JEE Main 13 April 2023 Shift 1 question paper with solutions

Let $$I = \int_0^\infty \left( \frac{6}{e^{3x} + 6e^{2x} + 11e^x + 6} \right) \, dx$$ which implies $$I = \int_0^\infty \frac{6}{(e^x + 1)(e^x + 2)(e^x + 3)} \, dx$$ Using partial fraction, we can write $$\Rightarrow I = 6 \int_0^\infty \left[ \frac{1}{2(e^x + 1)} - \frac{1}{(e^x + 2)} + \frac{1}{2(e^x + 3)} \right] \, dx$$ $$\Rightarrow I = 6 \int_0^\infty \left[ \frac{1}{2(e^x + 1)} - \frac{1}{(1 + 2e^{-x})} + \frac{1}{2(1 + 3e^{-x})} \right] e^{-x} \, dx$$ $$\Rightarrow I = 6 \left[ -\frac{1}{2} \log_e (e^{-x} + 1) + \frac{1}{2} \log_e (2e^{-x} + 1) - \frac{1}{6} \log_e (3e^{-x} + 1) \right]_0^\infty$$ $$\Rightarrow I = 6 \left[ 0 - \left\{ -\frac{1}{2} \log_e 2 + \frac{1}{2} \log_e 3 - \frac{1}{6} \log_e 4 \right\} \right]$$ $$\Rightarrow I = 3 \log_e 2 - 3 \log_e 3 + \log_e 4$$ $$\Rightarrow I = 5 \log_e 2 - 3 \log_e 3$$ $$\Rightarrow I = \log_e \left( \frac{32}{27} \right)$$

Given, $x|x - 1| + |x + 2| + a = 0$. Now taking, Case I: $x < -2$, we get $$-x^2 + x - x - 2 + a = 0$$ $$\Rightarrow a = x^2 + 2$$ And $y = x^2 + 2$ is decreasing $\forall x \in (-\infty, -2)$. Now taking Case II: $-2 \leq x < 1$ we get, $$-x^2 + x + x + 2 + a = 0$$ $$\Rightarrow a = x^2 - 2x - 2$$ And $y = x^2 - 2x - 2$. So, $$\frac{dy}{dx} = 2x - 1 \leq 0 \forall x \in \left[-2, 1\right)$$ Hence, $y$ is decreasing $\forall x \in \left[-2, 1\right)$. Now taking Case III: $x \geq 1$ we get, $$x^2 - x + x + 2 + a = 0$$ $$\Rightarrow a = -(x^2 + 2)$$ And $y = -(x^2 + 2)$ is decreasing $\forall x \in \left[1, \infty\right)$. Hence, from all the cases we can say that nature of function is continuously decreasing so, it will cut $x$-axis only one time, $\therefore$ Exactly one real root $\forall a \in \mathbb{R}$.

Given, $$((A \land (B \lor C)) \Rightarrow (A \lor B)) \Rightarrow A$$ We know that, $X \Rightarrow Y \equiv \sim X \lor Y$, Now using the above formula we get, $$\equiv (\sim (A \land (B \lor C)) \lor (A \lor B)) \Rightarrow A$$ $$\equiv \sim (\sim (A \land (B \lor C)) \lor (A \lor B)) \lor A$$ $$\equiv (A \land (B \lor C)) \land \sim (A \lor B) \lor A$$ Now to solve $(A \land (B \lor C)) \land \sim (A \lor B)$ let us assume a universal set $\{1, 2, 3, 4, 5, 6, 7, 8\}$, set $A = \{1, 2, 3\}$, set $B = \{3, 4, 5\}$ and set $C = \{5, 6, 7\}$ So, $A \land (B \lor C)$ will be $\{3\}$ and $\sim (A \lor B)$ will be $\{6, 7, 8\}$, So, $(A \land (B \lor C)) \land \sim (A \lor B) = \emptyset$ $$\equiv F \lor A$$ $$\equiv A$$ Therefore, negation of statement $= \sim A$

The direction ratio of line parallel to the line of the shortest distance between the lines $\vec{r} = \left( \hat{i} - \hat{j} \right) + \lambda \left( 2\hat{i} + \hat{k} \right)$ and $\vec{r} = \left( 2\hat{i} - \hat{j} \right) + \mu \left( \hat{i} - \hat{j} + \hat{k} \right)$ is given by, $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i} - \hat{j} - 2\hat{k}$$ Now given line passing through $A(-1, 2, 3)$ And having direction ratio $(1, -1, -2)$, So, the equation is given by, $$\frac{x+1}{1} = \frac{y-2}{-1} = \frac{z-3}{-2} = \lambda$$ Let the line intersect the plane at $P$ $$P \equiv (\lambda - 1, -\lambda + 2, -2\lambda + 3)$$ $P$ lies on the plane $x - 2y + 3z = 10$ $$\therefore (\lambda - 1) - 2(-\lambda + 2) + 3(-2\lambda + 3) = 10$$ $$\Rightarrow \lambda = -2$$ Hence, point $P \equiv (-3, 4, 7)$ So, the distance of point $P(-3, 4, 7)$ from point $A(-1, 2, 3)$ will be, $$AP = \sqrt{4 + 16 + 16} = 2\sqrt{6}$$

It is given that $P(H) = 3P(T)$. Therefore, $P(H) = \frac{3}{4}$, $P(T) = \frac{1}{4}$. Since the coin is tossed till either $1H$ or $3T$ occurs, the process will end in the maximum of 3 throws. $$\begin{array}{|c|c|c|c|} \hline X_i & 1 & 2 & 3 \\ \hline P(X_i) & \frac{3}{4} & \frac{1}{4} \times \frac{3}{4} & \frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} \\ \hline \end{array}$$ Now we know that, Mean is given by $\sum_{i=1}^{n} X_i P(X_i)$. Mean $= 1 \times \frac{3}{4} + 2 \times \frac{3}{16} + 3 \times \frac{1}{16} = \frac{21}{16}$. Hence, this is the correct option.

Given, System of linear equations, $$2x + 4y + 2az = b$$ $$x + 2y + 3z = 4$$ $$x + 2y + 3z = 4$$ Now finding, $$\Delta = \begin{vmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{vmatrix}$$ $$\Rightarrow \Delta = -18(a - 3)$$ Hence, for $a = 3$, $\Delta = 0$ Now finding, $$\Delta_3 = \begin{vmatrix} b & 2 & 4 \\ 4 & 1 & 2 \\ 8 & 2 & -5 \end{vmatrix}$$ $$\Rightarrow \Delta_3 = 9(b - 8)$$ So, for $b = 8$, $\Delta_3 = 0$ Now solving options we get, If $a = 3$ and $b = 8$ we'll have infinite solution For $a = 8, b = 8$, $\Delta \neq 0$, $\Delta_3 = 0 \Rightarrow$ unique solution For $a = b = 6$, $\Delta \neq 0$, $\Delta_3 \neq 0 \Rightarrow$ unique solution $\therefore$ For $a = 3, b = 6$ we'll have no solution

Given: $$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10 \ldots (1)$$ Replace $x \to \frac{1}{x}$ $$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10 \ldots (2)$$ Eliminating $f\left(\frac{1}{x}\right)$ from (1) and (2), we get $$\Rightarrow 5f(x) = \frac{3}{x} - 2x - 10$$ Let us replace $x$ with 3 $$\Rightarrow 5f(3) = \frac{3}{3} - 2(3) - 10$$ $$\Rightarrow f(3) = -3$$ $$\Rightarrow 5f'(x) = \frac{-3}{x^2} - 2$$ $$\Rightarrow 5f'\left(\frac{1}{4}\right) = -48 - 2$$ $$\Rightarrow f'\left(\frac{1}{4}\right) = -10$$ $$\Rightarrow f(3) + f'\left(\frac{1}{4}\right) = 13$$ Hence this is the correct option.

Given that $\frac{x^2}{36} + \frac{y^2}{4} = 1$. Observe that $\left( 3\sqrt{3}, 1 \right)$ lies on the given ellipse. Equation of tangent to the ellipse at the given point is $T : \frac{3\sqrt{3}x}{12} + \frac{y}{4} = 1$. $$\Rightarrow \frac{\sqrt{3}x}{12} + \frac{y}{4} = 1$$ Equation of normal passing through same point will be $N : \frac{x - 3\sqrt{3}}{3\sqrt{3}/36} = \frac{y - 1}{1/4}$. $$\Rightarrow \frac{12x - 36\sqrt{3}}{\sqrt{3}} = 4y - 4$$ $$\Rightarrow 3x - \sqrt{3}y = \sqrt{3}y - \sqrt{3}$$ $$\Rightarrow 3x - \sqrt{3}y = 8\sqrt{3}$$ Now tangent and normal intersects the $y$-axis at $A$ and $B$ respectively. $A(0, 4)$ $B(0, -8)$. Equation of circle with $AB$ as diameter will be $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$ $$\Rightarrow x^2 + (y - 4)(y + 8) = 0$$ Line $x = 2\sqrt{5}$ intersects the circle. $$20 + y^2 + 4y - 32 = 0$$ $$\Rightarrow y^2 + 4y - 12 = 0$$ $$\Rightarrow (y + 6)(y - 2) = 0$$ Hence $P(2\sqrt{5}, -6)$ $Q(2\sqrt{5}, 2)$. Now let us find the equation of tangents to the circle at $P, Q$. $$\Rightarrow T : xx_1 + yy_1 + 2y + 2y_1 - 32 = 0$$ $$\Rightarrow T_1 : 2\sqrt{5}x - 6y + 2y - 12 - 32 = 0$$ $$\Rightarrow 2\sqrt{5}x - 4y = 44$$ $$\Rightarrow T_1 : \sqrt{5}x - 2y = 22 \ldots \ldots \ldots (i)$$ $$\Rightarrow T_2 : 2\sqrt{5}x + 2y + 2y + 4 - 32 = 0$$ $$\Rightarrow 2\sqrt{5}x + 4y = 28$$ $$\Rightarrow T_2 : \sqrt{5}x + 2y = 14 \ldots \ldots \ldots (ii)$$ From (i) and (ii) $$\alpha = \frac{18}{\sqrt{5}} = \frac{18}{\sqrt{5}}$$ $$\Rightarrow \alpha^2 - \frac{9}{2} = \frac{304}{5}$$ Hence this is the required option.

