JEE Main 3 April 2025 Shift 1 question paper with solutions

Given $|2 adj(3A adj(2A))|$. We have $2^3 \cdot |3A adj(2A)|^2$. This becomes $2^3 \cdot (3^3)^2 \cdot |A|^2 \cdot |adj(2A)|^2$. Simplifying further, $2^3 \cdot 3^6 \cdot |A|^2 \cdot \left( |2A|^2 \right)^2$. This is $2^3 \cdot 3^6 \cdot |A|^2 \left[ (2^3)^2 \cdot |A|^2 \right]^2$. Simplifying, $2^3 \cdot 3^6 \cdot |A|^2 \cdot 2^{12} \cdot |A|^4$. This results in $2^{15} \cdot 3^6 \cdot |A|^6$. Finally, $2^{15} \cdot 3^6 \cdot 5^6 = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$. Given $\alpha = 15$, $\beta = 6$, $\gamma = 6$. Thus, $\alpha + \beta + \gamma = 27$.

Given the lines: $$L_1 = \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = p$$ $$L_2 = \frac{x-6}{1} = \frac{y}{1} = \frac{z-4}{-1} = q$$ Point $A$ is given by $(2p+1, 3p+2, 4p+3)$ and point $B$ by $(q+6, q, 4-q)$. The direction ratios of $PA$ are $2p-3, 3p+1, 4p+3$. The direction ratios of $PB$ are $q+2, q-1, 4-q$. Equating the direction ratios: $$2p-3 = 3p+1 = 4p+3$$ $$q+2 = q-1 = 4-q$$ Solving these equations: For $p$: $$2p - 3 = 3p + 1 \Rightarrow p = -4$$ $$2p - 3q + 3 = 3p + 6p + q + 2$$ $$pq + rp + 4q - 1 = 0$$ $$12p - 3pq + 4 - q = 4pq + 3q - 4p - 3$$ $$7pq - 16p + 4q - 7 = 0$$ $$8p - 2pq - 12 + 3q = 4pq + 8p + 3q + 6$$ $$6pq = -18 \Rightarrow pq = -3$$ For $q$: $$8p + 4q = 4 \Rightarrow 2p + q = 1$$ $$-21 - 16p + 4q - 7 \Rightarrow 4p - q = -7$$ $$16p - 4q = -28 \therefore p = -1, q = 3$$ Thus, $A(-1, -1, -1)$ and $B(9, 3, 1)$. The determinant is: $$\begin{vmatrix} 1 & 0 & 0 \\ 1 & -1 & 0 \\ 9 & 3 & 1 \end{vmatrix} = 1(-1 + 9) = 8$$

Given $x^2 + \sqrt{3}x - 16 = 0 < \beta$. For $P_n + \sqrt{3}P_{n-1} - 16P_{n-2} = 0$, $$P_{25} + \sqrt{3}P_{24} - 16P_{23} = 0$$ Thus, $$P_{25} + \sqrt{3}P_{24} = 8$$ $$2P_{23}$$ Similarly, $x^2 + 3x - 1 = 0 < \sum \gamma_\delta$ $Q_n = \gamma^n + \delta^n$ $$Q_{25} - Q_{23} = \gamma^{25} + \delta^{25} - \gamma^{23} - \delta^{23}$$ $$= \gamma^{23}(\gamma^2 - 1) + \delta^{23}(\delta^2 - 1)$$ $$= \gamma^{23}(-3\gamma) + \delta^{23}(-3\gamma)$$ $$= -3 \left[ \gamma^{24} + \delta^{24} \right]$$ $$= -3Q_{24}$$ Thus, $$Q_{25} - Q_{23} = -3$$ $$Q_{24}$$ $$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}} = 8 - 3 = 5$$

Given $S = (2 + \sqrt{3})^8$. For sum of rational terms: $$= \binom{8}{0} (2)^8 + \binom{8}{2} (2)^6 \cdot (\sqrt{3})^2 + \binom{8}{4} (2)^4 (\sqrt{3})^4$$ $$+ \binom{8}{6} (2)^2 (\sqrt{3})^6 + \binom{8}{8} (\sqrt{3})^8$$ $$= 2^8 + 28 \times 2^6 \cdot 3 + 70 \cdot 2^4 \cdot 9 + 28 \cdot 2^2 \cdot 27 + 81$$ $$= 256 + 5376 + 10080 + 3024 + 81$$ $$= 18817$$

By given data $\overrightarrow{AB} + \overrightarrow{AC} = \overrightarrow{CB}$. Let pv of $\overrightarrow{A}$ are $\overrightarrow{O}$ then $\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A}$. i.e. pv of $\overrightarrow{B} = -2\hat{i} - \hat{j} + \hat{k}$. $\overrightarrow{CA} = \overrightarrow{A} - \overrightarrow{C}$. i.e. pv of $\overrightarrow{C} = -(\hat{i} - 3\hat{j} - 5\hat{k})$. Now pv of centroid $\overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} = \frac{\overrightarrow{0} + (2, -1, 1) + (-1, 3, 5)}{3}$. $\overrightarrow{G} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k})$. Now $\overrightarrow{AG} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k})$. $\Rightarrow |\overrightarrow{AG}|^2 = \frac{1}{9} \times 41$. $\overrightarrow{BG} = \left(\frac{1}{3} - 2\right)\hat{i} + \left(\frac{2}{3} + 1\right)\hat{j} + (2 - 1)\hat{k}$. $\Rightarrow |\overrightarrow{BG}|^2 = \frac{59}{9}$. $\overrightarrow{CG} = \left(\frac{1}{3} + 1\right)\hat{i} + \left(\frac{2}{3} - 3\right)\hat{j} + (2 - 5)\hat{k}$. $\Rightarrow |\overrightarrow{CG}|^2 = \frac{146}{9}$. Now $6 \left[|\overrightarrow{AG}|^2 + |\overrightarrow{BG}|^2 + |\overrightarrow{CG}|^2\right] = 6 \times \left[\frac{41}{9} + \frac{59}{9} + \frac{146}{9}\right] = 6 \times \frac{246}{9} = 164$.

