81 + Interview Questions in SURFACE-AREAS-AND-VOLUMES IN GENERAL AWARENESS Page 2 InterviewSolution

51.	A cylindrical pipe has inner diameter of 4 cm and water flows through it at the rate of 20 metre per minute. How long would it take to fill a conical tank of radius 40 cm and depth 72cm?
Answer» `d= 4 cm` rate=`20m/(min) = 200 (cm)/(min)` `r= 2 cm` volume= `pir^2h ` `= pi * (2)^2 * 2000 (cm)/(min)` volume of cone = `1/3 pi r^2 h` `= 1/3 xx 22/7 xx (40)^2 xx 72` `= 1/3 xx pi xx 1600 xx 72 cm^3` time taken by the pipe=`(1/3 xx pi xx 1600 xx 72)/( pi xx 4 xx 2000)` `25/5 min` `4 min 48 sec` Answer

Discussion

52.	A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also find the rate of the canvas with 500 per `m^2`
Answer» area of canvas used=curved surface area of cone+curved surface area of cylinder. curved surface area of cone=>`pirl=22/7xx2xx2.8=17.6(m)^2` curved surface area of cylinder=>`2pirh=2xx22/7xx2xx2.1=26.4(m)^2` total surface area=`26.4+17.6=44(m)^2`. cost of canvas=`500xx44=Rs22000`

Discussion

53.	If two solid hemispheres of same base radius r are joined together along their bases, them curved surface area of the this new solid isA. `4pir^(2)`B. `6pir^(2)`C. `3pir^(2)`D. `8pir^(2)`
Answer» Correct Answer - A Because curved surfce area of a hemisphere is `2pir^(2)` and here, we joint two solid hemispheres along their bases of radius r, form which we get solid sphere. Hence, the curved surface area a new solid `= 2 pi r^(2) + 2pir^(2) = 4pir^(2)`

Discussion

54.	Water flows at the rate of 10 meter per minute through a cylindrical pipe having diameter as 5 mm. How much time will it take to fill a conical vessel whose diameter of base is 40 cm and depth 24 cm?
Answer» Let the time taken= t minutes `V_t=(pir^2)l` `=pi(25/100)^21000 t cm^3` `r=2.5mm` `V_t=V_(conical)` `pi25^2/100^21000t=1/3pi20^2*24` t=51 min 12 sec.

Discussion

55.	The curved surface area of cylinder is `264 m^2` and its volume is `924 m^3`.Find the ratio of its height to its diameter.
Answer» CSA=`pirl=264m^2` Volume=`pir^2h=924m^2` Ratio=`(2pirh)/(pir^2h)=264/924` `h/(2r)=(264/(pi2r))/((24*2)/(264))` `h/(2r)=264/(49pi)`.

Discussion

56.	The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of `R s 10 p e r m^2`is Rs 15000, find the height of the hall.[Hint : Area of the four walls = Lateral surface area.]
Answer» LSA=2(lb+bh) X=2(l+b)h X=250h-(1) `1m^2=Rs10` `Xm^2=Rs15000` `10X=Rs15000` `X=Rs1500` from equation 1 250h=1500`m^2` h=6m.

Discussion

57.	The diameter of the moon is approximately `1/4` of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon.
Answer» `D=2r` `D_m/D_e=(2rm)/(2re)=1/4` `r_m=r_e/4` `V_e=4/3pir_e^3` `V_m=4/3pir_m^3=4/3pi(r_e/4)^3=1/644/3pir_e^3` `V_m=V_e/64` Volume of noon is 1/64 part of volume of earth.

Discussion

58.	Find the volume of a sphere whose surface area is 154 `c m^2`
Answer» Here, we are given, the surface area of the sphere is `154cm^2` So, `4pir^2 = 154cm^2` `422/7r^2 = 154cm^2` `r^2 = 49/4=>r = 7/2 = 3.5cm` So, volume of sphere, `V = 4/3pir^3 = 4/322/73.53.53.5` `V = 179.67 m^3`

Discussion

59.	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?
Answer» `r_(mo on)=1/4r_(earth)` `Volume of mo on=4/3pir_(mo on)^3` `Volume of mo on=4/3pir_(earth)^3/64` `Volume of mo on=1/64(4/3pir_(earth)^3)` Volume of mo on=`1/64`Volume of earth

Discussion

60.	How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer» A hemispherical bowl can hold milk equal to its volume. Volume of hemisphere, `V = 2/3pir^3` Here,` r = d/2 = 10.5/2cm` So, `V = 2/322/710.5/210.5/210.5/2` `V = 303.1875cm^3` We know, `1m^3 = 1 litre` So, `1cm^3 = 1/1000 litre ` So, `V ~=0.303litres` That means the given bowl can hold milk upto `0.303` litres.

