1
Numerical

JEE Main 2021 (Online) 1st September Evening Shift

A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is 50 : x, then value of x is ____________.
Your Input ________

Answer

Correct Answer is 200

Explanation

Amax = AC + Am = 250 + 150 = 400

Amin = AC $$-$$ Am = 250 $$-$$ 150 = 100

$${{{A_{\min }}} \over {{A_{\max }}}} = {{100} \over {400}} = {1 \over 4} = {{50} \over {200}}$$

$$x = 200$$
2
Numerical

JEE Main 2021 (Online) 31st August Evening Shift

A bandwidth of 6 MHz is available for A.M. transmission. If the maximum audio signal frequency used for modulating the carrier wave is not to exceed 6 kHz. The number of stations that can be broadcasted within this band simultaneously without interfering with each other will be __________.
Your Input ________

Answer

Correct Answer is 500

Explanation

Signal bandwidth = 2 fm = 12 kHz

$$\therefore$$ N = $${{6MHZ} \over {12kHZ}} = {{6 \times {{10}^6}} \over {12 \times {{10}^3}}} = 500$$
3
Numerical

JEE Main 2021 (Online) 31st August Morning Shift

If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is ___________ km. (Take radius of Earth = 6400 km)
Your Input ________

Answer

Correct Answer is 64

Explanation

hT = hR = 160 ........ (i)

$$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $$

$$d = \sqrt {2R} \left[ {\sqrt {{h_T}} + \sqrt {{h_R}} } \right]$$

$$d = \sqrt {2R} \left[ {\sqrt x + \sqrt {160 - x} } \right]$$

$${{d(d)} \over {dx}} = 0$$

$${1 \over {2\sqrt x }} + {{1( - 1)} \over {2\sqrt {160 - x} }} = 0$$

$${1 \over {\sqrt x }} = {1 \over {\sqrt {160 - x} }}$$

x = 80 m

$${d_{\max }} = \sqrt {2 \times 6400} \left[ {\sqrt {{{80} \over {1000}}} + \sqrt {{{20} \over {1000}}} } \right]$$

$$ = {{80\sqrt 2 \times 2\sqrt {80} } \over {10\sqrt {10} }}$$

$$ = 8 \times 2 \times \sqrt 2 \times 2\sqrt 2 = 64$$ km
4
Numerical

JEE Main 2021 (Online) 27th August Morning Shift

A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is 'd'. The value of 'd' is ................. km.
Your Input ________

Answer

Correct Answer is 224

Explanation

$${d_m} = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $$

$${d_m} = \left( {\sqrt {2 \times 6400 \times {{10}^3} \times 320} + \sqrt {2 \times 6400 \times {{10}^3} \times 2000} } \right)$$m

dm = 224 km

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