Given ellipse is $\frac{x^2}{36} + \frac{y^2}{25} = 1$. Any point on line $AB$ can be assumed as $Q(\sqrt{5} + r \cos \theta, \sqrt{5} + r \sin \theta)$. Putting this in equation of ellipse, we get $$25(\sqrt{5} + r \cos \theta)^2 + 36(\sqrt{5} + r \sin \theta)^2 = 900$$ Simplifying, we get $$r^2 \left( 25 \cos^2 \theta + 36 \sin^2 \theta \right) + 2 \sqrt{5} r (25 \cos \theta + 36 \sin \theta) - 595 = 0$$ $|r| = \mathrm{PA}, \mathrm{PB}$. $$\mathrm{PA} \cdot \mathrm{PB} = 595 = \frac{595}{25 \cos^2 \theta + 36 \sin^2 \theta} = \frac{595}{25 + 11 \sin^2 \theta}$$ maximum, if $\sin^2 \theta = 0$. This means line $AB$ must be parallel to $x$-axis $\Rightarrow y_A = y_B = \sqrt{5}$. Putting $y = \sqrt{5}$ in equation of ellipse, we get $$\frac{x^2}{36} + \frac{1}{5} = 1 \Rightarrow x^2 = 36 \cdot \frac{4}{5}$$ Hence, $$\mathrm{PA}^2 + \mathrm{PB}^2 = \left( \sqrt{5} - \frac{12}{\sqrt{5}} \right)^2 + \left( \sqrt{5} + \frac{12}{\sqrt{5}} \right)^2 = 2 \left( 5 + \frac{144}{5} \right) = \frac{338}{5}$$ $$5 \left( \mathrm{PA}^2 + \mathrm{PB}^2 \right) = 338$$

Given sum is $$S_n = 1 + 3 + 11 + 25 + 45 + 71 + \ldots + T_n$$ First order differences are in A.P. Thus, we can assume that $$T_n = an^2 + bn + c$$ Solving $$\begin{cases} T_1 = 1 = a + b + c \\ T_2 = 3 = 4a + 2b + c \\ T_3 = 11 = 9a + 3b + c \end{cases}$$ we get $a = 3$, $b = -7$, $c = 5$ Hence, general term of given series is $$T_n = 3n^2 - 7n + 5$$ Hence, required sum equals $$\sum_{n=1}^{20} \left(3n^2 - 7n + 5\right) = 3 \left(\frac{20 \cdot 21 \cdot 41}{6}\right) - 7 \left(\frac{20 \cdot 21}{2}\right) + 5(20) = 7240$$

Given function is $f(x) = \log \left( \frac{2x-3}{5+4x} \right) + \sin^{-1} \left( \frac{4+3x}{2-x} \right)$. For domain, the conditions are $$\frac{2x-3}{5+4x} > 0 and \left| \frac{4+3x}{2-x} \right| \leq 1$$ Now, $$\frac{2x-3}{5+4x} > 0 \Rightarrow x \in \left( -\infty, -\frac{5}{4} \right) \cup \left[ \frac{3}{2}, \infty \right)$$ and $$-1 \leq \frac{4+3x}{2-x} \leq 1$$ $$\Rightarrow \left( -1 \leq \frac{4+3x}{2-x} \right) \cap \left( \frac{4+3x}{2-x} \leq 1 \right)$$ $$\Rightarrow \left( \frac{6+2x}{2-x} \geq 0 \right) \cap \left( \frac{2+4x}{2-x} \leq 0 \right)$$ $$\Rightarrow \frac{6+2x}{2-x} \cdot \frac{2+4x}{2-x} \leq 0$$ $$\Rightarrow x \in \left[ -3, -\frac{1}{2} \right]$$ Hence, we get the domain of $f$ as $x \in \left[ -3, -\frac{5}{4} \right)$. This means that $\alpha = -3, \beta = -\frac{5}{4}$. Thus, $\alpha^2 + 4\beta = 9 - 5 = 4$.

Given $\($ $\sum$_{r=1}^{9} $\left$( $\frac{r+3}{2^r}$ $\right$) $\cdot$ $\binom{9}{r}$ = $\alpha$ $\left$( $\frac{3}{2}$ $\right$)^9 - $\beta$, $\alpha$, $\beta$ $\in$ $\mathbb{N}$ $\)$. Now, $\[$ $\sum$_{r=1}^{9} $\left$( $\frac{r+3}{2^r}$ $\right$) $\cdot$ $\binom{9}{r}$ = $\sum$_{r=1}^{9} $\left$( $\frac{r}{2^r}$ $\right$) $\cdot$ $\binom{9}{r}$ + $\sum$_{r=1}^{9} $\left$( $\frac{3}{2^r}$ $\right$) $\cdot$ $\binom{9}{r}$ $\]$ $\[$ = $\sum$_{r=1}^{9} $\left$( $\frac{9}{2^r}$ $\right$) $\cdot$ $\binom{8}{r-1}$ + 3 $\sum$_{r=1}^{9} $\binom{9}{r}$ $\left$( $\frac{1}{2}$ $\right$)^r $\left$[ Using g $\binom{9}{r}$ = $\frac{9}{r}$ $\binom{8}{r-1}$ $\right$] $\]$ $\[$ = $\frac{9}{2}$ $\sum$_{r=1}^{9} $\binom{8}{r-1}$ $\left$( $\frac{1}{2}$ $\right$)^{r-1} + 3 $\left$( $\sum$_{r=0}^{9} $\binom{9}{r}$ $\left$( $\frac{1}{2}$ $\right$)^r - 1 $\right$) $\]$ $\[$ = $\frac{9}{2}$ $\left$( 1 + $\frac{1}{2}$ $\right$)^8 + 3 $\left$( $\left$( 1 + $\frac{1}{2}$ $\right$)^9 - 1 $\right$) $\]$ $\[$ = $\frac{9}{2}$ $\cdot$ $\left$( $\frac{3}{2}$ $\right$)^8 + 3 $\left$( $\frac{3}{2}$ $\right$)^9 - 3 = 6 $\cdot$ $\left$( $\frac{3}{2}$ $\right$)^9 - 3 $\]$ Hence, $\($ $\alpha$ = 6, $\beta$ = 3 $\)$ Thus $\($ ($\alpha$ + $\beta$)^2 = 81 $\)$

