Fluid Mechanics

Two large circular discs separated by a distance of

0.01%5C%2C%20m

are connected to a battery via a switch as shown in the figure Charged oil drops of density

900%5C%2C%20kg%20%5C%2Cm%20%5E%7B-3%7D

are released through a tiny hole at the center of the top disc Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of

200%5C%2C%20V

across the discs As a result, an oil drop of radius

8%20%5Ctimes%2010%5E%7B-7%7D%20m

stops moving vertically and floats between the discs The number of electrons present in this oil drop is (neglect the buoyancy force, take acceleration due to gravity

%3D10%20%5C%2Cms%20%5E%7B-2%7D

and charge on an electron

(e)%20%3D1.6%20%5Ctimes%2010%5E%7B-19%7D%20C

)

1. Work Done in Pushing the Ball:
The work done in pushing the ball is given by the formula:

$W%20%3D%20(%20%5Crho%20p%20g%20)%20d%20-%20(%20%5Csigma%20g%20g%20)%20d$

Where:
$%5Crho$ is the density of water and $%5Csigma$ is the density of the ball.

2. Calculating Work Done:
Substituting the given values into the formula:

$W%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20R%5E3%20%5Ctimes%2010%20%5Ctimes%200.7%20%5Cleft%5B1000%20-%20%5Cfrac%7B3%7D%7B4%7D%20%5Ctimes%2010%5E%7B-3%7D%20R%5E3%20%5Cright%5D$

After calculation, we get:

$W%20%3D%200.077%20%5C%2C%20%5Ctext%7BJ%7D$

3. When the Ball is Released:
The same work, i.e., 0.077 J, is done on the ball when it is released. Using the equation $%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%3D%200.077$ , we can solve for the velocity:

$v%20%3D%20%5Csqrt%7B%5Cfrac%7B0.077%20%5Ctimes%202%7D%7B22%20%5Ctimes%207%20%5Ctimes%2010%5E%7B-3%7D%7D%7D%20%3D%207%20%5C%2C%20%5Ctext%7Bm%2Fs%7D$

4. Kinetic Energy and Velocity:
The work done is converted into kinetic energy, and the ball reaches a velocity of:

$v%20%3D%207%20%5C%2C%20%5Ctext%7Bm%2Fs%7D$

Hence, option B is correct.

5. Calculating the Height:
Using the formula for height $H%20%3D%20%5Cfrac%7Bv%5E2%7D%7B2g%7D$ , we calculate:

$H%20%3D%20%5Cfrac%7B7%20%5Ctimes%207%7D%7B2%20%5Ctimes%2010%7D%20%3D%202.45%20%5C%2C%20%5Ctext%7Bm%7D$

This calculation is incorrect according to option C.

6. Net Force on the Ball:
The net force on the ball is given by:

$F_%7B%5Ctext%7Bnet%7D%7D%20%3D%20%5Crho%20%5Csigma%20g%20-%20%5Csigma%20g%20%3D%200.11%20%5C%2C%20%5Ctext%7BN%7D$

7. Viscous Force:
The maximum viscous force is calculated when $v%20%3D%207%20%5C%2C%20%5Ctext%7Bm%2Fs%7D$ using the formula for viscous force:

$F_%7Bv_%7B%5Ctext%7Bmax%7D%7D%7D%20%3D%206%20%5Cpi%20%5Ceta%20r%20v$

$F_%7Bv_%7B%5Ctext%7Bmax%7D%7D%7D%20%3D%206%20%5Ctimes%20%5Cfrac%7B22%7D%7B7%7D%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20%5Cleft(%20%5Cfrac%7B3%7D%7B2%7D%20%5Ctimes%2010%5E%7B-2%7D%20%5Cright)%20%5Ctimes%207$

$F_%7Bv_%7B%5Ctext%7Bmax%7D%7D%7D%20%3D%2018%20%5Ctimes%2011%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%20%5Ctext%7BN%7D$

8. Final Calculation:
Finally, using the net force $F_%7B%5Ctext%7Bnet%7D%7D%20%3D%20500$ , we get:

$%5Cfrac%7BF_%7B%5Ctext%7Bnet%7D%7D%7D%7BF_%7Bv_%7B%5Ctext%7Bmax%7D%7D%7D%7D%20%3D%20%5Cfrac%7B500%7D%7B9%7D$

Hence, option D is correct.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Fluid Mechanics

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs