- AThe electric field must be zero.
- BThe magnetic field must be zero.
- CThe electric field may or may not be zero.
- DThe magnetic field may or may not be zero.
Explanation:
Force on charged particle in an electric eld, $\text{F} = \text{qE} \ ...(1)$
Force on charged particle in a magnetic eld $\text{F} = \text{q} (\text{v}\times\text{b}) = \text{qvB} \sin\theta \ ...(2) $
Where boldface letter represent vector nature of that quantity, q is charge of the particle, v is the velocity of the particle( if any), and $\theta$ is the angle between velocity and magnetic eld.
From (1), FE = 0 only when either q = 0 or E = 0.
Let q ≠ 0, and F ≠ 0, then we must have E ≠ 0
From (2), if q ≠ 0, v ≠ 0 and B ≠ 0 even then FB can be 'zero' because of θ = 0° or 180°
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(a) 2.50 gm |
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(c) 6.30 gm |
(d) 1.35 gm |
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(a) |
(b) |
(c) |
(d) |
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(a) Work function decreases |
|
(b) Work function increases |
|
(c) Conductivity of cathode increases |
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(d) Cathode can be heated to high temperature |
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(a) Maximum in situation (A) |
(b) Maximum in situation (B) |
|
(c) Maximum in situation (C) |
(d) The same in all situations |
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(a) 0.01 mm |
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(d) 10 mm |
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→
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(a) 10α, 6β |
(b) 4 protons, 8 neutrons |
|
(c) 6 electrons, 8 protons |
(d) 6β, 8α |
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is perpendicular to the coil. The field is reduced to zero in 0.1 second. The induced e.m.f. in the coil is
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(a) 1 V |
(b) 5 V |
(c) 50 V |
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(a) |
(b) |
(c) |
(d) |