A cylindrical can with a horizontal base and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the radius of the base of the cylinder be $r = 3.5 \, cm$.Since the sphere fits exactly into the can,the radius of the sphere is also $r = 3.5

Vedclass · Accepted Answer

$\frac{7}{3} \, cm$

Vedclass · Answer

(C) Let the radius of the base of the cylinder be $r = 3.5 \, cm$.Since the sphere fits exactly into the can,the radius of the sphere is also $r = 3.5 \, cm$,and the height of the water level after placing the sphere is $H = 2r = 7 \, cm$.The volume of the water in the can is equal to the total volume of the cylinder up to height $H$ minus the volume of the sphere.Volume of water $= \pi r^2 (2r) - \frac{4}{3} \pi r^3 = 2 \pi r^3 - \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3$.Let the initial depth of the water be $h$. The volume of this water is $\pi r^2 h$.Equating the volumes: $\pi r^2 h = \frac{2}{3} \pi r^3$.Solving for $h$: $h = \frac{2}{3} r = \frac{2}{3} 	imes 3.5 = \frac{7}{3} \, cm$.

$A$ cylindrical can with a horizontal base and an internal radius of $3.5 \, cm$ contains sufficient water so that when a solid sphere is placed inside,the water just covers the sphere. The sphere fits exactly into the can. The depth of the water in the can before the sphere was added is:

Similar Questions

The circumference of one end of a frustum of a right circular cone is $48\, cm$ and of the other end is $34\, cm$. If the height of the frustum is $10\, cm$,its volume (in $cm^3$) is

If the side of a cube is increased by $100 \%$,its volume is increased by (in $\%$)

From a solid right circular cylinder with height $10 \ cm$ and radius of the base $6 \ cm$,a right circular cone of the same height and base is removed. The volume (in $cm^3$) of the remaining solid is

The height and base radius of a cone are each increased by $100 \%$. The volume of the cone now becomes

$A$ right circular cylinder is circumscribing a hemisphere such that their bases are common. The ratio of their volumes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module