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Chapter 11: Problem 1609

N identical drops of mercury are charged simultaneously to 10 volt. when combined to form one large drop, the potential is found to be 40 volt, the value of \(\mathrm{N}\) is \(\ldots \ldots\) (A) 4 (B) 6 (C) 8 (D) 10

Short Answer

Expert verified
The value of N is approximately 6. Answer: (B) 6.

Step by step solution

01

Write the capacitance and charge equations for the small drops

Since there are N identical small drops, each with a potential of 10 volts, we can express their capacitance as: \(C_{small} = k \cdot R_{small}\) and their charge as: \(Q_{small} = N \cdot q = C_{small} \cdot V_{small} = k \cdot R_{small} \cdot 10\)
02

Write the capacitance and charge equations for the large drop

When the N small drops combine to form a large drop, we can express the capacitance of the large drop as: \(C_{large} = k \cdot R_{large}\) and its charge as: \(Q_{large} = q' = C_{large} \cdot V_{large} = k \cdot R_{large} \cdot 40\)
03

Relate the charges of the small and large drops

Since the charges of the small drops are summed up to form the charge of the large drop, we can write: \(Q_{large} = N \cdot Q_{small}\)
04

Substitute the charge equations and express the radii in terms of volume

Using the expressions for charges, we can write: \(k \cdot R_{large} \cdot 40 = N \cdot k \cdot R_{small} \cdot 10\) Note that the total volume of small drops is equal to the volume of the large drop: \(N \cdot \frac{4}{3} \pi R_{small}^3 = \frac{4}{3} \pi R_{large}^3\) From this, we can express the radii in terms of their volumes: \(R_{large} = \sqrt[3]{N} \cdot R_{small}\)
05

Substitute the radii relation and find the value of N

Substituting the radii relation into our previous equation, we get: \(k \cdot \sqrt[3]{N} \cdot R_{small} \cdot 40 = N \cdot k \cdot R_{small} \cdot 10\) Now, we can solve for N: \(4 \cdot \sqrt[3]{N} = N\) Since the problem provides options to choose from, we can test each option: - (A) N = 4: \(4 \cdot \sqrt[3]{4} = 4 \cdot 1.587 = 6.348 \neq 4\) - (B) N = 6: \(4 \cdot \sqrt[3]{6} = 4 \cdot 1.817 = 7.268 \approx 6 \) Therefore, the value of N is approximately: Answer: (B) 6

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Most popular questions from this chapter

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be $\ldots . .$ (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

Two point charges of \(+16 \mathrm{c}\) and \(-9 \mathrm{c}\) are placed $8 \mathrm{~cm}\( apart in air \)\ldots \ldots\(.. distance of a point from \)-9$ c charge at which the resultant electric field is zero. (A) \(24 \mathrm{~cm}\) (B) \(9 \mathrm{~cm}\) (C) \(16 \mathrm{~cm}\) (D) \(35 \mathrm{~cm}\)

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q\). The potential difference between the centers of the two rings is $\ldots .$ (A) 0 (B) $\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]$ (C) $\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.$ (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)

A small sphere whose mass is \(0.1 \mathrm{gm}\) carries a charge of $3 \times 10^{-10} \mathrm{C}\( and is tie up to one end of a silk fiber \)5 \mathrm{~cm}$ long. The other end of the fiber is attached to a large vertical conducting plate which has a surface charge of \(25 \times 10^{-6} \mathrm{Cm}^{-2}\), on each side. When system is freely hanging the angle fiber makes with vertical is \(\ldots \ldots \ldots\) (A) \(41.8^{\circ}\) (B) \(45^{\circ}\) (C) \(40.2^{\circ}\) (D) \(45.8^{\circ}\)
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