Moving Charges and Magnetism - Result Question 80

«««< HEAD:content/english/neet-pyq-chapterwise/physics/moving-charges-and-magnetism/moving-charges-and-magnetism-result-question-80.md

82. A galvanometer of resistance $50 \Omega$ is connected to battery of $3 V$ along with a resistance of $2950 \Omega$ 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

======= ####82. A galvanometer of resistance $50 \Omega$ is connected to battery of $3 V$ along with a resistance of $2950 \Omega$ 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

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/moving-charges-and-magnetism/moving-charges-and-magnetism—result-question-80.md (a) $5050 \Omega$

(b) $5550 \Omega$

(c) $6050 \Omega$

(d) $4450 \Omega$

[2008]

Show Answer

Answer:

Correct Answer: 82. (d)

Solution:

  1. (d) Total internal resistance $=(50+2950) \Omega$ $=3000 \Omega$

Emf of the cell, $\varepsilon=3 V$

$\therefore$ Current $=\frac{\varepsilon}{R}=\frac{3}{3000}=1 \times 10^{-3} A=1.0 mA$

$\therefore$ Current for full scale deflection of 30 divisions is $1.0 mA$.

$\therefore$ Current for a deflection of 20 divisions,

$I=(\frac{20}{30} \times 1) mA$ or $I=\frac{2}{3} mA$

Let the resistance be $x \Omega$. Then

$ x=\frac{\varepsilon}{I}=\frac{3 V}{(\frac{2}{3} \times 10^{-3} A)}=\frac{3 \times 3 \times 10^{3}}{2} \Omega $

$=4500 \Omega$

But the resistance of the galvanometer is $50 \Omega$.

$\therefore$ Resistance to be added

$=(4500-50) \Omega=4450 \Omega$