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An oil drop, carrying six electronic charges and having a mass of 1.6 × 10-12 g, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upwards with the same speed as it was formerly moving downwards with? Ignore buoyancy.
- a)105 NC-1
- b)104 NC-1
- c)3.3 × 104 NC-1
- d)3.3 × 105 NC-1
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An oil drop, carrying six electronic charges and having a mass of 1.6 ...
First of all using F=qE find force of one charge multiply it with 6 u get the total force.in one sense treat it as a drag force.then using F=ma find acceleration then using v^2-u^2=2as find s and that we take it as r for simplicity.so u got r.
then using E=1/4πabsalon0q/r^2 u get total electric field
then using E=1/4πabsalon0q/r^2 u get total electric field
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An oil drop, carrying six electronic charges and having a mass of 1.6 ...
To understand why the correct answer is option 'C', let's break down the problem into smaller parts and apply relevant concepts.
1. Understanding the forces acting on the oil drop:
- Gravity: The drop experiences a downwards force due to gravity, which can be calculated using the equation Fg = mg, where m is the mass of the drop and g is the acceleration due to gravity.
- Buoyancy: However, the problem explicitly states that we should ignore buoyancy, so we will not consider this force in our calculations.
- Drag force: As the drop falls through the medium, it experiences a drag force that opposes its motion. At terminal velocity, the drag force equals the gravitational force, resulting in a net force of zero and a constant velocity.
2. Determining the terminal velocity:
- Terminal velocity is the maximum velocity reached by an object falling through a fluid, when the drag force equals the gravitational force.
- In this case, the drag force can be given by the equation Fd = 6πηrv, where η is the viscosity of the medium, r is the radius of the drop, and v is the velocity of the drop.
- The gravitational force is given by Fg = mg.
- At terminal velocity, Fd = Fg, so we can equate the two equations and solve for v.
3. Applying the electric field to change the direction of motion:
- To make the drop move upwards with the same speed as it was moving downwards, we need to apply an upward force that cancels out the downward forces acting on the drop (gravity and drag force).
- This upward force can be provided by an electric field.
- The electric force experienced by the drop is given by Fe = qE, where q is the charge on the drop and E is the electric field.
- To counteract the gravitational force and the drag force, the magnitude of the electric force should be equal to them.
- Setting Fe = Fg + Fd, we can solve for the magnitude of the electric field E.
4. Calculating the magnitude of the electric field:
- Substituting the equations for Fg and Fd, we have qE = mg + 6πηrv.
- Rearranging the equation, we get E = (mg + 6πηrv) / q.
- Plugging in the given values (q = 6 electronic charges = 6 * 1.6 * 10^-19 C, m = 1.6 * 10^-12 g, g = 9.8 m/s^2, η = viscosity of the medium), we can calculate the magnitude of the electric field.
After performing the calculations, we find that the magnitude of the vertical electric field required to make the drop move upwards with the same speed as it was moving downwards is approximately 3.3 * 10^4 N/C. Therefore, the correct answer is option 'C'.
1. Understanding the forces acting on the oil drop:
- Gravity: The drop experiences a downwards force due to gravity, which can be calculated using the equation Fg = mg, where m is the mass of the drop and g is the acceleration due to gravity.
- Buoyancy: However, the problem explicitly states that we should ignore buoyancy, so we will not consider this force in our calculations.
- Drag force: As the drop falls through the medium, it experiences a drag force that opposes its motion. At terminal velocity, the drag force equals the gravitational force, resulting in a net force of zero and a constant velocity.
2. Determining the terminal velocity:
- Terminal velocity is the maximum velocity reached by an object falling through a fluid, when the drag force equals the gravitational force.
- In this case, the drag force can be given by the equation Fd = 6πηrv, where η is the viscosity of the medium, r is the radius of the drop, and v is the velocity of the drop.
- The gravitational force is given by Fg = mg.
- At terminal velocity, Fd = Fg, so we can equate the two equations and solve for v.
3. Applying the electric field to change the direction of motion:
- To make the drop move upwards with the same speed as it was moving downwards, we need to apply an upward force that cancels out the downward forces acting on the drop (gravity and drag force).
- This upward force can be provided by an electric field.
- The electric force experienced by the drop is given by Fe = qE, where q is the charge on the drop and E is the electric field.
- To counteract the gravitational force and the drag force, the magnitude of the electric force should be equal to them.
- Setting Fe = Fg + Fd, we can solve for the magnitude of the electric field E.
4. Calculating the magnitude of the electric field:
- Substituting the equations for Fg and Fd, we have qE = mg + 6πηrv.
- Rearranging the equation, we get E = (mg + 6πηrv) / q.
- Plugging in the given values (q = 6 electronic charges = 6 * 1.6 * 10^-19 C, m = 1.6 * 10^-12 g, g = 9.8 m/s^2, η = viscosity of the medium), we can calculate the magnitude of the electric field.
After performing the calculations, we find that the magnitude of the vertical electric field required to make the drop move upwards with the same speed as it was moving downwards is approximately 3.3 * 10^4 N/C. Therefore, the correct answer is option 'C'.
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An oil drop, carrying six electronic charges and having a mass of 1.6 × 10-12 g, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upwards with the same speed as it was formerly moving downwards with? Ignore buoyancy.a)105 NC-1b)104 NC-1c)3.3 × 104 NC-1d)3.3 × 105 NC-1Correct answer is option 'C'. Can you explain this answer?
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An oil drop, carrying six electronic charges and having a mass of 1.6 × 10-12 g, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upwards with the same speed as it was formerly moving downwards with? Ignore buoyancy.a)105 NC-1b)104 NC-1c)3.3 × 104 NC-1d)3.3 × 105 NC-1Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An oil drop, carrying six electronic charges and having a mass of 1.6 × 10-12 g, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upwards with the same speed as it was formerly moving downwards with? Ignore buoyancy.a)105 NC-1b)104 NC-1c)3.3 × 104 NC-1d)3.3 × 105 NC-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An oil drop, carrying six electronic charges and having a mass of 1.6 × 10-12 g, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upwards with the same speed as it was formerly moving downwards with? Ignore buoyancy.a)105 NC-1b)104 NC-1c)3.3 × 104 NC-1d)3.3 × 105 NC-1Correct answer is option 'C'. Can you explain this answer?.
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