JEE Main 8 April 2023 Shift 1 question paper with solutions

Given, $$I(x) = \int \frac{x+1}{x(1+xe^x)^2} \, dx$$ Now let $1 + xe^x = t$ which implies $$e^x(x+1)dx = dt$$ So, $$I(x) = \int \frac{1}{(t-1)t^2} \, dt$$ $$= \int \frac{(1-t^2) + t^2}{(t-1)t^2} \, dt$$ $$= \int \frac{- (t+1)}{(t-1)t^2} \, dt$$ $$= \int \frac{t^2}{t} - \frac{1}{t^2} + \frac{1}{t-1} \, dt$$ $$= \int \frac{-1}{t} - \frac{1}{t^2} + \frac{1}{t-1} \, dt$$ $$= - \ln t + \frac{1}{t} + \ln(t-1) + C$$ $$I(x) = \ln \left( \frac{xe^x}{xe^x+1} \right) + \frac{1}{xe^x+1} + C$$ Also given, $$\lim_{x \to \infty} I(x) = 0$$ $$\Rightarrow \lim_{x \to \infty} I(x) = \lim_{x \to \infty} \left[ \ln \left( 1 - \frac{1}{xe^x+1} \right) + \frac{1}{xe^x+1} + C \right]$$ $$\Rightarrow \lim_{x \to \infty} I(x) = [\ln(1-0)+0+C]$$ $$\Rightarrow C = 0$$ Now finding, $$I(1) = \ln \left( \frac{e}{e+1} \right) + \frac{1}{e+1} = 1 + \frac{1}{e+1} - \ln(e+1)$$ $$I(1) = \frac{e+2}{e+1} - \ln(e+1)$$

Given, the equation of the plane containing the line $x + 2y + 3z - 4 = 0 = 2x + y - z + 5$ and perpendicular to the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda (\hat{i} + \hat{j} + \hat{k}) + \mu (\hat{i} - 2\hat{j} + 3\hat{k})$ is $ax + by + cz = 4$. Now let the equation of the required plane be, $$P : (x + 2y + 3z - 4) + \alpha (2x + y - z + 5) = 0$$ $$\Rightarrow (2\alpha + 1) + (\alpha + 2)y + (3 - \alpha)z = 4 - 5\alpha$$ So, the normal vector is given by, $$\vec{n}_1 = (2\alpha + 1) \hat{i} + (\alpha + 2) \hat{j} + (3 - \alpha) \hat{k}$$ And the normal of the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda (\hat{i} + \hat{j} + \hat{k}) + \mu (\hat{i} - 2\hat{j} + 3\hat{k})$ will be, $$\vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = 5\hat{j} - 2\hat{j} - 3\hat{k}$$ Now given they are perpendicular, so by perpendicular condition we get, $$\vec{n}_1 \cdot \vec{n}_2 = 0$$ $$\Rightarrow 5(2\alpha + 1) - 2(\alpha + 2) - 3(3 - \alpha) = 0$$ $$\Rightarrow 11\alpha + (-8) = 0$$ $$\Rightarrow \alpha = \frac{8}{11}$$ Hence, the equation of the required plane will be, $$P : \frac{27}{11}x + \frac{30}{11}y + \frac{25}{11}z - \frac{4}{11} = 0$$ $$\Rightarrow 27x + 30y + 25z = 4$$ Now comparing with $ax + by + cz = 4$ we get, $a = 27$, $b = 30$, $c = 25$. Hence, $a - b + c = 22$.

Given, $\nR$ be the focus of the parabola y^2 = 20x and the line y = mx + c intersect the parabola at two points P and Q, $\nAnd$ the points G(10, 10) be the centroid of the triangle PQR, $\nNow$ focus of the parabola y^2 = 20x will be, R(5, 0) $\nAnd$ parametric points of PQ be (5t^2, 10t), $\nNow$ plotting the diagram we get, $\nNow$ finding the centroid of the triangle we get, $\nFor$ x--coordinate we get, $\n$$\Rightarrow$ $\frac{5t_1^2 + 5t_2^2 + 5}{3}$ = 10 $\n$$\Rightarrow$ t_1^2 + t_2^2 = 5 $\cdots$ (i) $\nNow$ for y--coordinate we get, $\n$$\frac{10(t_1 + t_2)}{3}$ = 10 $\n$$\Rightarrow$ t_1 + t_2 = 3 $\cdots$ (ii) $\nNow$ solving both equations we get, t_1 = 1, $\ $t_2 = 2 $\nSo$, points will be, P $\equiv$ (5, 10) and Q $\equiv$ (20, 20) $\nHence$, equation of PQ = y - 10 = $\frac{10}{15}$(x - 5) $\n$$\Rightarrow$ 3y - 30 = 2x - 10 $\n$$\Rightarrow$ y = $\frac{2}{3}$x + $\frac{20}{3}$, so on comparing with y = mx + c, we get c - m = 6 $\nHence$, PQ^2 = 225 + 100 = 325

Given, $C(\alpha, \beta)$ be the circumcentre of the triangle formed by the lines $4x + 3y = 69$, $4y - 3x = 17$, and $x + 7y = 61$. Now plotting the diagram of the above equations and finding the intersection point we get, Now from the diagram we can see that it is a right-angled triangle, so the circumcentre is given by the midpoint of the hypotenuse. Hence, circumcentre will be, $$C\left( \frac{12+5}{2}, \frac{7+8}{2} \right)$$ $$\Rightarrow C\left( \frac{17}{2}, \frac{15}{2} \right) = (\alpha, \beta)$$ Hence, $$(\alpha - \beta)^2 + \alpha + \beta$$ $$= \left( \frac{17}{2} - \frac{15}{2} \right)^2 + \frac{17}{2} + \frac{15}{2}$$ $$= 1 + 16 = 17$$

Given: $$P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$$ Therefore, $$PP^T = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ Thus, $$PP^T = I$$ Now, $$P^T (PAP^T)^{2007} P = P^T (PAP^T)(PAP^T)(PAP^T) \ldots (PAP^T)P$$ $$\Rightarrow P^T (PAP^T)^{2007} P = A^{2007}$$ Now, $$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ $$A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$$ $$A^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$$ $\[$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\]$ $$A^{2007} = \begin{bmatrix} 1 & 2007 \\ 0 & 1 \end{bmatrix}$$ So, $$\Rightarrow P^T (PAP^T)^{2007} P = \begin{bmatrix} 1 & 2007 \\ 0 & 1 \end{bmatrix}$$ $$\Rightarrow P^T Q P = \begin{bmatrix} 1 & 2007 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ So, $a = 1$, $b = 2007$, $c = 0$, $d = 1$ $2a + b + 3c - 4d = 2 + 2007 - 4 = 2005$

Given that roots of $x^3 + bx + c = 0$ are $\alpha$, $\beta$, $\gamma$. Also given that $\beta \gamma = 1 = -\alpha$ implies $\alpha = -1$. (i) Now let us apply the relation between the roots of the cubic equation with the coefficients. $\Rightarrow \alpha + \beta + \gamma = 0$. (ii) $\Rightarrow \alpha \beta \gamma = -c$. (iii) $\Rightarrow (-1)(1) = -c$ Therefore, $c = 1$. (iv) On substituting the value of $\alpha = -1$ in eq (ii) we get, $\beta + \gamma = 1$. (v) Therefore, $\alpha \beta + \beta \gamma + \gamma \alpha = b$ implies $\alpha(\beta + \gamma) + \beta \gamma = b$. Therefore, $b = 0$. (vi) Hence, equation will be, $$x^3 + 1 = 0$$ Whose roots will be, $-1$, $-\omega$, $-\omega^2$. We know that $1 + \omega + \omega^2 = 0$ implies $\alpha + \beta + \gamma = 0$ implies $-1 + \beta + \gamma = 0$. Therefore, $\beta = -\omega$, $\gamma = -\omega^2$. (vii) $\Rightarrow \beta^3 = -\omega^3 = -1$ and $\Rightarrow \gamma^3 = -\omega^6 = -1$. Hence the value of $b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3$ is $$= 0 + 2 + 3 + 6 + 8 = 19$$ Hence this is the correct option.