Given $\tan x = \frac{\pi}{3} - \frac{2x}{3}$. The graph shows the function and its intersections with the line $y = \frac{\pi}{3} - \frac{2x}{3}$. There are 5 solutions where the line intersects the tangent function.

Given the transformation $C_3 \to C_3 - C_1$, we have: $$y(x) = \begin{vmatrix} \sin x & \cos x & 1 + \cos x \\ 27 & 28 & 0 \\ 1 & 1 & 0 \end{vmatrix}$$ $$y(x) = -(1 + \cos x)$$ Differentiating $y$ with respect to $x$: $$\frac{dy}{dx} = \sin x$$ $$\frac{d^2y}{dx^2} = \cos x$$ The differential equation is: $$\frac{d^2y}{dx^2} + y = -1$$

Differentiate with respect to $x$. Given $g(x) = 1 - xg(x)$, we have $$g(x) = \frac{1}{1 + x}$$ so $$\frac{dy}{dx} - y \tan x = 2 \sec x$$ The integrating factor is $$IF = e^{-\int \tan dx} = e^{\log \cos x} = \cos x$$ The solution of the differential equation is $$y \cos x = \int 2 dx + c$$ $$y \cos x = 2x + c$$ Given $y(0) = 0$, we find $$c = 0$$ Thus, $$y = \frac{2x}{\cos x}$$ $$y = 2x \sec x$$ Finally, $$y\left(\frac{\pi}{3}\right) = 2 \cdot \frac{\pi}{3} \cdot 2 = \frac{4\pi}{3}$$

Line is $y = x$. $m_1 = 1$, $m_2 = -2$. So $\tan \theta = \left| \frac{1 + 2}{1 - 2} \right|$. $\tan \theta = \frac{AM}{BM} = 3$.

Given $z^2 + 3i = z(2 + 3i) - 7 - 4i$. We have $$z^2 - z(2 + 3i) + 7 + 7i = 0$$ Then, $$z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1z_2$$ Calculating, $$= 4 - 9 + 12i - 14 - 14i$$ $$= -19 - 2i$$

Let $3 - x^2 = t^2$. Then $x \, dx = -t \, dt$. $$f(x) = \int (3 - t^2) \cdot t (-t \, dt) + c$$ $$= \int (t^4 - 3t^2) \, dt + c$$ $$= \frac{t^5}{5} - t^3 + c$$ $$f(x) = \frac{(3 - x^2)^{5/2}}{5} - (3 - x^2)^{3/2} + c$$ $$f(\sqrt{2}) = \frac{1}{5} - 1 + c = -\frac{4}{5}$$ Thus, $c = 0$. $$f(1) = \frac{2^{5/2}}{5} - 2^{3/2}$$ $$= 2^{1/2} \left( \frac{4}{5} - 2 \right)$$ $$f(1) = -\frac{6\sqrt{2}}{5}$$

Let the 1st term of G.P. be $a$ and common ratio be $r$. $$a_3 a_5 = ar^2 \cdot ar^4 = 729$$ $$= a^2 r^6 = 729$$ $$= ar^3 = 27 \ldots (i)$$ $$a_2 + a_4 = ar + ar^3 = \frac{111}{4}$$ $$= ar = \frac{3}{4} \ldots (ii)$$ Dividing (i) by (ii): $$ar^3 = 27$$ $$ar = \frac{3}{4}$$ $$r^2 = 36$$ $$r = 6$$ From (ii): $$a(6) = \frac{3}{4} \implies a = \frac{1}{8}$$ Now, $24 (a_1 + a_2 + a_3)$ $$= 24 \left(a + ar + a^2\right)$$ $$= 24a \left(1 + r + r^2\right)$$ $$= 24 \times \frac{1}{8} \left(1 + 6 + 36\right)$$ $$= 3(43)$$ $$= 129$$

Given $\log_4 \log_6 (3 + 4x - x^2) > 0$. (1) $\log_6 (3 + 4x - x^2) > 1$ $3 + 4x - x^2 > 6$ $x^2 - 4x + 3 < 0$ $(x - 1)(x - 3) < 0$ $x \in (1, 3)$ So $a = 1$ and $b = 3$. Therefore, $\int_0^2 \lfloor x^2 \rfloor \, dx = ?$ $I = \int_0^1 \lfloor x^2 \rfloor \, dx + \int_1^{\sqrt{2}} \lfloor x^2 \rfloor \, dx + \int_{\sqrt{2}}^{\sqrt{3}} \lfloor x^2 \rfloor \, dx + \int_{\sqrt{3}}^{\sqrt{4}} \lfloor x^2 \rfloor \, dx$ $= 0 + \lfloor x \rfloor \big|_1^{\sqrt{2}} + 2 \lfloor x \rfloor \big|_{\sqrt{3}}^{\sqrt{2}} + 3 \lfloor x \rfloor \big|_{\sqrt{4}}^{\sqrt{3}}$ $= (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$ $= 5 - \sqrt{2} - \sqrt{3} \Rightarrow p + q + r = 10$