Plotting the required diagram we get, Now from diagram we can say that, $\sin x$ is maximum for $x \in \left( -\pi, -\frac{3\pi}{4} \right) \cup \left( \frac{\pi}{4}, \pi \right)$ and $\cos x$ is maximum for $x \in \left( -\frac{3\pi}{4}, -\frac{\pi}{2} \right) \cup \left( -\frac{\pi}{2}, \frac{\pi}{4} \right)$. So, area of the region is given by, $$A = \left| \int_{-\pi}^{-\frac{3\pi}{4}} \sin x \, dx \right| + \left| \int_{\frac{\pi}{4}}^{\pi} \sin x \, dx \right| + \left| \int_{-\frac{\pi}{2}}^{-\frac{3\pi}{4}} \cos x \, dx \right| + \left| \int_{\frac{\pi}{4}}^{-\frac{\pi}{2}} \cos x \, dx \right|$$ $$\Rightarrow A = \left| \frac{1}{\sqrt{2}} - 1 \right| + 1 + \frac{1}{\sqrt{2}} + \left| -1 + \frac{1}{\sqrt{2}} \right| + \left| \frac{1}{\sqrt{2}} + 1 \right|$$ $$\Rightarrow A = 2 + \frac{2}{\sqrt{2}} + 2 - \frac{2}{\sqrt{2}} = 4$$ Hence this is the required option.

Given that $A = \{0, 1, 2, 3, \ldots 8, 9\}$. We need to find the number of symmetric matrices. We know that a matrix is symmetric if $A^T = A$. $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ Now $a_{11}, a_{22}, a_{33}$ can have any digit from 0 to 9. Number of ways this can be done is $10 \times 10 \times 10 = 10^3$. Now $a_{12} = a_{21}$. So number of ways this can be done is $10 \times 1$. Similarly $a_{13} = a_{31}$. So number of ways this can be done is $10 \times 1$. Also, $a_{23} = a_{32}$. So number of ways this can be done is $10 \times 1$. $\therefore$ Total symmetric matrix $= 10^3 \times 10 \times 10 \times 10 = 10^6$. Hence this is the correct option.

For $S_1$: $\lim_{n \to \infty} \frac{1}{n^2} \left[ 2 + 4 + 6 + \ldots + 2n \right]$. $$\lim_{n \to \infty} \frac{2n(n+1)}{2n^2} = 1$$ For $S_2$: $\lim_{n \to \infty} \frac{1}{n^{16}} \left( 1^{15} + 2^{15} + 3^{15} + \ldots + n^{15} \right)$. This is limit of sum. We know that $\lim_{n \to \infty} \sum_{r=1}^{n} f\left( \frac{r}{n} \right) \frac{1}{n} = \int_{0}^{1} f(x) \, dx$. Therefore, $\lim_{n \to \infty} \frac{\sum_{r=1}^{n} r^k}{n^{k+1}} = \lim_{n \to \infty} \frac{\sum_{r=1}^{n} \left( \frac{r}{n} \right)^k}{n}$. $$= \int_{0}^{1} x^k \, dx = \frac{1}{k+1}$$ Here $k = 15$. Therefore, $\lim_{n \to \infty} \frac{\sum_{r=1}^{n} r^{15}}{n^{16}} = \frac{1}{16}$. Therefore, both $S_1$ and $S_2$ are correct. Hence this is the required option.

We have been given the length of focal chord $PQ = 100$. Length of focal chord at $(t) = a \left( t + \frac{1}{t} \right)^2 = 100$ Where $a = 9$ $$t + \frac{1}{t} = \pm \frac{10}{3}$$ $$\therefore t = 3, \frac{1}{3}, -3, -\frac{1}{3}$$ Since ordinate of $P$ is $+\mathrm{ve}$ $$\therefore t = 3 or \frac{1}{3}$$ $(t = 3 as slope of PQ is positive)$ $$\therefore M(21, 9)$$ Required line: $y - 9 = \frac{-80}{60} \left( x - 21 \right)$ $$\Rightarrow 4x + 3y = 111$$ $(-3, 43)$ does not lie on line. Hence this is the correct option.

We have been given that $$f(g(x)) = x + 3 - \sqrt{x}$$ $$\Rightarrow f(\sqrt{x} + 1) = x + 3 - \sqrt{x}$$ Put $\sqrt{x} + 1 = t$ $$\sqrt{x} = t - 1$$ $$x = (t - 1)^2$$ $$\Rightarrow f(t) = (t - 1)^2 + 3 - (t - 1)$$ $$\Rightarrow f(0) = 1 + 3 + 1 = 5$$ Hence this is the correct option.

Given that $\frac{4^{2022}}{15}$. We know that $(a + x)^n = {}^{n}C_{0} a^n x^0 + {}^{n}C_{1} a^{n-1} x^1 + \ldots + {}^{n}C_{n} a^0 x^n$. $$\Rightarrow \frac{4^{2022}}{15} = \frac{16^{1011}}{15} = \frac{(15+1)^{1011}}{15}$$ $$= \frac{{}^{1011}C_{0} (15)^{1011} + {}^{1011}C_{1} (15)^{1010} + \ldots + {}^{1011}C_{1011} (15)^0}{15}$$ $$= {}^{1011}C_{0} (15)^{1010} + {}^{1011}C_{1} (15)^{1009} + \ldots + {}^{1011}C_{1010} (15)^1 + \frac{1}{15}$$ Therefore, the fractional part of $\frac{4^{2022}}{15} = \frac{1}{15}$. Hence this is the required option.

We have been given that $\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$. Therefore, $\left( \vec{d} - \vec{c} \right) \times \vec{b} = 0$. This implies $\vec{d} - \vec{c} = \lambda \vec{b}$ (since $\vec{d} \neq \vec{c}$). Thus, $\vec{d} = \vec{c} + \lambda \vec{b}$. It is given that $\vec{a} \cdot \vec{d} = 24$. Therefore, $\vec{a} \cdot \vec{c} + \lambda \vec{b} \cdot \vec{a} = 24$. This implies $6 + \lambda (3 - 8 + 14) = 24$. Therefore, $9 \lambda = 18$. Solving gives $\lambda = 2$. Thus, $\vec{d} = \vec{c} + 2 \vec{b}$. Therefore, $\vec{d} = 8 \hat{i} - 5 \hat{j} + 18 \hat{k}$. The magnitude of $\vec{d}$ is $|\vec{d}|^2 = 64 + 25 + 324 = 413$. Hence, this is the correct option.

We have been given that $B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}$, $\alpha > 2$. And $adj(A) = B$, $|A| = 2$ implies $|adj(A)| = |B|$. We know that $|adjA| = |A|^{n-1}$ implies $2^2 = (8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(-\alpha)$ implies $\alpha^2 - 6\alpha + 8 = 0$ implies $(\alpha - 4)(\alpha - 2) = 0$. $\alpha = 4, 2$ but $\alpha > 2$ so $\alpha = 4$. Now $\begin{bmatrix} \alpha & -2\alpha & \alpha \end{bmatrix} B = \begin{bmatrix} 4 & -8 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{bmatrix} = \begin{bmatrix} 4 \\ -8 \\ 4 \end{bmatrix} = [12 12 8] = 48 - 96 + 32 = -16$. Hence this is the correct option.

Given, first term of A.P are 1, 2, 3, $\ldots$ 10, so general term of the first term will be $i$. And common difference are 1, 3, 5, $\ldots$, so general term of common difference is given by $2i - 1$. Now sum of the A.P is given by, $$S_i = \frac{12}{2} [2 \times i + (12 - 1)(2i - 1)]$$ $$\Rightarrow S_i = 6 [2 \times i + 11(2i - 1)]$$ $$\Rightarrow S_i = 144i - 66$$ So, $$\sum_{i=1}^{10} S_i = \sum_{i=1}^{10} 144i - 66 \sum_{i=1}^{10} 1$$ $$\Rightarrow \sum_{i=1}^{10} S_i = 144 \left( \frac{10 \times 11}{2} \right) - 66 \times 10$$ $$\Rightarrow \sum_{i=1}^{10} S_i = 792(10) - 66 \times 10$$ $$\Rightarrow \sum_{i=1}^{10} S_i = 7260$$ Hence this is the correct option.

