Two metal wires of identical dimensions are connected in series if σ1a nd σ2 are the conductivity of the metal wires respectively, the conductivity of the combination is
A series R -C circuit is connected to an alternating voltage source. Consider tow situations:
1. When the capacitor is air filled.
2. When the capacitor is mica filled.
Current through resistor is i and voltage across capacitor is V then
Va < Vb
Va > Vb
ia >ib
ia >ib
A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 Ω all connected in series. If the ammeter has a coil of resistance 480 Ω and a shunt 20 Ω then reading in the ammeter will be
0.5 A
0.25 A
2 A
2 A
An electron moves on a straight line path XY as shown. The abcd is a coil adjacent in the path of the electron. What will be the direction of the current, if any induced in the coil?
abcd
adcb
the current will reverse its direction as the electron goes past the coil
the current will reverse its direction as the electron goes past the coil
A potentiometer wire of length L and a resistance r are connected in series with battery of e.m.f. Eo and a resistance r1. An unknown e.m.f is balanced at a length l of the potentiometer wire. The e.m.f E will be given by
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
4/9
9/4
27/5
27/5
A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is
(h = planck's constant, c= speed of light)
hc/ 2λ
hc/λ
2 hc/λ
2 hc/λ
A.
hc/ 2λ
According to Einstein photoelectric equation,
E = Kmax + Φ
Where Kmax is the maximum kinetic energy of emitted electron and Φ is work function of electrons.
Kmax = E - Φ = hv - Φ
Kmax =
Similarly, in the second case, maximum kinetic energy of emitted electron is 3 times that in the first case, we get
3Kmax
solving EQs (i) and (ii), we get work function of an emitted electron from a metal surface.
Φ = hc/2λ
Light of wavelength 500 nm is incident on metal with work function 2.28 eV. The de - Broglie wavelength of the emitted electrons is
<2.8 x 10-10 m
<2.8 x 10-9 m
> equal to 2.8 x 10-9 m
> equal to 2.8 x 10-9 m
The input signal given to a CE amplifier having a voltage gain of 150 is The corresponding output signal will be
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then,
the helium nucleus has more kinetic energy than the thorium nucleus
the helium nucleus has less momentum than the thorium nucleus
the helium nucleus has more momentum than the thorium nucleus
the helium nucleus has more momentum than the thorium nucleus