Volume Calculation - Online Quiz


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Following quiz provides Multiple Choice Questions (MCQs) related to Volume Calculation. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Questions and Answers

Q 1 - The entire surface range of a cuboid 24 cm long, 14 cm wide and 7.5 cm high is:

A - 2520 cm2

B - 1260 cm2

C - 1242 cm2

D - 621 cm2

Answer : C

Explanation

Zone of the entire surface = 2(Lb+ bh +Lh)
= 2 (24*14 + 14*15/2 + 24* 15/2) cm2
= 2(336+105+180) cm2= (621*2) cm2
= 1242 cm2

Q 2 - A rectangular water store contains 42000 liters of water. On the off chance that the length of store is 6 m and its broadness is 3.5m, then the profundity of the supply is:

A - 2 m

B - 5 m

C - 6 m

D - 8 m

Answer : A

Explanation

Volume of reservoir = 42000 ltr. = 42 cubic mtr.
Let the depth be x mtr. Then,
6*7/2*x =42 ⇒ x = 2 m
∴ Depth =2 m

Q 3 - The measurements of a cuboid are a, b,c units, its volume is V cubic units and its entire surface zone is S sq. units. At that point, 1/V=?

A - S/2(a+b+c)

B - 2/S(1/a + 1/b+ 1/c)

C - 2S(a+b+c)

D - 2S/(a+b+c)

Answer : B

Explanation

1/V =(1/S*S/V) = 2(ab+bc+ca)/s*abc=   2/S(1/a+1/b+1/c)

Q 4 - The result of the regions of three neighboring countenances of a rectangular box is equivalent to:

A - The volume of the container

B - twice of the volume of the crate

C - The square of the volume of the container

D - block foundation of the volume of the crate

Answer : C

Explanation

Product of areas of 3 adjacent faces = (Lb*bh*Lh)
=(L2*b2*h2) = (L*b*h) 2= V2

Q 5 - The aggregate surface zone of a solid shape is 150 cm2. Its volume is:

A - 64 cm3

B - 125 cm3

C - 150 cm3

D - 216 cm3

Answer : B

Explanation

6a2=150 ⇒a2= 25 ⇒a =5cm
Volume of the cube = a3= (5*5*5) cm3= 125cm3

Q 6 - A metal 3D square of edge 12cm is liquefied and framed into three littler 3D shape. In the event that the edge of two littler 3D shapes is 6cm and 8 cm. the edge of the third littler solid shape is:

A - 10 cm

B - 14 cm

C - 12 cm

D - 16 cm

Answer : A

Explanation

Let the edge of third smaller cube be a cm. Then,
(6) 3+ (8) 3+a3= (12) 3 ⇒ (216+512) +a3=1728
⇒a3= (1728-728) = 1000= (10) 3⇒a= 10cm
∴ Edge of third smaller cube = 10 cm

Q 7 - The measurement of the base of a tube shaped drum is 35dm and its tallness is 24 dm. It is brimming with lamp oil. What number of tins each of size 25cm *22cm* 35 cm can be loaded with lamp fuel from the drum?

A - 120

B - 600

C - 1020

D - 1200

Answer : D

Explanation

r= 35/2 dm=(35/2*10)cm= 175 cm , h=24 dm = 240cm
Volume of drum = (22/7*175*175*240) cm3
=(22*25*175*240) cm3
Volume of a tin = (25*22*35) cm3
Number of tin = (22*25*175*240)/ (25*22*35) = 1200

Q 8 - If the tallness of a barrel is expanded by 15% and the span of the base is diminished by 10%, then by what percent will its bended surface region change?

A - 3.5 % diminished

B - 3.5% expansion

C - 5% diminish

D - 5% expansion

Answer : B

Explanation

Let the original radius =r and height = h
Then curved surface area = 2πrh
New height = 115% of h = (115/100*h) = 23h/20
New radius = 90% of r = (90/100*r) = 9r/10
New curved surface area = (2π*9r/10*23h/20) = 207πrh/100
Increase = (207πrh/100-2 πrh) = 7 πrh/100
Increase %= (7 πrh/100*1/2 πrh*100) %= 3.5%

Q 9 - A empty greenery enclosure roller 63cm wide with a circumference of 440 cm is made of iron 4 cm thick. The volume of the iron utilized is:

A - 56372 cm3

B - 58752 cm3

C - 54982 cm3

D - 57636 cm3

Answer : B

Explanation

2πr= 440 ⇒2*22/7*r= 440 ⇒r= (440*7/44) = 70 cm
Outer radius = 70cm, inner radius = (70-4) = 66cm
Volume of iron = π [(70)2-(66)2*63cm3= (22/7*136*4*63)
= 58752 cm3

Q 10 - If 1 cubic cm of cast iron weight 21 gms. At that point the heaviness of a cast iron funnel of lenght1 m with a drag of 3 cm and in which the thickness of the metal is 1cm, is:

A - 21 kg

B - 24.2 kg

C - 18.6 kg

D - 26.4 kg

Answer : D

Explanation

External radius = 2.5cm, length= 100cm
External volume = [π*(2.5)2*100] cm3
Internal radius = 1.5 cm
Internal volume = [π*(1.5)2*100] cm3
Volume of metal =[π*(2.5)2*100-π*(1.5)2*100] cm3
=π*100*{(2.5)2-(1.5)2} = (22/7*100*4*1) cm3
Weight of metal = (8800/7 -21/1000) kg = 26.4 kg


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