We have been given that $\frac{dy}{dx} = y + 7 \implies \frac{dy}{y+7} = dx$. $$\implies \ln |y + 7| = x + c$$ $$\implies |y + 7| = k \cdot e^x$$ $$\implies y = k \cdot e^x - 7$$ $$\implies y_1(0) = 0 \implies k = 7 \implies y_1(x) = 7(e^x - 1)$$ $$\implies y_2(0) = 1 \implies 1 = k - 7 \implies 8 = k$$ $$\implies y_2(x) = 8e^x - 7$$ $$\implies y_1(x) = y_2(x) \implies 8e^x - 7 = 7e^x - 7$$ Hence, No point of intersection. This is the required option.

Let the equation of plane passing through the intersection of two given planes: $$x + 2y + az - 2 + \lambda(x - y + z - 3) = 0$$ $$\Rightarrow x(\lambda + 1) + y(2 - \lambda) + z(a + \lambda) - 2 - 3\lambda = 0$$ This is same as $5x - 11y + bz = 6a - 1$ $$\frac{\lambda + 1}{5} = \frac{2 - \lambda}{-11} = \frac{a + \lambda}{b} = \frac{2 + 3\lambda}{6a - 1}$$ $$\Rightarrow -11\lambda - 11 = 10 - 5\lambda$$ $$\Rightarrow 6\lambda = -21 \Rightarrow \lambda = \frac{-7}{2}$$ $$\frac{2 - \lambda}{-11} = \frac{2 + 3\lambda}{6a - 1} \Rightarrow \frac{2 + \frac{7}{2}}{-11} = \frac{2 - \frac{21}{2}}{6a - 1} \Rightarrow \frac{1}{2} = \frac{-17}{2(6a - 1)} \Rightarrow 6a - 1 = 17 \Rightarrow a = 3$$ $$\frac{2 - \lambda}{-11} = \frac{a + \lambda}{b} \Rightarrow \frac{1}{2} = \frac{3 - \frac{7}{2}}{b}$$ $$\Rightarrow \frac{-b}{2} = -\frac{1}{2} \Rightarrow b = 1$$ Therefore, point $(a, -c, c) \equiv (3, -c, c)$ Distance $= \frac{2}{\sqrt{a}} = \frac{2}{\sqrt{3}}$ Plane: $5x - 11y + z = 17$ The distance of the plane from $(3, -c, c)$ is $$\left| \frac{15 + 11c + c - 17}{\sqrt{147}} \right| = \frac{2}{\sqrt{3}}$$ $$\Rightarrow |12c - 2| = 14$$ $$\Rightarrow c = -1, \frac{4}{3}$$ $$c = -1 (c \in \mathbb{Z})$$ Therefore, $\frac{a + b}{c} = \frac{3 + 1}{-1} = -4$ Hence this is the required option.

Given, $$\left( \sqrt{x} - \frac{6}{x^{\frac{3}{2}}} \right)^n, \; n \leq 15$$ Now the $r^{th}$ term is given by $$T_{r+1} = nC_r \left( x^{\frac{1}{2}} \right)^{n-r} \left( -6x^{-\frac{3}{2}} \right)^r$$ $$\Rightarrow T_{r+1} = (-6)^r \cdot nC_r \cdot x^{\frac{n-r}{2}} \cdot x^{-\frac{3r}{2}}$$ $$\Rightarrow T_{r+1} = (-6)^r \cdot nC_r \cdot x^{\frac{n-4r}{2}}$$ Now we have to take, $n = 4r$ for constant term so $n$ is divisible by 4. So, $$\alpha = nC_{\frac{n}{4}} (-6)^{\frac{n}{4}}$$ And sum of all coefficients will be $$\left( \sqrt{1} - \frac{6}{3^{\frac{1}{2}}} \right)^n = (-5)^n$$ Also given, $$(-5)^n - nC_{\frac{n}{4}} (-6)^{\frac{n}{4}} = 649$$ Now by hit and trial we get, $n = 4$ Hence, $$\alpha = 4C_1 (-6)^{\frac{4}{4}} = -24$$ Now for coefficient of $x^{-4}$ $$\frac{n-4r}{2} = -4$$ $$\Rightarrow n = 4r - 8 \Rightarrow r = 3$$ So, $$\lambda(\alpha) = (-6)^3 \cdot 4C_3$$ $$\Rightarrow \lambda(-24) = (-6)^3 \cdot 4C_3$$ $$\Rightarrow \lambda = 36$$

Given, $\sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right)-\sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)=\frac{\pi}{4}$ $\Rightarrow \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right)=\frac{\pi}{4}+\sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)$ $\Rightarrow \sin\left(\sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right)\right)=\sin\left(\frac{\pi}{4}+\sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)\right)$ $\Rightarrow \left(\frac{x+1}{\sqrt{x^2+2x+2}}\right)=\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{2}}\times\frac{x}{\sqrt{x^2+1}}$ $\Rightarrow \frac{x+1}{\sqrt{x^2+2x+2}}=\frac{x+1}{\sqrt{2}\sqrt{x^2+1}}$ $\Rightarrow (x+1)\left(\sqrt{2}\sqrt{x^2+1}-\sqrt{x^2+2x+2}\right)=0$ $\Rightarrow x=-1 \text{ or } \sqrt{x^2+2x+2}=\sqrt{2}\cdot\sqrt{x^2+1}$ Now solving, $\sqrt{x^2+2x+2}=\sqrt{2}\cdot\sqrt{x^2+1}$ we get, $x^2+2x+2=2(x^2+1)$ $\Rightarrow x^2-2x=0$ $\Rightarrow x=0,\;x=2$ Hence, $S=\{0,2\}$. Now solving, $\sum_{x\in S}\left(\sin\left(\frac{(x^2+x+5)\pi}{2}\right)-\cos\left((x^2+x+5)\pi\right)\right)$ $=\left(\sin\frac{5\pi}{2}-\cos5\pi\right)+\left(\sin\frac{11\pi}{2}-\cos11\pi\right)$ $=1-(-1)+1-(-1)=4$

Given, $$w = \bar{z}z + k_1 z + k_2 i z + \lambda (1+i), \; k_1, k_2 \in \mathbb{R}.$$ And $\operatorname{Re}(w) = 0$, So, taking $z = x + iy$ and $\bar{z} = x - iy$ We get, $$w = x^2 + y^2 + k_1 (x + iy) + k_2 i (x + iy) + \lambda (1+i)$$ $$\Rightarrow \operatorname{Re}(w) = 0 \Rightarrow x^2 + y^2 + k_1 x - k_2 y + \lambda = 0$$ And, $\operatorname{Im}(w) = 0 \Rightarrow k_1 y + k_2 x + \lambda = 0$ Now given $\operatorname{Re}(w) = 0$ circle is touching $y = 1$ and $y$-axis, with radius 1, so we have equation of circle $$x^2 + y^2 - 2x - 4y + 4 = 0$$ So, on comparing with $\operatorname{Re}(w) = 0$ we get, $$k_1 = -2, \; k_2 = 4, \; \lambda = 4$$ So, $\operatorname{Im}(w) = 0 \Rightarrow -2y + 4x + 4 = 0$ And given $-2y + 4x + 4 = 0$ intersects the circle, So, solving $x^2 + y^2 - 2x - 4y + 4 = 0$ and $-2y + 4x + 4 = 0$ we get, Intersecting point as $A(0, 2)$ and $B\left(\frac{2}{5}, \frac{14}{5}\right)$ Hence, by distance formula we get, $$30 \left(AB\right)^2 = 30 \left( \left(0 - \frac{2}{5}\right)^2 + \left(2 - \frac{14}{5}\right)^2 \right) = 24$$

Given, $S_0(x) = x$, $S_k(x) = C_k x + k \int_0^x S_{k-1}(t) dt$. Now for $S_0(x) = x$, $C_0 = 1$. Now solving, $C_k = 1 - \int_0^1 S_{k-1}(x) dx$ for $k = 1$ we get, $$C_1 = 1 - \int_0^1 x dx = \frac{1}{2}$$ Now putting $k = 1$ in $S_k(x) = C_k x + k \int_0^x S_{k-1}(t) dt$ we get, $$\Rightarrow S_1(x) = \frac{x}{2} + 1 \cdot \int_0^x t dt = \frac{x}{2} + \frac{x^2}{2}$$ Now for $k = 2$ we get, $$C_2 = 1 - \int_0^1 \left( \frac{x}{2} + \frac{x^2}{2} \right) dx = \frac{7}{12}$$ And $$S_2(x) = \frac{7}{12} x + 2 \int_0^x \left( \frac{t}{2} + \frac{t^2}{2} \right) dt$$ $$\Rightarrow S_2(x) = \frac{7x}{12} + \frac{x^2}{2} + \frac{x^3}{3}$$ Now taking $k = 3$ we get, $$C_3 = 1 - \int_0^1 \left( \frac{7x}{12} + \frac{x^2}{2} + \frac{x^3}{3} \right) dx = \frac{11}{24}$$ Now finding the value of $S_2(3) + 6 \cdot C_3$ we get, $$S_2(3) + 6 \cdot C_3 = \frac{7 \times 3}{12} + \frac{3^2}{2} + \frac{3^3}{3} + 6 \cdot \frac{11}{24}$$ $$S_2(3) + 6 \cdot C_3 = \frac{61}{4} + \frac{11}{4} = 18$$

