NTA JEE Main 9th April 2014 Online

Modern vacuum pumps can evacuate a vessel down to a pressure of $$4.0 \times 10^{-15}$$ atm. at room temperature (300 K). Taking R = 8.3 JK$$^{-1}$$ mole$$^{-1}$$, 1 atm = $$10^5$$ Pa and $$N_{Avogadro} = 6 \times 10^{23}$$ mole$$^{-1}$$, the mean distance between molecules of gas in an evacuated vessel will be of the order of:

Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad s$$^{-1}$$, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take $$g = 10$$ m s$$^{-2}$$).

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbondioxide will be close to (ln 5 = 1.601, ln 2 = 0.693).

A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; $$x = a_1 \cos \omega t$$ and $$y = a_2 \cos 2\omega t$$ traces a curve given by:

A transverse wave is represented by: $$y = \frac{10}{\pi} \sin\left(\frac{2\pi}{T}t - \frac{2\pi}{\lambda}x\right)$$. For what value of the wavelength the wave velocity is twice the maximum particle velocity?

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be: [Given: $$\epsilon_0 = 8.85 \times 10^{-12}$$ C$$^2$$/N-m$$^2$$, $$R_E = 6.37 \times 10^6$$ m]

Three capacitances, each of 3 $$\mu$$F, are provided. These cannot be combined to provide the resultant capacitance of:

A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1 $$\Omega$$. The battery terminals are connected to external resistance $$R$$. The minimum value of $$R$$, so that a current passes through the battery to charge it is:

The mid points of two small magnetic dipoles of length $$d$$ in end-on positions, are separated by a distance $$x$$ ($$x \gg d$$). The magnitude of force between them is proportional to $$x^{-n}$$ where n is:

The magnetic field of earth at the equator is approximately $$4 \times 10^{-5}$$ T. The radius of earth is $$6.4 \times 10^6$$ m. Then the dipole moment of the earth will be nearly of the order of: