JEE Main 13 April 2023 Shift 2 question paper with solutions

Given system of equations. $$2x + y - z = 5$$ $$2x - 5y + \lambda z = \mu$$ $$x + 2y - 5z = 7$$ For infinite solution $\Delta = 0 = \Delta_1 = \Delta_2 = \Delta_3$ $$\Rightarrow \Delta = \begin{vmatrix} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{vmatrix} = 0$$ $$\Rightarrow 51 - 3\lambda = 0$$ $$\Rightarrow \lambda = 17$$ $$\Rightarrow \Delta_3 = \begin{vmatrix} 5 & 2 & 1 \\ \mu & 2 & -5 \\ 7 & 1 & 2 \end{vmatrix} = 0$$ $$\Rightarrow -3(\mu + 13) = 0$$ $$\Rightarrow \mu = -13$$ Now $$(\lambda + \mu)^2 + (\lambda - \mu)^2$$ $$= (17 + 13)^2 + (17 - 13)^2$$ $$= 900 + 16$$ $$= 916$$ Hence this is the correct option.

General term of the binomial expansion of $\left(2x^3 - \frac{1}{3x^2}\right)^5$ is $$T_{r+1} = \binom{5}{r} \left(2x^3\right)^{5-r} \left(-\frac{1}{3x^2}\right)^r$$ $$\Rightarrow T_{r+1} = \binom{5}{r} (2)^{5-r} \left(-\frac{1}{3}\right)^r x^{15-5r}$$ For coefficient of $x^5$, we must have $$15 - 5r = 5 \Rightarrow r = 2$$ So, required coefficient is $$= \binom{5}{2} (2)^3 \left(\frac{1}{9}\right)$$ $$= 10 \times 8 \times \frac{1}{9}$$ $$= \frac{80}{9}$$

Let $A(0, -1, 2)$ and $B(-1, 2, 1)$ and given plane is parallel to line passing through $(5, 1, -7)$ and $(1, -1, -1)$. So, normal vector to the plane is given by, $$ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 - 0 & 2 + 1 & 1 - 2 \\ 1 - 5 & -1 - 1 & -1 + 7 \end{vmatrix} $$ $$ \Rightarrow \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & -1 \\ -4 & -2 & 6 \end{vmatrix} = \hat{i}(16) - \hat{j}(-10) + \hat{k}(14) $$ $$ \Rightarrow \vec{n} = 16\hat{i} + 10\hat{j} + 14\hat{k} $$ Therefore, the equation of the plane is $$ 16(x - 0) + 10(y + 1) + 14(z - 2) = 0 $$ $$ \Rightarrow 8x + 5y + 7z = 9 $$ Now satisfying the points from option in the plane we get, $(-2, 5, 0)$ only point which satisfy the plane.

Given, $$x^2 - \sqrt{2}x + 2 = 0$$ $$\Rightarrow x = \frac{\sqrt{2} \pm \sqrt{6}i}{2}$$ $$\Rightarrow x = \sqrt{2} \left( \frac{1 \pm \sqrt{3}i}{2} \right)$$ $$\Rightarrow x = \sqrt{2} \left( \frac{1}{2} \pm \frac{\sqrt{3}i}{2} \right)$$ $$\Rightarrow x = \sqrt{2} \left( \cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3} \right)$$ So, $$\alpha = \sqrt{2} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = \sqrt{2} e^{i \frac{\pi}{3}}$$ And $$\beta = \sqrt{2} \left( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} \right) = \sqrt{2} e^{-i \frac{\pi}{3}}$$ Now $$\alpha^{14} + \beta^{14} = \left( \sqrt{2} e^{i \frac{\pi}{3}} \right)^{14} + \left( \sqrt{2} e^{-i \frac{\pi}{3}} \right)^{14}$$ $$\Rightarrow \alpha^{14} + \beta^{14} = \left( \sqrt{2} \right)^{14} \left( e^{i 14 \frac{\pi}{3}} + e^{-i 14 \frac{\pi}{3}} \right)$$ $$\Rightarrow \alpha^{14} + \beta^{14} = \left( \sqrt{2} \right)^{14} \left( \cos \frac{14\pi}{3} + i \sin \frac{14\pi}{3} + \cos \frac{14\pi}{3} - i \sin \frac{14\pi}{3} \right)$$ $$\Rightarrow \alpha^{14} + \beta^{14} = 2^7 \left( 2 \cos \frac{14\pi}{3} \right)$$ $$\Rightarrow \alpha^{14} + \beta^{14} = 2^8 \left( \cos \left( 4\pi + \frac{2\pi}{3} \right) \right)$$ $$\Rightarrow \alpha^{14} + \beta^{14} = 2^8 \left( \cos \frac{2\pi}{3} \right)$$ $$\Rightarrow \alpha^{14} + \beta^{14} = -2^8 \times \frac{1}{2} = -128$$

