Physics Test 58

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Physics Test 58

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Weekly Quiz Competition

Question 1
4 / -1

A junction diode has a resistance of 25Ω when forward-biased and 2500Ω when reverse-biased. The current in the arrangement shown in the figure will be

A
1/10 A

B
1/25 A

C
1/520 A

D
1/480 A

Solution

The junction diode is forward biased.

Therefore, the effective resistance

= 25 + 100

= 2500Ω

∴ Current in diode

5v/ 100 Ω = 1/10A

Hence, the correct option is (A).
Question 2
4 / -1

The motor of an electric train can give it an acceleration of 1 m/s² and its brakes can give it a negative acceleration of 3 m/s². The shortest time in which the train can make a trip between two stations 1350 m apart is

A
14.5 s

B
28.4 s

C
60 s

D
113.6 s

Solution
Question 3
4 / -1

A message signal of frequency 10 kHz and peak voltage 10 V is used to modulate a carrier of frequency 1 MHz and peak voltage 20 V. Determine the modulation index and side bands produced

A
0.5 and 103 ± 10kHz

B
5 and 103 ± 2kHz

C
8 and 102 ± 10kHz

D
4 and 105 ± 10kHz

Solution
Question 4
4 / -1

A
θ = cos^-1 (1/√3)

B
θ = sin^-1 (1/√3)

C
θ = cos^-1 (√3/2)

D
θ = sin^-1 (√3/2)

Solution
Question 5
4 / -1

The depletion layer in a p−n junction diode is 10⁻⁶ m wide and its knee potential is 0.5 V. What is the inner electric field in the depletion region?

A
5 × (10)⁶ V/m

B
5 × (10)^-7 V/m

C
5 × (10)⁵ V/m

D
None of these

Solution

In both forward biasing and reverse biasing, applied potential establishes an internal electric field which acts against or towards the potential barrier. This internal electric field is weakened or stronger at the junction. In forward biasing knee voltage is the forwards voltage at which the current through the junction starts to increase rapidly. Once the applied forward voltage exceeds the knee voltage, the current starts increasing rapidly.

In forward biasing condition, the inner electric field is given by E = − ΔV / Δr

or
Question 6
4 / -1

110 J joules of heat is added to a gaseous system, whose internal energy is 40 J; then the amount of external work done is

A
150J

B
70J

C
110J

D
40J

Solution

According to the first law of thermodynamics

ΔQ = ΔU + ΔW

⇒ ΔW = ΔQ − ΔU

⇒110−40 = 70 J

Hence, the correct option is (B).
Question 7
4 / -1

Two uniform metal rods of lengths l₁ and l₂, and linear coefficients of expansion α1 and α2, respectively are connected to form a single rod of length (l₁+l₂). When the temperature of this rod is raised by t^∘C, the length of each rod increases by the same amount. Then, (α₂ / α₁+α₂) is

A
l₁ / ( l₁ + l₂)

B
( l₁ + l₂) / l₁

C
l₂ / ( l₁ + l₂)

D
(l₁ + l₂) / l₂

Solution
Question 8
4 / -1

In Young's experiment, light of wavelength 6000A⁰ is used to produce fringes of width 0.8 mm at a distance of 2.5m. If the whole apparatus is dipped in a liquid of refractive index 1.6 , the fringe width will be

A
0.2 mm

B
0.4 mm

C
0.6 mm

D
0.5 mm

Solution

The Young's double slit experiment belongs to a general class of "double path" experiments, in which a wave is split into two separate waves that later combine into a single wave.
Question 9
4 / -1

what is the angle between A and B?

A
Zero

B
30^o

C
60^o

D
90^o

Solution

Using Dot Product principle: A×B=ABcos⁡($)

Here, $, means the angle between vector A and B

Angle of arbitrary vectors A,B, is defined in range (0^∘,180^∘)

Then, factor AB dissolved on both sides of equation. We will get cos⁡($) = 1/2

Finally referring to the above range, you can get only one answer:

60^∘

Hence, the correct option is (C).
Question 10
4 / -1

If g is the acceleration due to gravity on the surface of the Earth, then the gain in potential energy of an object of mass m raised from the Earth's surface to a height equal to the radius R of the earth is

A
mg R/4

B
mg R/2

C
mg R

D
2mg R

Solution