A) 12
B) 24
C) 36
D) 48
Correct Answer: B
Solution :
| Let the radius of ball = r |
| \[\therefore \] Radius of base of cylinder = 4r |
| and height of cylinder = 4r |
| \[\therefore \] Volume of spherical ball \[=\frac{4}{3}\pi {{r}^{3}}\] |
| and volume of water \[=\pi \,\,{{(4r)}^{2}}(2r)=32\pi \,\,{{r}^{3}}\] |
| Also, volume of remaining portion of cylinder\[=32\pi \,\,{{r}^{3}}\] |
| Let number of spherical balls = n |
| \[\therefore \] \[32\pi \,\,{{r}^{3}}=n\times \frac{4}{3}\pi \,\,{{r}^{3}}\]\[\Rightarrow n=8\times 3=24\] |
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