Given, 7 boys and 5 girls are to be seated around a circular such that no two girls to be seated together, Now we know that $n$ objects can be arranged in a circle in $(n - 1)!$ ways. Let us first arrange 7 boys in circular arrangement in $(7 - 1)!$ ways. Now there will be 7 gaps. So let us select any 5 gaps out of 7 gaps and arrange 5 girls in the chosen gaps. This can be done in $^7C_5 \times 5!$ ways. Hence, required arrangements are $6! \times \binom{7 \times 6}{2} \times 5!$ $$= 6 \times 5! \times \frac{7 \times 6}{2} \times 5!$$ $$= 126(5!)^2.$$ Therefore, required arrangements are $126(5!)^2$

Let $X \equiv$ Event that product is defective, then $$P\left( \frac{X}{A} \right) = \frac{3}{100}$$ $$P\left( \frac{X}{B} \right) = \frac{4}{100}$$ $$P\left( \frac{X}{C} \right) = \frac{2}{100}$$ Now, $$P\left( \frac{C}{X} \right) = \frac{P(C) P\left( \frac{X}{C} \right)}{P(A) P\left( \frac{X}{A} \right) + P(B) P\left( \frac{X}{B} \right) + P(C) P\left( \frac{X}{C} \right)}$$ $$\Rightarrow P\left( \frac{C}{X} \right) = \frac{\frac{20}{100} \times \frac{2}{100}}{\frac{30}{100} \times \frac{3}{100} + \frac{40}{100} \times \frac{4}{100} + \frac{50}{100} \times \frac{2}{100}}$$ $$\Rightarrow P\left( \frac{C}{X} \right) = \frac{\frac{100}{100}}{60 + 120 + 100}$$ $$\Rightarrow P\left( \frac{C}{X} \right) = \frac{5}{14}$$ Hence this is the correct option.

There are 5 vowels in the given word which are 4 $E$'s and 1 $I$. Since they have to always occur together we take them as a single object $E E E E I$ for the time being. This single object together with 7 remaining objects will account for 8 objects. There are 8 objects in which there are 3 $N$'s and 2 $D$'s can be arranged in $$\frac{8!}{3!2!}$$ ways. Corresponding to each of their arrangements the 5 vowels $E, E, E, E$ and $I$ which can be arranged in $$\frac{5!}{4!}$$. Hence, required number of arrangements. $$= \frac{8!}{3!2!} \times \frac{5!}{4!} = 16800$$ Hence this is the correct option.

Give that: $$f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$$ Convert the numerator and denominator in the form of $\sin(A \pm B)$. $$\Rightarrow f(x) = \frac{\sqrt{2} \sin \left( x + \frac{\pi}{4} \right) - \sqrt{2}}{\sqrt{2} \left( \sin \left( x - \frac{\pi}{4} \right) \right)}$$ $$\Rightarrow f(x) = \frac{\sin \left( x + \frac{\pi}{4} \right) - 1}{\sin \left( x - \frac{\pi}{4} \right)}$$ $$\Rightarrow f \left( x + \frac{\pi}{4} \right) = \frac{\cos x - 1}{\sin x}$$ $$\Rightarrow f \left( x + \frac{\pi}{4} \right) = -\tan \frac{x}{2}$$ $$\Rightarrow f(x) = -\tan \left( \frac{x}{2} - \frac{\pi}{8} \right)$$ $$\Rightarrow f'(x) = -\frac{1}{2} \sec^2 \left( \frac{x}{2} - \frac{\pi}{8} \right)$$ $$\Rightarrow f''(x) = -\frac{1}{2} \left( \sec^2 \left( \frac{x}{2} - \frac{\pi}{8} \right) \right) \tan \left( \frac{x}{2} - \frac{\pi}{8} \right)$$ $$\Rightarrow f \left( \frac{7\pi}{12} \right) = -\tan \left( \frac{7\pi}{24} - \frac{\pi}{8} \right) = -\tan \left( \frac{4\pi}{24} \right)$$ $$\Rightarrow -\tan \left( \frac{\pi}{6} \right) = -\frac{1}{\sqrt{3}}$$ Also, $$\Rightarrow f'' \left( \frac{7\pi}{12} \right) = -\frac{1}{2} \left( \frac{2}{\sqrt{3}} \right)^2 \times \frac{1}{\sqrt{3}} = -\frac{2}{3\sqrt{3}}$$ $$\Rightarrow f \left( \frac{7\pi}{12} \right) f'' \left( \frac{7\pi}{12} \right) = \frac{2}{9}$$ Hence this is the correct option.

The given position vectors can be written as - $A(\alpha, 10, 13)$ - $B(6, 11, 11)$ - $C\left(\frac{9}{2}, \beta, -8\right)$ Also given that these three points are collinear. Let us assume that the point $B$ divides $AB$, $BC$ in the ratio $k : 1$. Since, $A$, $B$, $C$ are collinear On applying section formula for $z$ coordinate we get, $$11 = \frac{-8k + 13}{k + 1}$$ $$\Rightarrow 11k + 11 = -8k + 13$$ $$\Rightarrow 19k = 2$$ $$\Rightarrow k = \frac{2}{19}$$ Therefore, Ratio $= 2 : 19$ $$\frac{\alpha \times 19 + \left(\frac{9}{2}\right) \times 2}{2 + 19} = 6$$ Now, $$\Rightarrow 19\alpha = 117$$ $$\Rightarrow \alpha = \frac{117}{19}$$ Now similarly, $$\frac{2\beta + 190}{21} = 11$$ $$\Rightarrow \beta = \frac{41}{2}$$ Therefore, $$(19\alpha - 6\beta)^2 = (117 - 123)^2 = 36$$ Hence this is the correct option.

Given, $nC_{r-1} : nC_r : nC_{r+1} = 1 : 5 : 20$. Now using the formula $\frac{nC_r}{nC_{r-1}} = \frac{n-r+1}{r}$, we get, $$\frac{nC_r}{nC_{r-1}} = \frac{n-r+1}{r} = \frac{5}{1}$$ $$\Rightarrow n - 6r + 1 = 0 \ldots \ldots \ldots (1)$$ And $$\frac{nC_{r+1}}{nC_r} = \frac{n-r}{r+1} = \frac{20}{5}$$ $$\Rightarrow n - 5r - 4 = 0 \ldots \ldots \ldots (2)$$ Solving above equations we get, $n = 29$ and $r = 5$. So, the coefficient of forth term will be $nC_3 = \binom{29}{3} = \frac{29 \times 28 \times 27}{3 \times 2 \times 1} = 3654$.

Given that $S_K = \frac{1+2+\ldots+K}{K}$. We know that $1 + 2 + 3 \ldots + n = \frac{n(n+1)}{2}$. Therefore, $$S_K = \frac{K(K+1)}{2K} = \frac{K+1}{2}.$$ Now, $$\sum_{j=1}^{n} (S_j)^2 = \sum_{j=1}^{n} \frac{1}{4} \left( 2^2 + 3^2 + 4^2 \ldots (n+1)^2 \right)$$ $$= \sum_{j=1}^{n} \frac{1}{4} \left( 1^2 + 2^2 + 3^2 + 4^2 \ldots (n+1)^2 - 1^2 \right)$$ $$= \frac{1}{4} \left( \frac{(n+1)(n+2)(2n+3)}{6} - 1 \right)$$ $$= \frac{1}{4} \left( \frac{(n^2+3n+2)(2n+3)-6}{6} \right)$$ $$= \frac{1}{4} \left( \frac{2n^3+6n^2+4n+3n^2+9n+6-6}{6} \right)$$ $$= \frac{1}{4} \left( \frac{2n^3+9n^2+13n}{6} \right)$$ $$= \frac{n}{24} (2n^2 + 9n + 13)$$ On comparing this with $\sum_{j=1}^{n} S_j^2 = \frac{n}{A} (Bn^2 + Cn + D)$ we get, $A = 24$, $B = 2$, $C = 9$, $D = 13$. Thus, $$\frac{A+B}{D} = \frac{26}{13} = 2.$$ That means, $A + B$ is divisible by $D$. Hence this is the correct option.

Given that $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$ and $|adj(adj^2 A)| = 16^n$. We know that $|adj A| = |A|^{n-1}$ and $|adj(adj A)| = |A|^{(n-1)^2}$. Similarly, $|adj(adj adj A)| = |A|^{(n-1)^3}$ where $n$ is the order of the square matrix. Let us find $|A|$. $$|A| = 2(2 \times 2 - (-1)(-1)) - 1(1 \times 2 - 0(-1)) + 0(1 \times (-1) - 0 \times 2)$$ $$= 2(3) - (2) + 0 = 4$$ $$\Rightarrow |2A|^{(n-1)^3} = |2A|^{(3-1)^3}$$ $$= |2A|^8 = (2^8)^3 |A|^8$$ $$\Rightarrow 2^{24} \times 4^8 = 16^n$$ $$\Rightarrow 16^6 \times 16^4 = 16^n$$ $$\Rightarrow 16^{10} = 16^n$$ $$\Rightarrow n = 10$$ Therefore, the value of $n$ is 10.

