Solution:
Key Concept: Power loss due to transmisssion lines having resistance (R) and Irms current flowing in the circuit is I2rms R.
The power is to be transmitted over the large distances ar high alternating voltages, so current flowing through the wires will be low because of given power (P).
For a given power level, we find that
P = ErmsIrms (Irms is low when Erms is high)
Powor loss = I2rms R = low ($\because$ Irms is low)
Now at the receiving end high voltage is reduced by using step-down transformers.
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In case of linearly polarized light, the magnitude of the electric field vector :
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(a) Does not change with time |
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(b) Varies periodically with time |
|
(c) Increases and decreases linearly with time |
|
(d) Is parallel to the direction of propagation |
The potential gradient along the length of a uniform wire is 10 volt/metre. B and C are the two points at 30 cm and 60 cm point on a meter scale fitted along the wire. The potential difference between B and C will be
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(a) 3 volt |
(b) 0.4 volt |
(c) 7 volt |
(d) 4 volt |
In a triode amplifier, 
kilo ohm and load resistance 
kilo ohm. If the input signal voltage is 0.5 volt, then output signal voltage will be
|
(a) 1.25 volt |
(b) 5 volt |
(c) 2.5 volt |
(d) 10 volt |
The path executed by a charged particle whose motion is perpendicular to magnetic field i
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(a) A straight line |
(b) An ellipse |
(c) A circle |
(d) A helix |
A cell of e.m.f. E is connected with an external resistance R, then p.d. across cell is V. The internal resistance of cell will be
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(a) |
(b) |
(c) |
(d) |
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(a) Frequency |
(b) Wave length |
|
(c) Velocity of the source of light |
(d) None of these |
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A denotes
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(a) Electron |
(b) Positron |
(c) Proton |
(d) Neutron |