Tangents at A and B must be parallel to the line $y = x$, so the slope of the tangents is $1$. $$\left( \frac{dy}{dx} \right)_{\min A} = 1 = \left( \frac{dy}{dx} \right)_{\min B}$$ For point B, $y = x^2 + 2$. $$\frac{dy}{dx} = 2x = 1$$ $$x = \frac{1}{2} \Rightarrow y = \frac{9}{4}$$ Therefore, $$Point B = \left( \frac{1}{2}, \frac{9}{4} \right) \Rightarrow Point A = \left( \frac{9}{4}, \frac{1}{2} \right)$$ $$AB = \sqrt{\left( \frac{1}{2} - \frac{9}{4} \right)^2 + \left( \frac{9}{4} - \frac{1}{2} \right)^2}$$ $$= \sqrt{\frac{98}{16}} = \frac{7\sqrt{2}}{4}$$ Radius $= \frac{7\sqrt{2}}{8}$

Given $f(0^-) = e^{\lim_{x \to 0} \frac{ax}{x}} = e^a$. $f(0) = 1 + b$ $f(0^+) = \frac{1}{3}(x + c)^{-\frac{2}{3}} = \frac{1}{3} \cdot c^{-\frac{2}{3}} = \frac{3}{4} c^{\frac{2}{3}}$ Also at $x = 0$; $c^{\frac{1}{3}} = 2 \Rightarrow c = 8$ So $f(0^+) = \frac{3}{4} (8)^{\frac{2}{3}} = 3$ Now, $e^a = b + 1 = 3$ $e^a \cdot b \cdot c = 3 \cdot 2 \cdot 8 = 48$

Given the lines: $$L_1 \equiv \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$$ $$L_2 \equiv \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}$$ The symmetric difference (SD) is given by: $$SD = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{vmatrix}$$ This simplifies to: $$= |\lambda - 1| = 3$$ Solving gives: $$\lambda = 4, -2$$ Letting $\lambda_1 = 4$ and $\lambda_2 = -2$. Let the foot of the perpendicular from $P(4, -2, 7)$ be $Q(1, 2, t + 3)$. So, $\vec{PQ} = (3, -4, 4-t) \cdot (0, 0, 1) = 0$. Solving gives $t = 4$. Thus, $Q(1, 2, 7)$. The distance $PQ^2 = 9 + 16$. Therefore, $PQ^2 = 25$.

Let $P(W_1) = x$. $$\sum_{i=1}^{120} P(W_i) = 1$$ $$x + 2x + 2^2 x + 2^3 x + \ldots + 2^{119} x = 1$$ $$x (2^{120} - 1) = 1 \Rightarrow x = \frac{1}{2^{120} - 1} \ldots (i)$$ Rank of CDBEA $$A\_\_\_\_ = \underline{4} = 24$$ $$B\_\_\_\_ = \underline{4} = 24$$ $$CA\_\_\_ = \underline{3} = 6$$ $$CB\_\_\_ = \underline{3} = 6$$ $$CDA\_\_ = \underline{2} = 2$$ $$64$$ CDBAE = 1 CDBEA = 1 So, $P(W_{64}) = 2P(W_{63}) = \ldots = 2^{63} P(W_1)$ $$= \frac{2^{63}}{2^{120} - 1}$$ $$\alpha + \beta = 63 + 120 = 183$$

Given $\($ $\frac{x^2}{a^2}$ - $\frac{y^2}{b^2}$ = 1 $\)$ $\ldots$ (1) Point $\($ P(4, 2$\sqrt{3}$) $\)$ $\($ PS_1 $\cdot$ PS_2 = 32 $\)$ $\($|PS_1 - PS_2| = 2a $\)$ Point $\($ P(4, 2$\sqrt{3}$) $\)$ lies on $\($ H $\)$ $\($ $\frac{16}{a^2}$ - $\frac{12}{b^2}$ = 1 $\)$ $\($ 16b^2 - 12a^2 = a^2b^2 $\)$ $\ldots$ (2) $\($|PS_1 - PS_2|^2 = 4a^2 $\)$ $\($ PS_1^2 + PS_2^2 - 2PS_1 $\cdot$ PS_2 = 4a^2 $\)$ $\($ (ae - 4)^2 + 12 + (ae + 4)^2 + 12 - 64 = 4a^2 $\)$ $\($ 2a^2e^2 - 8 = 4a^2 $\)$ $\($ a^2 + b^2 - 4 = 4a^2 $\)$ $\($ b^2 - a^2 = 4 $\)$ From (2) and (3) $\($ $\Rightarrow$ 16(a^2 + 4) - 12a^2 = a^2(a^2 + 4) $\)$ $\($ $\Rightarrow$ 16a^2 + 64 - 12a^2 = a^4 + 4a^2 $\)$ $\($ $\Rightarrow$ a^4 = 64 $\)$ $\($ $\Rightarrow$ a^2 = 8 $\)$ Thus, $\($ b^2 = 12 $\)$ $\($ p^2 + q^2 = 4b^2 + $\frac{4b^4}{a^2}$ $\)$ $\($ = 120 $\)$

Given $\vec{a} \times \vec{d} - \vec{b} \times \vec{d} = 0$. $\left( \vec{a} - \vec{b} \right) \times \vec{d} = 0$. $\vec{d} = t(\vec{a} - \vec{b})$. $\vec{d} = t(-2\hat{i} - \hat{j} + 2\hat{k})$. $|\vec{d}| = 1$. $|t| = \frac{1}{3}$. $\vec{c} \cdot \vec{a} = 0$. $\lambda + \mu = 0$. $\mu = -\lambda$. $\vec{c} = \lambda(\hat{j} - \hat{k}), \ |\vec{c}|^2 = 2\lambda^2$. $\vec{c} \cdot \hat{d} = 1$. $t(-2, -1, 2) \cdot \lambda(0, 1, -1) = 1$. $\lambda t = \frac{-1}{3} \Rightarrow \lambda^2 = 1$. $|3\lambda \hat{d} + \mu \vec{c}|^2 = 9\lambda^2 |\hat{d}|^2 + \mu^2 |\vec{c}|^2 + 6\lambda \mu (\hat{d} \cdot \vec{c})$. $= 3\lambda^2 + 2\lambda^4$. $= 5$.

