JEE Main 5 April 2026 Shift 1 question paper with solutions

Given $\beta - \alpha = \sqrt{11} \ldots$ (1) and $\beta^2 - \alpha^2 = 3i \sqrt{11}$ $$\Rightarrow (\beta + \alpha)(\beta - \alpha) = 3i \sqrt{11}$$ $$\Rightarrow \beta + \alpha = 3i \ldots (2)$$ From (1) and (2) $$\beta = \frac{\sqrt{11} + 3i}{2}, \alpha = \frac{3i - \sqrt{11}}{2}$$ $$\Rightarrow \alpha \beta = \frac{-9 - 11}{4} \Rightarrow \alpha \beta = -5$$ Now $\left(\beta^3 - \alpha^3\right)^2 = (\beta - \alpha)^2(\beta^2 + \alpha^2 + \alpha \beta)^2$ $$= (\beta - \alpha)^2((\beta + \alpha)^2 - \alpha \beta)^2$$ $$= (11)(-9 + 5)^2 = 176$$

Given $S_n = 3n^2 + 5n$. $T_n = S_n - S_{n-1}$. Therefore, $T_n = (3n^2 + 5n) - (3(n-1)^2) + 5(n-1))$. This simplifies to $T_n = 6n + 2$. The sum $\sum_{n=1}^{10} T_n^2 = \sum_{n=1}^{10} (6n + 2)^2$. This equals $36 \sum_{n=1}^{10} n^2 + 24 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 4$. The final result is $15220$.

Let $A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$. Now $A = \begin{pmatrix} \alpha_1 & \beta_1 & \gamma_1 \\ \alpha_2 & \beta_2 & \gamma_2 \\ \alpha_3 & \beta_3 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$. $$\Rightarrow \begin{pmatrix} \gamma_1 \\ \gamma_2 \\ \gamma_3 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} ...(1)$$ Now $\begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$. $$\Rightarrow \begin{pmatrix} \alpha_3 \\ \beta_3 \\ \gamma_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} ...(2)$$ $A^T \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - A^T \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$. $$\Rightarrow A^T \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$$ $$\begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} ...(3)$$ Now $A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - A \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \\ 4 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix}$. $$\Rightarrow A \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$$ $$\Rightarrow \begin{pmatrix} \alpha_1 \\ \beta_1 \\ \gamma_1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} ...(4)$$ $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & \beta_2 & 3 \\ 3 & 1 & 1 \end{pmatrix}$. $|A| = 2(\beta_2 - 3) - 1(1 - 9) + 1(1 - 3\beta_2) = 1$. $$\Rightarrow 2\beta_2 - 6 + 8 + 1 - 3\beta_2 = 1$$ $$\Rightarrow \beta_2 = 2$$ So $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}$. $A^2 = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 8 & 5 & 6 \\ 13 & 8 & 10 \\ 10 & 6 & 7 \end{pmatrix}$. $A^2 + A = \begin{pmatrix} 10 & 6 & 7 \\ 14 & 10 & 13 \\ 13 & 7 & 8 \end{pmatrix} \Rightarrow \det(A^2 + A) = 8$. $$\Rightarrow Det(adj(A^2 + A)) = 8^2 = 64$$

The determinant is given by $$ \begin{vmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & t^2 & f(t) \end{vmatrix} = 0 $$ which implies $$ 1(f(t) - 5t^3) - 2(6f(t) - 15t) + t(6t^2 - 3) = 0. $$ Solving for $f(t)$, we get $$ f(t) = \frac{t^3 + 12t}{11}. $$ The derivative is $$ f'(t) = \frac{1}{11}(3t^2 + 12) > 0 \forall \ t \in \mathbb{R}. $$ Therefore, $f(t)$ is strictly increasing on $\mathbb{R}$.

Let $\sum_{n=1}^{10} \frac{528}{n(n+1)(n+2)} = \lambda$. Let $T_n = \frac{1}{n(n+1)(n+2)}$. $$T_n = \frac{1}{2} \left\{ \frac{(n+2)-n}{n(n+1)(n+2)} \right\}$$ $$T_n = \frac{1}{2} \left\{ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right\}$$ $$T_1 = \frac{1}{2} \left\{ \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} \right\}$$ $$T_2 = \frac{1}{2} \left\{ \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} \right\}$$ $$\vdots$$ $$T_{10} = \frac{1}{2} \left\{ \frac{1}{10 \cdot 11} - \frac{1}{11 \cdot 12} \right\}$$ $$S_n = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{132} \right]$$ $$S_n = \frac{1}{2} \left[ \frac{66 - 1}{132} \right] = \frac{65}{264}$$ So, $\lambda = \frac{65}{264} \times 528$ $\lambda = 130$

Given $x^2 - 2x - 5 = 0$. $\tan A + \tan B = 2$; $\tan A \tan B = -5$. Therefore, $\tan(A + B) = \frac{2}{1 - (-5)} = \frac{1}{3}$. Thus, $\cos(A + B) = \frac{3}{\sqrt{10}}$. Therefore, $20 \left( \sin^2 \left( \frac{A + B}{2} \right) \right) = \frac{10}{2} (1 - \cos(A + B))$. This simplifies to $10 \left( 1 - \frac{3}{\sqrt{10}} \right) = (10 - 3 \sqrt{10})$.

