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10 Jun 2020, 03:30
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C
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55% (03:02) correct
45%
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In a rectangular box of dimensions 20 centimetres by 20 centimetres by 100 centimetres, identical spherical balls are arranged in layers such that each layer has exactly 4 balls. The top view of this arrangement is shown in the figure above. What is the approximate percentage of space left empty inside the box, if the box contains the maximum possible number of layers of such spherical balls? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius of the sphere)
A. 33%
B. 40%
C. 50%
D. 60%
E. 67%
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Updated on: 10 Jun 2020, 04:04
Originally posted by RaghavKhanna on 10 Jun 2020, 03:44.
Last edited by RaghavKhanna on 10 Jun 2020, 04:04, edited 2 times in total.
Last edited by RaghavKhanna on 10 Jun 2020, 04:04, edited 2 times in total.
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Let r be the radius of a sphere
=> 4r = 20
=> r = 5cm
Let the number of layers be : N
=> Height of the box = N*2r
=> N = 100/10 = 10
Volume of 1 layer = Volume of 4 spheres = (16/3)*π*(5)^3
=> Volume of 10 layers = (160/3)* π*(5)^3
Volume of box = 20*20*100
Fraction of Empty space = [ (20*20*100) - (160/3)*π*125] / 20*20*100
=> Empty space % = 47% = 50% approx
Answer: C
Posted from my mobile device
=> 4r = 20
=> r = 5cm
Let the number of layers be : N
=> Height of the box = N*2r
=> N = 100/10 = 10
Volume of 1 layer = Volume of 4 spheres = (16/3)*π*(5)^3
=> Volume of 10 layers = (160/3)* π*(5)^3
Volume of box = 20*20*100
Fraction of Empty space = [ (20*20*100) - (160/3)*π*125] / 20*20*100
=> Empty space % = 47% = 50% approx
Answer: C
Posted from my mobile device
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10 Jun 2020, 03:58
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Radius = 5 cm {20/4 = r}
A level consisting 4 identical spheres whose diameter is 10 = 10
Height of box = 100cm
Total no. of level = 100/10 = 10 levels
1 level = 4 balls.
10 level = 40 balls.
Volume of cuboid = l*b*h =20*20*100 = 40,000
Volume of 1 ball(sphere) = 4/3πr^2 = 4/3*22/7*5*5*5
Volume of 40 spheres = 40*4/3*22/7*5*5*5 = 20952.38 or 20952.4
Space left inside box = 40,000-20952.4 = 19047.6
Approximate percentage left = 19047.6*100/40,000
=> 47.619 or 50% approximate.
IMO C
Posted from my mobile device
A level consisting 4 identical spheres whose diameter is 10 = 10
Height of box = 100cm
Total no. of level = 100/10 = 10 levels
1 level = 4 balls.
10 level = 40 balls.
Volume of cuboid = l*b*h =20*20*100 = 40,000
Volume of 1 ball(sphere) = 4/3πr^2 = 4/3*22/7*5*5*5
Volume of 40 spheres = 40*4/3*22/7*5*5*5 = 20952.38 or 20952.4
Space left inside box = 40,000-20952.4 = 19047.6
Approximate percentage left = 19047.6*100/40,000
=> 47.619 or 50% approximate.
IMO C
Posted from my mobile device