Given, $$2 \cdot 2^2 - 3^2 + 2 \cdot 4^2 - 5^2 + 2 \cdot 6^2 - \cdots 20 terms,$$ $$= \sum_{r=1}^{10} \left( 2 \cdot (2r)^2 - (2r+1)^2 \right)$$ $$= \sum_{r=1}^{10} (8r^2 - 4r^2 - 4r - 1)$$ $$= \sum_{r=1}^{10} (4r^2 - 4r - 1)$$ $$= 4 \sum_{r=1}^{10} r^2 - 4 \sum_{r=1}^{10} r - \sum_{r=1}^{10} 1$$ $$= \frac{4 \cdot 10 \cdot 11 \cdot 21}{6} - 4 \cdot \frac{10 \cdot 11}{2} - 10$$ $$= 44 \cdot 35 - 220 - 10 = 1540 - 230 = 1310$$

The given numbers are 1, 2, 3, 4 and sum of digits should be 12. Case 1: $\underline{4 \ 3 \ 1 \ 1 \ 1 \ 1 \ 1}$. Here the sum of the digits is 12. The number of ways arranging them is $$\frac{7!}{5!} = 42.$$ Case 2: $\underline{4 \ 2 \ 2 \ 1 \ 1 \ 1 \ 1}$. The number of ways arranging them is $$\frac{7!}{2!4!} = 105.$$ Case 3: $\underline{3 \ 2 \ 2 \ 2 \ 1 \ 1 \ 1}$. The number of ways arranging them is $$\frac{7!}{3!3!} = 140.$$ Case 4: $\underline{3 \ 3 \ 2 \ 1 \ 1 \ 1 \ 1}$. The number of ways arranging them is $$\frac{7!}{2!4!} = 105.$$ Case 5: $\underline{2 \ 2 \ 2 \ 2 \ 2 \ 1 \ 1}$. The number of ways arranging them is $$\frac{7!}{5!2!} = 21.$$ The total number of ways are $$42 + 105 + 140 + 105 + 21 = 413.$$ Therefore, the required answer is 413.

Given that $H : \frac{y^2}{25} - \frac{x^2}{16} = 1$. Equation of tangent to the hyperbola is: $y = mx + \sqrt{25 - 16m^2}$. Passes through $(4, 1)$. $$\Rightarrow y = mx + \sqrt{25 - 16m^2}$$ $$\Rightarrow 1 = 4m + \sqrt{25 - 16m^2}$$ $$\Rightarrow 4m^2 - m - 3$$ $$\Rightarrow m_1 = 1, \ m_2 = \frac{-3}{4}$$ $$\Rightarrow |m_1| = 1, \ |m_2| = \frac{3}{4}$$ Equation of tangents with slopes $1$ and $\frac{3}{4}$ $$\Rightarrow y = x - 3 \ldots \ldots (i)$$ $$\Rightarrow y = \frac{3}{4}x - 4 \ldots \ldots (ii)$$ $(i)$ and $(ii)$ $\Rightarrow Q \equiv (-4, -7)$ $P \equiv (4, 1)$ $$\Rightarrow (PQ)^2 = 8^2 + 8^2 = 128$$ On converting $(i)$ and $(ii)$ in intercept form we get, $(i) \Rightarrow \alpha = 3, \ (ii) \Rightarrow \beta = \frac{16}{3}$ $$\Rightarrow \frac{(PQ)^2}{\alpha \beta} = \frac{128}{3 \times \frac{16}{3}} = 8$$ Hence this is the required option.

The image of the point $\left( \frac{5}{3}, \frac{5}{3}, \frac{8}{3} \right)$ with respect to the plane will be, $$\frac{x - \frac{5}{3}}{1} = \frac{y - \frac{5}{3}}{-2} = \frac{z - \frac{8}{3}}{1} = \frac{-2 \left( 1 \times \frac{5}{3} - 2 \times \frac{5}{3} + \frac{8}{3} - 2 \right)}{1^2 + 2^2 + 1^2}$$ $$\Rightarrow \frac{x - \frac{5}{3}}{1} = \frac{y - \frac{5}{3}}{-2} = \frac{z - \frac{8}{3}}{1} = \frac{-2(-1)}{6}$$ $$\Rightarrow P(x, y, z) = (2, 1, 3)$$ But $PQ = 13$ $$4^2 + 3^2 + (\alpha - 3)^2 = 169$$ $$|\alpha - 3| = 12$$ $$\alpha = 15 Since (\alpha > 0)$$ Hence this is the required answer.

Given, $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}$. And $\vec{b}$ is a vector such that $\vec{a} = \vec{b} \times \vec{c}$ and $|\vec{b}|^2 = 50$, Now solving, $|\vec{a}| = |\vec{b} \times \vec{c}|$ $$\Rightarrow \sqrt{3^2 + 1^2 + 1^2} = |\vec{b}| \cdot \sqrt{2^2 + 3^2 + 3^2} \cdot \sin \theta$$ $$\Rightarrow \sqrt{11} = \sqrt{50} \cdot \sqrt{22} \cdot \sin \theta$$ $$\therefore \sin \theta = \frac{1}{10} or \cos \theta = \frac{\sqrt{99}}{10}$$ Now solving, $|72 - |\vec{b} + \vec{c}|^2|$ we get, $$|72 - |\vec{b} + \vec{c}|^2|$$ $$= |72 - \left( |\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}| \cos \theta \right)|$$ $$= |72 - (50 + 22 + 2 \times 5 \sqrt{2} \cdot \sqrt{22} \frac{\sqrt{99}}{10})|$$ $$= |72 - (72 + \frac{2 \times 5 \times 2 \times 11 \times 3}{10})|$$ $$= |66| = 66$$

Given, the mean of the below data is 5. Now finding mean we get, $$\frac{4 + 72 + 28 \times 5 + 7\alpha + 72}{64 + \alpha} = 5$$ $$\Rightarrow \alpha = 16$$ So, total number of observation will be, $\sum f_i = 80$. Now finding, mean deviation we get, $$M. D = \frac{\sum f_i |x_i - 5|}{\sum f_i}$$ $$= \frac{4 \times 4 + 24 \times 2 + 0 + 16 \times 2 + 8 \times 4}{80} = \frac{8}{5}$$ So, $m = \frac{8}{5}$. Now finding, variance we get, $$\sigma^2 = \frac{\sum f_i |x_i - 5|^2}{\sum f_i}$$ $$= \frac{4 \times 16 + 24 \times 4 + 0 + 16 \times 4 + 8 \times 16}{80}$$ $$\Rightarrow \sigma^2 = \frac{22}{5}$$ Hence, the value of $\frac{3\alpha}{m + \sigma^2} = \frac{3 \times 16}{\frac{8}{5} + \frac{22}{5}} = 8$.

Maxwell's four equations describe the electric and magnetic fields arising from distributions of electric charges and currents, and how those fields change in time. The equations are as follows: $$\oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon_0} (1)$$ $$\oint \vec{B} \cdot d\vec{A} = 0 (2)$$ $$\oint \vec{E} \cdot d\vec{l} = -\frac{\partial \phi_B}{\partial t} (3)$$ $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I (4)$$ From the above set of equations, it is clear that only the third equation is valid for time varying field. Hence, this is the correct option.

The resistances can be calculated as $$R_A = R + \frac{R}{2} = \frac{3R}{2}$$ For second circuit $$\frac{1}{R_B} = \frac{1}{2R} + \frac{1}{R} = \frac{3}{2R}$$ $$\Rightarrow R_B = \frac{2R}{3}$$ For third circuit $$R_C = \frac{R}{3}$$ For fourth circuit $$R_D = 3R$$ The formula for power is given by $P = I^2 R$. Clearly, $$P_D > P_A > P_B > P_C$$

The apparent depth formula is given by $n = \frac{Real depth}{Apparent depth}$. The apparent depth in this case is given by $$d' = \frac{d_1}{n_1} + \frac{d_2}{n_2}$$ $$\Rightarrow d' = \frac{d}{2} \left( \frac{1}{n_1} + \frac{1}{n_2} \right)$$ $$\Rightarrow d' = \frac{d}{2} \left( \frac{n_1 + n_2}{n_1 n_2} \right)$$

An electric field varying linearly with time will not produce variable magnetic field as current will be constant. A decelerating charged particle can result in a time varying magnetic field by emitting radiation. Similarly a permanent magnet has a constant magnetic field and a direct current does not change its direction and cannot produce a magnetic field. An antenna can modulate the digital signals. So, the right option is the statement D.