Total 7 digit numbers = 9000000. 7 digit numbers having sum of digits even = 4500000 = 9.5 $\cdot$ 10^5. $m = 9$, $n = 5$. $m + n = 14$.

Required Area = $\frac{1}{2}$ $\times$ 2 $\times$ 2 + $\frac{1}{2}$ $\times$ 3 $\times$ 3 + $\frac{1}{2}$ $\times$ 1 $\times$ 11 = 12

Volume decreases during melting of ice so positive work is done on ice water system by atmosphere. Heat absorbed by ice water so $\Delta Q$ is positive, work done by ice water system is negative. Hence by first law of thermodynamics $$\Delta U = \Delta Q + \Delta W = Positive$$ So internal energy increases.

Apply Bernoulli equation between points 1 and 2 $$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + 0$$ $$P_0 + \frac{mg}{A} + \rho g \frac{70}{100} = P_0 + \frac{1}{2} \rho v_2^2$$ $$5000 + 0.5 \times 10^3 \times 10 \times \frac{70}{100} = \frac{1}{2} \times 10^3 v_2^2$$ $$10^3 + 10^3 \times 7 = \frac{1}{2} v_2^2$$ $$v_2^2 = 16$$ $$v_2 = 4 \, \mathrm{m/s}$$ As the tank area is large $v_1$ is negligible compared to $v_2$.

Q5. Only option (2) matches with the truth table.

Given $P_{rad} = \frac{2I}{C}$. Where $I =$ intensity at surface. $C =$ Speed of light. $$I = \frac{Power}{Area} = \frac{450}{4\pi^2}$$ $$= \frac{450}{4\pi \times 4} = \frac{450}{16\pi}$$ $$= \frac{2 \times 450}{16\pi \times 3 \times 10^8} = \frac{150}{8\pi \times 10^8}$$ $$= 5.97 \times 10^{-8} \approx 6 \times 10^{-8} Pascals$$

Given $L = 25 \, \mathrm{m}$, $A = 5 \, \mathrm{mm^2} = 5 \times 10^{-6} \, \mathrm{m^2}$, $\rho = 2 \times 10^{-6} \, \Omega \, \mathrm{m}$. $R_{wire} = \frac{\rho L}{A} = \frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10$. $R_{eq} = \frac{R}{4} = \frac{10}{4} = 2.5 \, \Omega$. Answer does not match with NTA option.

(A) As both blocks moving together so Time period $= 2\pi \sqrt{\frac{M+m}{K}}$; where $m = M + m$. (1) $T = 2\pi \sqrt{\frac{M+m}{K}}$ (B) Let block is displaced by $x$ in $(+ve)$ direction so force on block will be in $(-ve)$ direction $F = -Kx$ $(M+m)a = -Kx$ $a = -\frac{Kx}{(M+m)}$ (C) As upper block is moving due to friction thus $f = ma = \frac{mKx}{(M+m)}$ (D) This option is like two block problem in friction for maximum amplitude, force on block is also maximum, for which both blocks are moving together. $KA = (M+m)a$ $a = \frac{KA}{(M+m)}$ $f = ma = \frac{mKA}{(M+m)}$ $f_{max} = f_{L} = \mu mg$ $f = \mu mg$ $\frac{mKA}{(M+m)} = \mu mg$ $A = \frac{\mu (M+m)g}{K}$ (E) Maximum friction can be $\mu mg$ as force is acting between blocks $\&$ normal force here is $mg$.

For option (A) $\phi = kt + C$ it can be 1D motion. eg $\rightarrow x = A \sin \phi$ (SHM). For option (B) $v^2 + x^2 = constant$ yes 1D. For option (C) time can't be negative Not possible. For option (D) Possible. A, B $\&$ D only.

Area of plate is $A$. Then $$C = \frac{\varepsilon_2 \varepsilon_0 A}{d/2} = \frac{2 \varepsilon_2 \varepsilon_0 A}{d}$$ $$C' = \frac{\varepsilon_1 \varepsilon_0 A}{d/2} = \frac{2 \varepsilon_1 \varepsilon_0 A}{d}$$ Let $C_0 = \frac{\varepsilon_0 A}{d}$. $$C = 2 \varepsilon_2 C_0$$ $$C' = 2 \varepsilon_1 C_0$$ $C$ and $C'$ are in series. $$C_1 = \frac{CC'}{C + C'} = \frac{4 \varepsilon_2 \varepsilon_1 C_0^2}{2 C_0 (\varepsilon_2 + \varepsilon_1)}$$ Here $C = \frac{\varepsilon_1 \varepsilon_0 A}{2d} = \frac{\varepsilon_1 C_0}{2}$. $$C' = \frac{\varepsilon_2 C_0}{2}$$ $C$ and $C'$ are in parallel. $$C_2 = C' + C = (\varepsilon_1 + \varepsilon_2) \frac{C_0}{2}$$ Thus $$\frac{C_1}{C_2} = \frac{2 \varepsilon_2 \varepsilon_1 C_0}{(\varepsilon_2 + \varepsilon_1) (\varepsilon_1 + \varepsilon_2) C_0}$$ $$= \frac{4 \varepsilon_2 \varepsilon_1}{(\varepsilon_2 + \varepsilon_1)^2}$$

