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Chapter 12: Problem 1848

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at $500 \mathrm{~cm}$ on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The voltmeter reading will be. (A) \(1.96 \mathrm{~V}\) (B) \(1.8 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(0.96 \mathrm{~V}\)

Short Answer

Expert verified
The voltmeter reading is approximately 1.61 V, which is closest to option (C) 1.64 V.

Step by step solution

01

Calculate voltage across the entire potentiometer wire

First, we need to calculate the voltage across the entire 600 cm potentiometer wire. To do this, we can use Ohm's Law: \(V = IR\) Here, V is voltage, I is current, and R is resistance. We are given I = 40 mA (or 0.04 A) and the total resistance of the potentiometer wire can be found by: \(R_{total} = \frac{rL}{l}\) Where r is the internal resistance of the cell (10 Ω), L is the total length of the potentiometer wire (600 cm), and l is the null point (500 cm).
02

Calculate the total resistance of the potentiometer wire

Now we will calculate the total resistance of the potentiometer wire: \(R_{total} = \frac{(10 Ω)(600 cm)}{500 cm}\) \(R_{total} = 12 Ω\)
03

Calculate the potential difference across the entire wire

Now that we have the total resistance of the potentiometer wire, we can calculate the potential difference across the entire wire: \(V_{total} = IR_{total}\) \(V_{total} = (0.04 A)(12 Ω)\) \(V_{total} = 0.48 V\)
04

Calculate the potential difference across the new balancing length

Since the balancing length decreased by 10 cm, the new balancing length is 490 cm. Now we can calculate the potential difference across the new balancing length using the proportion of the original null point to the new balancing length, and multiplying it by the potential difference across the entire wire: \(V_{new} = \frac{490 cm}{600 cm} \times V_{total}\) \(V_{new} = \frac{490 cm}{600 cm} \times 0.48 V\) \(V_{new} = 0.392 V\)
05

Calculate the voltmeter reading

Finally, we can calculate the voltmeter reading. The voltmeter reading will be equal to the emf of the cell minus the potential difference across the new balancing length: \(V_{voltmeter} = E - V_{new}\) \(V_{voltmeter} = 2 V - 0.392 V\) \(V_{voltmeter} ≈ 1.61 V\) The voltmeter reading is approximately 1.61 V, which is closest to option (C) 1.64 V.

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Most popular questions from this chapter

Match the physical quantities given in column I with their dimensional formulae given in column II - I stands for the dimension of current. \(\begin{array}{ll}\text { Column I } & \text { Column II }\end{array}\) (a) Electromotive force (emf) (p) \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (b) Resistance (q) \(\mathrm{ML}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (c) Resistivity (r) \(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (d) Conductivity (s) \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\) (A) \(a-s, b-p, c-q, d-r\) (B) \(a-p, b-s, c-r, d-p\) (C) \(a-p, b-s, c-r, d-q\) (D) \(a-r, b-p, c-q, d-s\)

Match the following two columns. Column I \(\quad\) Column II (a) Electrical resistance (p) \(\left[\mathrm{MLT}^{-2} \mathrm{~A}^{2}\right]\) (b) Electric potential (q) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\) (c) Specific resistance (r) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]\) (d) Specific conductance (s) None of there (A) \(a-q, b-s, c-r, d-p\) (B) \(a-q, b-r, c-s, d-s\) (C) \(a-p, b-q, c-s, d-r\) (D) \(a-p, b-r, c-q, d-s\)

Two batteries each of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series to a resistor \(R\). Maximum Possible power consumed by the resistor \(=\ldots .\) (A) \(3.2 \mathrm{~W}\) (B) \((16 / 9) \mathrm{W}\) (C) \((8 / 9) \mathrm{W}\) (D) \(2 \mathrm{~W}\)

Length of a wire of resistance \(R \Omega\) is increased to 10 times, so its resistance becomes \(1000 \Omega\), therefore \(R=\ldots .\) (The volume of the wire remains same during increase in length) (A) \(0.01 \Omega\) (B) \(0.1 \Omega\) (C) \(1 \Omega\) (D) \(10 \Omega\)

How would you arrange 48 cells each of e.m.f \(2 \mathrm{~V}\) and internal resistance \(1.5 \Omega\) so as to pass maximum current through the external resistance of \(2 \Omega\) ? (A) 2 cells in 24 groups (B) 4 cells in 12 groups (C) 8 cells in 6 groups (D) 3 cells in 16 groups
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