The time taken for $A$ $$\Delta t_A = \frac{l+L}{V_A}$$ The time taken for $B$ $$\Delta t_B = \frac{4l+L}{V_B}$$ It is given that $B$ takes more time than $A$. So, $$\Rightarrow \frac{4l+L}{20} - \frac{l+L}{30} = 35$$ $$\Rightarrow \frac{64L}{20} - \frac{61L}{30} = 35$$ $$\Rightarrow l = \frac{35 \times 600}{700} = 30 \, \mathrm{m}$$ Therefore, $$L = 1800 \, \mathrm{m}.$$

The work done in raising the water is $W = mgh$. Power is $P = \frac{W}{t}$. Let the powers of both the motors be $P_1$ and $P_2$ respectively. Hence, the ratio of the powers is, $$\frac{P_1}{P_2} = \frac{300 \times g \times 100}{5 \times 60} \div \frac{50 \times g \times 100}{2 \times 60} = \frac{60}{25}$$ $$\Rightarrow \frac{3\sqrt{x}}{\sqrt{x+1}} = 2.4$$ $$\Rightarrow 3\sqrt{x} = 2.4\sqrt{x} + 2.4$$ $$\Rightarrow x = 16$$

The escape velocity on earth is given by $V_e = \sqrt{\frac{2GM_e}{R_e}}$. Escape velocity for the planet is $$V'_{esc} = \sqrt{\frac{2G(9M_e)}{4R_e}} \cdots (i)$$ The escape velocity of earth is $V_e = 11.2 \times 10^3 \, \mathrm{m \, s^{-1}}$. From equation (i) $$V'_{esc} = \sqrt{\frac{9}{4} \left( \frac{2GM_e}{R_e} \right)} = \frac{3}{2} \times 11.2 \times 10^3 \, \mathrm{m \, s^{-1}}$$ $$\Rightarrow V'_{esc} = 16.8 \, \mathrm{km \, s^{-1}}$$

The work function is given by $$\frac{hc}{\lambda} = \phi$$ which implies $$\lambda = \frac{hc}{\phi}$$ By the given data, $$\lambda_A = \frac{1242}{9} \, \mathrm{nm}, \lambda_B = \frac{1242}{4.5} \, \mathrm{nm}.$$ The difference between the wavelengths, $$\lambda_B - \lambda_A = \frac{1242}{4.5} \, \mathrm{nm} - \frac{1242}{9} \, \mathrm{nm} = \frac{1242}{9} \, \mathrm{nm} = 138 \, \mathrm{nm}$$

The formula to calculate the impulse ($J$) on the gun is given by $$J = m(v - u) \ldots(1)$$ where, $m$ is the mass, $u$ is the initial velocity and $v$ is the final velocity of the bullet. Substitute the values of the known parameters into equation (1) to calculate the required impulse. $$J = 10 \, \mathrm{g} \times \frac{1 \, \mathrm{kg}}{1000 \, \mathrm{g}} \times (600 - 0) \, \mathrm{m \, s^{-1}}$$ $$= 6 \, \mathrm{N \, s}$$

The formula to calculate the dipole moment is given by $$|\vec{p}| = q |\vec{d}| \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the dipole moment. $$|\vec{p}| = 0.01 \, \mathrm{C} \times 0.4 \, \mathrm{mm} \times \frac{1 \, \mathrm{m}}{10^3 \, \mathrm{mm}}$$ $$= 4.0 \times 10^{-6} \, \mathrm{C} \cdot \mathrm{m}$$ The formula to calculate the torque on the dipole in the electric field is given by $$|\vec{\tau}| = |\vec{p} \times \vec{E}| \ldots (2)$$ Substitute the values of the known parameters into equation (2) to calculate the required torque. $$|\vec{\tau}| = 4.0 \times 10^{-6} \, \mathrm{C} \cdot \mathrm{m} \times 10 \, \mathrm{dyne} \cdot \mathrm{C}^{-1} \times \frac{1 \, \mathrm{N}}{10^5 \, \mathrm{dyne}} \times \sin 30^\circ$$ $$= 2 \times 10^{-10} \, \mathrm{N} \cdot \mathrm{m}$$

The displacement is the hypotenuse of the diameter and the distance travelled for half the rotation $\pi R$. By using Pythagoras theorem, Displacement of $A = \sqrt{(2R)^2 + (\pi R)^2}$ $$= R \sqrt{\pi^2 + 4}$$

The $F_1$ layer is above the E layer and the ionization is less compared to E layer. It ranges to $170 - 190 \, \mathrm{km}$. The D layer is located above the stratosphere. It ranges from $60 - 90 \, \mathrm{km}$ above the earth's surface. The troposphere starts from the ground and ranges to $100 \, \mathrm{km}$. The E layer is located above the D layer. The intensity of ionization is less than the D layer. It ranges from $100 - 125 \, \mathrm{km}$ above the earth's surface.

The RMS speed is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ The average speed is given by $$v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$ The ratio is $$\Rightarrow \frac{v_{rms}}{v_{avg}} = \sqrt{\frac{\frac{3RT}{M}}{\frac{8RT}{\pi M}}} = \sqrt{\frac{3\pi}{8}} = \sqrt{\frac{3}{8} \times \frac{22}{7}}$$ $$\Rightarrow \frac{v_{rms}}{v_{avg}} = \sqrt{\frac{33}{28}}$$ Comparing it with the given value $\left(1 + \frac{5}{x}\right)^{\frac{1}{2}}$, $$\frac{v_{rms}}{v_{avg}} = \sqrt{1 + \frac{5}{28}}$$ $$\Rightarrow x = 28$$

The kinetic energy is given by, $$K = \frac{1}{2} mv^2$$ Differentiating the above equation, $$dK = \frac{1}{2} \left( v^2 dm + 2mvdv \right)$$ Dividing by $K$, $$\Rightarrow \frac{dK}{K} = \frac{dm}{m} + \frac{2dv}{v}$$ $$\Rightarrow \frac{dK}{1000} = \frac{0.5}{5} + \frac{2 \times 0.4}{20}$$ $$\Rightarrow dK = 100 + 40 = 140$$ $$\Rightarrow KE = (1000 \pm 140) \mathrm{J}$$

The kinetic energy and the momentum of a moving object are related by $$K = \frac{p^2}{2m} \ldots (1)$$ Hence, the ratio of the kinetic energies of two objects having masses $m$ and $m'$ can be calculated as follows: $$\frac{K}{K'} = \frac{\frac{p^2}{2m}}{\frac{p^2}{2m'}}$$ $$= \frac{m'}{m} \ldots (2)$$ Substitute the ratio of the kinetic energies into equation (2) and simplify to obtain the ratio of their masses. $$\Rightarrow \frac{16}{9} = \frac{m'}{m}$$ $$\Rightarrow m : m' = 9 : 16$$

Using Bernoulli's theorem, $$P_A + \frac{1}{2} \rho V_A^2 = P_B + \frac{1}{2} \rho V_B^2 \cdots (i)$$ Using equation of continuity, $$A_A V_A = A_B V_B$$ So, $1.5 \times 10^{-4} \times V_A = 25 \times 10^{-6} \times V_B$ $$V_A = \frac{25 \times 10^{-6}}{1.5 \times 10^{-4}} \times 0.6$$ $$= \frac{1}{10} \, \mathrm{m \, s^{-1}}$$ Substituting in equation (i) $$\Rightarrow P_A - P_B = \frac{1000}{2} \left[ 0.6^2 - 0.1^2 \right] = \frac{1000}{2} \times 0.7 \times 0.5$$ $$= 175 \, \mathrm{Pa}$$

The data given is $$P = aV^{-3}$$ Bulk modulus is given by, $$B = -\frac{dP}{dV} \cdot \frac{V}{P} = -V \frac{dP}{dV}$$ $$= -V \frac{d}{dV} \left\{ \frac{a}{V^3} \right\}$$ $$= -V \left[ \frac{-3a}{V^4} \right]$$ $$= \frac{3a}{V^3} = 3P$$

The signal $A$ is given to a NOR gate and $B$ is given to an OR gate. The output of these two signals is given to a NAND gate. Using the laws of Boolean algebra, $$Y = \overline{\overline{A} \cdot B} = A + \overline{B}.$$ Analyzing from the input graphs, one can see, $$t_1 - t_2: A = 0, \overline{B} = 0, Y = 0$$ $$t_2 - t_3: A = 1, \overline{B} = 0, Y = 1$$ $$t_3 - t_4: A = 0, \overline{B} = 1, Y = 1$$ $$t_4 - t_5: A = 1, \overline{B} = 0, Y = 1$$ $$t_5 - t_6: A = 1, \overline{B} = 1, Y = 1$$

In a SHM, mechanical energy is conserved. So, the sum of kinetic energy and potential energy remains constant. For a SHM, $$T.E. - U = K.E$$ $$\Rightarrow T.E. - U = \frac{1}{2} mv^2 = \frac{1}{2} m \omega^2 \left( A^2 - x^2 \right)$$ This is the equation of a downward parabola.

