1
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1 A uniform electric field $$\mathrm{E}=(8 \mathrm{~m} / \mathrm{e}) \,\mathrm{V} / \mathrm{m}$$ is created between two parallel plates of length $$1 \mathrm{~m}$$ as shown in figure, (where $$\mathrm{m}=$$ mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of $$2 \mathrm{~m} / \mathrm{s}$$. The angle of the deviation $$(\theta)$$ of the path of the electron as it comes out of the field will be _________. A
$$\tan ^{-1}(4)$$
B
$$\tan ^{-1}(2)$$
C
$$\tan ^{-1}\left(\frac{1}{3}\right)$$
D
$$\tan ^{-1}(3)$$
2
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1 A charge of $$4 \,\mu \mathrm{C}$$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be :

A
$$1 \,\mu \mathrm{C}$$ and $$3 \,\mu\mathrm{C}$$
B
$$2 \,\mu \mathrm{C}$$ and $$2\, \mu \mathrm{C}$$
C
0 and $$4\, \mu\, \mathrm{C}$$
D
$$1.5 \,\mu \mathrm{C}$$ and $$2.5\, \mu \mathrm{C}$$
3
JEE Main 2022 (Online) 27th July Morning Shift
+4
-1 Two identical positive charges $$Q$$ each are fixed at a distance of '2a' apart from each other. Another point charge $$q_{0}$$ with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge $$\mathrm{q}_{0}$$ executes $$\mathrm{SHM}$$. The time period of oscillation of charge $$\mathrm{q}_{0}$$ will be :

A
$$\sqrt{\frac{4 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}$$
B
$$\sqrt{\frac{q_{0} Q}{4 \pi^{3} \varepsilon_{0} m a^{3}}}$$
C
$$\sqrt{\frac{2 \pi^{2} \varepsilon_{0} m a^{3}}{q_{0} Q}}$$
D
$$\sqrt{\frac{8 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}$$
4
JEE Main 2022 (Online) 27th July Morning Shift
+4
-1 An electron (mass $$\mathrm{m}$$) with an initial velocity $$\vec{v}=v_{0} i\left(v_{0}>0\right)$$ is moving in an electric field $$\vec{E}=-E_{0} \hat{i}\left(E_{0}>0\right)$$ where $$E_{0}$$ is constant. If at $$\mathrm{t}=0$$ de Broglie wavelength is $$\lambda_{0}=\frac{h}{m v_{0}}$$, then its de Broglie wavelength after time t is given by

A
$$\lambda_{0}$$
B
$$\lambda_{0}\left(1+\frac{e E_{0} t}{m v_{0}}\right)$$
C
$$\lambda_{0} t$$
D
$$\frac{\lambda_{0}}{\left(1+\frac{e E_{0} t}{m v_{0}}\right)}$$
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Modern Physics
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