We know that, \[ A \to B \equiv \neg A \vee B \] Now using the above formula we will solve, \[ (p \to q)\to (q \to p) \] \[ = (\neg p \vee q)\vee(\neg q \vee p) \] \[ = (p \wedge \neg q)\vee(\neg q \vee p) \] \[ = q \vee p \] Now finding the negation of \[ (p \vee \neg q) \] we get, \[ \neg(p \vee \neg q) \] \[ = \neg p \wedge q \]

Given: $$\frac{x-4}{4} = \frac{y+2}{5} = \frac{z+3}{3}$$ $$\frac{x-1}{3} = \frac{y-3}{4} = \frac{z-4}{2}$$ So, $$\vec{a}_1 = 4\hat{i} - 2\hat{j} - 3\hat{k}$$ $$\vec{a}_2 = \hat{i} + 3\hat{j} + 4\hat{k}$$ So, $$\vec{a}_2 - \vec{a}_1 = -3\hat{i} + 5\hat{j} + 7\hat{k}$$ And, $$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix} = -2\hat{i} + \hat{j} + \hat{k}$$ $$\Rightarrow \left|\vec{b}_1 \times \vec{b}_2\right| = \sqrt{(-2)^2 + (1)^2 + (1)^2} = \sqrt{6}$$ Shortest distance between the lines $$= \left|\frac{\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|$$ $$= \left|\frac{\left(-3\hat{i} + 5\hat{j} + 7\hat{k}\right) \cdot \left(-2\hat{i} + \hat{j} + \hat{k}\right)}{\sqrt{6}}\right|$$ $$= \left|\frac{6 + 5 + 7}{\sqrt{6}}\right| = \frac{6 \times 3}{\sqrt{6}} \text{ units}$$ $$= 3\sqrt{6} \text{ units}$$ Hence this is the correct option.

The given curves are $x^2 \leq y$, $y \leq 8 - x^2$ and $y \leq 7$. The point of intersection of the curves $x^2 \leq y$ and $y \leq 8 - x^2$ is obtained by, $$x^2 = 8 - x^2$$ $$\Rightarrow x = \pm 2 and y = 4.$$ Hence the points are $(2, 4), (-2, 4)$. The points of intersection of the curves $y \leq 8 - x^2$ and $y \leq 7$ is obtained by, $$8 - x^2 = 7$$ $$\Rightarrow x = \pm 1 and y = 7.$$ Hence the points are $(1, 7), (-1, 7)$. The required graph is The required area is symmetrical about $y$-axis. Hence required area is $$A = 2 \left[ \int_0^4 \sqrt{y} \, dy + \int_4^7 \sqrt{8 - y} \, dy \right]$$ $$A = 2 \left[ \left[ \frac{y^{3/2}}{3/2} \right]_0^4 + \left[ \frac{(8-y)^{3/2}}{-3/2} \right]_4^7 \right]$$ $$= \frac{4}{3} \left[ (4)^{3/2} - (0)^{3/2} - \left\{ (8 - 7)^{3/2} - (8 - 4)^{3/2} \right\} \right]$$ $$= \frac{4}{3} (8 - 0 - 1 + 8)$$ $$= 20 sq units.$$ Hence, the required area is 20 sq units.

Given, the number of elements in sets $A$ and $B$ be five and two respectively. So, $n(A) = 5$ and $n(B) = 2$. Now Cartesian product will be, $n(A \times B) = 10$. Number of subsets having three elements $= \binom{10}{3}$. Number of subsets having four elements $= \binom{10}{4}$. Number of subsets having five elements $= \binom{10}{5}$. Number of subsets having six elements $= \binom{10}{6}$. So, number of subsets having at least 3 and at most 6 elements will be $= \binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6}$. $$= 120 + 210 + 252 + 210$$ $$= 792$$

Given, $$ \lim_{x \to 0} \left( \frac{\left(1 - \cos^2(3x)\right)}{\cos^3(4x)} \right) \left( \frac{\sin^3(4x)}{(\log_e(2x+1))^5} \right) $$ Now we know that, $$ \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \& \lim_{x \to 0} \frac{\log(1+x)}{x} = 1 $$ Now using the above formula we get, $$ \lim_{x \to 0} \left( \frac{1 - \cos^2 3x}{\cos^3 4x} \right) \left( \frac{\sin^3 4x}{(\log_e(2x+1))^5} \right) $$ $$ = \lim_{x \to 0} \frac{(1 - \cos 3x) \left(1 + \cos 3x\right) 9x^2 \left(\sin 4x\right)^3 \left(2x\right)^5}{(\cos^3 4x) 9x^2 \left(64x^3\right) \left(\log_e(2x+1)\right)^5 \left(2x\right)^5} $$ $$ = \lim_{x \to 0} \frac{\left(\frac{1 - \cos 3x}{9x^2}\right) \left(1 + \cos 3x\right) \left(\frac{\sin 4x}{4x}\right)^3 \left(\frac{\log_e(2x+1)}{2x}\right)^5 \times \frac{9 \times 64}{32}}{(\cos^3 4x)} $$ $$ = \left(\frac{1}{2}\right) \times (2) \left(1\right)^3 \left(1\right)^5 \times \frac{9 \times 64}{32} $$ $$ = \frac{9 \times 2}{1} = 18 $$

Given, $z = \alpha + i \beta$ and $|z + 2| = z + 4(1 + i)$ Now putting the value of $z = \alpha + i \beta$ in $|z + 2| = z + 4(1 + i)$ we get, $$|z + 2| = z + 4(1 + i)$$ $$\Rightarrow \sqrt{(\alpha + 2)^2 + \beta^2} = (\alpha + 4) + i(\beta + 4)$$ Now on comparing real and imaginary part, we get $$\sqrt{(\alpha + 2)^2 + \beta^2} = \alpha + 4 \ldots (i)$$ and $$\beta + 4 = 0 \Rightarrow \beta = -4 \ldots (ii)$$ Now solving, $$\sqrt{(\alpha + 2)^2 + 16} = \alpha + 4$$ $$\Rightarrow \alpha^2 + 4\alpha + 20 = \alpha^2 + 8\alpha + 16$$ $$\Rightarrow \alpha = 1$$ So, $\alpha + \beta = -3$, $\alpha \beta = -4$ We know that quadratic equation is given by, $$x^2 - (sum of roots)x + (product of roots) = 0$$ So, equation with roots $\alpha$ and $\beta$ will be, $$x^2 + 3x + 4 = 0$$

Let, $$I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left( 8[\csc x] - 5[\cot x] \right) dx$$ $$\Rightarrow I = 8 \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [\csc x] dx - 5 \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [\cot x] dx$$ $$\Rightarrow I = 8I_1 - 5I_2, where I_1 = 8 \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [\csc x] dx and I_2 = 5 \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [\cot x] dx$$ Now solving, $$I_1 = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [\csc x] dx$$ $$\Rightarrow I_1 = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 1 dx = \frac{2\pi}{3}$$ As when $x \in \left( \frac{\pi}{6}, \frac{5\pi}{6} \right)$, $\csc x \in \left[ 1, 2 \right)$, So, $[\csc x] = 1$ Now solving, $$I_2 = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [\cot x] dx$$ $$\Rightarrow I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 1 dx + \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 0 dx + \int_{\frac{3\pi}{4}}^{\frac{5\pi}{6}} (-1) dx + \int_{\frac{5\pi}{6}}^{\frac{3\pi}{4}} (-2) dx$$ $$\Rightarrow I_2 = \left( \frac{\pi}{4} - \frac{\pi}{6} \right) - \left( \frac{3\pi}{4} - \frac{\pi}{2} \right) - 2 \left( \frac{5\pi}{6} - \frac{3\pi}{4} \right)$$ $$\Rightarrow I_2 = -\frac{\pi}{3}$$ Required value will be, $$\frac{2}{\pi} I = \frac{2}{\pi} \left[ 8 \times \frac{2\pi}{3} + 5 \times \frac{\pi}{3} \right] = 14$$

The given expansion is $\left(3x^2 - \frac{1}{2x^5}\right)^7$. The general term in the binomial expansion of $(x + a)^n$ is given by $T_{r+1} = {}^{n}C_{r} x^{n-r} a^r$. Therefore, $$T_{r+1} = {}^{7}C_{r} (3x^2)^{7-r} \left(-\frac{1}{2x^5}\right)^r$$ $$= {}^{7}C_{r} (-1)^r \left(\frac{3^{7-r}}{2^r}\right) x^{14-2r-5r}$$ Now for term independent of $x$ implies $14 - 2r - 5r = 0$. Therefore, $r = 2$. Coefficient of $x^0$ is ${}^{7}C_{2} (-1)^2 \times \left(\frac{3^{7-2}}{2^2}\right)$. $$= \frac{7 \times 6}{2} \times \frac{3^5}{2^2}.$$ Therefore, $\alpha = \frac{5103}{4}$. Hence $[\alpha] = 1275$. Hence this is the required answer.

