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1

JEE Main 2021 (Online) 27th August Morning Shift

Numerical
A circuit is arranged as shown in figure. The output voltage V0 is equal to ................... V.

Your Input ________

Answer

Correct Answer is 5

Explanation

As diodes D1 and D2 are in forward bias, so they acted as negligible resistances

$$\Rightarrow$$ Input voltage become zero


$$\Rightarrow$$ Input current is zero

$$\Rightarrow$$ Output current is zero

$$\Rightarrow$$ V0 = 5 volt
2

JEE Main 2021 (Online) 26th August Evening Shift

Numerical
For the given circuit, the power across Zener diode is .............. mW.

Your Input ________

Answer

Correct Answer is 120

Explanation



$$i = {{10V} \over {5k\Omega }}$$ = 2mA

$$I = {{14V} \over {1k\Omega }}$$ = 14 mA

$$\therefore$$ Iz = 12 mA

$$\therefore$$ P = IzVz = 120 mW
3

JEE Main 2021 (Online) 26th August Evening Shift

Numerical
If the maximum value of accelerating potential provided by a ratio frequency oscillator is 12 kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is ...............

[mp = 1.67 $$\times$$ 10$$-$$27 kg, e = 1.6 $$\times$$ 10$$-$$19C, Speed of light = 3 $$\times$$ 108 m/s]
Your Input ________

Answer

Correct Answer is 543

Explanation

V = 12 kV

Number of revolution = n

$$n[2 \times {q_P} \times V] = {1 \over 2}{m_P} \times v_P^2$$

$$n[2 \times 1.6 \times {10^{ - 19}} \times 12 \times {10^3}]$$

$$ = {1 \over 2} \times 1.67 \times {10^{ - 27}} \times {\left[ {{{3 \times {{10}^8}} \over 6}} \right]^2}$$

n(38.4 $$\times$$ 10$$-$$16) = 0.2087 $$\times$$ 10$$-$$11

n = 543.4
4

JEE Main 2021 (Online) 25th July Evening Shift

Numerical
In a semiconductor, the number density of intrinsic charge carries at 27$$^\circ$$C is 1.5 $$\times$$ 1016/m3. If the semiconductor is doped with impurity atom, the hole density increases to 4.5 $$\times$$ 1022/m3. The electron density in the doped semiconductor is ___________ $$\times$$ 109/m3.
Your Input ________

Answer

Correct Answer is 5

Explanation

$${n_e}{n_h} = {n_i}^2$$

$${n_e} = {{{n_i}^2} \over {{n_h}}} = {{{{(1.5 \times {{10}^{16}})}^2}} \over {4.5 \times {{10}^{22}}}}$$

$$ = {{1.5 \times 1.5 \times {{10}^{32}}} \over {4.5 \times {{10}^{22}}}}$$

$$ = 5 \times {10^9}$$/m3

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