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)

A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30^o from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s^$$-$$1) will be (g = 10 ms^$$-$$2)

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :

Two point charges q₁$$\left( {\sqrt {10} \mu C} \right)$$ and q₂($$-$$ 25 $$\mu $$C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
[take $${1 \over {4\pi { \in _0}}}$$ = 9 $$ \times $$ 10⁹ Nm²C^$$-$$2]