JEE Main 2010 — Electromagnetism

Let $C$ be the capacitance of a capacitor discharging through a resistor R. Suppose $t_1$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_2$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $t_1 / t_2$ will be

Two long parallel wires are at a distance $2 \mathrm{~d}$ apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line $\mathrm{XX}$ ' is given by

A rectangular loop has a sliding connector PQ of length $\ell$ and resistance $\mathrm{R} \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are 

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $30^{\circ}$ with each other. When suspended in a liquid of density $0.8 \mathrm{~g} \mathrm{~cm}^{-3}$, the angle remains the same. If density of the material of the sphere is $16 \mathrm{~g} \mathrm{~cm}^{-3}$, the dielectric constant of the liquid is

In a series LCR circuit $R=200 \Omega$ and the voltage and the frequency of the main supply is $220 \mathrm{~V}$ and $50 \mathrm{~Hz}$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $30^{\circ}$. On taking out the inductor from the circuit the current leads the voltage by $30^{\circ}$. The power dissipated in the LCR circuit is

In the circuit shown below, the key $\mathrm{K}$ is closed at $\mathrm{t}=0$. The current through the battery is 

A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec{E}$ at the centre $O$ is 

If a source of power $4 \mathrm{~kW}$ produces $10^{20}$ photons/second, the radiation belong to a part of the spectrum called

Let there be a spherically symmetric charge distribution with charge density varying as $\rho(r)=\rho_0\left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r=R$, and $\rho(r)=0$ for $r>R$, where $r$ is the distance from the origin. The electric field at a distance $r(r < R)$ from the origin is given by