P(I) : $\frac{1}{2}$ P(A/I) $\rightarrow \frac{1}{5}$ P(II) : $\frac{1}{2}$ P(A/II) $\rightarrow \frac{2}{7}$ $\therefore$ Reqd. Prob = $\frac{\frac{1}{2} \times \frac{2}{7}}{\frac{1}{2} \times \frac{1}{5} + \frac{1}{2} \times \frac{2}{7}}$ = $\frac{\frac{2}{7} \times 5}{\frac{17}{7}}$ = $\frac{10}{17}$

The mean deviation (M.D.) is given by the formula: $$ M.D. = \frac{\sum f_i \left| x_i - \bar{x} \right|}{\sum f_i} $$ First, calculate the mean $\bar{x}$: $$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{40 + 42 + 18 + 20 + 24 + 90}{26} $$ This simplifies to: $$ \frac{234}{26} = 9 $$ Now, calculate the mean deviation: $$ M.D. = \frac{8(4) + 6(2) + 2(0) + 2(1) + 2(3) + 6(6)}{26} $$ This simplifies to: $$ \frac{84}{26} = \frac{44}{13} $$

Given $a > b$. Focus $(ae, 0) = (4, 0)$. Therefore, $ae = 4$. $$a \left( \frac{4}{5} \right) = 4 \implies a = 5$$ Thus, $b^2 = a^2 (1 - e^2)$. $$b^2 = 25 \left( 1 - \frac{16}{25} \right)$$ Therefore, $E : \frac{x^2}{25} + \frac{y^2}{9} = 1$. Thus, $P$ lies on the ellipse $E$. $$\frac{9}{25} + \frac{\alpha^2}{9} = 1$$ Therefore, $\alpha = \pm \frac{12}{5}$. Thus, the area of $\Delta_{POS} = \frac{1}{2} (OS)(PN)$. $$= \frac{1}{2} \times 4 \times \frac{12}{5} = \frac{24}{5}$$

Centre: $C(3,4)$ and $r = 2$. Parabola $(x + 3)^2 = -(y + 4)$. Vertex is $A(-3,-4)$. $$AP_{max} = AC + r = \sqrt{36 + 64} + 2 = 12$$

Given the triangle with vertices $P(3,5)$, $Q(1,3)$, and $R$. The line $x+y=4$ is given. For point $Q$: $$\frac{x-3}{1} = \frac{y-5}{1} = \frac{(3+5-4)}{1+1}$$ This gives $Q(1, 3)$. So the orthocenter is $$\left( \frac{2+3}{3}, \frac{6+5}{3} \right)$$ Calculating further: $$\left( \frac{5}{3}, \frac{11}{3} \right)$$ So $9(\alpha + \beta) = 3(16) = 48$

Given $P = 12 \cos x - 3 \cos^2 x$. $P = -3 (\cos^2 x - 4 \cos x)$ $P = -3 ((\cos x - 2)^2 - 2)$ Put $\cos x = -1$ implies $P = -15$. Put $\cos x = 1$ implies $P = 9$. $-15 \leq P \leq 9$. Sum of all integers $= -(10 + 11 + \ldots + 15) = -75$.

Given $\overrightarrow{PQ} \cdot \overrightarrow{A} = 0$. Therefore, $$(2\lambda - 3)2 + (3\lambda - 1)3 + (4\lambda - 5)4 = 0$$ which implies $\lambda = 1$. The point $P$ is $(5, 6, 7)$ and the vector $\overrightarrow{A}$ is $\langle 2, 3, 4 \rangle$. The point $Q$ is given by $(2\lambda + 2, 3\lambda + 5, 4\lambda + 2)$. Substituting $\lambda = 1$, we find Point $Q = (4, 8, 6)$. Therefore, $PQ^2 = 6$.

Given $\vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0}$. $$(\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$$ $$\vec{r} - \vec{b} = \lambda \vec{a}$$ $$\vec{r} = \vec{b} + \lambda \vec{a}$$ $$\vec{r} \cdot \vec{a} = 0 \implies \vec{a} \cdot \vec{b} + \lambda |\vec{a}|^2 = 0$$ $$\lambda = -\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} = -\frac{(1-2)}{9} = \frac{1}{9}$$ $$\vec{r} = \vec{b} + \frac{\vec{a}}{9}$$ $$|3\vec{r}|^2 = 9|\vec{r}|^2 = 9 \left( b^2 + \frac{a^2}{81} + \frac{2(\vec{a} \cdot \vec{b})}{9} \right) = 44$$

Equating both the equations $$ (\hat{i} + \hat{j} - \hat{k}) + \lambda (a\hat{i} - \hat{j}) = (4\hat{i} - \hat{k}) + \mu (2\hat{i} + a\hat{k}) $$ $$ 1 + a\lambda = 4 + 2\mu $$ $$ 1 - \lambda = 0 \Rightarrow \lambda = 1 $$ $$ -1 = -1 + a\mu $$ $$ a\mu = 0 \Rightarrow \mu = 0 $$ $$ a = 3 $$ point of intersection $$ = \hat{i} + \hat{j} - \hat{k} + 3\hat{i} - \hat{j} = 4\hat{i} - \hat{k} $$

The area is given by the integral: $$\int_{1}^{3} (x^2 - 1) \, dx + \int_{3}^{27} \left( \frac{27}{x} - 1 \right) \, dx$$ This simplifies to: $$\left[ \frac{x^3}{3} - x \right]_{1}^{3} + \left[ 27 \ln x - x \right]_{3}^{27}$$ Evaluating the integrals, we get: $$= \frac{26}{3} - 2 + 27 \ln 9 - 24$$ Simplifying further: $$= 27 \ln 9 - \frac{26 \times 2}{3}$$ Finally, we have: $$= 54 \ln 3 - \frac{52}{3}$$