(A) $G = \frac{Fr^2}{m^2}$ $$[G] = \left[ \mathrm{MLT^{-2}} \right] \left[ \mathrm{L^2} \right] \left[ \mathrm{M^2} \right]^{-1} = \left[ \mathrm{M^{-1} L^3 T^{-2}} \right] (\mathrm{IV})$$ (B) P.E. $= mgh = \left[ \mathrm{MLT^{-2} L} \right]$ $$= \left[ \mathrm{ML^2 T^{-2}} \right] (\mathrm{III})$$ (C) Gravitational Potential $= \frac{GM}{r}$ $$= \left[ \mathrm{M^{-1} L^3 T^{-2}} \right] \left[ \mathrm{M} \right] \left[ \mathrm{L} \right]^{-1} = \left[ \mathrm{M^0 L^2 T^{-2}} \right] (\mathrm{II})$$ (D) Acceleration due to gravity $= [g] = \left[ \mathrm{LT^{-2}} \right] (\mathrm{I})$$

Torque about bottom point $$F \times 2r = I \alpha$$ $$49 \times 2r = \frac{7}{5} m r^2 \alpha$$ $$14 = 4r \alpha$$ As sphere rolls without slipping $$a = r \alpha$$ $$a = \frac{14}{4} = \frac{7}{2} = 3.5 \, \mathrm{m/s^2}$$

Using WET, total energy supplied equals gravitational potential energy plus spring potential energy plus work done by gas. $$Mg \left( L_1 - L_0 \right) + \int_{L_0}^{L_1} k^3 \, dx + nRT \ln$$ $\[$ $\begin{bmatrix}$ L_1 & A $\\$ L_0 & A $\end{bmatrix}$ $\]$ plus $\($ W_{ext} = 0 $\)$ $$\frac{K}{4} \left[ x^4 \right]_{L_0}^{L_1} + Mg(L_1 - L_0) + \int_{L_0}^{L_1} kx^3 \, dx + nRT \ln$$ $\[$ $\begin{bmatrix}$ L_1 $\\$ L_0 $\end{bmatrix}$ $\]$ plus $\($ W_{ext} = 0 $\)$ $$\frac{k}{4} \left( L_1^4 - L_0^4 \right) + Mg \left( L_1 - L_0 \right) + nRT \ln$$ $\[$ $\begin{bmatrix}$ L_1 $\\$ L_0 $\end{bmatrix}$ $\]$ plus $\($ W_{ext} = 0 $\)$ $$W_{ext} = \frac{k}{4} \left( L_1^4 - L_0^4 \right) + Mg(L_1 - L_0) + nRT \ln$$ $\[$ $\begin{bmatrix}$ L_1 $\\$ L_0 $\end{bmatrix}$ $\]$

Given $V_1 = 800 \, \mathrm{cm}^3$, $V_2 = 200 \, \mathrm{cm}^3$, $T_1 = 300 \, \mathrm{K}$. For adiabatic process, $TV^{\gamma - 1} = const.$ $$(300)(800)^{1.5 - 1} = T_2(200)^{1.5 - 1}$$ $$T_2 = 300 \left[ \frac{800}{200} \right]^{0.5} = 300 \times (2^2)^{1/2}$$ $$T_2 = 600 \, \mathrm{K}$$ $$\Delta T = 600 - 300 = 300 \, \mathrm{K}$$

Conceptual (2)

Potential inside shell is equal to potential on surface. $$V_{in} = V_{surface} = \frac{kQ}{R} = 120 \, \mathrm{V}$$ at $r = 15 \, \mathrm{cm}$ $$V = \frac{kQ}{r} = \frac{120 \times 10}{15} = 80 \, \mathrm{V}$$

Given $(KE)_{\max} = \frac{hc}{\lambda} - \phi$. For emission, $\frac{hc}{\lambda} > \phi$. This implies $\lambda < \frac{hc}{\phi} \Rightarrow \lambda < \frac{1242}{3} \, \mathrm{nm}$. So blue light option (B) is correct.

Given $V^2 = 0 + 2g(S - x)$. $V^2 = 2g(S - x)$. At B, potential energy $= mgx$. $$mgx = 3 \times \frac{1}{2} mv^2$$ $$gx = \frac{3}{2} \times 2g(S - x)$$ $$4x = S$$ $$x = \frac{S}{4}$$ Therefore, $$V = \sqrt{2g \times \frac{3S}{4}} = \sqrt{\frac{3gS}{2}}$$

Given $m_1 + m_2 + m_3 = 435.42 + 226.3 + 0.125$. According to least significant digits $m = 661.8 \, \mathrm{g}$.

Given $\($ $\frac{1}{f}$ = ($\mu$ - 1) $\left$( $\frac{1}{R_1}$ - $\frac{1}{R_2}$ $\right$) $\)$ $\($ $\frac{1}{12}$ = ($\mu$ - 1) $\left$( $\frac{1}{10}$ - $\frac{1}{-15}$ $\right$) $\)$ $\($ $\frac{1}{12}$ = ($\mu$ - 1) $\left$( $\frac{3 + 2}{30}$ $\right$) $\)$ $\($ $\mu$ = $\frac{3}{2}$ $\)$

The maximum height $H_{max}$ of a projectile launched at an angle $\theta$ with initial velocity $u$ is given by: $$H_{max} = \frac{u^2 \cos^2 \phi}{2g}.$$

Given $\delta = I + e - A$. For $\delta_{\min} \Rightarrow I = e$ and refracted ray is parallel to base. A, C, D are correct.

Given $T \propto m^x G^y a^z$. $T \propto M^x \left[ M^{-1} L^3 T^{-2} \right]^y \left[ L \right]^z$. $T \propto M^{x-y} L^{3y+z} T^{-2y}$. $x - y = 0 \Rightarrow x = y$. $-2y = 1 \Rightarrow y = -\frac{1}{2}, x = -\frac{1}{2}$. $\Rightarrow 3y + z = 0$. $z = -3y = \frac{3}{2}$. Hence $T \propto m^{-1/2} G^{-1/2} a^{3/2}$. $T \propto \left( \frac{a^3}{m} \right)^{1/2}$. $T = 4 \times \left( \frac{2^3}{2} \right)^{1/2} = 8 \, \mathrm{s}$.

