Answer: (b) Electric dipole

Explanation: The dipole moment has two charges with the separation distance between them. The two charges are +q and -q, while the separation distance is 2l. It means that the net charge on an electric dipole is 0.

Answer: (d) Electrostatic field lines cannot originate from the positive charge.

Explanation: Electrostatic field lines can originate only from the positive charge and terminates at the negative charge. It cannot originate from the negative charge.

Answer: (c) Electric potential

Explanation: Electric potential is a scalar quantity, which is calculated as:

Electric potential = Work done/Charge

The unit of work done is Joule, and the charge of a conductor is Coulomb. Thus, the S.I. unit of the electric potential is:

Joule/Coulomb = J/C.

Answer: (a) 150V

Explanation:

Here, we will first solve the capacitance network to find the network’s equivalence capacitance.

Step 1: The two capacitors are connected in series. After solving, we get:

C = 1/200 + 1/200 = 100pF

Step 2: The two capacitors of 100pF are now in parallel. The equivalent capacitance will be 200pF.

Step 3: The two capacitors are now in series. After solving we get:

C = 1/200 + 1/200 = 100pF

The equivalent circuit appears as:

Ceq = 100pF

Q = Ceq x V

Q=100×10^-12×300=3×10^-8

C4 = Q/C4

C4 = 150V

Answer: (c) 17.52uC

Explanation: Consider the below figure:

The flux through S’ = 4 x flux through S

By Gauss theorem,

Net flux through the surface is equal to:

Flux through S’ is equal to:

Flux through S is equal to:

Equating both equations, we get:

The value of charge Q is:

= 3 (1 – 4 +8.84)

= 17.52

Answer: (c) Increases

Explanation: The earth conductor has low electric potential due to its connection to the earth. It increases the capacitance of the conductor.

Answer: (b) High resistivity, low melting point

Explanation: During the high current, the fuse melts and breaks the circuit. It prevents the circuit from any damage. Thus, the fuse has a low melting point and high resistivity.

Answer: (d) None of the above

Explanation: Internal resistance of the cell is defined as the hindrance offered by the electrolyte of the cell to the flow of current. It depends on the area of the immersed part of the electrodes in the electrolyte solution, nature, and concentration of the electrolyte. It also depends on the distance between the two electrodes (anode and cathode) of a cell.

Answer: (b) Curved

Explanation: The path of the electrons in the presence of an electric field is curved. But, in the absence of the electric field, the path of electrons is straight.

Answer: (a) 2.3MΩ±10%

Explanation: The above resistor has four color bands. The fourth color band is used to find the tolerance. The table is shown below:

The corresponding digits for the first three colors are 2, 3, and 5. The fourth digit corresponds to the tolerance of 10%.

The format to find the resistance value is:

ab×Multiplier ±Tolerance

Where,

a, b, and n are the digits of the first three colors.

a = 2, b = 3, and n = 5

The resistance value of the given resistor is:

=2.3MΩ

The resistance value with the tolerance can be written as:

=2.3MΩ±10%

Answer: (d) 10 Ohms, 6V

Explanation:

In series combination, the resistance can be calculated as:

R = 2 + 3 + 5 = 10 Ohms

We know that the voltage is 12V. The current can be calculated as I = V/R

I = 12/10 = 1.2V

The potential difference across the resistor of 5 Ohms is:

V = IR3 = 1.2 x 5 = 6V

Answer: (c) Both (a) and (b)

Explanation: A semiconductor consists of both positive charge carriers known as holes and negative charge carriers electrons. Both these charge carriers are responsible for the current generation in a semiconductor.

Answer: (d) Magnetic susceptibility

Explanation: Magnetic susceptibility is defined as a substance’s ability to permit the passage of magnetic field lines through it.

Answer: (c) Magnetic field lines are completely straight

Explanation: Magnetic field lines can be curved but not straight.