Discussion

61.	A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Answer» Volume of hemispherical bowl=`2/3 xx pi xx r^3` given `r=3.5cm` volume of water it can contain`=2/3 xx pixx(3.5)^3=179.66(cm)^3`

Discussion

62.	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?
Answer» Total cardboard required=35(CSA+BA) =35(2pirh+BA) =35(222/73+(21+3) =5223*24 `cm^2` =7920`cm^2`

Discussion

63.	A dome of a building is in the form of a hemisphere. From inside, it was white-washedat the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, findthe(i) inside surface area of the dome, (ii) volume of the air inside the dome.
Answer» Cost of white wash(CSA)=Rs498.46 `1m^2=Rs2` CSA of hemisphere=249.48`m^2` `2pir^2=249.48` `r^2=36.69` `r=6.057m` Volume of air inside=Volume of hemisphere =`2/3pir^3` =`2/322/7(6.05)^2` =`465.627m^3`

Discussion

64.	A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand
Answer» Total Volume= Volume of cuboid-Total volume of d =`15103.5-(41/3pi (.5)^214cm)` =`523.53cm^3`

Discussion

65.	Combination of cylinder and hemisphere : A solid is in the form of a cylinder with hemispherical ends. The the whole length of the solid is 108 cm and the diameter of the hemispherical ends is 36 cm; Find the cost of polishing the surface of the solids at the rate of 7 paise per sq. cm.
Answer» two hemisphere together make 1 sphere so surafec area of 1 sphere=`4pir^2=4xx22/7xx18xx18=4073.14(cm)^2` Surface area of cylinder=`2pirh=2xx22/7xx18xx72=8146.28(cm)^2` total surface area=`12219.4(8146.28+4073.14)(cm)^2` Cost of polishing=`7xx12219.4=85536`paise.

Discussion

66.	A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of watercan it hold? (`1 m^3= 1000 l`)
Answer» Water tank can hold water equal to its volume. As tank is a cuboid in shape, its volume will be ,`V = lbh` Here, Length,`l = 6m` Breadth,`b = 5m` Height, `h = 4.5m` So, `V = 654.5 = 135m^3` Also, `1m^3 = 1000l` So, `V = 135000l` So, the water tank can hold `135000` `litres` of water.

Discussion

67.	A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cmwide 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?
Answer» 1) TSA of cube=`4a^2=4*10^2=400cm^2` TSA of Cuboid=`2(l+b)h=2(12+10)8=352cm^2` difference=400-352=48`cm^2` 2)TSA of cube=`6a^2=6(10)^2=600m^2` TSA of cuboid=`2(lb+bh+hl)=352+2(120)=592m^2` difference=600-592=8`cm^2`.

Discussion

68.	A bucket isin the form of a frustum of a cone and holds 28.490 litres of water. Theradii of the top and bottom are 28 cm and 21 cm respectively. Find the heightof the bucket.
Answer» Given volume of the frustum = 28.49 L `= 28.49 xx 1000 cm^(3)" "[therefore 1 L = 1000 cm^(3)] ` `= 28490 cm^(3)` and radius of the top `(r_(1) )= 28 cm` radius of the bottom `(r_(2)) = 21 cm` Let height of the bucket = h cm Now, volume of the bucket `= (1)/(3) pi h(r_(1)^(2) + r_(2)^(2) + r_(1)r_(2))= 28490 ` `rArr " " (1)/(3) xx (22)/(7) xx h (28^(2) + 21^(2) + 28xx 21) =28490` `rArr " "h (784 + 441 +588) = (28490xx3xx7)/(22)` `rArr " " 1813 h = 1925 xx 12` `therefore" "h=(1295 xx 21)/(1813) = (27915)/(1813) = 15 cm`

Discussion

69.	How many shots each having diameter 3 cm can be made form a cuboidal lead solid of dimensions `9 cm xx 11 cm xx 12 cm`?
Answer» Given , dimesion of cuboidal `= 9 cm xx 11 cm xx 12 cm` `therefore ` Volume of cuboidal `= 9 xx 11 xx 12 = 1188cm^(3)` and diameter of short = 3 cm `therefore` " "Radius of shot , `r= (3)/(2) = 1.5 cm` " "volume of shot `= (4)/(3)pir^(3) = (4)/(3) xx (22)/(7) xx (1.5)^(3)` ` " "= (297)/(21) = 14.143 cm^(3)` `therefore ` Required number of shots `= (1188)/( 14.143) = 84` (approx)`

Discussion

70.	Mearbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.
Answer» Given , diameter of a marble = 1.4 cm `therefore` Radius of marble `= (1.4)/(2) = 0.7 cm` So, volume of one marble `= (4)/(3 pi (0.7)^(3)` `=(4)/(3) pi xx 0.343= (1.372)/(3) pi cm^(3)` Also, given diameter of beaker = 7 cm `therefore ` Radius of beaker `=(7)/(2) = 3.5 cm` Height of water level raised water in beaker `= pi (3.5)^(2) xx 5.6 = 68.6 pi cm^(3)` Now, required number of marbles `= ("Volume of the raised water in beaker")/("Volume of one spherical marble")` `=(68.6pi)/(1.372 pi) xx 3=150`

Discussion

71.	Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker, so that the water level rises by 5.6 cm.
Answer» Volume of water=`pir^2h` Change in volume=Volume of marbles `pir^2(h+5.6)-pir^2h=n4/3pi(0.7)^3` `7^2/2^25.6=n4/37/107/107/10` `n=150`.

Discussion

72.	If a hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that `(1)/(8)` space of the cube remains unfilled. Then, the number of marbles that the cube can accommodate isA. 142244B. 142344C. 142444D. 142544
Answer» Correct Answer - A Given , edge of the cube = 22 cm `therefore` Volume of the cube `= (22)^(3) = 10648 cm^(3) " "[because "Volume of cube " = ("Side")^(3)]` Also, given diameter of marble = 0.5 cm `therefore` Radius of a marble, `r = (0.5)/(2) = 0.25 cm" " [because diameter = 2 xx radius]` volume of one marble `= (4)/(3) pi r^(3) = (4)/(3) xx (22)/(7) xx (0.25)^(3)" "[therefore " Volume of sphere " (4)/(3) xx pi xx ("radius")^(3)]` `= (1.375)/(21) = 0.0655 cm^(3)` Filled space of cube = Volume of the cube `(1)/(8) xx ` volume of cube `= 10648-10648 xx (1)/(8)` `= 10648 xx (7)/(8) = 9317 cm^(3)`

Discussion

73.	How manyspherical bullets can be made out of a solid cube of lead whose edge measures44 cm, each being 4 cm in diameter.
Answer» Given that,lost of spherical lead shots made out a soild cube of lead `therefore` Number of spherical shots `= ("Volumeof a solid cube of lead")/("Volume of a spherical lead shot") ` Given that, diameter of a spherical lead shot i.e, sphere = 4 cm `rArr" ""Radius of a spheical lead shot "(r)=(4)/(2)` `r = 2cm " "[therefore "diameter " = 2xx "radius"]` So Volume of a spherical lead shot i..e, sphere `" "= (4)/(3) pi r^(3)` `" "(4)/(3) xx (22)/(7) xx (2)^(3) ` `" "= (4 xx 22 xx 8)/(21) cm^(3)` Now, sine edge of a solid cube (a) = 44 cm so, volume of a soild `= (a)^(3) = (44)^(3) = 44 xx 44 cm^(3)` From Eq. (i) Number of spherical lead shots `= (44 xx 44 xx 44)/(4 xx 22 xx 8) xx 21` `" " = 11 xx 21 xx 11 = 121 xx 21` `" " =2541` Hence, the required number of spherical lead shots is 2541.

Discussion

74.	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
Answer» Actual needed value `= l(b + 20)` `= lb + 20l` `lb = csa + 20l` `b = ( pi r l + 20l)/l` `= (22/7 xx 6 xx 10 + 20 xx 3)/3 ` `= (1320 + 420)/21 = 83m` `b = 83 m - 0.143m` `= 82.857m` now for the cone, `3l = pi xx 6 xx sqrt(8^2 + 6^2)` `l = 22/7 xx 2 xx 10= 62m + 6/7 m` `L= l+ 20cm` `= 62m + (6/7 xx100cm)= ` `~~ 63m ` answer

Discussion

75.	A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Answer» Volume of a cone can be given as, `V = 1/3piR^2h` Here, `R = d/2 = 3.5/2m` `h = 12m` So, `V = 1/322/73.5/23.5/212 = 11**3.5= 38.5m^3` Now, we know, `1m^3 = 1000litre=>1m^3 = 1 kilolitre` so, capacity of given come will be `38.5` `kilolitres.`

Discussion

76.	A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity.
Answer» Apparent Capacity=`pir^2h` =`22/7(5/2)^210` =`1.57125` actual Capacity=approve capacity-`2/3pir^3` =`1.57125-2/33.14(5/2)^3` =`163.54 cm^2`.

Discussion

77.	The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pin is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one-fifth of a litre?
Answer» Given, length of the barrel of a fountain pen = 7 cm and daimeter = 5 mm`= (5)/(10) cm = (1)/(c) cm` `therefore` Radius of the barrel `= (1)/(2 xx2) = 0.25 cm` Volume of the barrel `= pir^(2)h" "["Since, its shape is cylindrical"]` `= (22)/(7) xx (0.25)^(2) xx 7` `22 xx 0.0625 = 1.375 CM^(3)` Also , given volume of ink in the bottole `= (1)/(5)` of liter `=(1)/(5) xx 1000 cm^(3)= 200 cm^(3)` Now, `1.375 cm^(3)` ink used for writing number of wards = 3300 `therefore 1 cm^(3)` ink is used for writing number of words `= (3300)/(1.375)` `therefore 200 cm^(3)` ink used for witing number of words `= (3300)/(1.375) xx 200= 480000`

Discussion

78.	A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Answer» Volume of embankment=Volume of earth dig out `h(pi(r+4)^2-pir^2)=pir^2h` `h=(r^2h)/((r+h)^2-r)` `h=((3/2)^2 14)/(((3/2)+4)^2-3/2)` h=1.125m

Discussion

79.	A well with 10m inside diameter meter is dug 14m deep earth taken out of of its spread all around to a width of 5 m to form an enbankment. Find the hight of the enbankment. Also show me how does an enbankment look like ?
Answer» volume of wall = volume of earth dug out `pi r^2h=` volume of earth `V= 22/7 xx (10/2)^2 xx 14` `= 1100 m^3` `V = pi(10^2 - 5^2)h = 1100 m^3` `22/7 (75) xx h= 110` `h = (50 xx 7)/(75) = 7 xx 2/3 = 14/3m` Answer

Discussion

80.	A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Answer» Inner surface area ofvessel=`4pir^2+2pirh` here r=7cm and h=6cm subtituting values=>`4xx22/7xx49+2xx22/7xx7xx6` =>`880(cm)^2`

Discussion

81.	A solid is in the form of a right circular cone mounted on ahemisphere. The radius of the hemisphere is 3.5cm and the height of the coneis 4cm. The solid is placed in a cylindrical tub, full of water, in such away that the whole solid is submerged in water. If the radius of the cylinderis 5cm and its height is 10.5cm, find the volume of water left in thecylindrical tub. `(U s epi=(22)/7)`
Answer» Given, `r=3.5cm` `h=4cm` `R=5cm` `H=10.5cm` Volume of water left in the cylinder=Volume of cylinder-(volume of cone + volume of hemisphere)=`piR^2H-((pi/3)r^2h + 2/3pir^3)`=`(825-(51.33+89.83)) cm^3` (put (`pi=22/7)`) =`683.84cm^3`

Discussion

Explore topic-wise InterviewSolutions in .

A cylindrical pipe has inner diameter of 4 cm and water flows through it at the rate of 20 metre per minute. How long would it take to fill a conical tank of radius 40 cm and depth 72cm?

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also find the rate of the canvas with 500 per `m^2`

If two solid hemispheres of same base radius r are joined together along their bases, them curved surface area of the this new solid isA. `4pir^(2)`B. `6pir^(2)`C. `3pir^(2)`D. `8pir^(2)`

Water flows at the rate of 10 meter per minute through a cylindrical pipe having diameter as 5 mm. How much time will it take to fill a conical vessel whose diameter of base is 40 cm and depth 24 cm?

The curved surface area of cylinder is `264 m^2` and its volume is `924 m^3`.Find the ratio of its height to its diameter.

The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of `R s 10 p e r m^2`is Rs 15000, find the height of the hall.[Hint : Area of the four walls = Lateral surface area.]

The diameter of the moon is approximately `1/4` of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon.

Find the volume of a sphere whose surface area is 154 `c m^2`

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?

How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?

A dome of a building is in the form of a hemisphere. From inside, it was white-washedat the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, findthe(i) inside surface area of the dome, (ii) volume of the air inside the dome.

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand

Combination of cylinder and hemisphere : A solid is in the form of a cylinder with hemispherical ends. The the whole length of the solid is 108 cm and the diameter of the hemispherical ends is 36 cm; Find the cost of polishing the surface of the solids at the rate of 7 paise per sq. cm.

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

A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cmwide 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?

A bucket isin the form of a frustum of a cone and holds 28.490 litres of water. Theradii of the top and bottom are 28 cm and 21 cm respectively. Find the heightof the bucket.

How many shots each having diameter 3 cm can be made form a cuboidal lead solid of dimensions `9 cm xx 11 cm xx 12 cm`?

Mearbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker, so that the water level rises by 5.6 cm.

If a hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that `(1)/(8)` space of the cube remains unfilled. Then, the number of marbles that the cube can accommodate isA. 142244B. 142344C. 142444D. 142544

How manyspherical bullets can be made out of a solid cube of lead whose edge measures44 cm, each being 4 cm in diameter.

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

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pin is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one-fifth of a litre?

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

A well with 10m inside diameter meter is dug 14m deep earth taken out of of its spread all around to a width of 5 m to form an enbankment. Find the hight of the enbankment. Also show me how does an enbankment look like ?

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.