JEE Main 2026 (Online) 4th April Evening Shift

A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is 6.4 m . The speed of the bullets from the gun is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$.

$$\text { (take } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 \text { ) }$$

$$\text { For a first order reaction } \mathrm{A} \rightarrow \mathrm{~B}$$

$$\begin{array}{|l|l|} \hline \mathrm{t} / \min & {[\mathrm{A}] / \mathrm{M}} \\ \hline 0 & 0.6500 \\ \hline x & 0.0650 \\ \hline 20 & 0.00065 \\ \hline \end{array}$$

$x=$ $\_\_\_\_$ min. (Nearest integer)

Let $f(x)=\left\{\begin{array}{cc}e^{x-1} & , x<0 \\ x^2-5 x+6 & , x \geq 0\end{array}\right.$ and $g(x)=f(|x|)+|f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha+\beta$ is equal to $\_\_\_\_$

Two identical small bar magnets each of dipole moment $3 \sqrt{5} \mathrm{~J} / \mathrm{T}$ are placed at a center to center separation of 10 cm , with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is $\alpha \times 10^{-3} \mathrm{~T}$. The value of $\alpha$ is $\_\_\_\_$

$$\left(\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}\right)$$

An electrochemical cell, consist of the following two redox couples, $\mathrm{M}^{x+} (\mathrm{aq}) / \mathrm{M}(\mathrm{s})\left[\mathrm{E}_{\text {red }}^{\Theta}=+0.15 \mathrm{~V}\right]$ and $\mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}(\mathrm{s})\left[\mathrm{E}_{\text {red }}^{\Theta}=-0.036 \mathrm{~V}\right]$. The cell EMF $\left(\mathrm{E}_{\text {cell }}\right)$ is recorded to be 0.2057 V . If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of $x$ is

$\_\_\_\_$ .(Nearest integer)

[Given : M is a p-block metal and $\frac{2.303 R T}{F}=0.059 \mathrm{~V}$ ]

Let $\mathrm{A}, \mathrm{B}$ be points on the two half-lines $x-\sqrt{3}|y|=\alpha, \alpha>0$ at a distance of $\alpha$ from their point of intersection $P$. The line segment $A B$ meets the angle bisector of the given half-lines at the point $Q$. If $P Q=\frac{9}{2}$ and $R$ is the radius of the circumcircle of $\triangle \mathrm{PAB}$, then $\frac{\alpha^2}{R}$ is equal to $\_\_\_\_$