(ii) Water is filled in a right cylindrical tank with base radius 14 cm, such that water
level is 3 cm below the top. When an iron ball is dropped in the tank, 3003 cm' of
water flows out. Find the radius of the ball.
Answers
Answer:
radius of the ball will be 9.75 or 10 cm.
Answer:
Radius of water tank is 14 cm and water level is 3 cm below the top.
When iron ball is dropped it water 3003 cm³ of water flows out.
Volume of empty tank = \pi r^2 hπr
2
h
The volume of ball = Volume of empty tank + Volume of water came out
The volume of ball = \pi r^2 hπr
2
h + Volume of water came out
The volume of ball = \frac{22}{7}\times14^2\times3
7
22
×14
2
×3 + 3003 cm³
The volume of ball = 1848 cm³ + 3003 cm³
The volume of ball = 4851 cm³
The volume of sphere is: V=\frac{4}{3}\pi r^3V=
3
4
πr
3
Substitute the respective values in the above formula.
4851 =\dfrac{4}{3}\times\dfrac{22}{7}\times r^34851=
3
4
×
7
22
×r
3
r^3= \dfrac{4851\times3\times7}{22\times4}r
3
=
22×4
4851×3×7
r^3=1157.625r
3
=1157.625
r=10.5r=10.5
Hence, the radius of the ball is 10.5 cm³.