A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance

⇦

⇨

A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

Options

(a) 6050 Ω
(b) 4450 Ω
(c) 5050 Ω
(d) 5550 Ω

Correct Answer:

4450 Ω

Explanation:

I=3 / (50+2950) = 10⁻³ A

Current for 30 divisions=10⁻³ A

Current for 20 divisions=10⁻³ * 20/30

=2*10⁻³ / 3 Amperes

For the same deflection to obtain for 20 divisions let resistance added be R, therefore,

2*10⁻³ / 3 = 3 / (50+R)

(50+R) = 9 *10⁻³ / 2

(50+R) = 4500

Therefore R=4450Ω

Related Questions:

1. Which of the following is true for number of spectral lines in going from Lyman series
2. A wave travelling in the +ve x-direction having displacement along y-direction as 1m
3. If the feedback voltage is increased in a negative feedback amplifier, then
4. The electric resistance of a certain wire of iron is R. If its length and radius
5. Dimensional formula of magnetic field in terms of mass M, length L

Topics: Alternating Current (96)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends