In the given circuit the the internal resistance of the 18 V cell is negligible. If R₁ = 400 $$\Omega $$, R₃ = 100 $$\Omega $$ and R₄ = 500 $$\Omega $$ and the reading of an ideal voltmeter across R₄ is 5V, then the value of R₂ will be :

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the radio of the magnetic field at the central of the loop (B_L) to that at the center of the coil (B_C), i.e. $${{{B_L}} \over {{B_C}}}$$ will be :

The magnetic field associated with a light wave is given, at the origin, by B = B₀ [sin(3.14 $$ \times $$ 10⁷)ct + sin(6.28 $$ \times $$ 10⁷)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(Take c = 3 $$ \times $$ 10⁸ ms^$$-$$1, h = 6.6 $$ \times $$ 10^$$-$$34J-s)