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Volume Calculation - Online Quiz
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.
Q 1 - An iron shaft is 9 m long, 40 cm wide and 20 cm high. In the event that 1 cubic meter of iron measures 50 kg. What is the heaviness of the bar?
Answer : C
Explanation
Volume of beam = (9*40/100*20/100)m3= 18/25m3 Weight of the bean = (18/25*50) kg = 36 kg
Q 2 - Half cubic meter of gold sheet is stretched out by pounding in order to spread a region of 1 hectare. The thickness of the sheet is:
Answer : C
Explanation
Area = 1 hectare= 10000m2, Volume = 0.5 m3 Thickness =Volume/Area = (0.5/10000)m =(0.5*100/10000)cm=0.005cm
Q 3 - A rectangular tank measuring (5m*4.5 m*2.1 m) is dove in the focal point of field measuring 13.5m*2.5 m. The earth burrow is spread equitably over the remaining part of the field. What amount is the level of the field raised?
Answer : C
Explanation
Volume of the earth dugout = (5*9/2*21/10) m3= 189/4 m3 Area of the remaining field = [(27/2*5/2) - (5*9/2)] m2 = (135/4-45/2) m2= 45/4m2 Let the level of the field be raised h meter. Then, 45/4*h = 189/4 ⇒h= 189/45 m =21/5 m =4.2 m
Q 4 - A rectangular square 6 cm*12 cm *15 cm is sliced into careful no. of equivalent solid shapes. The slightest conceivable number of blocks will be:
Answer : D
Explanation
Each side of required cube = HCF of 6cm, 12cm, 15cm = 3cm Volume of given block = (6*12*15) cm3= 1080 cm3 Volume of each cube = (3*3*3) cm3=27cm3 ∴ required number of cubes = 1080/27= 40
Q 5 - The aggregate surface zone of a solid shape of side 27 cm is:
Answer : C
Explanation
Surface area =6a2= (6*27*27) cm2= 4374cm2
Q 6 - The rate increment in the surface zone of a 3D square (cube) when every side is multiplied is:
Answer : C
Explanation
Let each side be a, Then its surface area =6a2 New side =2a, new surface area = 6(2a) 2=24a2 Increase % = (18a2/6a2*100) %= 300%
Q 7 - The tallness of the barrel is 14 cm and its bended surface range is 264 cm2. The volume of the barrel is:
Answer : B
Explanation
Given h= 14cm and 2πrh= 264 ∴ 2*22/7*r *14= 264 ⇒88r= 264 ⇒r =3 Volume = πr2h = (22/7*3*3*14) cm3= 396 cm3
Q 8 - A divider with 10 m inside diameter is burrowed 14 m profound. Earth taken out of it is spread all around to a width of 5 m to shape a dike. The tallness of the bank is:
Answer : C
Explanation
Volume of the earth dugout = πr2h = (22/7* 5*5*14) m3= 1100m3 Area of embankment = π (R2- r2) = 22/7*[(10)2-(5)2] = (22/7*75) m2 Let the height of the embankment be h meters. Then, 22/7*75*h = 1100 ⇒h = (1100*7/22*1/75) m = 14/3 m = 4.66m
Q 9 - The bended surface region and the aggregate surface zone of a barrel are in the proportion 1:2. In the event that the aggregate surface territory of the barrel is 616 cm2, then its volume:
Answer : A
Explanation
2πrh/2πr (h+r) =1/2⇒ h/(h+r)=1/2 ⇒2h= h+r⇒h=r 2πr (h+r) = 616 ⇒2πr*2r= 616 ⇒πr2= 154 ⇒22/7*r2=154 ⇒ r2 = (154*7/22) =49 ⇒ r = 7cm and h= 7cm ∴ Volume of cylinder = πr2h = (22/7*7*7*7) cm3= 1078 cm3
Q 10 - The bended surface region of a circle is 5544 cm2. Its volume is:
Answer : A
Explanation
4πr2= 5544 ⇒4*22/7* r2 =5544 ⇒ r2= (5544*7/88)=(63*7)= (72*32)⇒ r =(7*3)=21 Volume= 4/3πr3= (4/3*22/7*21*21*21) cm3= 38808cm3