Answer: (b) 2800Hz

Explanation: The frequency range for speech signals lies between 300Hz and 3100Hz. Thus, the bandwidth of the speech signal or the telephonic communication can be calculated as:

3100Hz – 300Hz = 2800Hz

Answer: (d) All of the above

Explanation: The electromagnetic waves through space can be transmitted using antennas and satellites. The three processes used for such transmission are space wave propagation, ground wave, and sky wave propagation.

Answer: (b) Increase the size of the antenna.

Explanation: The purpose of modulating a signal is to reduce the size of antenna. The reduced antenna size will improve its efficiency. It will also reduce the cost of the antenna.

Answer: (b) 0.125

Explanation: The length of the dipole antenna is:

= ¼ (c/v)

=¼ (3 x 10^8)/ 6 x 10^8)

c is the speed of light, which is equal to 3 x 10^8 m/s.

v is the carrier wave frequency, which is given as 6 x 10^8 Hz.

= ¼(1/2)

= 1/8

=0.125m

Answer: (b) It is expensive

Explanation: Amplitude modulation is cheaper because it is simple to manufacture.

Answer: (a) the top of the valence band

Explanation: The majority of carriers in the p-type semiconductor are holes. It means that the p-type semiconductor is deficient in electrons. The movement of electrons is from one hole to another. Hence, the acceptor level lies on the top of the valence band.

Answer: (d) Both (a) and (b)

Explanation: The Light Emitting diodes have low operational voltage, fast action, and longer life as compared to other light sources.

Answer: (b) Solar cell

Explanation: The solar cell uses sunlight as the incident energy and gives rise to the photocurrent, which is used to recharge the batteries in satellites.

Answer: (a) 2V

Explanation: Av = A1 x A2 = 10 x 20 = 200

Voltage gain = Output voltage/ Input signal voltage

Av = Vo/Vi

Vo = Av x Vi

Vo = 200 x 0.01

Vo = 2V

Thus, the output voltage is 2V.

Answer: (d) AND

Explanation: Let consider the output of the two gates be Y1 and Y2. We can write it as:

The output of the two gates is again fed to the NOR gate. Let the output of the NOR gate be Y.

Y = (Y1 + Y2)’

We can write it as:

Hence, the logic gate output of the complete circuit is AND gate.

Answer: (c) Stability of an atom

Explanation: Rutherford model was unable to explain the stability of an atom.

Answer: (b) visible region

Explanation: The short and longer wavelength of the Balmer series lies in the range of the visible region.

Answer: (b) Gamma rays have the least penetration power.

Explanation: Gamma rays have the highest penetration power. It is because these rays do not possess any electric charge. The interaction of the gamma rays with the matter is not strong enough as compared to the charged particles. Hence, it has the highest penetration power.

Answer: (a) 1/4

Explanation: Let the two radii be R1 and R2. We know,

R1/R2 = (A1/A2)^(1/3)

Where,

A1 and A2 are the mass number of the two nuclei.

R1/R2 = (1/64)^(1/3)

R1/R2 = 1/4

Answer: (c) Increases

Explanation: After the emission, alpha particles pick up the electrons from the atoms. The emission also increases the atomic number. Hence, the neutron to proton ratio increases.

Answer: (b) Ultraviolet radiations

Explanation: Ultraviolet radiations are high-frequency radiations. The electrons are emitted when the U.V. radiations fall on the surface of the metal. Visible light is suitable for alkali metals or sensitive metals.

Answer: (c) Increases

Explanation: The increase of frequency results in the rise of the maximum kinetic energy of the photoelectrons. It causes an increase in the stopping potential.

Answer: (d) 5V

Explanation: The stooping potential can be calculated as:

Ek = eVo = eVo = Vo = 5V

Answer: (c) Energy of photon of red light is less the work function of the zinc.

Explanation: The electrons are ejected from the metal surface if the incident light frequency is greater or equal to the threshold frequency. We can also say that the incident light’s energy should be greater than the metal’s work function.

Red light has low energy and low frequency. Hence, it cannot produce electrons from a clean zinc metal surface.

