| The radius of a spherical lead ball is 6 cm. The radius and height of a cylindrical vessel are 24 cm and 14 cm, respectively. If the lead ball is dropped in the cylindrical vessel, half filled with water, what is the rise in the water level? |
A) 0.5 cm
B) 0.25 cm
C) 0.45 cm
D) 0.9 cm
Correct Answer: A
Solution :
| Volume of lead ball = Volume of the water content |
| That has risen from the original height. |
| Let the rise in water level \[=\,\,h\] |
| \[\frac{4}{3}\times \pi \times 6\times 6\times 6=\pi \times 24\times 24\times h\] |
| \[\Rightarrow \] \[h=\frac{4\times 6\times 6\times 6}{3\times 24\times 24}=\frac{1}{2}=0.5\,\,cm\] |
You need to login to perform this action.
You will be redirected in
3 sec
