1
JEE Main 2025 (Online) 29th January Morning Shift
Numerical
+4
-1

Two light beams fall on a transparent material block at point 1 and 2 with angle $\theta_1$ and $\theta_2$, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given : the distance between 1 and 2, $\mathrm{d}=4 \sqrt{3} \mathrm{~cm}$ and $\theta_1=\theta_2=\cos ^{-1}\left(\frac{n_2}{2 n_1}\right)$. where refractive index of the block $n_2>$ refractive index of the outside medium $\mathrm{n}_1$, then the thickness of the block is ______ cm .

JEE Main 2025 (Online) 29th January Morning Shift Physics - Geometrical Optics Question 4 English
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2
JEE Main 2025 (Online) 29th January Morning Shift
Numerical
+4
-1

In a hydraulic lift, the surface area of the input piston is 6 cm2 and that of the output piston is 1500 cm2. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is _______ kJ.

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3
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance.

Reason (R) : A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Both (A) and (R) are true and (R) is the correct explanation of (A).

B

Both (A) and (R) are true but (R) is not the correct explanation of (A).

C

(A) is false but (R) is true

D

(A) is true but (R) is false

4
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
At the interface between two materials having refractive indices $n_1$ and $n_2$, the critical angle for reflection of an em wave is $\theta_{1C}$. The $\mathrm{n}_2$ material is replaced by another material having refractive index $n_3$ such that the critical angle at the interface between $n_1$ and $n_1$ materials is $\theta_{2 C}$. If $n_3>n_2>n_1 ; \frac{n_2}{n_3}=\frac{2}{5}$ and $\sin \theta_{2 C}-\sin \theta_{1 C}=\frac{1}{2}$, then $\theta_{1 C}$ is :
A

$ \sin^{-1}\left( \frac{1}{6n_1} \right) $

B

$ \sin^{-1}\left( \frac{1}{3n_1} \right) $

C

$ \sin^{-1}\left( \frac{5}{6n_1} \right) $

D

$ \sin^{-1}\left( \frac{2}{3n_1} \right) $

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