In an interference experiment,the third bright fringe is obtained at a point on the screen with light of wavelength $700 \, nm$. What should be the wavelength of the light source in order to obtain the $5$th bright fringe at the same point (in $, nm$)?

The smallest positive root of the equation $\tan x - x = 0$ lies in the interval:

In a region of uniform magnetic induction $B = 10^{-2} \, T$,a circular coil of radius $30 \, cm$ and resistance $\pi^2 \, \Omega$ is rotated about an axis which is perpendicular to the direction of $B$ and which forms a diameter of the coil. If the coil rotates at $200 \, rpm$,the amplitude of the alternating current induced in the coil is ...... $mA$.

In an experiment on photoelectric emission from a metallic surface,the wavelength of incident light is $2 \times 10^{-7} \ m$ and the stopping potential is $2.5 \ V$. The threshold frequency of the metal (in $Hz$) is approximately (charge of electron $e = 1.6 \times 10^{-19} \ C$,Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$):

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$ and if $px^2+qx+r=0$ has roots $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$,then $r$ is equal to