By using L.H. Rule we get $$\Rightarrow \frac{1}{2}(\alpha + 1)^2 [\alpha^2 + (\alpha + 1)^2 + (\alpha + 2)^2] = 2$$ $$\alpha^2 + (\alpha + 1)^2 + (\alpha + 2)^2 = 4(\alpha + 1)^2$$ $$\alpha^2 + (\alpha + 2)^2 = 3(\alpha + 1)^2$$ $$\Rightarrow \alpha^2 + 2\alpha - 1 = 0$$ Product $= -1$

Put $x = 2t \Rightarrow dx = 2dt$ $$I = \int_0^\infty \frac{\ln 2t}{4t^2 + 4} (2dt) = \frac{1}{2} \int_0^\infty \frac{\ln 2 + \ln t}{t^2 + 1} dt$$ $$= \frac{1}{2} \int_0^\infty \frac{\ln 2}{t^2 + 1} dt + \frac{1}{2} \int_0^\infty \frac{\ln t}{t^2 + 1} dt$$ $$= \left[ \frac{\ln 2}{2} \tan^{-1} t \right]_0^\infty + I_1$$ $$= \frac{\ln 2}{2} \cdot \frac{\pi}{2} + I_1$$ $$I_2 = \frac{1}{2} \int_0^\infty \frac{\ln t}{t^2 + 1} dt t = \frac{1}{u} \Rightarrow dt = -\frac{du}{u^2}$$ $$= \frac{1}{2} \int_\infty^0 \frac{\ln (1/u)}{1/u^2 + 1} \left( -\frac{du}{u^2} \right)$$ Add $\Rightarrow I_1 = 0$ $$I = \frac{\pi \ln 2}{4}$$

Given $f\left( \frac{x+y}{3} \right) = \frac{f(x) + f(y)}{3}$, put $x = y = 0$. $f(0) = \frac{2f(0)}{3}$ $\Rightarrow f(0) = 0 \ldots (1)$ $f'\left( \frac{x+y}{3} \right) \cdot \frac{1}{3} = \frac{1}{3} f'(x)$ Put $x = 0$ $f'\left( \frac{y}{3} \right) \frac{1}{3} = \frac{1}{3} \times 3$ $f'\left( \frac{y}{3} \right) = 3$ Put $y = 3x$ $f'(x) = 3$ Integrate both sides $f(x) = 3x + c$ From eq. (1) $f(0) = 0$ $\Rightarrow f(x) = 3x$ Now $g(x) = 3 + e^x \cdot 3x$ $g'(x) = 3[e^x + x \cdot e^x]$ $= 3e^x(x + 1)$ $g'(x) = 0$, at $x = -1$ $(g(x))_{\min} = 3 + e^{-1}(-3)$ $= 3 \left[ 1 - \frac{1}{e} \right] = \frac{3(e-1)}{e}$

The solution is given by the integral: $$ \int_{\pi/6}^{\pi/3} \frac{4}{\cos^4 x} \, dx - \int_{\pi/6}^{\pi/3} \frac{\csc^2 x}{\cos^4 x} \, dx $$ This can be rewritten as: $$ \int_{\pi/6}^{\pi/3} \frac{4}{\cos^4 x} \, dx - \left[ -\frac{\cot x}{\cos^4 x} \right]_{\pi/6}^{\pi/3} - \int_{\pi/6}^{\pi/3} (-\cot x) \frac{-4}{\cos^5 x} (-\sin x) \, dx $$ Simplifying further: $$ = \frac{\cot x}{\cos^4 x} \bigg|_{\pi/6}^{\pi/3} $$ Evaluating the expression: $$ = \frac{1/\sqrt{3}}{(1/2)^4} - \frac{\sqrt{3}}{\left(\frac{\sqrt{3}}{2}\right)^4} $$ This simplifies to: $$ = \frac{16}{\sqrt{3}} - \frac{\sqrt{3} \cdot 16}{9} = \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{16}{\sqrt{3}} \left(1 - \frac{1}{3}\right) $$ Finally, we have: $$ \frac{32}{3\sqrt{3}} = \frac{32\sqrt{3}}{9} $$

Given $f(2) + f(3) = 5$. C-I: $f(2) = 4$, $f(3) = 1$ So, $f(1) = 3, 5, 6 \rightarrow 3$ choices. $f(4) \rightarrow 3$ choices. $f(5) \rightarrow 2$ choices. $f(6) \rightarrow 1$ choice. So total function for C-I: $3 \times 3 \times 2 \times 1 = 18$. Similarly C-II: $f(2) = 1$, $f(3) = 4 \rightarrow 18$ function. C-III: $f(2) = 3$, $f(3) = 2 \rightarrow 18$ function. C-IV: $f(2) = 2$, $f(3) = 2 \rightarrow 18$ function. Answer total 72 function.

Maximum number of games will be 9. A must win 5 games so, Number of ways = $$^9C_5 = 126$$ Answer = 126

Given $T_{r+1} = \binom{n}{r} \left( \frac{1}{x^3} \right)^{n-r} (-x^4)^r$. $T_{r+1} = \binom{n}{r} x^{7r-3n} (-1)^r$. For the coefficient of $x^7$, $7r_1 - 3n = 7$. $$\Rightarrow r_1 = \frac{7 + 3n}{7}$$ Coefficient of $x^7 \Rightarrow \binom{n}{r_1} (-1)^{r_1}$. For the coefficient of $x^{14}$, $7r_2 - 3n = 14$. $$\Rightarrow r_2 = \frac{14 + 3n}{7} = r_1 + 1$$ Coefficient of $x^{14} \Rightarrow \binom{n}{r_1+1} (-1)^{r_1+1}$. $$\Rightarrow \binom{n}{r_1} (-1)^{r_1} + \binom{n}{r_1+1} (-1)^{r_1+1} = 0$$ $$\Rightarrow \binom{n}{r_1} = \binom{n}{r_1+1} \Rightarrow r_1 + r_1 + 1 = n$$ $$\Rightarrow 2r_1 + 1 = n \Rightarrow 2 \left( \frac{7 + 3n}{7} \right) + 1 = n$$ Ans. $\Rightarrow n = 21$