Given, $\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}$, $\vec{b} = \alpha \hat{i} + 11 \hat{j} - 2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$, $\vec{a} \times \vec{c} - \vec{a} \times \vec{b} = 0$. Therefore, $\vec{a} \times \left( \vec{c} - \vec{b} \right) = 0$. So, $\vec{a}$ and $\left( \vec{c} - \vec{b} \right)$ are parallel vectors. Hence, $\lambda \vec{a} = \vec{c} - \vec{b}$. Therefore, $\vec{c} = \vec{b} + \lambda \vec{a}$. Thus, $\vec{a} \cdot \vec{c} = \vec{a} \cdot \vec{b} + \lambda \left| \vec{a} \right|^2$. This implies $-12 = (6\alpha + 75) + \lambda (261)$. Therefore, $2\alpha + 87\lambda = -29 \ldots (i)$. Now again using $\vec{c} = \vec{b} + \lambda \vec{a}$ we get, $$\vec{c} = \hat{i}(\alpha + 6\lambda) + \hat{j}(11 + 9\lambda) + \hat{k}(-2 + 12\lambda)$$ Also given $\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$. This implies $(\alpha + 6\lambda) - 2(11 + 9\lambda) + (-2 + 12\lambda) = 5$. Therefore, $\alpha = 29$. So, $2\alpha + 87\lambda = -29$. Thus, $\lambda = -1$. Hence, $\vec{c} = 23\hat{i} + 2\hat{j} - 14\hat{k}$. So, the value of $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11$

We know that, maximum value of $n$ for $p^n$ where $p$ is prime divides the number $a!$ is given by, $$\left\lfloor \frac{a}{p} \right\rfloor + \left\lfloor \frac{a}{p^2} \right\rfloor + \left\lfloor \frac{a}{p^3} \right\rfloor + \left\lfloor \frac{a}{p^4} \right\rfloor + \ldots$$ Since, 3 is a prime number, so we have $$= \left\lfloor \frac{66}{3} \right\rfloor + \left\lfloor \frac{66}{3^2} \right\rfloor + \left\lfloor \frac{66}{3^3} \right\rfloor + \left\lfloor \frac{66}{3^4} \right\rfloor + \ldots$$ $$= 22 + 7 + 2 + 0$$ $$= 31$$ So, maximum value of $n$ is 31.

Let, $$y = \frac{x^3}{x^4 + 147} = f(x)$$ We know that, for increasing function $\frac{dy}{dx} > 0$ So, differentiating the given function $f(x) = \frac{x^3}{x^4 + 147}$ we get, $$\frac{dy}{dx} = \frac{-x^2 \left(x^4 - 441\right)}{\left(x^4 + 147\right)^2}$$ $$\Rightarrow \frac{-x^2 \left(x^4 - 441\right)}{\left(x^4 + 147\right)^2} > 0$$ $$\Rightarrow \frac{x^2 \left(x^4 - 441\right)}{\left(x^4 + 147\right)^2} f(4)$$ $$\Rightarrow \alpha = 5$$

Given, Set $A = \{10, 9, 8, 7, 6, 4, 3, 0\}$ Now relation $x - y$ is odd or $x - y = 2$ can be given by, $$R = \{(10, 9), (10, 8), (10, 7), (10, 3), (9, 7), (9, 6), (9, 4), (9, 0), (8, 7), (8, 6), (8, 3), (7, 6), (7, 4), (7, 0), (6, 4), (6, 3), (4, 3), (3, 0)\}$$ So, total there are 19 elements and all the elements of $R$, $(a, b)$ are of type $a > b$. Hence, we need to add total of 19 more elements to $R$ to make it symmetric.

Given, $C_1 : x^2 + y^2 - 4x - 2y + (5 - \alpha) = 0$ So, its centre will be, $O_1 = (2, 1)$ and radius $= \sqrt{\alpha}$ And $C_2 : 5x^2 + 5y^2 - 10fx - 10gy + 36 = 0$ $$\Rightarrow C_2 : x^2 + y^2 - 2fx - 2gy + \frac{36}{5} = 0$$ So, Centre $O_2 = (f, g)$ and radius $r = \sqrt{f^2 + g^2 - \frac{36}{5}}$ Also given $O_2$ is reflection of $O_1$ in $2x - y + 1 = 0$, so image formula we get, $$\Rightarrow \frac{f - 2}{2} = \frac{g - 1}{-1} = -2 \cdot \left(\frac{2 \times 2 - 1 + 1}{2^2 + 1^2}\right)$$ $$\Rightarrow f = \frac{-6}{5} and g = \frac{13}{5}$$ So, radius $r = \sqrt{\left(\frac{-6}{5}\right)^2 + \left(\frac{13}{5}\right)^2 - \frac{36}{5}} = \sqrt{\frac{25}{25}} = 1$ $$\Rightarrow$ $r = 1$ and $\alpha = 1$ as they both are same radius circle, Hence, $r + \alpha = 2.$

Given, $\left( y - 2 \log_e x \right) dx + \left( x \log_e x^2 \right) dy = 0, \; x > 1$ $\[$$\Rightarrow$ 2x $\ln$ x $\frac{dy}{dx}$ + y = 2 $\ln$ x $\;$ $\{$where $\log$_e x = $\ln$ x$\}$$\]$ $\[$$\Rightarrow$ $\frac{dy}{dx}$ + $\frac{y}{2x \ln x}$ = $\frac{1}{x}$$\]$ Which is linear differential equation, So, $I.F = e^{\int \frac{1}{2x \ln x}} = \sqrt{\ln x}$ Now, solution of the equation is given by, $\[$y $\cdot$ $\sqrt{\ln x}$ = $\int$ $\frac{\sqrt{\ln x}}{x}$ $\,$ dx$\]$ $\[$$\Rightarrow$ y $\cdot$ $\sqrt{\ln x}$ = $\frac{2}{3}$ ($\ln$ x)^{$\frac{3}{2}$} + C $\cdots$ (i)$\]$ Given, eq. (i) passes through point $\left( e, \frac{4}{3} \right)$ So, $C = \frac{2}{3}$ Hence, solution of differential equation will be, $\[$y $\sqrt{\ln x}$ = $\frac{2}{3}$ ($\ln$ x)^{$\frac{3}{2}$} + $\frac{2}{3}$$\]$ Also given, above equation passes through point $\left( e^4, \alpha \right)$ $\[$$\alpha$ $\sqrt{\ln e^4}$ = $\frac{2}{3}$ ($\ln$ e^4)^{$\frac{3}{2}$} + $\frac{2}{3}$$\]$ $\[$$\Rightarrow$ 2 $\alpha$ = $\frac{2}{3}$ (4)^{$\frac{3}{2}$} + $\frac{2}{3}$$\]$ $\[$$\Rightarrow$ $\alpha$ = $\frac{1}{3}$ $\times$ 8 + $\frac{1}{3}$$\]$ $\[$$\Rightarrow$ $\alpha$ = 3$\]$

Given, $\lambda_1$, $\lambda_2$ be the values of $\lambda$ for which the points $\left( \frac{5}{2}, 1, \lambda \right)$ and $(-2, 0, 1)$ are at equal distance from the plane $2x + 3y - 6z + 7$, So, by using distance formula of a point from a plane we get, $$\left| \frac{-4 \cdot 0 - 6 \cdot 7 + \lambda}{\sqrt{2^2 + 3^2 + 6^2}} \right| = \left| \frac{5 + 3 - 6 \cdot \lambda + 7}{\sqrt{2^2 + 3^2 + 6^2}} \right|$$ $$\Rightarrow \left| \frac{4 + 0 - 6 \cdot 7}{7} \right| = \left| \frac{5 + 3 - 6 \cdot \lambda + 7}{7} \right|$$ $$\Rightarrow \frac{3}{7} = \frac{15 - 6 \lambda}{7}$$ $$\Rightarrow \lambda = 2 or 3$$ $$\Rightarrow \lambda_1 = 3, \lambda_2 = 2 as \lambda_1 > \lambda_2$$ So, $(\lambda_1 - \lambda_2, \lambda_2, \lambda_1) = (1, 2, 3)$ Now finding the distance of point $(1, 2, 3)$ from the line $\frac{x - 5}{1} = \frac{y - 1}{2} = \frac{z + 7}{2} = \lambda$ we get, Now from diagram we can see that, $\overrightarrow{PM}$ is perpendicular to given line, So, by perpendicular condition we get, $$\overrightarrow{PM} \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k}) = 0$$ $$\Rightarrow (\lambda + 4) + 2(2\lambda - 1) + 2(2\lambda - 10) = 0$$ $$\Rightarrow 9 \lambda = 18 or \lambda = 2$$ Hence, the point $M$ will be $M(7, 5, -3)$ So, the distance $\left| \overrightarrow{PM} \right| = \sqrt{6^2 + 3^2 + 6^2} = 9$

