The correct circuit diagram is D with galvanometer resistance G=R−SRS
In the circuit diagrams (A, B, C and D) shown below, R is a high resistance and S is a resistance of the order of galvanometer resistance G. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as: (a) 
(b)
(C)
(d)
Held on 11 Apr 2014 · Verified 6 Jul 2026.
Circuit A with G=(R−S)RS
Circuit B with G=S
Circuit C with G=S
Circuit D with G=(R−SRS)
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A solenoid has a core made of material with relative permeability $400$. The magnetic field produced in the interior of solenoid is $1.0$ T. The magnetic intensity in SI units is $\alpha \times 10^5$. The value of $\alpha$ is ______. (Free space permeability $\mu_0=4\pi \times 10^{-7}$ SI units.)
Two identical small bar magnets each of dipole moment $3\sqrt{5}$ J/T are placed at a center to center separation of $10$ cm, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is $\alpha \times 10^{-3}$ T. The value of $\alpha$ is ______. ($\mu_0=4\pi \times 10^{-7}$ Tm/A) 
The heat generated in 1 minute between points $A$ and $B$ in the given circuit, when a battery of 9 V with internal resistance of $1 \Omega$ is connected across these points is $\_\_\_\_$ J. 
Two known resistances of $R \Omega$ and $2 R \Omega$ and one unknown resistance $X \Omega$ are connected in a circuit as shown in the figure. If the equivalent resistance between points $A$ and $B$ in the circuit is $X \Omega$, then the value of $X$ is $\_\_\_\_$ $\Omega$. 
A cylindrical conductor of length 2 m and area of cross-section $0.2 \mathrm{~mm}^{2}$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \mathrm{~m}^{2} / V. s$. The value of $\alpha$ is : (electron concentration $=5 \times 10^{28} / \mathrm{m}^{3}$ and electron charge $\left.=1.6 \times 10^{-19} \mathrm{C}\right)$
Work through every JEE Main Electromagnetism PYQ, year by year.