A solid which is not transparent to visible light and whose conductivity increases with temperature is formed by (a) ionic bonding (b) covalent bonding (c) van der Waals' bonding (d) metallic bonding

The electron emitted from an electron gun are having

2 eV

of kinetic energy, then they are accelerated by applying a potential of

V

volts and are allowed to fall on atomic hydrogen gas. It has been observed that red light of wavelength

656.3 nm

has been emitted by the hydrogen. Assume that total energy of electron has been given, then

V

has the value given by
(a)

2 V

0.11 V

- 2 V

- 0.11 V

Topic:

Electric Charges and Fields

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2006

View solution

The voltage gain of an amplifier without feedback is 100 . If a negative feedback is introduced with a feedback fraction

β = 0.1

, then the gain of the feedback amplifier is (a)

9.09

(b) 10 (c)

100.1

90.0

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2006

View 2 solutions

A long solenoid has 200 turns per

cm

and carries a current

I

. The magnetic field at its centre is

6.28 \times 1 0^{- 2} Wb / m^{2}

. Another long solenoid has 100 turns per

cm

and it carries a current

I /3

. The value of the magnetic field at its centre is (a)

1.05 \times 1 0^{- 2} Wb / m^{2}

1.05 \times 1 0^{- 5} Wb / m^{2}

1.05 \times 1 0^{- 3} Wb / m^{2}

1.05 \times 1 0^{- 4} Wb / m^{2}

Topic:

Moving Charges and Magnetism

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2006

View 3 solutions

The time taken by a photoelectron to come out after the photon strikes is approximately
(a)

1 0^{- 4} s

1 0^{- 10} s

1 0^{- 16} s

1 0^{- 1} s

Topic:

Dual Nature of Radiation and Matter

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2006

View 3 solutions

Practice questions from Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

A zener diode is operating in its normal region i.e., the breakdown region for which the circuit diagram is as shown in the figure.

Here, take

V_{Z} = 7 V

and

R = 10 k Ω

. For potential difference equal to

8 V

across

A B

, what is the current through microammeter?
(a)

1000 μ A

1 mA

10 μ A

100 μ A

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

View solution

The logic circuit shown below has the input waveforms A and B as shown. Pick out the correct output waveform.

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2009

View 2 solutions

transistor operates as a common emitter amplifier, with a power gain of 60

n - p - n

dB

. The input circuit resistance is

100Ω

and the output load resistance is

10 k Ω

. The common emitter current gain

β

is (a)

1 0^{2}

6 \times 1 0^{2}

1 0^{4}

(d) 60

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

View 3 solutions

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is

6 V

and the load resistance is

R_{L} = 4 k Ω

. The series resistance of the circuit is

R_{i} = 1 k Ω

. If the battery voltage

V_{B}

varies from

8 V

16 V

, what are the minimum and maximum values of the current through Zener diode?

$1.5 mA, 8.5 mA$
$1 mA, 8.5 mA$
$0.5 mA, 8.5 mA$
$0.5 mA, 6 mA$

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

View 3 solutions

Practice questions from Semiconductor Electronics: Materials, Devices and Simple Circuits in the same exam

Carbon, silicon and germanium have four valence electrons each. At room temperature, which one of the following statements is most appropriate?
(a) The number of free conduction electrons is significant in

C

but small in

Si

and Ge
(b) The number of free conduction electrons is negligibly small in all the three
(c) The number of free electrons for conduction is significant in all the three
(d) The number of free electrons for conduction is significant only in Si and Ge but small is

C

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2007

View solution

A TV transmission tower antenna is at a height of

20 m

. Suppose that the receiving antenna is at.
(i) ground level
(ii) a height of

5 m

.
The increase in antenna range in case (ii) relative to case (i) is

n %

The value of n, to the nearest integer, is

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Exam:

JEE Mains 2021 Mar 18 Shift 2

View 4 solutions

The maximum and minimum amplitude of an amplitude modulated wave is

16 V

and

8 V

respectively. The modulation index for this amplitude modulated wave is

x \times 1 0^{- 2} .

The value of

x

is___________.

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Exam:

JEE Mains 2021 Feb 26 Shift 1

View solution

A telephonic communication service is working at carrier frequency of

10 GHz

. Only

10%

of it is utilised for transmission. How many telephonic channels can be transmitted simultaneously, if each channel requires a bandwidth of

5 kHz

?
(a)

2 \times 1 0^{3}

2 \times 1 0^{4}

2 \times 1 0^{5}

2 \times 1 0^{6}

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2018

View 2 solutions

Practice more questions from Semiconductor Electronics: Materials, Devices and Simple Circuits

The following configuration of gate is equivalent to

-

-

NAND
XOR
OR
None of these

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

View 6 solutions

A waveform shown when applied to the following circuit will produce which of the following output waveform? Assuming ideal diode configuration and

Hard

R_{1} = R_{2}

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

View 2 solutions

In an n-type semiconductor, the fermi energy level lies

In the forbidden energy gap nearer to the conduction band
In the forbidden energy gap nearer to the valence band
In the middle of forbidden energy gap
Outside the forbidden energy gap

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

View 8 solutions

Hole in semiconductor is

an anti-particle of electron
a vacancy created when an electron leaves a covalent bond
absence of free electrons
an artificially created particle

Topic:

Semiconductor Electronics: Materials, Devices and Simple Circuits

View solution

Practice questions on similar concepts asked by Filo students

parameters for the circuit in Fig. P16.43.

Find the

A BC D

Topic:

Physics

View solution

The de-Broglie wavelength of an electron moving with a speed of

Question

2.

6.6 \times 1 0^{15} ms^{- 1}

is nearly equal to

$1 0^{- 11} m$
$1 0^{- 9} m$
$1 0^{- 7} m$
$1 0^{- 5} m$

Topic:

Atoms, Nuclei & Dual nature

View solution

F. physics (a) FIGURE 9.3 Focus of a concave and convex mirror. The distance between the focus

F

and the pole

P

of the mirror is

Galk

the focal length of the mirror, denoted by

f

. We now show that

f = p

geometh of reflection of an incident ray is shown in Fig. 9.4. ray parallel to the principal axis striking the of incidence, and MD be the perpendicular from

M

on the principal axis. Then,

∠ MCP = θ

and

∠ MFP = 2 θ

Now,

tan θ = \frac{M D}{C D} and tan 2 θ = \frac{M D}{F D}

For small

θ

, which is true for paraxial rays,

tan θ = θ

tan 2 θ \approx 2 θ

. Therefore, Eq. (9.1) gives

\frac{M D}{F D} = 2 \frac{M D}{C D} or, F D = \frac{C D}{2} Now, for

Topic:

Optics

View solution