Given, the data $x, y, 10, 12, 4, 6, 8, 12$. So, Mean $$= \frac{x+y+10+12+4+6+8+12}{8}$$ $$\Rightarrow 9 = \frac{x+y+52}{8}$$ $$\Rightarrow x+y+52 = 72$$ $$\Rightarrow x+y = 20$$ And, Variance $$= \left( \frac{\sum x_i^2}{n} \right) - \left( \frac{\sum x_i}{n} \right)^2$$ $$\Rightarrow 9.25 = \left( \frac{x^2+y^2+100+144+16+36+64+144}{8} \right) - (9)^2$$ $$\Rightarrow x^2 + y^2 = 218$$ $$\Rightarrow (x+y)^2 - 2xy = 218$$ $$\Rightarrow 20^2 - 2xy = 218$$ $$\Rightarrow 2xy = 182$$ $$\Rightarrow xy = 91$$ Now solving $x+y = 20$ and $xy = 91$ we get, $$x = 13, \; y = 7 \; or \; y = 3x - 2y = 25$$

The formula to calculate the radius of the path followed by the charged particle under an external magnetic field can be written as $$r = \frac{mv}{Bq} \ldots (1)$$ The perpendicular component of velocity results in circular motion, while the parallel component results in linear motion. Hence, the path of the charged particle is helical in nature, the axis of which lies along the direction of the magnetic field.

The range $\($(R_1)$\)$ of the first projectile is given by $$R_1 = \frac{u_1^2 \sin 2\theta_1}{g} \cdots (1)$$ The range $\($(R_2)$\)$ of the second projectile is given by $$R_2 = \frac{u_2^2 \sin 2\theta_2}{g} \cdots (2)$$ Divide equation $\($(1)$\)$ by equation $\($(2)$\)$ and simplify to obtain the ratio of the ranges. $$\frac{R_1}{R_2} = \frac{\frac{u_1^2 \sin 2\theta_1}{g}}{\frac{u_2^2 \sin 2\theta_2}{g}}$$ $$= \frac{u_1^2 \sin 2\theta_1}{u_2^2 \sin 2\theta_2} \cdots (3)$$ Substitute the values of the known parameters into equation $\($(3)$\)$ to calculate the required ratio. $$\frac{R_1}{R_2} = \frac{40^2 \times \sin 60^\circ}{60^2 \times \sin 120^\circ}$$ $$= \frac{4}{9}$$

Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly. Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field.

The given data is $H = 98 \, \mathrm{m}$ $R = 6400 \, \mathrm{km}$ The line of sight formula for an antenna is $$d = \sqrt{2RH}$$ Therefore, the area is given by $$Area = \pi d^2 = \pi \times 2 \times R \times h = 2 \times 6400 \times 98 \times 10^{-3} \times \pi = 3940 \, \mathrm{km}^2$$

Mirrors are used in reflecting telescopes, which allow astronomers to see distant objects in space more clearly. A mirror forms an image by gathering light from spatial objects. A second mirror receives the picture that was reflected by the first mirror, which may be rather wide. The light is reflected by this tiny mirror onto an eyepiece lens, which magnifies or enlarges the image of the object. Hence, the secondary mirror is used to move the eyepiece outside the telescopic tube.

If heat energy is added to a thermodynamic system, one part increases the internal energy and the other part performs some work. If heat is added to a system, the system can do work without changing its internal energy. Hence, statement I is false. The formula to calculate the work done is given by $$W = \int PdV$$ From the above equation, it can be concluded that if $W > 0$, i.e., work is done by the system, its volume must increase. Hence, statement II is true.

The acceleration due to gravity changes with depth from the surface, which is given by $$g' = g \left( 1 - \frac{d}{R} \right) \cdots (i)$$ The given data is $W = 400 \, \mathrm{N}$ $d = \frac{R}{2}$ Multiplying equation (i) with the mass we get the weight. $$mg' = mg \left( 1 - \frac{R}{2R} \right) = mg \left( \frac{1}{2} \right)$$ $$\Rightarrow W' = \frac{400 \, \mathrm{N}}{2} = 200 \, \mathrm{N}$$

The data given is $Y = 7 \times 10^{10} \, \mathrm{N \, m^{-2}}$ and $\frac{\Delta l}{l} = \frac{0.04}{100}$. Young's modulus is given by the ratio of stress and strain, $$Y = \frac{F}{A} \frac{\Delta l}{l} = \frac{F l}{A \Delta l}$$ The energy stored is given as $$E = \left( \frac{Y A}{2 l} \right) \Delta l^2$$ $$\Rightarrow E = \left( \frac{Y A}{2} \right) \left( \frac{\Delta l}{l} \right)^2 \times l \cdots (i)$$ The energy per unit volume can be written from equation (i) $$\frac{E}{V} = \frac{Y}{2} \times \left( \frac{\Delta l}{l} \right)^2 \cdots (ii)$$ Substituting the values in equation (ii) $$\frac{E}{V} = \frac{1}{2} \times 7 \times 10^{10} \times \frac{0.04 \times 0.04}{10^4} = 56 \times 10^2$$

The equation for velocity is $$\vec{v} = \left( 2t \, \hat{i} + 3t^2 \, \hat{j} \right)$$ Thus, the acceleration is $$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d(2t \, \hat{i} + 3t^2 \, \hat{j})}{dt} = \left( 2 \, \hat{i} + 6t \, \hat{j} \right)$$ At $t = 1 \, \mathrm{s}$, $$\vec{a} = 2 \, \hat{i} + 6 \, \hat{j}$$ The force is written as $$\vec{F} = m \vec{a} = \frac{2 \, \hat{i} + 6 \, \hat{j}}{2} = \hat{i} + 3 \, \hat{j}$$ Comparing it with the given value of force, $x = 3$.

All the gates used in the diagram represent NAND gates. Thus, the output $Y$ can be calculated as follows $$Y = \overline{\overline{A} \cdot \overline{B}}$$ $$= \overline{\overline{A}} + \overline{\overline{B}}$$ $$= A + B$$ The above equation suggests that the output of the given combination yields the same result as an OR gate. The truth table can be written as follows

The formula to calculate the radius of a nucleus is given by $$R = R_0 A^{\frac{1}{3}} \ldots (1)$$ Hence, the volume energy can be written as $$b_v = C_1 R^3$$ $$= a_3 \left( A^{\frac{1}{3}} \right)^3$$ $$= a_3 A$$ The nuclear binding energy is the minimum energy that is required to disassemble the nucleus of an atom into its constituent protons and neutrons. Also, the energy of the nucleus is negative with regard to the energy of the particles pulled apart to infinite distance, because energy must be utilised to split a nucleus into its individual protons and neutrons. It is proportional to the surface area. Hence, statements C and D are most appropriate.

Let $m$ be the mass of the satellite. The potential energy of the satellite is $$U = -\frac{GMm}{r}$$ The velocity of the satellite is given by $$v = \sqrt{\frac{GM}{r}}$$ Kinetic energy is given by $$K = \frac{mv^2}{2} = \frac{m}{2} \frac{GM}{r} = \frac{GMm}{2r}$$ The total mechanical energy is the sum of potential and kinetic energy. So, $$T = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$$ Hence, both the statements are false.

The speed of light can be written as $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \cdots (1)$$ Rearrange equation (1) to obtain the required quantity. $$\frac{1}{\mu_0 \varepsilon_0} = c^2 \cdots (2)$$ Use the method of dimensional analysis to obtain the dimension of the required quantity. $$\left[ \frac{1}{\mu_0 \varepsilon_0} \right] = \left[ L T^{-1} \right]^2$$ $$= L^2 \ T^{-2}$$

Current flowing through both capacitors will be zero at steady state. The potential difference across capacitor $C_1$ will be the sum of the potential difference across the resistor $6 \, \Omega$ and the galvanometer resistance. Mathematically, $$V_{C_1} = i[6 \, \Omega + R_G] \ldots (1)$$ Similarly, the potential difference across capacitor $C_2$ will be the sum of the potential difference across the resistor $8 \, \Omega$ and the galvanometer resistance. Mathematically, $$V_{C_2} = i[R_G + 8 \, \Omega] \ldots (2)$$ Divide equation (1) by equation (2) to obtain the required ratio. $$\frac{V_{C_1}}{V_{C_2}} = \frac{i[6 \, \Omega + R_G]}{i[R_G + 8 \, \Omega]}$$ $$= \frac{6 \, \Omega + 2 \, \Omega}{2 \, \Omega + 8 \, \Omega}$$ $$= \frac{4}{5}$$

The electric field inside a solid insulated sphere is given by $$E_{inside} = \frac{\rho r}{3 \varepsilon_0}$$ for $r R$ is given by $$E_{outside} = \frac{Q}{4 \pi \varepsilon_0 r^2}$$ $$E_{outside} \propto \frac{1}{r^2}$$ Thus, the graph starts decreasing as a power law graph as shown in the figure for the region $r > R$. Hence, this graph is the right option.

