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UGC NET Entrance exams Science Electronic Science TEST : 50

UGC NET Entrance exams Science # Electronic Science Moving Charges and Magnetism Prepare / Learn

Q1. A wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire in the plane of the paper with center of ring being the center of the wire .Initially there is no current in the wire. What happens after a steady dc current is established in the wire?

A. The needles become parallel to each other

B. The needles become tangential to the ring of which they are a part

C. some needles become parallel to each other while others are uneffected

D. nothing happens

Answer : Option B
Explaination / Solution:

The current carrying wire has a magnetic field around it and the lines of force are in the form of concentric circles with their centers on the wire.

UGC NET Entrance exams Science # Electronic Science Electronic Devices Prepare / Learn

Q2. In P-N- P transistor operating as an amplifier with common emitter configurations a change in base current from

100 μ A

200 μ A

produces a change in collector current from 10 mA to 18 mA. The current gain of transistor is

A. 12.5

\times 1 0^{- 3}

B. 25.0

\times 1 0^{- 3}

Answer : Option D
Explaination / Solution:

Current gain of transistor

$β = \frac{△ I_{c}}{△ I_{b}} = \frac{(18 - 10) \times 10^{- 3} A}{(200 - 100) 10^{- 6} A} = \frac{8 \times 10^{- 3}}{100 \times 10^{- 6}} \frac{8 \times 10^{- 3}}{10^{- 4}} = 80$

UGC NET Entrance exams Science # Electronic Science Current Electricity Prepare / Learn

Q3. The resistance of a metallic conductor increases due to

A. Change in carrier density

B. increase in the number of collisions between the carriers

C. Change in dimensions of the conductor

D. increase in the rate of collisions between the carriers and vibrating atoms of the conductor

Answer : Option C
Explaination / Solution:

Resistance of a metal of length l , area A is

R = \frac{m l}{n e^{2} τ A}

where n is the carrier density, m the mass of the electron , e the charge of the electron and

τ

, the average time between two collisions. at constant temperature, n and

τ

remain constant.

\frac{m}{n e^{2} τ}

is constant at constant temperature and is called

ρ

∴ R = \frac{ρ l}{A}

The resistance increases if the length increases or the area decreases.

UGC NET Entrance exams Science # Electronic Science Electrostatic Potential and Capacitance Prepare / Learn

Q4. A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is then disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

A. zero

B. 9/2CV2

C. 25/6CV2

D. 3/2CV2

Answer : Option D
Explaination / Solution:

The charges Q1 and Q2 on the two capacitors

Q_{1} = C V; Q_{2} = (2 C) (2 V) = 4 C V

.The capacitors are connected in parallel in such a way that the positive plate of one is connected to the negative plate of the other. The common potential

V = \frac{Q_{2} - Q_{1}}{C + 2 C} = \frac{4 C V - C V}{3 C} = V

.The final energy

U_{f} = \frac{1}{2} C V^{2} + \frac{1}{2} (2 C) V^{2} = \frac{3}{2} C V^{2}

UGC NET Entrance exams Science # Electronic Science Electronic Devices Prepare / Learn

Q5. The peak electric field in the device is

A. 0.15 MV . cm-1, directed from p -region to n -region

B. 0.15 MV . cm-1, directed from n -region to p -region

C. 1.80 MV . cm-1, directed from p-retion to n -region

D. 1.80 MV . cm-1, directed from n -region to p -region

Answer : Option B
Explaination / Solution:

The peak electric field in device is directed from p to n and is

UGC NET Entrance exams Science # Electronic Science Digital Circuits Prepare / Learn

Q6. Which of the following Boolean Expressions correctly represents the relation between P,Q,R and M₁

A. M1 = (PORQ) XOR R

B. M1 = (PANDQ) XOR R

C. M1 = (PNORQ) XOR R

D. M1 = (PXORQ) XOR R

Answer : Option D
Explaination / Solution:

UGC NET Entrance exams Science # Electronic Science Magnetism and Matter Prepare / Learn

Q7. A closely wound solenoid of 2000 turns and area of cross-section

1.6 \times 1 0^{- 4} m^{2}

, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. What is the magnetic moment associated with the solenoid?

A. 1.28Am2

B. 1.08Am2

C. 1.38Am2

D. 1.18Am2

Answer : Option A
Explaination / Solution:
No Explaination.

UGC NET Entrance exams Science # Electronic Science Moving Charges and Magnetism Prepare / Learn

Q8. A wire of length L carrying current i is placed perpendicular to the magnetic induction B. The total force on the wire is

A. LB/i

B. iB/L

C. iL/B

D. iLB

Answer : Option D
Explaination / Solution:

Magnitude of the Force experienced by a current carrying conductor placed in a magnetic field is . If the angle between the directions of the current and the magnetic field is 90⁰, F=iLB

UGC NET Entrance exams Science # Electronic Science Electronic Devices Prepare / Learn

Q9. A p- type semiconductor can be obtained by adding

A. antimony to pure germanium

B. phosphorus to pure germanium

C. gallium to pure silicon

D. arsenic to pure silicon

Answer : Option C
Explaination / Solution:

In P-type doping, boron or gallium is the dopant. Boron and gallium each have only three outer electrons. When mixed into the silicon lattice, they form "holes" in the lattice where a silicon electron has nothing to bond to. The absence of an electron creates the effect of a positive charge, hence the name P-type. Holes can conduct current. A hole happily accepts an electron from a neighbor, moving the hole over a space. P-type silicon is a good conductor.

UGC NET Entrance exams Science # Electronic Science Electric Circuits Prepare / Learn

Q10. The L-C circuit shown in the figure has an inductance L = 1mH and a capacitance C = 10µF.

The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t = 0. The current i through the circuit is:

A. 5cos(5×10³t)A

B. 5sin(10⁴t)A

C. 10cos(5×10³t)A

D. 10sin(10⁴t)A

Answer : Option D
Explaination / Solution:
No Explaination.

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UGC NET Entrance exams Science Electronic Science TEST : 50