Explanation:
The plate current varies directly with the plate voltage. Therefore, if the plate voltage is zero, the plate current is also zero. Due to the same reason, if the plate voltage is negative, the plate current will be zero. Now, if the plate is slightly positive, then it may be the reason that the plate voltage is not able to reduce the effect of space charge. Hence, the current may be zero. Now, as the temperature of the filament is low, it will not be able to emit electrons and the resulting plate current will be zero. Hence, all the options are possible.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
In Thomson's method of determining e/m of electrons
|
(a) Electric and magnetic fields are parallel to electrons beam |
(b) Electric and magnetic fields are perpendicular to each other and perpendicular to electrons beam |
|
(c) Magnetic field is parallel to the electrons beam |
(d) Electric field is parallel to the electrons beam |
A diver in a swimming pool wants to signal his distress to a person lying on the edge of the pool by flashing his water proof flash light
|
(a) He must direct the beam vertically upwards |
|
(b) He has to direct the beam horizontally |
|
(c) He has to direct the beam at an angle to the vertical which is slightly less than the critical angle of incidence for total internal reflection |
|
(d) He has to direct the beam at an angle to the vertical which is slightly more than the critical angle of incidence for the total internal reflection |
Starting with a sample of pure 
of it decays into Zn in 15 min. The corresponding half-life is
| (a) 5 min |
(b) 7  min
|
(c) 10 min |
(d) 15 min |
The minimum wavelength of X-ray emitted by X-rays tube is 0.4125 Å. The accelerating voltage is
|
(a) 30 kV |
(b) 50 kV |
(c) 80 kV |
(d) 60 kV |
When a dielectric material is introduced between the plates of a charged condenser then electric field between the plates
|
(a) Decreases |
(b) Increases |
(c) Remain constant |
(d) First (b) then (a) |
For the transistor circuit shown below, if b = 100, voltage drop between emitter and base is 0.7 V then value of VCE will be
|
(a) 10 V |
(b) 5 V |
(c) 13 V |
(d) 0 V |
$\text{zero}$
$\frac{1}{8}\omega\text{Bl}^2$
$\frac{1}{2}\omega\text{Bl}^2$
$\text{B}\omega\text{l}^2$