Given $F = I \ell B$. $$F = 8 \times \frac{4}{100} \times 0.15$$ $$= 48 \times 10^{-3} \, \mathrm{N} = 48 \, \mathrm{mN}$$

Given $$I_{max} = \left( \sqrt{I_1} + \sqrt{I_2} \right)^2$$ $$= (\sqrt{4I} + \sqrt{9I})^2 = 25I$$ $$I_{min} = \left( \sqrt{I_1} - \sqrt{I_2} \right)^2$$ $$= (\sqrt{4I} - \sqrt{9I})^2 = I$$ $$I_{max} - I_{min} = 24I$$ $$x = 24$$

Magnetic field due to AB & CD $= 0$ $$B_0 = |B_{R_1} - B_{R_2}|$$ $$= \frac{\mu_0 I}{4R_2} - \frac{\mu_0 I}{4R_1}$$ $$= \frac{4\pi \times 10^{-7} \times 12}{4} \left( \frac{1}{4\pi} - \frac{1}{6\pi} \right)$$ $$= 12\pi \times 10^{-7} \left( \frac{1}{12\pi} \right)$$ $$= 1 \times 10^{-7}$$ $$K = 1$$

The equivalent resistance is given by $R_{eq} = R_1 + R_2$. Calculating $R_{eq}$: $$R_{eq} = 150.4 + \frac{240 \times 10}{240 + 10}$$ Simplifying, $$R_{eq} = 150.4 + 9.6 = 160 \Omega$$ The current $I_1$ is given by $$I_1 = \frac{IR_2}{240}$$ Substituting the values, $$I_1 = \frac{20}{160} \times \frac{9.6}{2400} = \frac{1}{200} = 5 \times 10^{-3} \, A = 5 \, mA$$

The electron in a H-atom's stationary state moves in a spherical path.

The $\mathrm{N} - \mathrm{N}$ single bond is weaker than $\mathrm{P} - \mathrm{P}$ due to more $\ell_p - \ell_p$ repulsion. Bond length implies $d_{p-p} > d_{N-N}$ (size increases, bond length increases). In group 15 elements, only N and P show disproportionation in the +3 oxidation state, while As, Sb, and Bi are almost inert for disproportionation in the +3 oxidation state. So both statements are false.

A catalyst can change equilibrium composition if it is added at constant pressure, but it cannot change equilibrium constant.

Given magnetic moment = 4.9 B.M. We know M.M $=\sqrt{n(n+2)}$ B.M. Where, $n \rightarrow$ No. of unpaired $e^-$ $4.9=\sqrt{n(n+2)}$ We get $n=4$ (A) $_{24}\mathrm{Cr}^{2+}\Rightarrow[\mathrm{Ar}]\,3d^4$ (4 unpaired $e^-$) (B) $_{26}\mathrm{Fe}^{2+}\Rightarrow[\mathrm{Ar}]\,3d^6$ (4 unpaired $e^-$) $(C)$ $_{26}\mathrm{Fe}^{3+}\Rightarrow[\mathrm{Ar}]\,3d^5$ (5 unpaired $e^-$) (D) $_{27}\mathrm{Co}^{3+}\Rightarrow[\mathrm{Ar}]\,3d^6$ (3 unpaired $e^-$) (E) $_{25}\mathrm{Mn}^{3+}\Rightarrow[\mathrm{Ar}]\,3d^4$ (4 unpaired $e^-)$

Given: $[A]_0 = 8[B]_0$. $[t_{1/2}]_A = 10 \, min$. $[t_{1/2}]_B = 40 \, min$. First order kinetics. $t = ?$ $[A]_t = [B]_t$ $-k_A \times t = -k_B \times t$ $\Rightarrow [A]_0 e = [B]_0 e$ $\Rightarrow \frac{[A]_0}{[B]_0} = e^{(k_A - k_B)t}$ $\Rightarrow 8 = e^{(k_A - k_B) \times t}$ $\Rightarrow \ln 8 = (k_A - k_B) \times t$ $\Rightarrow \ln 8 = \ln 2 \left( \frac{1}{(t_{a2})_A} - \frac{1}{(t_{a2})_B} \right) \times t$

The structures shown are of D-Fructose and L-Fructose. The D-Fructose has the hydroxyl group on the right side, while the L-Fructose has the hydroxyl group on the left side.

In option C are homologues to each other and option D are only organic molecules not isomers.

No. of atoms = $\frac{Mass in g}{Molar Mass (g/mol)}$ $\times$ N_A. Therefore for the same mass, the element having the least molar mass will have the higher number of atoms. - M_{$\mathrm{Po}$} = 209 - M_{$\mathrm{Pr}$} = 141 - M_{$\mathrm{Pb}$} = 207 - M_{$\mathrm{Pt}$} = 195

The process of adding an $e^-$ to a neutral gaseous atom is not always exothermic; it may be exothermic or endothermic. Be B $$1s^22s^2 1s^22s^22p^1$$ In Be, the 2s subshell is fully filled. So, high energy is needed to remove $e^-$ as compared to B. In $\mathrm{CCl_4}$: $$\delta^- -Cl \delta^+ -C \delta^- -Cl$$ Due to partially positive charge $z_{eff} \uparrow$, EN $\uparrow$. So, EN of C $\Rightarrow \mathrm{CCl_4} > \mathrm{CH_4}$. Cs is most electropositive.

Both solutions are having same composition, which is 1 mole of 'x' in 1 'l' water, so all the intensive properties will remain same, but as total amount is greater in solution '1' compared to solution '2'. So extensive properties will be different hence Gibbs free energy will be different.