Given $$\alpha = \frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right)$$ This simplifies to $$= \frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^p - 2^{p-1}}{1 + 2^{2p-1}} \right)$$ Further simplifying, we have $$= \frac{\pi}{4} + \sum_{p=1}^{11} \left( \tan^{-1} \left( 2^p \right) - \tan^{-1} \left( 2^{p-1} \right) \right)$$ Finally, $$= \frac{\pi}{4} + \tan^{-1} \left( 2^{11} \right) - \tan^{-1} \left( 2^0 \right) = 2^{11}$$ The answer is 2048.

Given $\sin \frac{y}{x} \frac{dy}{dx} = \frac{y}{x} \sin \frac{y}{x} - 1$. Put $y = tx$ which implies $\frac{dy}{dx} = t + x \frac{dt}{dx}$. Then, $\sin t \left( t + x \frac{dt}{dx} \right) = t \sin t - 1$. This gives $x \sin t \frac{dt}{dx} + 1 = 0$. Therefore, $\sin t \, dt + \frac{dx}{x} = 0$. Integrating, we get $-\cos t + \ln x = C$. Substituting back, $-\cos \left( \frac{y}{x} \right) + \ln x = C$. Given $y(1) = \frac{\pi}{2}$, we find $C = 0$. Thus, $\cos \left( \frac{y}{x} \right) = \ln x$. Solving $\csc \left( \frac{y(e^{12})}{e^{12}} \right) = 12$, we find $\alpha = 12$. For the equation $x^2 + y^2 - 2px + 2py + 14 = 0$, we have $r = \sqrt{p^2 + p^2 - 14}$. This implies $r \leq 6 \Rightarrow r^2 \leq 36$. Therefore, $2p^2 - 14 \leq 36$, which simplifies to $p^2 \leq 25$. Hence, $p \in [-5, 5]$. The number of integral values of $p$ is 11.

Scale reading = main scale + Least count $\times$ V.S.R. $$= 0 + \frac{1}{10} \times 7 = 0.7 \, \mathrm{mm}$$ $$= 0.07 \, \mathrm{cm}$$

Given $R = \frac{V}{I} = \frac{w}{qI} = \frac{ML^2T^{-2}}{I^2T} = ML^2T^{-3}I^{-2}$ $U = \frac{1}{2}LI^2 \rightarrow ML^2T^{-2} = LA^2$ $L = ML^2T^{-2}A^{-2}$ $Q = CV$ $C = \frac{IT \times q}{ML^2T^{-2}} = \frac{A^2T^2}{ML^2T^{-2}}$ $C = M^{-1}L^{-2}A^{-2}T^4$ $LC = ML^2T^{-2}A^{-2} \times M^{-1}L^{-1}A^2T^4$ $LC = T^2$ $\frac{R}{\sqrt{LC}} = \frac{ML^2T^{-3}A^{-2}}{T} = (ML^2T^4A^{-2})$

Given $$g_{above} = g - \frac{2hg}{R} = 10 - \frac{2h}{R}$$ $$g_{below} = g - \frac{hg}{R}$$ $$\Delta g = g_{above} - g_{below} = g - \frac{2hg}{R} - g + \frac{hg}{R}$$ $$\Delta g = \frac{hg}{R}$$ $$\frac{\Delta g}{g} = \frac{h}{R} = \frac{16 \times 100}{6400} = 0.25\%$$

Given the equations: $$60 - T_2 = 6a$$ $$T_2 + 40 - T_1 = 4a$$ $$60 = 14a$$ Solving for $a$: $$a = \frac{60}{14} = \frac{30}{7}$$ Substituting $a$ into the equation for $T_1$: $$T_1 = 40 + 4 \times \frac{30}{7} = \frac{280 + 120}{7}$$ $$T_1 = \frac{400}{7}$$ Substituting $a$ into the equation for $T_2$: $$60 - T_2 = 6 \times \frac{30}{7}$$ $$60 - \frac{180}{7} = T_2$$ $$\frac{420 - 180}{7} = T_2 = \frac{240}{7}$$ Finally, the ratio of $T_1$ to $T_2$ is: $$\frac{T_1}{T_2} = \frac{5}{3}$$

Given the diagrams, we start with the equations: $$20 \sin 37^\circ - 2a_\omega \cos 37^\circ = 2a_{rel}$$ $$20 \cos 37^\circ - N + 2a_\omega \sin 37^\circ = 0$$ $$24 - N \sin 37^\circ = 10 \times a_\omega$$ Solving the equations: $$24 - (20 \cos 37^\circ + 2a_\omega \sin 37^\circ) \times \sin 37^\circ = 10a_\omega$$ $$24 - \left(16 + 2a_\omega \times \frac{3}{5}\right) \frac{3}{5} = 10a_\omega$$ $$24 - \frac{48}{5} - \frac{18}{25} a_\omega = 10a_\omega$$ $$\frac{72}{5} = \left(10 - \frac{18}{25}\right) a_\omega$$ $$\frac{72}{5} = a_\omega = \frac{180}{116} = \frac{90}{58}$$ $$a_\omega = \frac{45}{29}$$ Substituting back: $$20 \sin 37^\circ - 2 \times \frac{45}{29} \times \frac{4}{5} \times \cos 37^\circ = 2a_{rel}$$ $$12 - \frac{72}{29} = 2a_{rel}$$ $$\frac{348 - 72}{29} = 2a_{rel}$$ $$\frac{276}{29 \times 2} = a_{rel}; \frac{138}{29} = a_{rel}$$ Calculating time $t$: $$t = \sqrt{\frac{2 \times 8.8 \times 29}{138}}$$ $$t = \sqrt{\frac{2 \times 8.8 \times 29}{10 \times 138}}$$ $$t = \sqrt{3.69}$$ $$t \approx 2 sec$$ After rounding off.

Young's modulus is the property of material not depends on dimensions of material.

Both statements are correct but the second does not explain the first statement.

Statement A, B and D are correct

Loop-1: $$-4 \left(i_1 - i_3\right) - 2 \left(i_1 + i_2 - i_3\right) - i_1 + 10 = 0$$ $$-7i_1 - 2i_2 + 6i_3 + 10 = 0 \ldots(1)$$ Loop-2: $$-5 - 2 \left(i_3 - i_2\right) + 2 \left(i_1 + i_2 - i_3\right) = 0$$ $$2i_1 + 4i_2 - 4i_3 - 5 = 0 \ldots(2)$$ Loop-3: $$-i_1 + 10 - 4i_3 - 2 \left(i_3 - i_2\right) = 0$$ $$2i_2 - 6i_3 - i_1 + 10 = 0 \ldots(3)$$ On solving equation- (i) & (ii) & (iii) $$i_1 = 2.5 \, \mathrm{A}$$ $$i_2 = 1.875 \, \mathrm{A}$$ $$i_3 = 1.875 \, \mathrm{A}$$

Given $V = U + at$. $V = V_0 + \left( \frac{2eE_0}{m} \right)t$ Therefore, $a = \frac{(2E_0)e}{m}$ $\lambda = \frac{h}{mV}$ $\lambda = \frac{h}{m \left[ V_0 + \frac{2eE_0}{m}t \right]}$ Put $\lambda_0 = \frac{h}{4mV_0}$ which implies $h = 4\lambda_0 m V_0$ $\lambda = \frac{4m\lambda_0 V_0}{m \left[ V_0 + \frac{2eE_0}{m}t \right]}$ $\lambda = \frac{4\lambda_0}{1 + \frac{2eE_0}{mV_0}t}$

Given: $\frac{r_i}{r_f} = \frac{16}{4} = \frac{n_1^2}{n_f^2}$ $\frac{n_i}{n_f} = \frac{4}{2}$ $$\frac{1}{\lambda} = \mathrm{R} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = 1.0973 \times 10^7 \left( \frac{1}{(2)^2} - \frac{1}{(4)^2} \right)$$ $\lambda = 4.86 \times 10^{-7} \, \mathrm{m} \approx 486 \, \mathrm{nm}$

Given $I_d = C \frac{dV}{dt}$. $$4 = (6 \times 10^{-6})(\alpha \times 10^6)$$ $$\alpha = \frac{2}{3} \approx 0.67$$

Solving each branch as series $$\varepsilon_{eq} = \frac{15}{9} + \frac{15}{9} + \frac{15}{9} = 15 \, volt$$ and $$\frac{1}{r_{eq}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}$$ $$r_{eq} = 3 \, \Omega$$ $$I_{AB} = \frac{15}{12} = 1.25 \, A$$

Since Fringe width in a medium becomes $$\beta' = \frac{\beta}{\mu} \left( Here \beta = \frac{\lambda D}{d} \right)$$ $$\beta' = \frac{\beta}{\mu} = \frac{2.4}{1.2} = 2 \, \mu m$$

Given $\mu = \frac{\sin \left( \frac{\delta_{\min} + A}{2} \right)}{\sin \frac{A}{2}}$. $$\mu = \frac{C}{V} = \frac{3 \times 10^8}{2.12 \times 10^8} = \sqrt{2}$$ $$\sqrt{2} = \frac{\sin \left( \frac{\delta_{\min} + 60^\circ}{2} \right)}{\sin 30^\circ}$$ $$\sin \left( \frac{\delta_{\min} + 60^\circ}{2} \right) = \frac{1}{\sqrt{2}}$$ $\delta_{\min} + 60^\circ = 90^\circ \Rightarrow \delta_{\min} = 30^\circ$

Given $V_s = \frac{1}{e} \left( \frac{hc}{\lambda} - \phi \right)$. $\phi = \frac{hc}{\lambda} - eV_s$ $$= \left( \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}} \right) - (1.6 \times 10^{-19} \times 0.2)$$ $$= 5.68 \times 10^{-19} \, \mathrm{J}$$

M = $\frac{L}{F_0} \frac{D}{F_e} \newline \frac{1}{F_0}$ = (n-1) $\left( \frac{2}{R} \right) \newline$ But when cut $\newline \frac{1}{F_0'}$ = (n-1) $\left( \frac{1}{R} \right)$ = $\frac{1}{2F_0} \newline$ So, $\frac{L}{F_0}$ = $\frac{L'}{F_0'}$ = 2L

The tension $T_1 = mg$. The tension $T_2 = (10 + m)g$. The tension $T_3 = (30 + 10 + m)g = g(40 + m)$. Stress in upper wire $$\frac{(40 + m)g}{0.008 \, \mathrm{cm}^2}$$ $$= \frac{(40 + m) \times 10}{8 \times 10^{-7}}$$ Stress in lower wire $$= \frac{(10 + m)g}{0.004 \, \mathrm{cm}^2}$$ $$= \frac{(10 + m) \times 10}{4 \times 10^{-7}}$$ Upper wire breaks: $$\frac{(40 + m) \times 10}{8 \times 10^{-7}} = 12 \times 10^8$$ So, maximum mass, without breaking any wire is 38 kg.

As we know that current in RLC circuit is given by $$I = \frac{V}{Z} = \frac{V}{\left[R^2 + (X_L - X_C)^2\right]^{1/2}}$$ $$I = \frac{V_0}{\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}}$$ As circuit is RC $$I_0 = \frac{V_0}{(R^2 + X_C^2)^{1/2}}$$ $$\frac{I_0}{3} = \frac{V_0}{\left[R^2 + (4X_C)^2\right]^{1/2}}$$ $$R^2 + (X_C)^2 = \frac{(R^2 + 16X_C^2)}{9}$$ $$\frac{8R^2}{9} = \frac{7X_C^2}{9}$$ $$\frac{R}{X_C} = \sqrt{\frac{7}{8}}$$

The circuit consists of an OR gate and two AND gates. The inputs to the OR gate are 1 (A) and 0 (B), resulting in an output of 1. This output is fed into the first AND gate along with another input of 1, resulting in an output of 1. This output is connected to LED-1, which remains off (0). The output of the first AND gate is also fed into the second AND gate along with input 1 (C), resulting in an output of 1. This output is connected to LED-2, which also remains off (0).

The equation for viscosity is given by $$\eta = \frac{F/A}{\phi} = \frac{Fh}{Ax}.$$ Solving for $x$, we have $$x = \frac{Fh}{A\eta} = \frac{10 \times 5 \times 10^{-2}}{25 \times 10^{-4} \times 10^5} = \frac{1}{500} \, \mathrm{m}.$$ Therefore, $x = 2 \, \mathrm{mm}$.

At any time $t$ after it is released, velocity $V = gt$. Acceleration of the ball remains constant $= g$. Therefore, both will be equal at $t = 1 \, sec$. Thus, the position of the ball from the ground will be $(18 - 5) \, m$.

Least distance between successive crest = $\lambda$ $y = A \sin(\omega t + kx + \phi)$, $\omega = 2 \pi f$ and $k = \frac{2\pi}{\lambda}$ & $y = A \sin(36t + 0.018x + \pi/4)$ $\Rightarrow k = 0.018 \, \mathrm{cm}^{-1} \Rightarrow \frac{2\pi}{\lambda} = 0.018$ $$\lambda = \frac{2\pi}{0.018} = \frac{2 \times 3.14}{18 \times 10^{-3}} \, \mathrm{cm} = 348.88 \, \mathrm{cm}$$ $\approx 349 \, \mathrm{cm}$ (nearest integer)

The force $\vec{F}_B$ is proportional to $(\vec{V} \times \vec{B})$. This implies $\vec{F}_B \perp \vec{B}$. Given $\vec{a} \perp \vec{B}$, it follows that $\vec{a} \cdot \vec{B} = 0$. Solving $12 - x = 0$ gives $x = 12$.

Magnetic energy of inductor is $U = \frac{1}{2} LI^2$. So $U_t = \frac{1}{2} LI^2 + \frac{1}{2} L \left( \frac{I}{2} \right)^2 + \frac{1}{2} L \left( \frac{I}{2} \right)^2 = \frac{3}{4} LI^2$. $$U_\ell = \frac{1}{2} L \left( \frac{I}{2} \right)^2 \Rightarrow \frac{U_t}{U_\ell} = 6$$

The reaction is given by: $$\mathrm{Ag_2CO_3(s)} \xrightarrow{\Delta} 2\mathrm{Ag(s)} + \mathrm{CO_2(g)} + \frac{1}{2}\mathrm{O_2(g)}$$ Given $2.76 \, \mathrm{g}$ of $\mathrm{Ag_2CO_3}$: $$\frac{1}{100} \, \mathrm{mol}$$ Moles of $\mathrm{Ag(s)} = \frac{2}{100} \, \mathrm{mol}$ Mass of residue $= 2.16 \, \mathrm{g}$

(A) $n = 3$, $l = 2$, $m = +1$; $(n + l) = 5$ (B) $n = 4$, $l = 0$, $m = 0$; $(n + l) = 4$ (C) $n = 6$, $l = 1$, $m = 0$; $(n + l) = 7$ (D) $n = 5$, $l = 1$, $m = +1$; $(n + l) = 6$ (E) $n = 2$, $l = 1$, $m = +1$; $(n + l) = 3$ More the value of $(n + l)$, more will be the energy of that particular orbital. Ans. E < B < A < D < C

Mass ratio = mol ratio $\times$ GMM ratio $$\frac{10}{90} = \frac{n}{N} \times \frac{60}{18}$$ $$\frac{n}{N} = \frac{1}{30}$$ $$x_{\mathrm{H_2O}} = \frac{W}{n+N} = \frac{30}{31} = 0.967$$

The reaction is given by: $$\mathrm{M_3A_2 (s) \rightleftharpoons 3M^{+2} (aq.) + 2A^{-3} (aq.)}$$ Molarity is given as: $$\frac{x}{y} - - 3 \frac{x}{y} 2 \frac{x}{y}$$ The solubility product is: $$K_{sp} = [\mathrm{M^{+2}}]^3 [\mathrm{A^{-3}}]^2$$ Substituting the molarity values: $$K_{sp} = \left[\frac{3x}{y}\right]^3 \left[\frac{2x}{y}\right]^2$$ The ratio of $\frac{[\mathrm{A^{-3}}]}{K_{sp}}$ is: $$\frac{2x/y}{\left[\frac{3x}{y}\right]^3 \left[\frac{2x}{y}\right]^2}$$ Simplifying gives: $$= \frac{1}{54} \frac{y^4}{x^4}$$

(A) $\mathrm{H^+_{(aq)}} + \mathrm{OH^-_{(aq)}} \rightarrow \mathrm{H_2O(\ell)}$ At $t = 0$, $2.5 \, \mathrm{mmol}$ of $\mathrm{H^+}$ and $4 \, \mathrm{mmol}$ of $\mathrm{OH^-}$. At $t = \infty$, $0 \, \mathrm{mmol}$ of $\mathrm{H^+}$ and $1.5 \, \mathrm{mmol}$ of $\mathrm{OH^-}$. In the final solution, $[\mathrm{OH^-}] = \frac{1.5 \, \mathrm{mmol}}{35 \, \mathrm{ml}} = \frac{1.5}{35}$. pOH $= -\log \frac{1.5}{35}$. pH $= 14 + \log \left( \frac{1.5}{35} \right)$. (B) $\mathrm{H^+_{(aq)}} + \mathrm{OH^-_{(aq)}} \rightarrow \mathrm{H_2O(\ell)}$ At $t = 0$, $0.2 \, \mathrm{mmol}$ of $\mathrm{H^+}$ and $0.2 \, \mathrm{mmol}$ of $\mathrm{OH^-}$. Final pH $= 7$. (C) $\mathrm{H^+_{(aq)}} + \mathrm{OH^-_{(aq)}} \rightarrow \mathrm{H_2O(\ell)}$ At $t = 0$, $2 \, \mathrm{mmol}$ of $\mathrm{H^+}$ and $1 \, \mathrm{mmol}$ of $\mathrm{OH^-}$. At $t = \infty$, $1 \, \mathrm{mmol}$ of $\mathrm{H^+}$ and $0 \, \mathrm{mmol}$ of $\mathrm{OH^-}$. Final pH $< 7$. So, order of pH values: (C) < (B) < (A)

The reaction is given as A(g) $\rightarrow$ B(g) + C(g). At time $t = 0$, the pressure is $P_i$ for A, and $-$ for B and C. At time $t = t$, the pressure is $P_i - x$ for A, $x$ for B, and $x$ for C. As given, $(P_i - x) + x + x = P_t$. Therefore, $x = (P_t - P_i)$. The rate constant $K$ is given by: $$K = -\frac{1}{t} \ln \frac{P_i}{P_i - x} = -\frac{1}{t} \ln \frac{P_i}{P_i - (P_t - P_i)}$$ Simplifying, we get: $$K = \frac{1}{t} \ln \frac{P_i}{2P_i - P_t}$$

Statement-I $$(\mathrm{EN})_F = 4.0$$ $$(\mathrm{EN})_O = 3.5$$ on pauling scale $$(\mathrm{EN})_N = 3.0$$ $$(\mathrm{EN})_F > (\mathrm{EN})_O > (\mathrm{EN})_N$$ Statement-II Oxidation state of oxygen in $\mathrm{OF_2}$ $$\mathrm{OF_2} \Rightarrow x - 1 \times 2 = 0$$ $$x = +2$$ Oxidation state of oxygen in $\mathrm{Na_2O}$ $$\mathrm{Na_2O} \Rightarrow +1 \times 2 + x = 0$$ $$x = -2$$ Both the statement I and statement II are correct.

(A) In the case of nitrogen, all oxidation states from $+1$ to $+4$ tend to disproportionate in acid solution. For example. $$3\mathrm{HNO_2} \rightarrow \mathrm{HNO_3} + \mathrm{H_2O} + 2\mathrm{NO}$$ (B) Nitrogen form $p_\pi - p_\pi$ bonds with itself and other elements with small size and high electronegativity, $d_\pi - p_\pi$ is not possible with itself as it has no d-orbital. (C) The single N-N bond is weaker than the single P–P bond because of high interelectronic repulsion of the non-bonding electrons (lone pair), owing to the small bond length. (D) Nitrogen has lowest density in its group. (E) The maximum covalency of nitrogen is four. Correct answer are (A) and (E)

The physical and chemical characteristics of interstitial compounds are: (1) They have high melting points, higher than those of pure metals. (2) They are very hard, some borides approach diamond in hardness. (3) They retain metallic conductivity. (4) They are chemically inert and usually non-stoichiometric. Ans. (2)

(A) The metal-carbon bonds in metal carbonyls possess both $\sigma$ and $\pi$-character. (B) The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal i.e. metal carbon bond becomes strong. (C) The metal-carbon $\sigma$ is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of metal. (D) The metal-carbon $\pi$ bond is formed by the donation of electrons from filled d-orbital of metal into vacant $\pi^*$ orbital of CO. Correct answer are A, C, D

Statement-I: In the absence of ligands, the isolated 'd' subshell is shown. In the presence of ligands, the energy levels split into $e_x$ and $t_{2g}$ with $+0.6\Delta_0$ and $-0.4\Delta_0$ respectively. Statement-II: The degeneracy of the d-orbitals is removed due to ligand electron-metal electron repulsions according to crystal field theory. Both the statement I and II are correct.

Statement-I $\Theta \mathrm{CH_3} > \mathrm{CH_3} - \Theta \mathrm{CH_2} > (\mathrm{CH_3})_2 \Theta \mathrm{CH} > (\mathrm{CH_3})_3 \Theta \mathrm{C}$ It is correct order of anion stability and it is based on inductive effect. Statement-II $\mathrm{CH_2 = CH - \Theta CH_2}$ Allyl anion $\&$ Benzyl carbanion Both are stabilised by resonance. Statement-I : Correct Statement-II : Incorrect

The reaction sequence begins with ozonolysis of the alkene using $\mathrm{O_3}$ and $\mathrm{Zn/H_2O}$, resulting in the formation of a diketone. This diketone undergoes an aldol condensation in the presence of $\mathrm{OH^-}$ and heat ($\Delta$), forming a cyclic compound. The cyclic compound is then treated with $\mathrm{NaOH}$ and $\mathrm{I_2}$, leading to the formation of a carboxylate ion and $\mathrm{CHI_3}$ as products.

The reaction begins with the alkyl chloride reacting with $AlCl_3$ to form a carbocation. This carbocation then undergoes a rearrangement to form a more stable carbocation. The benzene ring attacks this carbocation, leading to the formation of the major product, which is an alkylated benzene.

(A) Benzyl bromide reacts with $\mathrm{AgCN}$ to form benzyl isocyanide. (B) Benzylamine reacts with $\mathrm{CHCl_3}$ and aqueous $\mathrm{NaOH}$ to form benzyl isocyanide. (C) Benzyl bromide with $\mathrm{AgCN}$ gives no reaction. (D) Benzylamine reacts with $\mathrm{CHCl_3}$ and aqueous $\mathrm{NaOH}$ to form benzyl isocyanide. (E) Benzyl bromide reacts with $\mathrm{KCN}$ to form benzyl cyanide. (A) and (B) give the correct product.

The reaction of compound (A) with concentrated $\mathrm{H_2SO_4}$ or $\mathrm{H_3PO_4}$ leads to the formation of compound (B). Similarly, compound (C) reacts with concentrated $\mathrm{H_2SO_4}$ or $\mathrm{H_3PO_4}$ in excess to form compound (E).

(A) Basic strength order: $C_6H_5CH_2NH_2 > C_6H_5NH_2$. (B) Cannot be synthesized by Gabriel phthalimide synthesis. (C) Hoffmann's bromamide reaction: $C_6H_5CONH_2 \xrightarrow{Br_2 + NaOH} C_6H_5CH_2NH_2$ (Aliphatic amine). (D) Reaction: $C_6H_4NO_2NH_2 \xrightarrow{NaNO_2 + HCl, \, 0^\circ C} C_6H_4NO_2N_2Cl \xrightarrow{\Delta, \, H_2O} C_6H_4NO_2OH$ (Soluble in NaOH). Only (B) and (C) are incorrect hence Ans. is (3).

It is based on theory

Na_2Cr_2O_7 is hygroscopic in nature and is not used as a primary standard. Phenolphthalein is a weak acid.

The species and their hybridisation are as follows: 1. $\mathrm{BrF_5}$: $\mathrm{sp^3d^2}$ 2. $\mathrm{XeF_5^-}$: $\mathrm{sp^3d^3}$ 3. $\mathrm{BF_4^-}$: $\mathrm{sp^3}$ 4. $\mathrm{ICl_4^-}$: $\mathrm{sp^3d^2}$ 5. $\mathrm{XeF_4}$: $\mathrm{sp^3d^2}$ 6. $\mathrm{SF_4}$: $\mathrm{sp^3d}$ 7. $\mathrm{NH_4^+}$: $\mathrm{sp^3}$ 8. $\mathrm{ClF_3}$: $\mathrm{sp^3d}$ 9. $\mathrm{XeF_2}$: $\mathrm{sp^3d}$ 10. $\mathrm{ICl_2^-}$: $\mathrm{sp^3d}$ Number of species having $\mathrm{sp^3d}$ hybridisation are 4.

Moles of $\mathrm{BaSO_4}$ formed $\frac{0.6 \, \mathrm{gm}}{231 \, \mathrm{gm}} = 0.0026$. Moles of $\mathrm{BaSO_4} = Moles of S$. Mass of sulphur $(W_s) = \frac{0.6}{231} \times 32$. Percentage of sulphur $(S) = \frac{0.6 \times 32}{231 \times 0.2} \times 100$. $$= \frac{1920}{46.2} = 41.558 \approx 42$$

% oxygen in phenol = $\frac{16}{94} \times$ 100 = 17.02$\%$ % oxygen in o-Nitrophenol ($C_6H_5NO_3$) Molecular mass = $C_6H_5NO_3 = 139 \, \mathrm{g/m}$ $$= \frac{48}{139} \times 100 = 34.53\%$$ % increase = (34.53 - 17.02) = 17.5 $$17.5 \times 10^{-1} = 175 Answer$$

Using formula $$\log K_{P_2} - \log K_{P_1} = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$$ $$3.5 - 2.5 = \frac{\Delta H}{2.303 R} (0.06 - 0.05)$$ $$\frac{\Delta H}{R} = \frac{2.303}{0.01}$$ $$\frac{\Delta H}{R} = 230.3 \approx 230$$

Given $t_{1/2} = \frac{\ln 2}{K} \implies K_1 = \frac{0.693}{6.93} \implies 0.1 \, min^{-1}$. $t_{99\%} = \frac{1}{K} \ln \left[ \frac{A_0}{A_t} \right]$ $$= \frac{1}{0.1} \ln \left[ \frac{100}{1} \right] \implies \frac{2 \ln 10}{0.1}$$ $$= \frac{2 \times 2.303}{0.1}$$ $$\implies 46.06 \, min \approx 46$$