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

Iwo small charged spheres repel each other with a force $2 \times 10^{-3} \mathrm{~N}$. The charge on one sphere is twice that of the other. When these two spheres displaced \(10 \mathrm{~cm}\) further apart the force is $5 \times 10^{-4} \mathrm{~N}$, then the charges on both the spheres are....... (A) \(1.6 \times 10^{-9} \mathrm{C}, 3.2 \times 10^{-9} \mathrm{C}\) (B) \(3.4 \times 10^{-9} \mathrm{C}, 11.56 \times 10^{-9} \mathrm{C}\) (C) \(33.33 \times 10^{9} \mathrm{C}, 66.66 \times 10^{-9} \mathrm{C}\) (D) \(2.1 \times 10^{-9} \mathrm{C}, 4.41 \times 10^{-9} \mathrm{C}\)

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases \(4.5\) times in comparison with the initial value. The ratio of the initial charges of the balls is ....... (A) \(4: 1\) (B) \(6: 1\) (C) \(3: 1\) (D) \(2: 1\)

An electric circuit requires a total capacitance of \(2 \mu \mathrm{F}\) across a potential of \(1000 \mathrm{~V}\). Large number of \(1 \mu \mathrm{F}\) capacitances are available each of which would breakdown if the potential is more then \(350 \mathrm{~V}\). How many capacitances are required to make the circuit? (A) 24 (B) 12 (C) 20 (D) 18

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Two point positive charges \(q\) each are placed at \((-a, 0)\) and \((a, 0)\). A third positive charge \(q_{0}\) is placed at \((0, y)\). For which value of \(\mathrm{y}\) the force at \(q_{0}\) is maximum \(\ldots \ldots \ldots\) (A) a (B) \(2 \mathrm{a}\) (C) \((\mathrm{a} / \sqrt{2})\) (D) \((\mathrm{a} / \sqrt{3})\)

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