/7 xx 2.5 xx 3.5 ` sq - metre = 55 sq- metre The radius of the conical upper part of the tent = 2.5 m and height = `(9.5 - 3.5) m = 6 ` m The slant height of the conical ` = sqrt(r^(2) + h^(2))` or , ` l = sqrt((2.5)^(2) + (6)^(2)) ` or , `l = sqrt(42.25 )` m = 6.5 m ` :. ` the curved surface area of this conical part ` = pirl ` ` = 22/7 xx 2.5 xx 6.5 ` sq- m ` :.` the tarpaulin required to make one tent = ` ( 55 + 22/7 xx 2.5 xx 6.5 ) `sq - metre ` :. ` the tarpaulin required to make 14 tent = ` 14 ( 55 + 22/7 xx 2.5 xx 6.5) ` sq - metre Hence the required quantity of tarpaulin = 1485 sq - metre |
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| 8. |
One part of an iron - pillar is cylindrical and the rest part is conical . The radii of the cylindrical and conical part 8 cm and their heights are 240 cm and 36 cm respectively , If the weight of 1 cc iron be 7.8 gm , what will be the total weight of the pillar ? |
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Answer» The radius of the cylindrical part = 8 cm and its height = 240 cm ` :.` the volume of the cylindrical part ` = 22/7 xx 8^(2) xx 240 ` cubic cm ` = 22/ 7 xx 64 xx 240 ` cc Again , the radius of the conical part = 8 cm and its height = 36 cm ` :. ` the volume of the conical part ` = 1/3 xx 22/7 x 8^(2) xx 36 ` cc ` = 22/7 xx 64 xx 2 ` cc So, the total volume of the pillar ` = ( 22/7 xx 64 xx 240 + 22/7 xx 64 xx 12) ` cc ` = 22/7 xx 64 xx( 240 + 12 )` cc Now, the weight of 1 cc pillar is 7.8 gm ` :.` the weight of 1 cc pillar is 7.8 gm ` :. ` the weight of 1 cc pillar is 7.8 gm ` :.` the weight of 1 cc pillar is 7.8 gm ` :. ` the weight of ` 22/7 xx 64 xx 252 ` cc is `7.8 xx 22/7 xx 64 xx 252 ` gm = 395366.4 gm = 395 . 3664 kg Hence the weight of the total pillar = 395.37 kg (approx ) |
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| 9. |
Half of a tank of length 21 dcm , breadth 11 dcm and depth 6 dcm and depth 6 dcm is full of water . If 100 spheres made of iron and of diameter 21 cm each are fully immeresed into this water of tank , then how many dcm of the water - level of the tank will be raised ? |
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Answer» The length of the tank = 21 dcm breadth of the tank = 11 dcm , and depth of the tank = 6 dcm Let the water - level of the tank rises x dcm after the complete immersion of 100 oiron - spheres ` :. ` the volume of the raised - water = `(21 xx 11 xx x)` cubic - dcm Again , diameter of each sphere = 21 cm ` :. ` radius of each sphere = `21/2 ` cm = `(21)/(2 xx 10) ` dcm = 1.05 dcm ` :.` the volume of 100 iro - spheres ` = 100 xx 4/3 xx 22/7xx (105)^(3) ` cubic - dcm As per question , ` 21 x 11 xx x = 100 xx 4/3 xx 22/ 7 xx (1.05)^(3)` ` rArr x = ( 100 xx 4 xx 22 xx 1.05 xx 1.05 xx 1.05 )/(3 xx 7 xx 21 xx 1)` ` = ( 100 xx 4 xx 22 xx 105 xx 105 xx 105 )/(3 xx 7 xx 21 xx 11 xx 100 xx 100 xx100 )` = 2.1 Hence the water - level of the tank will rise 2.1 dcm |
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| 10. |
Half of a tank of length 5 metre , breadth 4 metre and height 2.2 metre , is filled with water How many iron - bullets of radius 5 cm each should be completely immersed into the water of the tank so that the tank full to the brim ? |
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Answer» The radius of each iron - bullet = 5 cm ` :. ` the volume of each iron - bullet ` = 4/3 pi xx 5^(3) ` cc Let the required number of iron - bullets be x ` :.` the total volume of x bullets ` = x xx 4/3 pi xx 5^(3) ` cc Since half of the tank is filled with water , so the water - level of the tank will be increased ` (2.2)/2 ` m = 1.1 m to be filled to the brim So , the volume of the raised - water ` = 5 xx 4 xx 1.1 ` cubic - metre ` = 500 xx 400 xx 110 ` cc As per question , `4/3 xx 22/7 xx 5^(3) xx x = = 500 xx 400 xx 110` ` rArr x = (500 xx 400 xx 110 xx 3 xx 7 )/(4 xx 22 xx 125 ) rArr x = 42000` Hence the required number of iron - bullets = 42000 |
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| 11. |
What will be the ratio of the volumes of a solid cone , a solid hemisphere and a solid cylinder when their radii of bases and heights are equal ? |
