A) 25
B) 75
C) 50
D) 125
Correct Answer: D
Solution :
Radius of cone \[=20\,cm\] and height of cone\[=10\,cm\] \[\therefore \] Volume of cone \[=\frac{1}{3}\pi {{(20)}^{2}}\times 10\] \[=\frac{4000}{3}\pi \,c{{m}^{3}}\] and radius of spherical ball \[=2\,cm\] \[\therefore \] Volume of spherical ball \[=\frac{4}{3}\pi {{(2)}^{3}}=\frac{32\pi }{3}\] \[\therefore \] Required number of balls \[=\frac{4000}{32}=125\]
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