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1
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Refer to the figure given below. The values of $I_1, I_2$ and $I_3$ are $\_\_\_\_$ .

JEE Main 2026 (Online) 5th April Morning Shift Physics - Current Electricity Question 2 English
A

$I_1=2.5 \mathrm{~A}, I_2=1.875 \mathrm{~A}, I_3=1.875 \mathrm{~A}$

B
$I_1=1.875 \mathrm{~A}, I_2=2.5 \mathrm{~A}, I_3=1.875 \mathrm{~A}$
C
$I_1=1.875 \mathrm{~A}, I_2=1.875 \mathrm{~A}, I_3=2.5 \mathrm{~A}$

D

$I_1=2.5 \mathrm{~A}, I_2=2.5 \mathrm{~A}, I_3=1.875 \mathrm{~A}$

2
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An electron of mass $m$ is moving in an electric field $\vec{E}=-2 E_{\mathrm{o}} \hat{i}\left(E_{\mathrm{o}}=\right.$ constant $\left.>0\right)$, with an initial velocity $\vec{V}=v_{\mathrm{o}} \hat{i} \left(v_{\mathrm{o}}=\right.$ constant $\left.>0\right)$. If $\lambda_{\mathrm{o}}=\frac{h}{4 m v_{\mathrm{o}}}$, its de Broglie wavelength at time $t$ is

$\_\_\_\_$ .

( $e=$ charge of electron)

A

$$ \frac{4 \lambda_{\mathrm{o}}}{\left[1-\frac{E_{\mathrm{o}} e}{2 m} \frac{t}{v_{\mathrm{o}}}\right]} $$

B

$$ \frac{4 \lambda_{\mathrm{o}}}{\left[1+\frac{E_{\mathrm{o}} e}{2 m} \frac{t}{v_{\mathrm{o}}}\right]} $$

C

$$ \frac{4 \lambda_{\mathrm{o}}}{\left[1+\frac{2 E_{\mathrm{o}} e}{m} \frac{t}{v_{\mathrm{o}}}\right]} $$

D

$$ \frac{4 \lambda_{\mathrm{o}}}{\left[1-\frac{2 E_{\mathrm{o}} e}{m} \frac{t}{v_{\mathrm{o}}}\right]} $$

3
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit $(f)$. The ratio of the radius of the orbits in given by $r_i: r_f=16: 4$. The wavelength of photon emitted due to this transition is $\_\_\_\_$ nm.

(Given Rydberg constant $=1.0973 \times 10^7 / \mathrm{m}$ )

A

121

B

242

C

486

D

974

4
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A displacement current of 4.0 A can be set up in the space between two parallel plates of $6 \mu \mathrm{~F}$ capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $\alpha \times 10^6 \mathrm{~V} / \mathrm{s}$. The value of $\alpha$ is $\_\_\_\_$ .

A

0.58

B

0.67

C

0.82

D

0.75

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