NTA JEE Main 9th April 2019 Shift 1

A rigid square loop of side 'a' and carrying current $$I_2$$ is lying on a horizontal surface near a long current $$I_1$$ carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be:

A rectangular coil (Dimension 5 cm $$\times$$ 2.5 cm) with 100 turns, carrying a current of 3 A in the clock-wise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45° about Z-axis, then the torque on the coil is:

The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to:

The magnetic field of a plane electromagnetic wave is given by $$\vec{B} = B_0[\hat{i}\cos(kz - \omega t)] + B_1[\hat{j}\cos(kz + \omega t)]$$, where $$B_0 = 3 \times 10^{-5}$$ T and $$B_1 = 2 \times 10^{-6}$$ T. The RMS value of the force experienced by a stationary charge $$Q = 10^{-4}$$ C at z = 0 is closest to:

A concave mirror for face viewing has a focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is:

The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index $$\mu$$ is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe width. If the wavelength of light used is $$\lambda$$ then t will be:

The electric field of light wave is given as $$\vec{E} = 10^{-3} \cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14}t\right) \hat{x} \frac{N}{C}$$. This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is: Given, E (in eV) = $$\frac{12375}{\lambda(\text{in } \mathring{A})}$$

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:

An NPN transistor is used in common emitter configuration as an amplifier with 1 k$$\Omega$$ load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 $$\mu$$A change in the base current of the amplifier. The input resistance and voltage gain are:

A signal A cos$$\omega$$t is transmitted using $$v_0 \sin\omega_0 t$$ as carrier wave. The correct amplitude modulated (AM) signal is: