MHT CET 2024 16th May Morning Shift | Semiconductor Devices and Logic Gates Question 4 | Physics

The input signal given to C.E. amplifier having a voltage gain of 126 is $V_i=2 \cos \left(12 t+\frac{\pi}{3}\right)$. The corresponding output signal will be

$252 \cos \left(12 t+\frac{4 \pi}{3}\right)$

$252 \cos \left(12 \mathrm{t}+\frac{\pi}{3}\right)$

$63 \cos \left(12 t+\frac{2 \pi}{3}\right)$

$2 \cos \left(12 t+\frac{5 \pi}{3}\right)$

MHT CET 2024 16th May Morning Shift

MCQ (Single Correct Answer)

-0

The truth table of the following circuit is

MHT CET 2024 16th May Morning Shift Physics - Semiconductor Devices and Logic Gates Question 1 English

A	B	Y
0	0	1
0	1	1
1	0	1
1	1	0

A	B	Y
0	0	0
0	1	0
1	0	1
1	1	1

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	1

A	B	Y
0	0	0
0	1	1
1	0	0
1	1	1

MHT CET 2024 16th May Morning Shift

MCQ (Single Correct Answer)

-0

If the $\mathrm{p}-\mathrm{n}$ junction diode is unbiased,

n-type side has lower potential than p-type side.

n-type side has same potential as p-type side.

An electric field is established at the junction such that n side is at positive voltage relative to p -side.

The potential across the junction is such that p -side is at positive voltage relative to n-side

MHT CET 2024 15th May Evening Shift

MCQ (Single Correct Answer)

-0

For detecting light intensity we use

photodiode in reverse bias.

photodiode in forwarded bias.

LED in reverse bias.

LED in forward bias.

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