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1

JEE Main 2021 (Online) 27th August Morning Shift

Numerical
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
2

JEE Main 2021 (Online) 26th August Morning Shift

Numerical
An amplitude modulated wave is represented by
Cm(t) = 10(1 + 0.2 cos 12560t) sin(111 $$\times$$ 104t) volts. The modulating frequency in kHz will be ................. .
Your Input ________

Answer

Correct Answer is 2

Explanation

Wm = 12560 = 2$$\pi$$fm

$${f_m} = {{12560} \over {2\pi }}$$

= 2000 Hz
3

JEE Main 2021 (Online) 27th July Evening Shift

Numerical
The maximum amplitude for an amplitude modulated wave is found to be 12V while the minimum amplitude is found to be 3V. The modulation index is 0.6x where x is ____________.
Your Input ________

Answer

Correct Answer is 1

Explanation

Amax = Ac + Am = 12

Amin = Ac $$-$$ Am = 3

$$\Rightarrow$$ Ac = $${{15} \over 2}$$ & Am = $${9 \over 2}$$

modulation index = $${{{A_m}} \over {{A_c}}} = {{9/2} \over {15/2}} = 0.6$$

$$\Rightarrow$$ x = 1
4

JEE Main 2021 (Online) 27th July Morning Shift

Numerical
The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5V amplitude are $${a \over {10}}V$$ and $${b \over {10}}V$$ respectively. Then the value of $${a \over b}$$ is ____________.
Your Input ________

Answer

Correct Answer is 1

Explanation



$${a \over {10}} = {b \over {10}} = {{\mu {A_C}} \over 2}$$

$$ \Rightarrow {a \over b} = 1$$

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