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.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Light of wavelength 5000 Å falls on a sensitive plate with photoelectric work function of 1.9 eV. The kinetic energy of the photoelectron emitted will be
|
(a) 0.58 eV |
(b) 2.48 eV |
(c) 1.24 eV |
(d) 1.16 eV |
A thin prism 
with angle 6° and made from glass of refractive index 1.54 is combined with another thin prism P2 of refractive index 1.72 to produce dispersion without deviation. The angle of prism 
will be
|
(a) 5° 24’ |
(b) 4° 30’ |
(c) 6° |
(d) 8° |
Young's experiment is performed in air and then performed in water, the fringe width
|
(a) Will remain same |
(b) Will decrease |
(c) Will increase |
(d) Will be infinite |
A 36 W galvanometer is shunted by resistance of 4W. The percentage of the total current, which passes through the galvanometer is
|
(a) 8 % |
(b) 9 % |
(c) 10 % |
(d) 91 % |
In a P-type semi-conductor, germanium is dopped with
|
(a) Gallium |
(b) Boron |
(c) Aluminium |
(d) All of these |
In a Young’s double slit experiment, the source illuminating the slits is changed from blue to violet. The width of the fringes
|
(a) Increases |
(b) Decreases |
(c) Becomes unequal |
(d) Remains constant |
Which is the correct relation for forbidden energy gap in conductor, semi conductor and insulator
|
(a) |
|
(b) |
|
(c) |
|
(d) |
The induction coil works on the principle of
|
(a) Self-induction |
(b) Mutual induction |
|
(c) Ampere's rule |
(d) Fleming's right hand rule |