The value of $Q$ is $$Q = (m_A - m_B - m_D) \times 931.5 \, \mathrm{MeV}$$ $$\Rightarrow Q = (238.05079 - 234.04363 - 4.00260) \times 931.5 \, \mathrm{MeV}$$ $$\Rightarrow Q = 4.25 \, \mathrm{MeV}$$

The formula to calculate the Young's modulus of the material is given by $$Y = \frac{stress}{strain} = \frac{stress}{\frac{\Delta l}{L}} \cdots (1)$$ The formula to calculate the potential energy $(U)$ stored into the wire is given by $$U = \frac{1}{2} \times stress \times strain \times volume$$ $$= \frac{1}{2} \times stress \times \frac{\Delta l}{L} \times A \times L$$ $$= \frac{1}{2} \times stress \times A \times \Delta l \cdots (2)$$ From equation (1) and equation (2), it can be written that $$U = \frac{1}{2} \times Y \times \left(\frac{\Delta l}{L}\right)^2 \times A \cdots (3)$$ Substitute the values of the known parameters into equation (3) and solve to calculate the required cross-sectional area of the wire. $$80 \, J = \frac{1}{2} \times 2.0 \times 10^{11} \, Nm^{-2} \times \left(\frac{0.02 \, m}{20 \, m}\right)^2 \times A$$ $$\Rightarrow A = \frac{2 \times 80 \, J \times 20}{2.0 \times 10^{11} \times 4 \times 10^{-4} \, Nm^{-1}}$$ $$= 4 \times 10^{-5} \, m^2 \times \frac{10^6}{1 \, m^2}$$ $$= 40 \, mm^2$$

The formula to calculate the power dissipated in the first case can be written as $$P_1 = \frac{V_0^2}{R} \ldots (1)$$ When the wire is halved, its resistance will also become half the previous value. Thus, the formula to calculate the power dissipated in the second case can be written as $$P_2 = \frac{V_0^2}{\frac{R}{2}} \times 2$$ $$= \frac{4V_0^2}{R} \ldots (2)$$ Divide equation (2) by equation (1) to calculate the required ratio. $$\frac{P_2}{P_1} = \frac{\frac{4V_0^2}{R}}{\frac{V_0^2}{R}}$$ $$= 4$$ $$= \sqrt{16}$$ Hence, $\($ z = 16 $\)$.

The formula to calculate the kinetic energy $(K)$ of a particle executing SHM is given by $$K = \frac{1}{2} k (A^2 - y^2) \ldots (1)$$ Substitute the values of the known parameters into equation (1) and solve to calculate the value of the unknown parameter. $$K = \frac{1}{2} \times k \times (A^2 - A^2 \cos^2 30^\circ)$$ $$= \frac{1}{2} \times k \times A^2 \sin^2 30^\circ$$ $$\Rightarrow 200 = \frac{1}{2} \times 1.0 \times 10^x \times (0.4)^2 \times \left(\frac{1}{2}\right)^2$$ $$= \frac{2 \times 10^x}{100}$$ $$\Rightarrow 10^2 = 10^{x-2}$$ $$\Rightarrow x - 2 = 2$$ $$\Rightarrow x = 4$$

The formula to calculate the maximum current through the galvanometer, when a shunt resistance $r_1$ is connected is given by $$(i_G)_{\max} = \frac{r_1}{r_1 + R_G} i_{\max} \ldots (1)$$ When a series resistance $r_2$ is connected, the potential difference across the galvanometer is given by $$V = (i_G)_{\max} [R_G + r_2] \ldots (2)$$ From equations (1) and (2), it can be written that $$V = \frac{r_1}{r_1 + R_G} i_{\max} (R_G + r_2)$$ $$\Rightarrow V r_1 + V R_G = i_{\max} r_1 R_G + i_{\max} r_1 r_2$$ $$\Rightarrow (V - i_{\max} r_1) R_G = i_{\max} r_1 r_2 - V r_1$$ $$\Rightarrow R_G = \frac{i_{\max} r_1 r_2 - V r_1}{V - i_{\max} r_1} \ldots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required galvanometer resistance. $$R_G = \frac{0.250 \, \mathrm{A} \times 5 \, \Omega \times 1050 \, \Omega - 25 \, \mathrm{V} \times 5 \, \Omega}{25 \, \mathrm{V} - 0.250 \, \mathrm{A} \times 5 \, \Omega}$$ $$= 50 \, \Omega$$

The formula to calculate the radius (r) of any nth Bohr's orbit is given by $$r = r_0 \frac{n^2}{Z} \ldots (1)$$ where, $r_0$ is Bohr's radius, $n$ is the principal quantum number and $Z$ is the atomic number. Thus, for Helium ion, $$r_1 = r_0 \frac{2^2}{2}$$ $$= 2r_0 \ldots (2)$$ And, for the Beryllium ion, $$r_2 = r_0 \frac{4^2}{4}$$ $$= 4r_0 \ldots (3)$$ Divide equation (3) by equation (2) to obtain the required ratio. $$\frac{r_2}{r_1} = \frac{4r_0}{2r_0}$$ $$= 2$$ Hence, $x = 2$.

Let's consider the following diagram: The formula to calculate the electric field due to an infinitely long charged sheet is given by $$E_s = \frac{\sigma}{2\varepsilon_0} \ldots (1)$$ The formula to calculate the electric field due to an infinitely long line charge is given by $$E_l = \frac{\lambda}{2\pi\varepsilon_0 r} \ldots (2)$$ Using equations (1) and (2), the electric field at $P$ can be calculated as follows: $$E_P = \left| \frac{\sigma}{2\varepsilon_0} - \frac{1}{4\pi\varepsilon_0} \frac{2\lambda}{3\pi} \right|$$ $$= \left| \frac{\sigma}{2\varepsilon_0} - \frac{\lambda}{6\varepsilon_0} \right|$$ $$= \frac{\sigma}{6\varepsilon_0} \ldots (3)$$ Similarly, the electric field at $Q$ is given by $$E_Q = \left| \frac{\sigma}{2\varepsilon_0} - \frac{1}{4\pi\varepsilon_0} \frac{2\lambda}{4\pi} \right|$$ $$= \left| \frac{\sigma}{2\varepsilon_0} - \frac{\lambda}{8\varepsilon_0} \right|$$ $$= \frac{\sigma}{4\varepsilon_0} \ldots (4)$$ Divide equation (3) by equation (4) to obtain the required ratio. $$\frac{E_P}{E_Q} = \frac{\frac{\sigma}{6\varepsilon_0}}{\frac{\sigma}{4\varepsilon_0}}$$ $$= \frac{4}{6}$$ Hence, $a = 6$.

The formula to calculate the current flowing through an L-R circuit is given by $$i = \frac{E}{R} \left[ 1 - e^{-\frac{Rt}{L}} \right] \ldots (1)$$ The energy stored within the inductor can be calculated as follows: $$E_{induction} = \frac{1}{2} L i^2$$ $$= \frac{1}{2} L \frac{E^2}{R^2} \left[ 1 - e^{-\frac{Rt}{L}} \right]^2 \ldots (2)$$ The rate of change of energy stored within the inductor is given by $$\frac{d}{dt} (E_{induction}) = 2 \left[ 1 - e^{-\frac{Rt}{L}} \right] \cdot e^{-\frac{Rt}{L}} \cdot \frac{R}{L} \cdot \frac{E^2 L}{2R^2}$$ $$= \frac{E^2}{R} \left[ e^{-\frac{Rt}{L}} - e^{-\frac{2Rt}{L}} \right] \ldots (3)$$ For maximum rate, $$\frac{d^2 E}{dt^2} = 0 \ldots (4)$$ From equation (3) and (4), it can be written that $$\frac{d}{dt} \left[ \frac{E^2}{R} \left[ e^{-\frac{Rt}{L}} - e^{-\frac{2Rt}{L}} \right] \right] = 0$$ $$\Rightarrow e^{-\frac{Rt}{L}} = 2 e^{-\frac{2Rt}{L}}$$ $$\Rightarrow e^{\frac{Rt}{L}} = 2 \ldots (5)$$ From equation (3) and (5), it can be written, under maximum energy stored condition for the circuit, that $$\frac{d}{dt} (E_{induction}) = \frac{E^2}{R} \left[ 2^{-1} - 2^{-2} \right]$$ $$= \frac{E^2}{4R}$$ Substitute the values of the resistance and compare with the given expression to obtain the values of the parameters. $$a = 2$$ $$b = 2R$$ $$= 50$$ Hence, the required ratio is given by $$\frac{b}{a} = 25$$

Let the fish be at a depth $x$ from the surface and let the bird be at height $y$ from the surface. So, with respect to fish $$y' = \, \mu y + x$$ $$\Rightarrow \, y' = \frac{4y}{3} + x$$ Differentiating with respect to time, $$\frac{dy'}{dt} = \frac{4dy}{3dt} + \frac{dx}{dt}$$ $$\Rightarrow \frac{4y}{3} + \frac{dx}{dt} = -12$$ $$\Rightarrow \frac{4}{3} \cdot \frac{dy}{dt} = -12 + 8 = -4$$ $$\Rightarrow \frac{dy}{dt} = -3 \, \mathrm{m \, s^{-1}}$$

