All categories

3 years ago

9. In Millikan's oil drop experiment an oil

1 answer:

6 0

From the figure, F
e

and F
g

represents electric and gravitational forces respectively.
We can observe that,

V∝
q
m

and in turn m∝r
3

given V = 400V is initial voltage and let V
1

substituting these values in the respective equations we get

V
V
1

=
r
3
(q)
(2r)
3
q

solving the above eqaution will give us
V
1

=8×V=3200 V
So, the answer is option (C).

You might be interested in

What type of wave is shown below?

Elena-2011 [213]

Answer:

B. Transverse wave.

Explanation:

Because it has troughs and crests.

5 0

3 years ago

A 2-kilogram object falls 3 meters. a. How much potential energy did the object have before it fell? b. How much work was accomp

stiks02 [169]

Below are the answers:

(a) Ep = mass x gravitational acceleration x height
= 2kg x 9.8ms-2 x 3m 
= 58.8J 

(b) F = mg 
= 2kg x 9.8ms-2 
= 19.6N 
W = Fd 
= 19.6N x 3m 
= 58.8J

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

6 0

3 years ago

Hey can someone please help answer this question?

yan [13]

Answer:

yes ............................................ks

Explanation: is good

4 0

3 years ago

Describe how charge is transferred from the ruler to the metal rod.

spin [16.1K]

When the ruler is broughı near the inetal knob, it repels electrons in the metal. Electrons move away froni the ruler and down the metal rod. The knob now has a positive charge. The thin pieces of metal foil at the bottom of the metal rod now have a negative charge.

3 0

3 years ago

Read 2 more answers

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If

horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

4 0

3 years ago