1
JEE Main 2023 (Online) 10th April Evening Shift
Numerical
+4
-1
Out of Syllabus
Change Language

In the figure, $$\theta_{1}+\theta_{2}=\frac{\pi}{2}$$ and $$\sqrt{3}(\mathrm{BE})=4(\mathrm{AB})$$. If the area of $$\triangle \mathrm{CAB}$$ is $$2 \sqrt{3}-3$$ unit $${ }^{2}$$, when $$\frac{\theta_{2}}{\theta_{1}}$$ is the largest, then the perimeter (in unit) of $$\triangle \mathrm{CED}$$ is equal to _________.

JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Properties of Triangle Question 2 English

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2
JEE Main 2023 (Online) 10th April Evening Shift
Numerical
+4
-1
Change Language

Let the equations of two adjacent sides of a parallelogram $$\mathrm{ABCD}$$ be $$2 x-3 y=-23$$ and $$5 x+4 y=23$$. If the equation of its one diagonal $$\mathrm{AC}$$ is $$3 x+7 y=23$$ and the distance of A from the other diagonal is $$\mathrm{d}$$, then $$50 \mathrm{~d}^{2}$$ is equal to ____________.

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3
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A person travels $$x$$ distance with velocity $$v_{1}$$ and then $$x$$ distance with velocity $$v_{2}$$ in the same direction. The average velocity of the person is $$\mathrm{v}$$, then the relation between $$v, v_{1}$$ and $$v_{2}$$ will be.

A
$$\mathbf{V}=\mathbf{V}_{1}+\mathbf{V}_{2}$$
B
$$V=\frac{v_{1}+V_{2}}{2}$$
C
$$\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{v}_{1}}+\frac{1}{\mathrm{v}_{2}}$$
D
$$\frac{2}{\mathrm{~V}}=\frac{1}{\mathrm{v}_{1}}+\frac{1}{\mathrm{v}_{2}}$$
4
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If each diode has a forward bias resistance of $$25 ~\Omega$$ in the below circuit,

JEE Main 2023 (Online) 10th April Evening Shift Physics - Semiconductor Question 17 English

Which of the following options is correct :

A
$$\frac{I_{3}}{I_{4}}=1$$
B
$$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=2$$
C
$$\frac{I_{2}}{\mathrm{I}_{3}}=1$$
D
$$\frac{I_{1}}{I_{2}}=1$$
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