The angular momentum is given by $$L = I_{CM} \omega = \frac{2MR^2 \omega}{5}$$ The total kinetic energy is $$E = \frac{1}{2} \left( Mv^2 + \frac{2MR^2 \omega^2}{5} \right)$$ The ratio of angular momentum and kinetic energy is $$\frac{L}{E} = \frac{I_{CM} \omega}{\frac{1}{2} Mv^2 + \frac{2}{5} MR^2 \omega^2} = \frac{\frac{2}{5} MR^2 \omega}{\frac{7}{10} MR^2 \omega^2}$$ $$\Rightarrow \frac{\pi}{22} = \frac{4}{7} \cdot \frac{1}{\omega} \Rightarrow \omega = 4 \, \mathrm{rad} \, \mathrm{s}^{-1}$$

The formula to calculate the current gain ($\beta$) in CE configuration is given by $$\beta = \frac{\Delta I_C}{\Delta I_B} \ldots(1)$$ Considering the third and the first points on the given graph, the current gain can be calculated as follows: $$\beta = \frac{(30 - 10) \times 10^{-3} \, \mathrm{A}}{(300 - 100) \times 10^{-6} \, \mathrm{A}}$$ $$= 100$$ The formula to calculate the power gain ($P$) in CE mode can be written as $$P = \beta^2 \cdot \frac{R_C}{R_B} \ldots(2)$$ Substitute the known values of the parameters into equation (2) and solve to calculate the required unknown parameter. $$10^x = 100^2 \times \frac{1 \, \mathrm{k}\Omega}{10 \, \mathrm{k}\Omega}$$ $$= 1000$$ $$= 10^3$$ $$\Rightarrow x = 3$$

The mechanism of basic amide hydrolysis begins with the nucleophilic addition of a hydroxide ion at the carbonyl carbon creating a tetrahedral alkoxide intermediate. In the given case ring opening reaction takes place. On acidification, we get a secondary amino acid.

The difference between the zeolite and ion exchange process is that the zeolite process uses the mineral zeolite as the exchanging resin for cations in the hard water whereas the ion exchange process includes several different resins for the ion exchange. Synthetic resins method is more efficient as it can exchange both cations and anions.

The chlorophyll molecule has a central magnesium atom engulfed by a nitrogen-containing structure called a porphyrin ring; attached to the ring is a long carbon-hydrogen side chain. Hardness of water is determined by titrating with a standard solution of ethylene diamine tetra acetic acid (EDTA) which is a complexing agent. Since EDTA is insoluble in water, the disodium salt of EDTA is taken for this experiment. EDTA can form four or six coordination bonds with a metal ion. Silver bromide is used in photography as a component of an emulsion that helps develop a photographic image. Silver bromide is sensitive to light, and when suspended in gelatin, silver bromide's grains create a photographic emulsion. Wilkinson's catalyst is the common name for chloridotris(triphenylphosphine)rhodium(I), a coordination complex of rhodium with the formula $[(\mathrm{Ph}_3 \mathrm{P})_3 \mathrm{RhCl}]$. D-Penicillamine is chelating ligand.

The table shows the process, change in magnetic nature, and bond order change for different molecules. For the process $\mathrm{N_2 \rightarrow N_2^+}$, the change is from diamagnetic to paramagnetic with a bond order change from 3 to 2.5. For $\mathrm{NO \rightarrow NO^+}$, the change is from paramagnetic to diamagnetic with a bond order change from 2.5 to 3. For $\mathrm{O_2 \rightarrow O_2^{2-}}$, the change is from paramagnetic to diamagnetic with a bond order change from 2 to 1. For $\mathrm{O_2 \rightarrow O_2^+}$, the change is from paramagnetic to paramagnetic with a bond order change from 2 to 2.5.

The statement "It contains a banana bond" is incorrect about Borazine. Borazine, also known as inorganic benzene or borazole, is a chemical compound with the formula $\mathrm{B_3N_3H_6}$. Borazine is a hexagonal planar molecule with alternating boron and nitrogen atoms, and it exhibits some structural and chemical similarities to benzene. Borazine is an aromatic compound with electron delocalization. It does not contain banana bond. On heating, borazine hydrolyses slowly. $\mathrm{B_3N_3H_6 + 9H_2O \rightarrow 3NH_3 + 3H_3BO_3 + 3H_2O}$

The compound which possess aromatic character will show highest dipole moment. From the given compounds the following compound will have the highest dipole moment. Because the positive and negative ends acquire aromaticity.

Nylon 6 or polycaprolactam is a polymer, in particular semicrystalline polyamide. Vulcanization is a chemical process in which the rubber is heated with sulphur, accelerator and activator at $140$–$160 \, ^\circ \mathrm{C}$. The process involves the formation of cross-links between long rubber molecules to achieve improved elasticity, resilience, tensile strength, viscosity, hardness and weather resistance. Polychloroprene is the polymer name for the synthetic rubber known as neoprene. Natural rubber is the only addition polymer found in nature and is known as Cis - 1,4 - polyisoprene.

The reaction proceeds through a carbocation intermediate. The carbocation formed undergoes rearrangement by ring expansion. Four-membered ring undergoes ring expansion to a five-membered ring and the five-membered ring undergoes ring expansion to a six-membered ring and finally it converts to alkene by losing a proton.

Chlorofluorocarbons are chemicals which contain carbon, chlorine and fluorine. Ozone layer breaks when these CFC's are released into the atmosphere. So, chlorine is one of the radicals which belongs to CFC's family, responsible for breakdown.

When n-Hexane is heated with anhydrous $AlCl_3$ and $HCl$ gas at $573\ K$ under a pressure of about $30-35\ atm$ it Isomerises to give a branched chain alkanes. $CH_3(CH_2)_4CH_3\xrightarrow[\Delta]{\text{Anhy.}\ AlCl_3,\ HCl}(CH_3)_2CHCH_2CH_2CH_3+3\text{-methylpentane}$ $2\text{-methylpentane}$ Major product in the above reaction is $2\text{-methylpentane}$.

The reaction between methyl propyl bromide and $\mathrm{C_2H_5O^-}$ proceeds via an $\mathrm{S_N2}$ (substitution nucleophilic bimolecular) mechanism. In this mechanism, the nucleophile attacks the carbon atom bearing the bromine (the electrophilic carbon) in a single step, leading to the displacement of the bromine atom. $$(i) Br \xrightarrow{\mathrm{C_2H_5O^-}} \mathrm{OC_2H_5}$$ $\mathrm{C_2H_5O^-}$ is a strong nucleophile. The reaction between methyl propyl bromide and $\mathrm{C_2H_5OH}$ can proceed via either an $\mathrm{S_N1}$ (substitution nucleophilic unimolecular). Here $\mathrm{C_2H_5OH}$ is a weak nucleophile. During the reaction 1,2-hydrogen shift will occur and tert-butyl ethyl ether is formed. $$(ii) Br \xrightarrow{\mathrm{C_2H_5OH}} \mathrm{CH_2^+} \xrightarrow{1,2 - H shift} (+)$$ $$\mathrm{C_2H_5OH} \xrightarrow{-\mathrm{H^+}} \mathrm{OC_2H_5}$$ $\mathrm{C_2H_5OH}$ is a weak nucleophile.

The aldehyde group of glyceraldehyde reacts with hydrogen cyanide to give cyanohydrin. Two isomers are formed in the above step because the aldehyde is planar and changed to a tetrahedral center. On hydrolysis, the cyanide group converts to a carboxylic acid. Upon reaction with nitric acid, the primary alcoholic group undergoes oxidation to a carboxylic acid group. One compound is a meso compound, optically inactive, and the other compound is optically active.

Mond's process involves the combination of carbon monoxide with nickel readily and reversibly and gives nickel tetracarbonyl. This process is used to extract and purify nickel. The Van Arkel technique is used to refine Zirconium or Titanium. Impurities found in Zirconium or Titanium in the form of nitrogen and oxygen are eliminated using this technique. Copper is purified by electrolysis. Electricity is passed through solutions containing copper compounds, such as copper sulphate. The anode (positive electrode) is made from impure copper and the cathode (negative electrode) is made from pure copper. Mercury and zinc are purified by distillation method. Liquation is a metallurgical method to remove lead containing silver from copper and remove antimony minerals from ore.

Chlorine pentafluoride is an inter halogen compound. $\mathrm{ClF_5}$ is a colourless liquid. The central chlorine atom is having five bond pairs and one lone pair on it. Hence, according to VSEPR theory it has square pyramidal geometry. It is toxic by inhalation and an irritant to skin, eyes and mucus membranes.

$\mathrm{Be(OH)_2}$ has amphoteric nature. $\mathrm{Sr(OH)_2}$ has basic nature. These two compounds undergo acid-base reaction to get salt, $$\mathrm{Be(OH)_2 + Sr(OH)_2 \rightarrow Sr^{+2}}[Be(OH)_4]^{-2}$$ As Be is present in anionic part.

