A right circular cylindrical tank having a height of 2 feet and perimeter of the base as 22 feet was filled with water up to 80% of its actual capacity. What is the minimum number of lead balls each having radius of 0.10 feet, that should be dropped into the tank so as to increase the volume to at least 90%?
- 800
- 900
- 919
- 1838
- 2757
Since the % increase is 10, the volume which should be filled with lead balls is 10 % of the cylindrical volume. (volume of sphere)*(no of spheres) = 10%(volume of cylinder)
On equating, I get ans in decimals.
Where am I going wrong?
perimeter of the base is $22= 2\pi r_c\implies r_c = \dfrac{11}{\pi}$
now with the actual problem ;
$n\times \dfrac43\pi r_s^3 = \dfrac{10}{100}\pi r_c^2h $
$n \cdot \dfrac43 \cdot (0.1)^3 = (0.1)\cdot (\dfrac{11}{\pi})^2\cdot 2$
$n = \dfrac{121\cdot 3}{2\cdot \pi^2\cdot 0.01}$
$n = 1838.979483$
The key word in the answer is "minimum" and "atleast 90%". So we take the lower bound of the answer which is $1838$