Using parallelogram law of vector addition, the formula can be written as $$A_{net} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \theta}$$ The data given is $$|\vec{A_1}| = A, |\vec{A_2}| = \frac{A}{2}$$ Since the vectors are perpendicular $\cos 90^\circ = 0$ $$\Rightarrow F_{net} = \sqrt{A^2 + \frac{A^2}{4}} = \frac{\sqrt{5}A}{2}$$

The formula to calculate the frequency of the sound as heard by the passenger inside the train can be written as $$f' = f_0 \left[ \frac{v - v_o}{v - v_s} \right] \ldots (1)$$ Since the passenger is inside the train, the speeds of the source and the observer are the same. Substitute the values of the known parameters into equation (1) to get the required frequency. $$f' = 400 \left[ \frac{330 - 10}{330 - 10} \right] \, \mathrm{Hz}$$ $$= 400 \, \mathrm{Hz}$$

The pressure at the bottom of the lake can be calculated as follows: $$P_{bottom} = P_0 + \rho gh$$ $$= (10^5 + 1000 \times 10 \times 40) \, Pa$$ $$= 5 \times 10^5 \, Pa$$ As the temperature of water remains constant, it can be written that $$P_{bottom} V_{bottom} = P_{top} V_{top} \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required volume of the bubble when it reaches the surface. $$5 \times 10^5 \times 1 = 1 \times 10^5 \times V_{top}$$ $$\Rightarrow V_{top} = \frac{5 \times 10^5 \times 1}{1 \times 10^5} \, cm^3$$ $$= 5 \, cm^3$$

The formula to calculate the density of the material of the wire is given by $$\rho = \frac{M}{\pi R^2 L} \cdots (1)$$ Hence, the formula to calculate the percentage error in calculating the density of the material can be written as $$\frac{d\rho}{\rho} \times 100 = \left[ \frac{dM}{M} + \frac{2dR}{R} + \frac{dL}{L} \right] \times 100 \cdots (2)$$ Substitute the values of the known parameters into equation (2) to calculate the required percentage error. $$= \left[ \frac{0.01}{0.4} + \frac{2 \times 0.03}{6} + \frac{0.04}{8} \right] \times 100$$ $$= 4\%$$

The De Broglie wavelength is $\lambda = \frac{h}{mv}$. The momentum is given by $p = mv$. The momentum of proton is $p_p = m_p v_p$. The momentum of electron is $p' = m_e v_e$. Since $h$ is the Planck's constant and it is given that both wavelengths are same, i.e. $\lambda_p = \lambda_e$, so $p_p = p'$.

The work done for rotating a dipole is given by the change in the potential energy. The potential energy is given by $$U = -\vec{p} \cdot \vec{E}$$ $$W = \Delta U = U_f - U_i$$ So, the work done is $$W = -pE \cos 180^\circ - \left(-pE \cos 0^\circ \right) = 2pE$$ $$= 2 \times 6 \times 10^{-6} \times 1.5 \times 10^3 = 18 \, \mathrm{mJ}$$

As can be seen from the image above, object distance is $8 \, \mathrm{cm}$ [For image by B] In a plane mirror the image distance is equal to the object distance. Thus, the required distance is $$8 + 8 + 2 \, \mathrm{cm} = 18 \, \mathrm{cm}$$

The relation of kinetic energy and momentum is given by $$K = \frac{p^2}{2m}$$ The new value of momentum is $p' = 1.5p$ The new value of kinetic energy is $$\Rightarrow K' = \frac{(1.5p)^2}{2m} = 2.25K$$ Hence, the percentage increase in kinetic energy is $$\frac{K' - K}{K} \times 100 = \frac{2.25K - K}{K} = 125\%$$

The moment of inertia of a semicircular ring about a line perpendicular to the plane of the ring and passing through its centre is given as $I = MR^2$, where $m$ and $R$ are the mass and radius of the ring. Comparing the given value $$I = MR^2 = \frac{1}{x} MR^2$$ implies $x = 1$.

The formula for the frequency in an open organ pipe is $$f = \frac{nv}{2L}$$ Since, it is a second harmonic, $n = 2$. The frequency of the second harmonic becomes $$f = \frac{2v}{2L} = \frac{v}{L}$$ The data given is $L = 0.4 \, \mathrm{m}$, $v = 360 \, \mathrm{m \, s^{-1}}$. Substituting the values in the frequency formula $$f = \frac{360 \, \mathrm{m \, s^{-1}}}{0.4 \, \mathrm{m}} = 900 \, \mathrm{Hz}$$

The bubble is moving with a constant velocity so the net force acting on the body is zero. Stokes' law defines the drag force which is given by $F = 6 \pi \eta r v$. The buoyant force is given by $B = V \rho g = \frac{4}{3} \pi r^3 \rho g$. Hence, it can be written that $$\frac{4}{3} \pi r^3 \rho g = 6 \pi \eta r v$$ Therefore, $$\Rightarrow \eta = \frac{2 r^2 \rho g}{9 v} = \frac{2 \times \left(3 \times 10^{-3}\right)^2 \times 1750 \times 10}{9 \times 0.35 \times 10^{-2}} = 10 Pas$$

When the capacitor is fully discharged, the current is maximum and all the energy resides in the inductor. Hence, the energy of the capacitor is equal to the energy of the inductor. $$\frac{Li_m^2}{2} = \frac{Q_m^2}{2C} \cdots (i)$$ where $i_m$ is the maximum current and $Q_m$ is the maximum charge. The given values are $$Q_m = 2.7 \times 10^{-6} \, \mathrm{C}$$ $$L = 75 \times 10^{-3} \, \mathrm{H}$$ $$C = 1.2 \times 10^{-6} \, \mathrm{F}$$ The maximum current is found from equation (i) $$i_m = \sqrt{\frac{1}{LC} Q_m}$$ $$\Rightarrow \; i_m = \frac{2.7 \times 10^{-6}}{\sqrt{75 \times 10^{-3} \times 1.2 \times 10^{-6}}} = 9 \, \mathrm{mA}$$

The magnetic intensity $H$ is defined as the product of the number of turns per unit length in a coil $n$ and the current that it carries $I$. The magnetic field of a solenoid is given by $B = \mu_0 n i$. $$H = \frac{B}{\mu_0} = \frac{\mu_0 n i}{\mu_0}$$ $$\Rightarrow H = n i \ldots (i)$$ The given data is $$H = 1.6 \times 10^3 \, \mathrm{A \, m^{-1}}$$ $$n = 8 per cm$$ Substituting the values in equation (i) $$\Rightarrow 1.6 \times 1000 = \frac{8}{\frac{1}{100}} \times i$$ $$\Rightarrow i = 2 \, \mathrm{A}$$

The drift velocity is given by $$v_d = \frac{I}{neA}$$ Thus, $$I = nAe v_d \ldots (i)$$ The given data is $$I = 2 \, \mathrm{A}$$ $$A = 25 \times 10^{-6} \, \mathrm{m^2}$$ $$n = 2 \times 10^{28}$$ $$e = 1.6 \times 10^{-19} \, \mathrm{C}$$ Substituting the given values in equation (i) $$2 = 2 \times 10^{28} \times 25 \times 10^{-6} \times 1.6 \times 10^{-19} \times v_d$$ Thus, $$v_d = \frac{2}{2 \times 10^{28} \times 25 \times 10^{-6} \times 1.6 \times 10^{-19}}$$ Thus, $$v_d = \frac{1}{25 \times 1.6 \times 1000} \, \mathrm{m \, s^{-1}} = 25 \times 10^{-6} \, \mathrm{m \, s^{-1}}$$

Binding energy is given by \[ E=\Delta m\,c^2 \] where \[ \Delta m \] is the mass defect. The energy per nucleon of the nucleus having mass number \[ 242 \] is \[ 7.6\ \mathrm{MeV}. \] The initial binding energy is, \[ BE=242\times 7.6\ \mathrm{MeV} \] The energy per nucleon of the nucleus with mass number \[ 121 \] is \[ 8.1\ \mathrm{MeV}. \] Therefore, binding energy is \[ BE'=2(121\times 8.1\ \mathrm{MeV}) \] The gain in the binding energy is \[ BE'-BE=(8.1-7.6)\times 242\ \mathrm{MeV} \] \[ =121\ \mathrm{MeV} \]

Assign the oxidation number to all the elements in the reaction, $$2 \mathrm{IO_3^-} + x \mathrm{I^-} + 12 \mathrm{H^+} \rightarrow 6 \mathrm{I_2} + 6 \mathrm{H_2O}$$ $$2 \mathrm{IO_3^-} + 10 \mathrm{e^-} \rightarrow \mathrm{I_2}$$ $$[2 \mathrm{I^-} \rightarrow \mathrm{I_2} + 2 \mathrm{e^-}] \times 5$$ $$2 \mathrm{IO_3^-} + 10 \mathrm{I^-} + 12 \mathrm{H^+} \rightarrow 6 \mathrm{I_2} + 6 \mathrm{H_2O}$$ Therefore, the value of $x = 10$ in the reaction.