In the first step, protonation of alcohol takes place which leads to the formation of carbocation. Ring expansion takes place in the next step. In the given process, the five-membered ring becomes a six-membered ring. Now, the carbocation loses a proton in the presence of potassium hydroxide to give alkene.

A is incorrect. F and Cl belong to the same group of the periodic table. The electron gain enthalpy usually becomes less negative on moving down a group. However, in this case, the value of the electron gain enthalpy of Cl is more negative than that of F. This is because the atomic size of F is smaller than that of Cl. B is correct (ionisation energy decreases down the group). C is incorrect (Electronegativity is a fixed property of an element and is typically measured on scales such as the Pauling scale or the Mulliken scale.). D is incorrect. The oxides which behave as both acidic and basic oxides are called amphoteric oxides. Here $\mathrm{Al_2 O_3}$ is amphoteric oxide and NO is neutral oxide. Hence, A, C and D are incorrect.

The energy of an electron in the $n^{th}$ Bohr orbit of a hydrogen atom can be calculated using the formula: $$E_n = \frac{-2.18 \times 10^{-18} Z^2}{n^2}$$ Where $n$ is the principal quantum number of the Bohr orbit. $$E_n \propto \frac{1}{n^2}$$ $$E_n \propto \frac{1}{9}$$ Therefore, the energy of the electron in the third Bohr orbit of a hydrogen atom is three times of this value.

Film of lyophilic sol is formed over lyophobic sol as lyophilic sol acts as a protective colloid. Lyophobic solutions are unstable in nature. When a lyophilic sol is added to lyophobic sol, stability of lyophobic sol increases due to a protective film of lyophilic sol present over the lyophobic sol.

Europium has an atomic number of 63 and a stable electronic configuration of $[\mathrm{Xe}] 4f^7 6s^2$. Ytterbium, on the other hand, has an atomic number of 70 and a stable electronic configuration of $[\mathrm{Xe}] 4f^{14} 6s^2$. Both have high 3rd ionisation energy due to half filled & fully filled configuration respectively.

There are two tertiary carbocations possible due to rearrangement of hydride or methyl groups. Totally four structural alkene products are formed, out of which one is having chiral centre. Hence, total number of products formed through the formation of carbocation intermediate are 5.

The pressure which applied externally in the solution side to prevent the osmosis process is known as osmotic pressure. It can be denoted by $\pi$. Since, $\left( \pi_1 \right)_{non electrolyte} = \left( \pi_2 \right)_{glucose}$ $$C_1 \, RT = C_2 \, RT$$ $$\frac{W_1}{M_1} = \frac{W_2}{M_2}$$ Empirical mass of $$\frac{12}{M_1} = 0.05$$ $$M_1 = \frac{12}{0.05} = 240$$

Given: Volume of $\mathrm{Ba(NO_3)_2}$ solution = $25.0 \, \mathrm{mL}$, Concentration of $\mathrm{Ba(NO_3)_2}$ solution = $0.050 \, \mathrm{M}$, Volume of $\mathrm{NaF}$ solution = $25.0 \, \mathrm{mL}$, Concentration of $\mathrm{NaF}$ solution = $0.020 \, \mathrm{M}$. $$[\mathrm{Ba^{2+}}] = \left( \frac{25 \times 0.05}{50} \right) = 0.025 \, \mathrm{M}$$ $$[\mathrm{F^-}] = \left( \frac{25 \times 0.02}{50} \right) = 0.01 \, \mathrm{M}$$ $$[\mathrm{Ba^{2+}}][\mathrm{F^-}] = 25 \times 10^{-7}$$ $$K_{sp} = 0.5 \times 10^{-6} = 5 \times 10^{-7}$$ $$Ratio = \frac{[\mathrm{Ba^{2+}}][\mathrm{F^-}]}{K_{sp}} = \frac{25 \times 10^{-7}}{5 \times 10^{-7}} = 5$$

First write balanced reaction for the formation of one mole of AB with the help of diatomic molecule: $$\frac{1}{2} \mathrm{A_2} + \frac{1}{2} \mathrm{B_2} \rightarrow \mathrm{AB} \Delta H_f = -200 \, \mathrm{kJ/mol}$$ Now, $$\Delta H_{reaction} = \frac{1}{2} \Delta H_{\mathrm{A-A}} + \frac{1}{2} \Delta H_{\mathrm{B-B}} - \Delta H_{\mathrm{AB}} (let, \Delta H_{\mathrm{AB}} = x)$$ $$\frac{1}{2} x + \frac{1}{2} \left(0.5x\right) - x = -200 \Rightarrow \frac{x}{2} + 0.25x - x = -200$$ $$\Rightarrow -0.25x = -200 \Rightarrow x = 800 \, \mathrm{kJ/mol}$$ Bond enthalpy = 800 kJ/mol.

Moles of $\mathrm{CO_2} = \frac{0.22}{44} = \frac{1}{200}$. Then, Mass of $\mathrm{C} = \frac{0.220}{44} \times 12 = 0.06 \, \mathrm{g}$. $\%$ of $\mathrm{C} = \frac{0.06}{W} \times 100 = 24$ (W = Wt. of organic compound) $W = 0.25$ Moles of $\mathrm{H_2O} = \frac{0.126}{18} = 0.007$. Moles of $\mathrm{H} = 0.007 \times 2 = 0.014 \, \mathrm{g}$. $\%$ of Hydrogen = $\frac{0.014 \times 1}{W} \times 100$ $$= \frac{0.014 \times 1}{0.25} \times 100$$ $$= 5.6$$ $$= 56 \times 10^{-1}$$

Titration principle is $N_1 V_1 = N_2 V_2$. $N =$ Normality $V =$ Volume of solution $M_1 V_1 n_1 = M_2 V_2 n_2$ $M =$ molarity $n =$ Valence factor of acid or base As pink colour is obtained on consumption of $10 \, \mathrm{ml}$ of $\mathrm{Ca(OH)_2}$, we have $$20 \times 0.5 \times 1 = 10 \times M \times 2$$ Molarity of $\mathrm{Ca(OH)_2} = \frac{10}{20} = \frac{1}{2} M$ Also, moles of $\mathrm{Ca(OH)_2} = \mathrm{mm}$ of $\mathrm{H_2SO_4}$ $$\frac{1}{2} \times 20 = 10 \times M$$ Molarity of $\mathrm{H_2SO_4} = 1 \, \mathrm{M}$

The compressibility factor (Z) is given as: $$Z = \frac{PV}{nRT}$$ $$n = \frac{PV}{ZRT}$$ Given $Z_1 = 1.07$, $P_1 = 100 \, atm$, $V_1 = 0.15 \, L$, $T_1 = 500 \, K$ $Z_2 = 1.4$, $P_1 = 300 \, atm$, $V_2 = ?$, $T_2 = 300 \, K$ The relation is: $$\frac{z_1}{z_2} = \frac{P_1 V_1 \times T_2}{T_1 \times P_2 V_2}$$ Substituting the values: $$\frac{1.07}{1.4} = \frac{100 \times 0.15 \times 300}{500 \times 300 \times V_2}$$ Solving for $V_2$: $$V_2 = 0.03925 \, dm^3$$

The time for completion of the reaction and concentration of reactant relation is as follows, $$A_t = A_0 \times \frac{12.5}{100} = \frac{A_0}{8} \left[ 87.5\% completed \right]$$ $$A_0 \xrightarrow{t_{1/2}} \frac{A_0}{2} \xrightarrow{t_{1/2}} \frac{A_0}{4} \xrightarrow{t_{1/2}} \frac{A_0}{8}$$ $$t_{87.5} = 3t_{1/2}$$ So, $x = 3$

By balancing redox reaction: 10[$FeSO_4$. $($NH_4$)_2$ $SO_4$. 6$H_2O$] + 2 $KMnO_4 + 8 $H_2SO_4 $\rightarrow$ 5 $Fe_2(SO_4)_3$ + 2 $MnSO_4$ + 10 $($NH_4$)_2$ $SO_4$ + $K_2SO_4$ + 68 $H_2O$ $$8\mathrm{H^+} + \mathrm{MnO_4^-} + 5 \mathrm{Fe^{+2}} \rightarrow 5 \mathrm{Fe^{+3}} + \mathrm{Mn^{+2}} + 4\mathrm{H_2O}$$ For 2 moles of $\mathrm{MnO_4^-}$, 8 moles of $\mathrm{H_2O}$ are released. 5 moles of Mohr's salt contain 30 moles of $\mathrm{H_2O}$. So, 10 moles of Mohr's salt contain 60 moles of $\mathrm{H_2O}$. So, total moles of $\mathrm{H_2O}$ produced = 68

One equivalent of substance deposited for 96500C (one Faraday). $i \times t$ Columbus of charge can be utilised to get $x$ g of substance. Millimoles of Ni coated $$= \frac{(2x)100}{96500 \times 2}$$ $$= \frac{10x}{965}$$ Mass of Ni $= \left(\frac{x}{965}\right) 600$ $$= \frac{600x}{965} mg$$ Volume $= \frac{60x}{965} \times 10^{-3} cm^3$ $$= 100 \times 10^{-4}$$ $$\Rightarrow x = 0.1608 \times 10^3$$ $$\approx 161$$