JEE Mains 2026 Exam Booster Series: 30-Day Rank Booster Plan

Day 1 (Optics, Coordination Compound and Coordination Geometry)

Physics:

Optics: Lens Maker’s Formula is a high-scoring topic of this Chapter

Question: A lens having a refractive index 1.6 has a focal length of 12 cm , when it is in air. Find the focal length of the lens when it is placed in water.
(Take the refractive index of water as 1.28 )

Solution:

As we know,

$\frac{1}{\mathrm{f}}=\left[\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{m}}}-1\right]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]$

For air $\mu_{\mathrm{m}}=1$

$\begin{aligned} & \frac{1}{12}=[1.6-1]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & \frac{1}{12}=\frac{6}{10}\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & {\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]=\frac{10}{72}}\end{aligned}$

For water

$\begin{aligned} & \frac{1}{\mathrm{f}}=\left[\frac{1.6}{1.28}-1\right]\left[\frac{10}{72}\right]=\frac{32}{128} \times \frac{10}{72} \\ & \frac{1}{\mathrm{f}}=\frac{1}{4} \times \frac{10}{72} \\ & \mathrm{f}=28.8 \mathrm{~cm} \\ & \mathrm{f}=288 \mathrm{~mm}\end{aligned}$

Hence, the answer is option (2).

Chemistry

Co-ordination Compounds: Crystal Field Splitting in an Octahedral Field is the most asked topic of this question

Question: Identify the homoleptic complex(es) that is/are low spin.
(A) $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}$
(B) $\left[\mathrm{CoF}_6\right]^{3-}$
(C) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
(D) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
(E) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Choose the correct answer from the options given below :

1. (B) and (E) only

2. (A) and (C) only

3. (C) and (D) only

4. (C) only

Solution:

(A) $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{-2} \rightarrow$ Heteroleptic, $\mathrm{Fe}^{+2}, 3 \mathrm{~d}^6$, $\mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{e}_{\mathrm{g}}{ }^0, \mathrm{~d}^2 \mathrm{sp}^3$, Low spin (3d series + SFL)
(B) $\left[\mathrm{CoF}_6\right]^{-3} \rightarrow$ Homoleptic, $\mathrm{sp}^3 \mathrm{~d}^2$, High spin, $\mathrm{Co}^{+3}, 3 \mathrm{~d}^6(3 \mathrm{~d}$ series + WFL)
(C) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-4} \rightarrow$ Homoleptic
$\mathrm{Fe}^{-2}, 3 \mathrm{~d}^6, \mathrm{~d}^2 \mathrm{sp}^3, \mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{eg}^0$ Low spin
$(3 \mathrm{~d}$ series +SFL$)$
(D) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{+3} \rightarrow$ Homoleptic, $\mathrm{Co}^{+3} 3 \mathrm{~d}^6, \mathrm{~d}^2 \mathrm{sp}^3$,
$\mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{eg}^0$, Low spin (3d series + SFL)
(E) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2} \rightarrow$ Homoleptic
$\mathrm{Cr}^{+2} 3 \mathrm{~d}^4, \mathrm{~d}^2 \mathrm{sp}^3$, High spin $\mathrm{t}_{2 \mathrm{~g}}{ }^3 \mathrm{e}_{\mathrm{g}}{ }^1$
(3d series + WFL)

Hence, the correct answer is option (3).

Maths

Coordinate geometry: Centroid is the most scoring topic of this chapter

Question: Let the triangle $P Q R$ be the image of the triangle with vertices $(1,3),(3,1)$ and $(2,4)$ in the line $x+2 y=2$. If the centroid of $\triangle P Q R$ is the point $(\alpha, \beta)$, then $15(\alpha-\beta)$ is equal to :

1) 24

2) 19

3) 21

4) 22

Solution:

Let ' $G$ ' be the centroid of $\Delta$ formed by $(1,3)(3,1)$ & $(2,4)$.
$G=\left(\frac{1+3+2}{3}, \frac{3+1+4}{3}\right)=\left(2, \frac{8}{3}\right)$.

Reflection of the centroid $G$ about the line $x+2 y-2=0$.

The formula for reflection of a point $\left(x_0, y_0\right)$ about the line $A x+B y+C=0$ is:
$\frac{\alpha-x_0}{A}=\frac{\beta-y_0}{B}=-\frac{2\left(A x_0+B y_0+C\right)}{A^2+B^2}$
Here, $A=1, B=2, C=-2$.

$
\begin{aligned}
& \frac{\alpha-2}{1}=\frac{\beta-\frac{8}{3}}{2}=-2 \frac{\left(2+\frac{16}{3}-2\right)}{1+4} =\frac{-2}{5}\left(\frac{16}{3}\right)=-\frac{32}{15}
\end{aligned}
$
Then,
$\alpha=2+1 \times\left(-\frac{32}{15}\right)=2-\frac{32}{15}=\frac{30-32}{15}=-\frac{2}{15}$,
$\beta=\frac{8}{3}+2 \times\left(-\frac{32}{15}\right)=\frac{8}{3}-\frac{64}{15}=\frac{40}{15}-\frac{64}{15}=-\frac{24}{15}=-\frac{8}{5}$
So, $\alpha-\beta=-\frac{2}{15}-\left(-\frac{8}{5}\right)=-\frac{2}{15}+\frac{24}{15}=\frac{22}{15}$
$\therefore 15(\alpha-\beta)=15\times\frac{22}{15}=22$

Hence, the correct answer is option (4).

Upcoming Plan For Tommorow

Maths: Integral Calculus

Physics: Electrostatics

Chemistry: Chemical Thermodynamics

How to Use This Series

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