Leaching is often used if the ore is soluble in some suitable solvent. The ore of tin metal is cassiterite ($\mathrm{SnO_2}$). Cassiterite is readily soluble in alkali. The reaction is shown below. $$\mathrm{SnO_2 + 2NaOH + 2H_2O \rightarrow Na_2[Sn(OH)_6]}$$

Saccharin is an artificial sweetener that is commonly used as a sugar substitute. It is about $300 - 400$ times sweeter than sugar and has a slightly bitter aftertaste. Aspartame use is limited to cold foods and soft drinks because it is unstable at cooking temperature. Alitame is high potency sweetener, which means that only a small amount is needed to achieve the same level of sweetness as sugar. Sucralose is stable at cooking temperature. So the correct option is, $A \to \mathrm{II}$, $B \to \mathrm{IV}$, $C \to \mathrm{I}$, $D \to \mathrm{III}$.

The equation $\frac{x}{m} = KP^{1/n}$ represents Freundlich adsorption isotherm. It is an empirical relationship between the amount of gas adsorbed by a given amount of solid adsorbent surface and pressure of the gas at a particular temperature. So, Freundlich adsorption isotherm equation is $$\frac{x}{m} = KP^{1/n}$$ $$\Rightarrow \log \frac{x}{m} = \log K + \frac{1}{n} \log P$$ Extent of adsorption $\left( \frac{x}{m} \right)$ decreases with increase in temperature. With the help of the above two equations, following plots are obtained.

The leaving group which results in the formation of more stable carbocation will be more reactive towards $S_N1$ reaction. In the above molecule, by leaving of $\mathrm{Br}_{(a)}$ benzyl carbocation is formed. Benzyl carbocation is more stable than aliphatic carbocation. By replacing $\mathrm{I}_{(a)}$ in the above molecule, allyl carbocation is formed. Allyl carbocation is more stable than aliphatic carbocation. In the above molecule, it is difficult to replace $\mathrm{Br}_{(a)}$, as carbocation formed at bridge head position is very difficult. Hence, $\mathrm{Br}_{(b)}$ is more reactive. $\mathrm{Br}_{(a)}$ is more reactive than $\mathrm{Br}_{(b)}$ in the above molecule, because tertiary carbocation is more stable than primary carbocation. Hence, the correct answer is option (3).

Isoleucine - Isoleucine is a neutral amino acid. Cysteine - Cysteine is a sulphur containing amino acid. It contains a thiol group. Lysine - Lysine is a basic amino acid. Methionine - Methionine is a sulphur containing amino acid containing a thioether group. Glutamic acid - Glutamic acid is an acidic amino acid.

Methanol can be produced from syn gas (a mixture of carbon monoxide and hydrogen) through a different process, mainly through synthesis reaction. In this process, syn gas is passed over a heterogeneous catalyst, at high temperatures and pressures. $$\underbrace{\mathrm{CO(g) + 2H_2(g)}}_{Water gas} \xrightarrow{Cobalt catalyst} \mathrm{CH_3OH(l)}$$

Lithium borohydride is commonly used for the selective reduction of esters and lactones to the corresponding alcohols in the presence of carboxylic acids, tertiary amides, and nitriles.

The coordination compound $\mathrm{K_3[Co(CN)_6]}$ contains a cobalt(III) ion, which has a $d^6$ electronic configuration. The cyanide ligands in this complex are indeed strong field ligands, which means they will cause a large splitting of the $d$ orbitals. So here cobalt is in $+3$ state having $3d^6$ configuration and also cyanide ligand which is a strong field ligand so, it is most stable and will have octahedral shape. The given compounds have the following nature: (1) $\mathrm{Na_3[CoCl_6]}$ – Paramagnetic (2) $\mathrm{[Ni(NH_3)_6]Cl_2}$ – Paramagnetic (3) $\mathrm{K_3[Co(CN)_6]}$ – Diamagnetic (4) $\mathrm{[Co(H_2O)_6]Cl_2}$ – Paramagnetic

Oxidation is loss of electron and in a galvanic cell it occurs at anode. Reduction is gain of electron and in a galvanic cell it occurs at cathode. At anode: $$\frac{1}{2} \mathrm{H_2} \left( \mathrm{g} \right) \rightarrow \mathrm{H^+} + \mathrm{e^-}$$ At cathode: $$\mathrm{AgCl} + \mathrm{e^-} \rightarrow \mathrm{Ag} + \mathrm{Cl^-} \left( \mathrm{aq} \right)$$ Overall reaction: $$\frac{1}{2} \mathrm{H_2} \left( \mathrm{g} \right) + \mathrm{AgCl} \left( \mathrm{s} \right) \rightarrow \mathrm{H^+} + \mathrm{Cl^-} + \mathrm{Ag} \left( \mathrm{s} \right)$$ Cell representation: Anode / Anode electrolyte || Cathodic electrolyte / Cathode. The cell representation for the above cell will be $$\mathrm{Pt|H_2} \left( \mathrm{g} \right)|\mathrm{HCl} \left( \mathrm{soln}^n \right)|\mathrm{AgCl(s)|Ag}$$

(A) Aldehydes give red precipitate with Fehling's solution. $$CHO + Cu^{+2} + OH^- \rightarrow COO^- + Cu_2O + H_3O^+$$ Copper(I) oxide (brick red ppt) (B) Phenols give violet colour with neutral ferric chloride solution. $$OH + FeCl_3 \rightarrow \left[ Fe \left( O \right) \right]_6^{3-}$$ Violet complex (C) Primary amines give isocyanides with chloroform in the presence of alkali. This reaction is known as carbylamine reaction. $$NH_2 + CHCl_3 + OH^- \rightarrow NC$$ Carbylamine reaction (D) 1-phenylethanol gives haloform reaction with sodium hypochlorite. The products are benzoate and chloroform. $$OH + NaOCl \rightarrow O^- + CHCl_3$$

Butan-1-ol can undergo hydrogen bonding due to the presence of an $-\mathrm{OH}$ group, ethoxyethane (also known as diethyl ether) can also participate in hydrogen bonding. Even though there are no hydrogen atoms directly bonded to a highly electronegative atom (like F, O, or N) in ethoxyethane, the molecule still has polar $\mathrm{C} - \mathrm{O}$ bonds that can lead to dipole-dipole interactions with other polar molecules. Owing to intermolecular hydrogen bonding in butanol, it has higher boiling point than ethoxyethane.

In chromyl chloride (CrO$_2$ Cl$_2$), the oxidation state of chromium (Cr) is $+6$. So, Cr (VI) $= [\mathrm{Ar}]^{18} 3d^0$. For the given elements the configuration is, Mn(VII) $= [\mathrm{Ar}]^{18} 3d^0$. $$\mathrm{Fe(III)} = [\mathrm{Ar}]^{18} 3d^5$$ $$\mathrm{Ti(III)} = [\mathrm{Ar}]^{18} 3d^1$$ $$\mathrm{V(IV)} = [\mathrm{Ar}]^{18} 3d^1$$ So among the given options Mn(VII) and Cr (VI) will have same number of d-electrons.

When mixed with water the setting of cement takes place to give a hard mass. This is due to the hydration of molecules of the constituents and their rearrangement. Gypsum is $\mathrm{CaSO_4} \cdot 2\mathrm{H_2O}$. The purpose of adding gypsum is only to slow down the process of setting of the cement so that it gets sufficiently hardened.

