<p><strong>Sixty four conducting drops each of radius 0.02m and each carrying a charge of <math><mrow><mn>5</mn><mi>μ</mi><mi>C</mi></mrow></math> are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:</strong></p>

Since, volume remain constant64×43πr3=43πR3R3 = r3 × 64R = 4rσbiggerσsmall drop=5×64×4π24πR2×5=64 (rR)2=64 (r4r)2=6416=4:1

Ask & Answer Question

Physics Class 12th Electric Charges and Fields physics ncert exemplar solutions class 12th chapter two

Sixty four conducting drops each of radius 0.02m and each carrying a charge of $5 μ C$ are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:

Option 1 -

1 : 4

Option 2 -

4 : 1

Option 3 -

1 : 8

Option 4 -

8 : 1

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

Since, volume remain constant

$64 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$

R³ = r³ × 64

R = 4r

$\frac{σ_{b i g g e r}}{σ_{s m a l l d r o p}} = \frac{5 \times 64 \times 4 π^{2}}{4 π R^{2} \times 5} = 64 {(\frac{r}{R})}^{2} = 64 {(\frac{r}{4 r})}^{2} = \frac{64}{16} = 4 : 1$

0 Upvotes 0 Downvotes

Similar Questions for you

Which are the main topics covered in the Electric Charges and Fields?

Pallavi Pathak

The following are the topics covered in this chapter: Electric Field and Field Lines, Gauss's Law, Electric Dipole, Conductors and Insulators, and Electric Flux.

In comparison to other Physics Class 12 chapters, is Electric Charges and Fields an easy chapter to read?

Pallavi Pathak

Indeed, it is an easy chapter of Class 12 Physics. In this chapter, you will learn about the foundational concepts like Gauss's law and electric fields.

A point charge of + $12 μ C$ is at a distance 6cm vertically above the centre of a squared of side 12cm as shown in figure. The magnitude of the electric flux through the square will be……………..× 10³ Nm²/C.

Vishal Baghel

Flux through the 6 sides of square (i.e. cube)

$?_{T} = \frac{q_{i n}}{ε_{0}} = \frac{12 \times 1 0^{- 6} C}{ε_{0}}$

Flux through a square

$? = \frac{?_{T}}{6} = \frac{12 \times 1 0^{- 6}}{ε_{0} \times 6} = \frac{2 \times 1 0^{- 6}}{ε_{0}}$

$\Rightarrow ? = 225.98 \times 1 0^{3} N m^{2} / C ? 226 N m^{2} / C$

Is Electric Charges and Fields chapter important for NEET examination?

Pallavi Pathak

In the medical entrance test NEET, this chapter has a moderate weightage. You can expect around 2-3 questions of this chapter that contributes to the 3-5% of the total marks in the Physics section.

The electric field inside a sphere having charge density related to the distance from the centre as r (a is a constant) is:

alok kumar singh

From Guass’s Law

$? \vec{E} . \vec{d A} = \frac{q_{i n}}{\in_{0}}$

$E . 4 π r^{2} = \frac{\int_{0}^{r} (4 π r^{2} . d r) (α r)}{\in_{0}}$

$E = (\frac{α r^{2}}{4 \in_{0}})$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Sixty four conducting drops each of radius 0.02m and each carrying a charge of $5 μ C$ are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Sixty four conducting drops each of radius 0.02m and each carrying a charge of 5μC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Sixty four conducting drops each of radius 0.02m and each carrying a charge of $5 μ C$ are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be: