JEE Main 2026 (Online) 6th April Evening Shift

A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is

$\_\_\_\_$ cm.

In a closed flask at 600 K , one mole of $\mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g})$ attains equilibrium as given below :

$$\mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}_2(\mathrm{~g})$$

At equilibrium, $75 \% \mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g})$ was dissociated and the total pressure is 1 atm . The magnitude of $\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}$ (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) at this temperature is $\_\_\_\_$ . (Nearest Integer)

(Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ; \ln 10=2.3, \log 2=0.3, \log 3=0.48, \log 5=0.69, \log 7=0.84$ )

If $\left(1-x^3\right)^{10}=\sum\limits_{\mathrm{r}=0}^{10} \mathrm{a}_{\mathrm{r}} x^{\mathrm{r}}(1-x)^{30-2 \mathrm{r}}$, then $\frac{9 \mathrm{a}_9}{\mathrm{a}_{10}}$ is equal to $\_\_\_\_$ .

Two cells of emfs 1 V and 2 V and internal resistance $2 \Omega$ and $1 \Omega$, respectively connected in parallel, gave a current of 1 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\frac{\alpha}{5} \mathrm{~A}$. The value of $\alpha$ is $\_\_\_\_$.

Decomposition of a hydrocarbon follows the equation $\mathrm{k}=\left(5.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{\frac{-28000 \mathrm{~K}}{\mathrm{~T}}}$. The activation energy of reaction is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)

Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Let $f(x)=\left\{\begin{array}{ll}x^3+8 ; & x<0, \\ x^2-4 ; & x \geq 0,\end{array}\right.$ and $g(x)= \begin{cases}(x-8)^{1 / 3} ; & x<0, \\ (x+4)^{1 / 2} ; & x \geq 0 .\end{cases}$

Then the number of points, where the function $g \circ f$ is discontinuous, is $\_\_\_\_$ .