Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

A train is moving on a straight track with speed 20 ms-1 .It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms-1 ) close to :

  • 6%

  • 12%

  • 18%

  • 18%


B.

12%

Apparent frequency heard by the person before crossing the train.

straight f subscript 1 space equals space open parentheses fraction numerator straight c over denominator straight c minus straight v subscript straight s end fraction close parentheses space straight f subscript straight o space equals space open parentheses fraction numerator 320 over denominator 320 minus 20 end fraction close parentheses space 1000
Similarly, apparent frequency heard, after crossing the trainsstraight f subscript 2 space equals space open parentheses fraction numerator straight c over denominator space straight c plus straight V subscript straight s end fraction close parentheses straight f subscript straight o space equals space open parentheses fraction numerator 320 over denominator 320 plus 20 end fraction close parentheses 1000
left square bracket straight c space equals speed space of space sound right square bracket
increment straight f space equals space straight f subscript 1 minus straight f subscript 2 space equals space open parentheses fraction numerator 2 cv over denominator straight c squared minus straight v subscript straight s superscript 2 end fraction close parentheses straight x space straight f subscript straight o
or space fraction numerator increment straight f over denominator straight f subscript straight o end fraction straight x 100 space equals space open parentheses fraction numerator 2 cv over denominator straight c squared minus straight v subscript straight s superscript 2 end fraction close parentheses straight x 100
space equals space fraction numerator 2 space straight x space 320 space straight x space 20 over denominator 300 space straight x space 340 end fraction straight x 100
equals space fraction numerator 2 space space straight x space 32 space straight x space 20 over denominator 3 space straight x 34 end fraction space equals space 12.54 space equals space 12 percent sign

472 Views

2.

In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be

  • 50 Hz

  • 25 Hz

  • 100 Hz

  • 100 Hz


C.

100 Hz

frequency = 2 (frequency of input signal).

285 Views

3.

Choose the correct statement:

  • In amplitude modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

  • In amplitude modulation, the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

  • In frequency modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

  • In frequency modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.


B.

In amplitude modulation, the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

As, we know, an amplitude modulated wave, the bandwidth is twice the frequency of modulating the signal. Therefore, amplitude modulation (AM), the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

214 Views

4.

This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements. 
Statement 1 - Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals.
Statement 2- The state of ionosphere varies from hour to hour, day to day and season to season.

  •  Statement 1 is true, statement 2 is false. 

  • Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1

  •  Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1

  •  Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1


D.

 Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1

233 Views

5.

A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are:

  • 2005 kHz, and 1995 kHz

  • 2005 kHz, 2000 kHz and 1995 kHz

  • 2000 kHz and 1995 kHz

  • 2000 kHz and 1995 kHz


C.

2000 kHz and 1995 kHz

Frequency associated with AM are
fc - fm, f, fc + fm
according to the question
fc = 2 MHz = 2000 kHz

fm = 5 kHz
Thus, frequency of the resultant signal is are carrier frequency fc = 2000 kHz, LSB frequency  fc-fm = 2000 kHz-5kHz = 1995 kHz and USB frequency fc+fm = 2005 kHz
397 Views

6.

A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

  • 1000

  • 10000

  • 10

  • 10


D.

10

straight B subscript 1 space equals space 10 space log space open parentheses straight I over straight I subscript 0 close parentheses
straight B subscript 2 space equals space log space space open parentheses fraction numerator straight I apostrophe over denominator straight I subscript 0 end fraction close parentheses
given space straight B subscript 2 minus straight B subscript 1
20 space equals space straight l 0 space log space space open parentheses fraction numerator straight I apostrophe over denominator straight I end fraction close parentheses
straight I apostrophe space equals space 100
491 Views

7.

Arrange the following electromagnetic radiations per quantum in the order of increasing energy:

A: Blue light
B: Yellow light
C: X-ray
D: Radiowave

  • D, B, A, C

  • A, B, D, C

  • C, A, B, D

  • C, A, B, D


A.

D, B, A, C

As, we know energy liberated, E = hc/λ

i.e straight E space proportional to space 1 over straight lambda
So, lesser the wavelength greater will be energy liberated by electromagnetic radiations per quantum.

As, order of wavelength is given by

X- rays, VIBGYOR, Radio waves
therefore, the order of electromagnetic radiations per quantum.

D<B<A<C

294 Views

8.

A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?

  • 2 x 106

  • 2 x 105

  • 2 x 104

  • 2 x 105


D.

2 x 105

If n = no. of channels

10% of 10 GHz = n x 5 KHz

or 101000x 10 x 109 = n x 5 x 103 n = 2 x 105


9.

A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it. 

  • 10.62 MHz

  • 10.62 kHz

  • 5.31 mHz

  • 5.31 mHz


B.

10.62 kHz

The frequency is given as
straight f space equals space fraction numerator 1 over denominator 2 πμτ end fraction
straight tau space equals space RC
therefore space straight f space equals open parentheses space fraction numerator 1 over denominator 2 πRC end fraction close parentheses
μ =0.6
R = 100 k = 100 x1000 Ω
c = 250 pico farad = 250 x 10-12 F
So,
straight f space equals space fraction numerator 1 over denominator 2 straight pi space straight x space 0.6 space straight x space 1000 space straight x 100 space straight x 250 space straight x 10 to the power of negative 12 end exponent end fraction
rightwards double arrow
straight f equals space fraction numerator 1 over denominator 9.42 space straight x 1 to the power of negative 5 end exponent end fraction
space equals space 0.1061 space straight x space 10 to the power of negative 5 end exponent
straight f space equals space 10.61 space kHz

313 Views

10.

A radar has a power of 1 Kw and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 x 106 m) is

  • 80 km

  • 16 km

  • 40km

  • 40km


A.

80 km

Maximum distance on earth where object can be detected is d, then

(h + R)2 = d2 + R2

⇒ d2 = h2 + 2Rh
Since, h<<R,
⇒d2 =2hR
straight d equals square root of 2 left parenthesis 500 right parenthesis left parenthesis 6.4 space straight x 10 to the power of 6 right parenthesis end root equals 80 space km

321 Views