Given the dimensions of cylindrical water tank, spherical solid balls and the amount of water present in the tank check if water tank will overflow when balls are dipped in the water tank.
input : H = 10, r = 5 h = 5 N = 2, R = 2 output : Not in overflow state Explanation : water tank capacity = 3.14 * r * r * H = 3.14 * 5 * 5 * 10 = 785 volume of water in tank = 3.14 * r * r * h = 3.14 * 5 * 5 * 5 = 392.5 Volume of balls = N * (4/3) * 3.14 * R * R * R = 2 * (4/3) * 3.14 * 2 * 2 * 2 = 67.02 Total volume of water + dip balls = 392.5 + 67.02 = 459.52 Total volume (459.02) < tank capacity (785) So, there is no overflow in tank input : H = 5, r = 3 h = 3 N = 3, R = 2 output : Overflow Explanation: water tank capacity = 3.14 * r * r * H = 3.14 * 3 * 3 * 5 = 141.3 volume of water in tank = 3.14 * r * r * h = 3.14 * 3 * 3 * 3 = 84.78 volume of balls = N * (4/3) * 3.14 * R * R * R = 3 * (4/3) * 3.14 * 2 * 2 * 2 = 100.48 Total volume of water + dip balls = 84.78 + 100.48 = 185.26 Total volume (185.26) > tank capacity (141.3) So, tank will overflow
When solid balls are dipped in water, level of water increases, hence volume of water will also increase.
Increasing in water volume = Total volume of dip balls
Volume of Cylinder = 3.14 * r * r * h
where: r: radius of tank
h: height of tank
Number of balls are n
Balls have shape of Sphere
Volume of Sphere = (4/3) * 3.14 * R * R * R
Where R: Sphere’s(solid ball) radius
After dipping all balls, if the total volume of water and all balls is less than or equal to the total volume of tank capacity then there will no overflow in tank, otherwise there will be overflow.
Below is the implementation of above approach:
Not in overflow state
- Find amount of water wasted after filling the tank
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Improved By : jit_t