1
JEE Main 2025 (Online) 23rd January Morning Shift
Numerical
+4
-1
Change Language

Consider the following sequence of reactions to produce major product (A)

JEE Main 2025 (Online) 23rd January Morning Shift Chemistry - Compounds Containing Nitrogen Question 14 English

$\begin{aligned} & \text { Molar mass of product }(\mathrm{A}) \text { is } \mathrm{g} \mathrm{~mol}^{-1} \text {. } \\ & \text { (Given molar mass in } \mathrm{g} \mathrm{~mol}^{-1} \text { of } \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{Br}: 80, \mathrm{~N}: 14, \mathrm{P}: 31 \text { ) }\end{aligned}$

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2
JEE Main 2025 (Online) 23rd January Morning Shift
Numerical
+4
-1
Change Language

During " S " estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is _________ %.

(Given molar mass in $\mathrm{g} \mathrm{~mol}^{-1}$ of $\mathrm{Ba}: 137, \mathrm{~S}: 32, {\mathrm{O}: 16}$)

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3
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
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Let a curve $y=f(x)$ pass through the points $(0,5)$ and $\left(\log _e 2, k\right)$. If the curve satisfies the differential equation $2(3+y) e^{2 x} d x-\left(7+e^{2 x}\right) d y=0$, then $k$ is equal to

A
32
B
8
C
4
D
16
4
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let P be the foot of the perpendicular from the point $\mathrm{Q}(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $(3,-2,1)$, is

A
 $\sqrt{30}$
B
$9 \sqrt{15}$
C
$3 \sqrt{30}$
D
$8 \sqrt{15}$
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