$[\mathrm{MnBr_4}]^{2-}$ is a tetrahedral complex. The electronic configuration of $\mathrm{Mn}^{2+}$ is $e^2t_2^3$ and it has five unpaired electrons. $\therefore\ \mu=\sqrt{5(5+2)}=\sqrt{35}\,\mathrm{B.M.}$ $[\mathrm{CoF_6}]^{3-}$ is an octahedral high-spin complex. The electronic configuration of $\mathrm{Co}^{3+}$ is $t_{2g}^4e_g^2$ and it has four unpaired electrons. $\therefore\ \mu=\sqrt{4(4+2)}=\sqrt{24}\,\mathrm{B.M.}$ $[\mathrm{Mn(CN)_6}]^{3-}$ is an octahedral low-spin complex. The electronic configuration of $\mathrm{Mn}^{3+}$ is $t_{2g}^4e_g^0$ and it has two unpaired electrons. $\therefore\ \mu=\sqrt{2(2+2)}=\sqrt{8}\,\mathrm{B.M.}$ $[\mathrm{Fe(CN)_6}]^{3-}$ is an octahedral low-spin complex. The electronic configuration of $\mathrm{Fe}^{3+}$ is $t_{2g}^5e_g^0$ and it has one unpaired electron. $\therefore\ \mu=\sqrt{1(1+2)}=\sqrt{3}\,\mathrm{B.M.}$ Hence, the correct answer is option (3).

All $\mathrm{Cu^{2+}}$ halides are known except the iodide. In this case, $\mathrm{Cu^{2+}}$ oxidises $\mathrm{I^-}$ to $\mathrm{I_2}$. $$2 \mathrm{Cu^{2+}} + 4 \mathrm{I^-} \rightarrow \mathrm{Cu_2I_2} + \mathrm{I_2}$$ However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation. $$2 \mathrm{Cu^+} \rightarrow \mathrm{Cu^{2+}} + \mathrm{Cu}$$

\begin{tabular}{|c|c|c|} \hline & Species & Maximum allowed concentration in ppm in drinking water \\ \hline A. & F$^-$ & 2 \\ \hline B. & SO$_4^{2-}$ & 500 \\ \hline C. & NO$_3^-$ & 50 \\ \hline D. & Zn & 5 \\ \hline \end{tabular} F⁻ ion concentration above 2 ppm causes brown mottling of teeth. At the same time, excess fluoride (over 10 ppm) causes harmful effect to bones and teeth. Excessive sulphate (>500 ppm) in drinking water causes laxative effect, otherwise at moderate levels it is harmless. The maximum limit of nitrate in drinking water is 50 ppm. Excess nitrate in drinking water can cause disease such as methemoglobinemia ('blue baby' syndrome).

On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group. As a result, the most electronegative elements are found on the top right of the periodic table, while the least electronegative elements are found on the bottom left. Element Electronegativity P 2.1 C 2.5 Br 3.0 At 2.2 Hence, the correct order is Br > C > At > P.

Coupling reaction between benzene diazonium chlorides with aniline in acidic medium takes place to give yellow colour. In this reaction the nitrogen $\mathrm{N_2}$ in the diazonium ion is lost and forms the bridge between two benzene rings forming an electrophilic substitution reaction. Fluoro benzene is prepared by treating benzene diazonium chloride with fluoro boric acid. This reaction produces diazonium fluoroborate which on heating produces fluorobenzene. This reaction is called Balz-Schiemann reaction. Gattermann reaction is used for obtaining chlorobenzene from benzenediazonium chloride by treating it with $\mathrm{Cu/HCl}$ respectively. Similarly, Benzene diazonium chloride solution on reaction with cuprous cyanide and potassium cyanide gives benzonitrile. This reaction is known as Sandmeyer's reaction.

The values of ionic radii of $\mathrm{Li}^+ = 0.74 \, Å$ and $\mathrm{Mg}^{+2} = 0.72 \, Å$ respectively. Thus, the lithium ion with $+1$ charge is only marginally larger than the magnesium ion having a charge of $+2$. The superoxide releases the most energy when formed, the superoxide is preferentially formed for the larger metals where the more complex anions are not polarised. So, $\mathrm{Li}$ and $\mathrm{Mg}$ do not form superoxides. Therefore both the given options were correct.

When 2-methyl butene-2 is subjected to ozonolysis and then reduced in presence of dimethyl sulphide or Zn dust, it yields acetone and acetaldehyde. Cycloaddition occurs first and then it is cleaved reductively to give aldehyde and ketone.

The percentage of covalent character in an ionic bond is determined by the polarization power of the cation, the polarizing ability of the anion, and the degree of distortion of the anion's electron cloud that occurs due to the electric field of the cation. The polarization power of a cation refers to its ability to distort the electron cloud of the anion in the ionic bond. So the options A, B, C are correct.

Heat capacity $= \frac{Heat absorbed}{change in temperature}$ 1 Watt $= 1$ Joule per second. Hence, heat absorbed $= 60 \times 100 \, \mathrm{J}$ Change in temperature $= 5^\circ \, \mathrm{C}$ Heat capacity $= \frac{60 \times 100 \, \mathrm{J}}{5}$ $= 1200 \, \mathrm{JK^{-1}}$

Rate constant is given by Arrhenius equation $$k = Ae^{-E_a/RT}$$ Higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant. The pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy. If the activation energy of reaction is zero, temperature will have no effect on the rate constant.

We know that the temperature at which vapour pressure of a liquid becomes 1 atm is called the boiling point of the liquid. From the figure, we can clearly see that Graph I is for pure solvent whereas graph II is for solution. Hence the boiling point of solvent and solution are $82^\circ \mathrm{C}$ and $83^\circ \mathrm{C}$ respectively.

Xenon tetrafluoride ($\mathrm{XeF_4}$) is square planar and acts as a fluoride donor with $\mathrm{SbF_5}$. $\mathrm{XeF_4}$ changes to the cation $[\mathrm{XeF_3}]^+$ and $\mathrm{SbF_5}$ changes to the anion $[\mathrm{SbF_6}]^-$. $$\mathrm{XeF_4} + \mathrm{SbF_5} \rightarrow [\mathrm{XeF_3}]^+ [\mathrm{SbF_6}]^-$$ $m = 3$, $n = 1$, $y = 6$, $z = 1$

To calculate the mass percent of an element in a compound, we divide the mass of the element in 1 mole of the compound by the compound's molar mass and multiply the result by 100. The combustion reaction is $\mathrm{C(s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}$. $$C\% = \frac{12}{44} \times \frac{\text{Wt. of } \mathrm{CO_2}}{\text{Wt. of organic compound}} \times 100$$ $$60 = \frac{12}{44} \times \frac{\text{Wt. of } \mathrm{CO_2}}{0.5} \times 100$$ Wt. of $\mathrm{CO}_2 = 1.1$

Phenolphthalein is an organic acid and can be represented as HPh. $$\mathrm{HPh \rightleftharpoons H^+ + Ph^-}$$ Using Henderson equation for phenolphthalein $$\mathrm{pH = pK_{In} + \log \left[ \frac{[Ph^-]}{[HPh]} \right]}$$ At equivalence point $[\mathrm{Ph^-}] = [\mathrm{HPh}]$ $$\therefore \mathrm{pH_2 = pK_{In} = -\log[4] + 10 = -0.6 + 10 = 9.4}$$ Hence at $(9.4 \pm 1) \mathrm{PH}$, phenolphthalein starts changing colour. Phenolphthalein is colourless in acidic medium and pink in basic medium. Phenolphthalein indicator distinguish the pH change between 8 to 10. Therefore, it is used for strong acid and strong base titration or weak acid and strong base titration. Hence, A and C are incorrect statements.

The total number of moles remains constant. $n_1 + n_2 + n_3 = n_f$ $n_1, n_2, n_3$ moles of gases in the individual containers. $n_f$ is resultant moles. Using ideal gas equation $PV = nRT$, we can write as $$\frac{P_1 V_1}{RT} + \frac{P_2 V_2}{RT} + \frac{P_3 V_3}{RT} = \frac{P_f V_f}{RT}$$ $P_1 V_1 + P_2 V_2 + P_3 V_3 = P_f V_f$ $2 \times 2 + 4 \times 3 + 3 \times 4 = P_f \times 9$ $$P_f = \frac{28}{9} \approx 3$$

The emission spectrum of an element is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. The energy of the emitted photon is equal to the energy difference between the two states. There are many possible electron transitions for each atom, and each transition has a specific energy difference. This collection of different transitions, leading to different radiated wavelengths, make up an emission spectrum. Each element's emission spectrum is unique. The emission spectra of atoms in the gaseous phase do not show a continuous spread of wavelength from red to violet, rather they emit light only at specific wavelengths with dark spaces between them. An absorption spectrum is like the photographic negative of an emission spectrum. The element helium was discovered in the sun by spectroscopic method.