Q7. Following will not give iodoform reaction/test. (1) Butanal (2) 2-Pentanone (3) Pentanal (4) 3-Pentanol

The reaction of the compound with $\mathrm{N_2Cl}$ in the presence of $\mathrm{EtOH}$ does not yield the expected $\mathrm{OEt}$ product. Instead, a deamination reaction occurs, resulting in the formation of benzene.

We know $E = h \nu = \frac{hC}{\lambda}$. $E \propto \frac{1}{\lambda}$ Here all Co in +3 oxidation state. So, as the ligand field strength increases, CFSE increases. Order of field strength of ligand: CN $>$ $\mathrm{NH_3}$ $>$ $\mathrm{H_2O}$ $>$ $\mathrm{Cl^-}$ CFSE order: C $>$ B $>$ A $>$ D Wavelength order: D $>$ A $>$ B $>$ C

On addition of inert gas at constant $P$ and $T$, reaction moves in the direction of greater number of moles so it will shift in forward direction, so $[\mathrm{PCl}_5]$ decrease and $[\mathrm{PCl}_3]$ and $[\mathrm{Cl}_2]$ will increase.

Given $\Delta T_b = i_1 \, m_1 k_b + i_2 \, m_2 k_b$. $$= 1 \times \frac{2}{0.5} \times 0.52 + \frac{1 \times 2}{0.5} \times 0.52 = 4.16$$ $$(T_b)_{solution} = 373.16 + 4.16 = 377.3 \, \mathrm{K}.$$

The reaction shown is ozonolysis of 1,3-dimethylcyclohexene. The ozonolysis process involves the cleavage of the double bond and formation of carbonyl compounds. The product formed is 3-Methyl-6-ketoheptanal.

For $PF_5$: $5\sigma + 0\ell p \rightarrow sp^3d$. For $SF_6$: $6\sigma + 0\ell p \rightarrow sp^3d^2$. For $Ni(CO)_4$: Ni is in the zero oxidation state. In the presence of a ligand field, Ni(0) has the electron configuration $[Ar]$ with $sp^3$ hybridisation. For $[PtCl_4]^{2-}$: Pt is in the $+2$ oxidation state. In the presence of a ligand field, $Pt^{2+}$ has the electron configuration $[Kr]$ with $dsp^2$ hybridisation.

C.B. of terminal alkyne will be sp hybridisation and localised. In other C.B. will be resonance stabilised.

Limiting Molar Conductivities of Ions: $\oplus H: 349.8 \, \mathrm{Scm^2 \, mol^{-1}}$ $Na^+: 50.11 \, \mathrm{Scm^2 \, mol^{-1}}$ $K^+: 73.52 \, \mathrm{Scm^2 \, mol^{-1}}$ $Ca^{+2}: 119 \, \mathrm{Scm^2 \, mol^{-1}}$ $Mg^{+2}: 106.12 \, \mathrm{Scm^2 \, mol^{-1}}$ Therefore correct order of limiting molar conductivity of cations will be - $\oplus H > Ca^{+2} > Mg^{+2} > K^+ > Na^+$

M.wt. of given compound = 86 Applying POAC on 'N' $n_X \times 2 = n_{N_2} \times 2$ $$\frac{0.42}{86} = n_{N_2}$$ Therefore, the volume of $N_2$ at STP is $$\frac{0.42}{86} \times 22.4 \, \mathrm{L}$$ $$= 0.1108 \, \mathrm{L} = 110.8 \, \mathrm{mL}$$

Applying POAC on 'C' (mole) of 'C' in compound = n_{$\mathrm{CO_2}$} $\times$ 1 So mass of 'C' in compound $$= \frac{1.46}{44} \times 12$$ So, $\%$ of 'C' in compound = $$\frac{1.46}{44} \times \frac{12}{0.5} \times 100$$ $$= 79.63$$

Given the reaction $\mathrm{FeCl_3} + \mathrm{KOH} + \mathrm{H_2SO_4} \rightarrow \mathrm{K_3[Fe(C_2O_4)_3]}$. The complex $[\mathrm{Fe(C_2O_4)_3}]^{3-}$ is of the $\mathrm{[M(AA)_3]}$ type. So the total optical isomers are $2$.

The reaction is given as follows: $$2\mathrm{C}_{(s)} + 2\mathrm{H}_2{(g)} \rightarrow \mathrm{CH}_2 = \mathrm{CH}_2{(g)} : [\Delta H_f^\circ]_{\mathrm{C}_2\mathrm{H}_4{(g)}}$$ The enthalpy change is calculated as: $$[\Delta H_f^\circ]_{\mathrm{C}_2\mathrm{H}_4{(g)}} = 2 \times [\Delta H_{sub}^\circ]_{\mathrm{C}_{(s)}} + 2 \times \Delta H_{H--H}^\circ - 1 \times \Delta H_{C=C}^\circ - 4 \times \Delta H_{CH}^\circ$$ Substituting the values: $$\Rightarrow [\Delta H_f^\circ]_{\mathrm{C}_2\mathrm{H}_4{(g)}} = (2 \times 710) + (2 \times 436) - 611 - 4 \times 414$$ Simplifying gives: $$\Rightarrow [\Delta H_f^\circ]_{\mathrm{C}_2\mathrm{H}_4{(g)}} = 25 \, \mathrm{kJ/mol}$$

$Cr_2O_7^{2-}+NaCl+H_2SO_4\rightarrow CrO_2Cl_2$ $CrO_2Cl_2\ \text{(Vapour)}+NaOH\rightarrow Na_2CrO_4+NaCl+H_2O$ $\mathrm{Na_2Cr_2O_7 \rightarrow 2Na^+ + Cr_2O_7^{2-}}$ No. of terminal "O" = 6