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Answer» Let radius of the cone , hemisphere and cylinder be r unit and their equal heights be h unit . Since the heights of the three objects are equal , ` :. ` height of the cone = height of the cylinder = height of the hemisphere = radius of the hemisphere : volume of the cylinder ` = 1/3 pir^(2) h : 2/3 pir^(3) : pir^(2) h ` ` = 1/3 pi r^(2) . r , 2/3 pir^(3) : pir^(2) r ` ` = 1/3 pi r^(3) : 2/3 pir^(3) : pir^(3) ` ` = 1/3 : 2/3 : 1 ` = 1: 2 : 3 Hence the required ratio is 1 : 2 : 3 |
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| 12. |
The upper part of a solid right circular cylindrical pillar is a hemisphere . What will be the volume of the pillar it the radius of base of it is 2 metres and its total length is 10 metres ? |
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Answer» Since the radius of base is 2 metres , so the radius of the hemisphere of the upper part of the pilar is 2 metres . ` :.` the height o the cylindrical part = `(10-2)` metres = 8 metres Now, the volume of the hemisphere ` = 1/2 . 4/3 xx pixx 2^(3)` cubic - metre ` = 16/3` pi cubic - metre and the volume of the cylindrical part ` = pi xx (2)^(2) xx 8 ` cubic - metre = `32 pi` cubic - metre ` :.` the total volume ` ( 16/3 pi + 32 pi )` cubic - metre ` = pi (16/3 + 32 ) ` cubic - metre ` pi xx 112/3 `cubic - metre ` = 22/7 xx 112 / 3` cubic - metre ` 352 / 3 ` cubic - metre = ` 117 1/3 ` cubic - metres Hence the total volume of the pillar = ` 117 1/3 ` cubic - metres . |
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| 13. |
The radius of base of a solid right circular iron - rod is 32 cm and the length of the rod is 35 cm . How many solid right circular cone of radius of base 8 cm and of height 28 cm can be made by melting this rod ? |
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Answer» The radius of the base of the rod = 32 cm and its length = 35 cm ` :.` volume of the rod ` = pi xx (32)^(2) xx 35 ` cc Again , the radius of base of each cone = 8 cm and height of each cone = 28 cm ` :. ` volume = ` 1/3 xx pi xx (8)^(2) xx 28 xx x = pi xx (32)^(2) xx 35 ` ` rArr x= ( 32 xx 32 xx 35 xx 3)/(8xx 8xx 28 ) ` ` rArr x = 60 ` Hence the required number of solid cone is 60 . |
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| 14. |
How many coins of diameter 1.5 cm and of thickness 0.2 cm can be made by melting a metal right circular cylinder of diameter 4.5 cm and height 10 cm ?A. 430B. 440C. 450D. 460 |
| Answer» Correct Answer - C | |
| 15. |
How many solid spheres of diameter 2.1 dcm each can be made by melting a solid rectangular parallelopiped copper piece of length 6.6 dcm , of breadth 4.2 dcm and of thickness 1.4 dcm ? How many cubic - dcm metal will possess in each sphere ? |
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Answer» The length of copper piecec = 6.6 dcm = 66 cm The breadth of copper piece = 4.2 dcm = 42 cm Height of copper piece = 1.4 dcm = 14 cm So , the volume of coper piece = ` ( 66 xx 42 xx 14) ` cc The diameter of each piece = `(66 xx 42 xx 14) ` cc The diameter of each sphere = 2.1 dcm = 21 cm ` :.` the radius of each sphere ` = 21/2 ` cm ` :. ` the volume of each sphere = `21/2 ` cm ` :.` the volume od each sphere =` 4/3 xx 22/7 xx (21 xx 21 xx 21 )/(2 xx 2 xx 2 ) ` ` = 11 xx 21 xx 21` cc Let the number of sphere = ` 11 xx 21 xx 21 ` cc ` = (11 xx 21 xx 21 )/(1000) ` cubic - dcm = 4.851 cubic dcm Hence , the required number of sphere is 8 and the volume of each sphere is 4.851 cubic - dcm |
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| 16. |
The external radius of a hollow sphere of thickness 1 cm, made of lead - plate is 6 cm If a solid right circular rod of radius 2 cm is made by melting this hollow sphere , then what will be the length of the rod ? |
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Answer» The external radius of the hollow sphere = 6 cm The thickness of the sphere = 1 cm ` :.` the inner radius of the hollow sphere |
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| 17. |
The total surface area of a solid hemisphere is 1848 sq. cm. A solid right circular cone is made by melting this hemisphere. If the radius of base of the cone is equal to the radius of the hemisphere , then what will be the height of the cone ? |