Answer: (b) Short radio waves

Explanation: The range of the wavelength of the radio waves lies between 0.1m and 0.0001m. Hence, the electromagnetic spectrum with a wavelength of 0.1m is short radio waves.

Answer: (b) Orange

Explanation: The approximate range of orange color of the visible spectrum has a wavelength between 600nm and 650 nm. Hence, the wavelength of 620nm lies in the visible orange spectrum.

Answer: (a) arises due to the time variation of electric flux

Explanation: The current flows due to the flow of electrons. The direction of the current is opposite to the flow of electrons. We can also say that the current direction is in the same direction as the flow of holes. The displacement current also exists when electrons flow at a uniform rate. It depends on the changing electric field. Hence, the origin of the displacement current arises due to the time variation of the electric flux.

Answer: (c) Ferromagnetic substances

Explanation: The ferromagnetic substances are strongly attracted to the magnet. Hence, the direction of such substances’ magnetism is in the same direction of the applied magnetic field.

Answer: (b) (n – 1)G

Explanation: The current across the Galvanometer can be calculated as:

Ig = Vo/G

Where,

V = nVo

From the given relation, we can write the above equation as:

Ig = Vo/(R + G)

Vo/G = nVo(R + G)

R + G = nG

R = nG – G

R = G(n – 1)

Answer: (a) Humming losses

Explanation: When the alternating current flows across the transformer’s core, it starts producing humming sounds and vibrations. Such losses are termed humming losses.

Answer: (b) 4- times

Explanation: Self-inductance of a solenoid is directly proportional to the square of the number of turns. Hence, if the number of turns in a solenoid is doubled, the self-inductance will be changed by 4 -times.

Answer: (d) 180 degrees

Explanation: The phase difference between the voltage across the inductance and a capacitor is 180 degrees.

Answer: (d) Both (a) and (c)

Explanation: The transformer is only used to step up or step down the ac voltage. It cannot step up or step down the dc voltage. If the dc voltage is applied to the primary coil of a transformer, the current will remain same. It may produce no change in magnetic flux. It may produce no output current.

Answer: (a) 0.32A

Explanation: Given: Power across the load PL = 30W, and IL = 0.64A

VL = PL/IL = 30/0.64 = 47V

The transformer given is a step-down. Hence, it has ½ input voltage.

Thus, the current across the primary winding Ip can be calculated as:

Ip = ½ IL

Ip = ½ 0.64 = 0.32A

Answer: (c) Simple telescope

Explanation: The simple telescope comprises a single convex lens. The compound and refracting telescope include two co-axial convex lenses. While reflecting the telescope, a concave mirror is used.

Answer: (c) Partially polarized light beams show no variation in intensity.

Explanation: The light beams that show no light intensity variations are known as unpolarized light beams.

Answer: (a) Wavelength

Explanation: Speed of sound depends directly on the wavelength. It is independent of other factors, such as nature, the direction of propagation, and intensity.

Answer: (b) Blue

Explanation: Blue deviates more than red, orange, and yellow. It is because the refractive index of blue is greater than these three colors. We can also say that the refractive index is inversely proportional to the wavelength. For example, violet has the shortest wavelength. Thus, it has the highest refractive index.

Answer: (d) 1.732

Explanation: The refractive index of a medium can be calculated as:

n = tan ip

Where, ip is the polarizing angle

n = tan 60 = 1.732

Answer: (b) -60cm

Explanation: Let the focal length of the convex and the concave lens be F1 and F2.

F1 = 30cm

F2 = -20cm

The focal length F of the combination can be calculated as:

F = 1/F1 + 1 /F2

F = 1/30 + 1/(-20)

F = F1.F2/ (F1 + F2)

F = 30.(-20) / (30 – 20)

F = -600 / 10 = -60cm

Answer: (c) -24

Explanation: The magnifying power can be calculated as:

M = – fo / fe

Where,

fo and fe are the focal lengths of the objective lens and eye piece.

M = – 144/6.0 = -24