Name: But What Are Vectors?
Uploaded: 2019-07-19
Duration: 10 min 42 s
Description: Anyone's movement can be written in many different ways, in the case of vectors you'll have to find the coordinates on the x, y, and z axes and learn another...

Question

A cylinder of radius, $R$ is moving down with speed, $v$ into water, placed in a cylinder of radius, $4 R$ as shown. Find the rate by which,

(a) water surface is rising.

(b) wet surface area of cylinder of radius, $R$ is increasing.

Accepted Answer

(a) Given that, a cylinder of radius, $4 R$ is moving down with a velocity, $v$ into the water in a cylinder of radius, $4 R$ . The rate of increase of the submerged length, $x$ of the moving cylinder with respect to the fixed cylinder is

$\frac{d x}{d t} = v . . . (1)$

Let, $h$ be the rate at which water rises in the fixed cylinder.

Then water displaced by the moving cylinder equals the volume due to the rise of the height of water.

Rate of volume displaced by the cylinder is

$\frac{d V}{d t} = \frac{d (π R^{2} x)}{d t} = π R^{2} \frac{d x}{d t} \Rightarrow \frac{d V}{d t} = π R^{2} v . . . (2)$

Rise of volume of water in the annular area is,

$V = h \{π {(4 R)}^{2} - π R^{2}\}$ .

Differentiating the above equation, the rate of rise of volume of the water due to rise in height is,

$\frac{d V}{d t} = \frac{d [h \{π {(4 R)}^{2} - π R^{2}\}]}{d t} = 15 π R^{2} \frac{d h}{d t} . . . (3)$

Equating equations $(2) and (3)$ ,

$π R^{2} v = 15 π R^{2} \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{π R^{2} v}{15 π R^{2}} = \frac{v}{15} \Rightarrow \frac{d h}{d t} = \frac{v}{15} . . . (4)$

(b) Wet surface area of the moving cylinder is

$A = π R^{2} + 2 π R (x + h)$ .

So, the rate of increase of the wet area is

$\frac{d A}{d t} = \frac{d \{π R^{2} + 2 π R (x + h)\}}{d t} = 2 π R (\frac{d x}{d t} + \frac{d h}{d t}) .$

Substituting equation $(1)$ and $(4)$ ,

$\frac{d A}{d t} = 2 π R (\frac{d x}{d t} + \frac{d h}{d t}) = 2 π R (v + \frac{v}{15}) = \frac{32 π R v}{15} .$

Therefore, the rate of increase of the wet area of the cylinder of radius, $R$ is $\frac{32 π R v}{15}$ .

A cylinder of radius, $R$ is moving down with speed, $v$ into water, placed in a cylinder of radius, $4 R$ as shown. Find the rate by which,

(a) water surface is rising.

(b) wet surface area of cylinder of radius, $R$ is increasing.

Important Questions on Mathematical Tools in Physics

Learn and Prepare for any exam you want !

A cylinder of radius, R is moving down with speed, v into water, placed in a cylinder of radius, 4R as shown. Find the rate by which, (a) water surface is rising. (b) wet surface area of cylinder of radius, R is increasing.

Important Questions on Mathematical Tools in Physics

Learn and Prepare for any exam you want !

A cylinder of radius, $R$ is moving down with speed, $v$ into water, placed in a cylinder of radius, $4 R$ as shown. Find the rate by which,

(a) water surface is rising.

(b) wet surface area of cylinder of radius, $R$ is increasing.