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Answer» Let the radius of the hemisphere be r cm So , the total surface area of the hemisphere = ` 3 pir^(2) ` sq - cm As per question , ` 3pir^(2) = 1848` or , ` 3 xx 22/7 xx r^(2) or , r^(2) = (1848 xx 7) (3 xx 22 )` or , `r^(2) = 196 or , r = 14` Also , the volume of the hemisphere ` = 2/3 pi r^(3) ` cc and the volume of the cone ` = 1/3 pir^(2) h` cc As per question , ` 2/3 pir^(3) = 1/3 pir^(2) h` ` rArr h = (2pir^(3))/(pir^(2)) rArr h = 2r rArr h = 2r rArr h = 2 xx 14 rArr h = 28 ` Hence the height of the cone = 28 cm |
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| 18. |
The length of radius of base of a solid right circular cone is equal to the length of radius of a solid sphere . If the volume of the sphere is 2 times the volume of the cone , then find the ratio of the height and radius of base of the cone . |
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Answer» Let the radius of the base of the cone = r unit ` :. ` the radius of the sphere = r unit Let the height of the cone = h unit ` :.` the volume of the cone ` = 1/2 pi r^(2) h` cubic - unit and the volume of the sphere `= 4/3 pi r^(3)` cubic - unit As per question , ` 2xx 1/3 pi r^(2) h = 4/3 pir^(3)` ` rArr h/r = 4/2 rArr h/r = 2/1 rArr h : r = 2 : 1 ` Hence the required ratio is 2:1 |
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| 19. |
If a solid sphere of radius r unit be melted to make a solid right circular cone of height r - unit , then find the radius of the base of the cone . |
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Answer» The radius of the sphere = r unit ` :.` the volume of the sphere `= 4/3 pi r^(3)` cubic - unit ` :.` The volume of the sphere ` = 4/3 pir^(3)` cubic - unit Again , the height of the cone = x unit ` :.` the volume of the cone ` = 1/3 pix^(2) r` cubic - unit As per question , ` 1/3 pi r^(2) r = 4/3 pir^(3) rArr x^(2) = 4r^(2) rArr x^(2) rArr x = 2r ` Hnece the radius of base of the cone is 2r unit |
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| 20. |
If a sphere is made by melting a solid right circular cylinder of radus 2x cm and height x cm , then the radius of the sphere will beA. ` root 3 (3x) ` cmB. ` root 3 (3x) ` cmC. x cmD. 2x cm |
| Answer» Correct Answer - B | |
| 21. |
If a solid right circular cone of height r unit is made by melting a solid sphere of radius r unit . Then the radius of the base of the cone will beA. 2r unitB. 3r unitC. r unitD. 4r unit |
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Answer» Correct Answer - A Radius of the sphere = r unit ` :.` Volume of the sphere ` = 4/3 pir^(3) ` cubic - unit Also , the height of the cone Let the radius of the base of the cone be `r_(1)` unit ` :. ` Let the radius of the base ` = r_(1)` unit ` :. ` Let the radius of the base = ` r_(1)` unit ` :. ` the volume of the cone ` = 1/3 pir_(1)^(2) xx r ` cubic - unit As per question , ` 1/3 pir_(1)^(2) xx r = 4/3 pir^(3) ` ` rArr r_(1)^(2) = 4r^(2) rArr r_(1)^(2) = (2r)^(2) rArr r_(1) = 2r` ` :. ` radius of the cone = 2r unit Hence (a ) is correct |
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| 22. |
The radius of a cone is r unit and its slant height is l unit . If the curved surface area of the cone be equal to the total surface area of a sphere of same radius , then the height of the cone will beA. `sqrt(15) ` r unitB. `sqrt(12)` r unitC. `sqrt(10)` r unitD. `sqrt(17)` r unit |
| Answer» Correct Answer - A | |
| 23. |
A right circular cone is made by cutting a cubical wooden block of side 9 cm each , then what will be the greatest volume of the cone ? |
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Answer» The length of each side of the cube = 9 cm i.e ., if the radius of the required cone r cm , then 2r = 9 `rArr r = 9/2` ` :.` height of the cone = 9 cm ` :.` colume of the cone `= 1/3 pir^(2)h ` cc ` 1/3 xx 22/7 xx (9/2)^(2) xx 9 ` cc ` = 2673/14` cc = 190 . 93 cc (approx.) Hence the greatest volume of the cone = 190.93 cc (approx) |
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