Given, $a_1,a_2,a_3,\ldots$ be a G.P. of increasing positive numbers, and the sum of its $6$th and $8$th terms be $2$. So, $a_6+a_8=2$ $\Rightarrow ar^5+ar^7=2 \qquad \ldots\ (1)$ And the product of its $3$rd and $5$th terms be $\frac{1}{9}$. So, $a_3\cdot a_5=\frac{1}{9}$ $\Rightarrow (ar^2)(ar^4)=\frac{1}{9}$ $\Rightarrow a^2r^6=\frac{1}{9}$ $\Rightarrow (ar^3)^2=\frac{1}{9}$ Since the terms are positive, $\Rightarrow ar^3=\frac{1}{3}$ Now putting the value of $ar^3=\frac{1}{3}$ in equation $(1)$, we get $\frac{r^2}{3}+\frac{r^4}{3}=2$ $\Rightarrow r^4+r^2=6$ $\Rightarrow (r^2+3)(r^2-2)=0$ $\Rightarrow r^2=2$ Also, $ar^3=\frac{1}{3}$ $\Rightarrow ar\cdot r^2=\frac{1}{3}$ $\Rightarrow ar\cdot 2=\frac{1}{3}$ $\Rightarrow ar=\frac{1}{6}$ Now finding the value of $6(a_2+a_4)(a_4+a_6)$, we get $=6(ar+ar^3)(ar^3+ar^5)$ $=6\left(\frac{1}{6}+\frac{1}{3}\right)\left(\frac{1}{3}+\frac{2}{3}\right)$ $=6\cdot\frac{1}{2}\cdot 1$ $=3$.

The given set of equations are: $$15x - y = 82$$ $$6x - 5y = -4$$ $$9x + 4y = 17$$ Now on solving the above equation by taking two at a time and plotting the diagram we get, $$A(1, 2)$$ $$B(5, -7)$$ $$C(6, 8)$$ $$(\alpha, \beta) \equiv \left( \frac{1+5+6}{3}, \frac{2-7+8}{3} \right) \equiv (4, 1)$$ Hence, $\alpha + 2\beta = 6$ and $2\alpha - \beta = 7$ The required equation is $(x - 6)(x - 7) = 0$ Equation $x^2 - 13x + 42 = 0$ Hence this is the required option.

Given, $|\vec{a}| = 2$, $|\vec{b}| = 3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Now solving, $$\left| (\vec{a} + 2\, \vec{b}) \times (2\, \vec{a} - 3\, \vec{b}) \right|^2$$ $$= \left| -3 \left( \vec{a} \times \vec{b} \right) + 4 \left( \vec{b} \times \vec{a} \right) \right|^2 \left\{ as \vec{a} \times \vec{a} = 0 and \vec{b} \times \vec{b} = 0 \right\}$$ $$= \left| 7 \left( \vec{b} \times \vec{a} \right) \right|^2 \left\{ as \vec{b} \times \vec{a} = -\vec{a} \times \vec{b} \right\}$$ $$= 49 |a|^2 |b|^2 \sin^2 \frac{\pi}{4}$$ $$= 49 \times 4 \times 9 \times \frac{1}{2}$$ $$= 882$$

Given, $N$ be the foot of perpendicular from the point $P(1, -2, 3)$ on the line passing through the points $(4, 5, 8)$ and $(1, -7, 5)$, Now finding direction ratio of line $L$ we get, $$(4 - 1, 5 + 7, 8 - 5) \equiv (3, 12, 3) \equiv (1, 4, 1)$$ Hence, the equation of Line will be, $$L : \frac{x-1}{1} = \frac{y+7}{4} = \frac{z-5}{1} = r$$ Now let point $N \equiv (r + 1, 4r - 7, r + 5)$ And given $P \equiv (1, -2, 3)$ So, direction ratio of $PN(r, 4r - 5, r + 2)$ Since, $PN \perp L$ so by perpendicular condition we get, $$r + 4(4r - 5) + (r + 2) = 0$$ $$\Rightarrow r = 1$$ Hence, $N \equiv (2, -3, 6)$ Now finding, distance of $N(2, -3, 6)$ from plane $2x - 2y + z + 5 = 0$ we get, $$Distance = \frac{|4 + 6 + 6 + 5|}{\sqrt{4 + 4 + 1}} = 7$$

Given that $$ \lim_{x \to 0} \frac{e^{ax} - \cos(bx) - \frac{cx}{2} e^{-cx}}{1 - \cos 2x} = 17 $$ We know that $e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots$ and $\cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots$ $$ \Rightarrow \lim_{x \to 0} \frac{\left(1 + ax + \frac{(ax)^2}{2!} + \ldots \right) - \left(1 - \frac{(bx)^2}{2!} + \ldots \right) - \frac{cx}{2} \left(1 - (cx) + \frac{(cx)^2}{2!} - \ldots \right)}{\frac{1 - \cos 2x}{(2x)^2} \times 4x^2} = 17 $$ $$ \Rightarrow \lim_{x \to 0} \frac{\left(a - \frac{c}{2} \right)x + \left(\frac{a^2 + b^2 + c^2}{2} \right)x^2 + \ldots}{\frac{1}{2} \times 4x^2} = 17 $$ Now for limit to exist, $a - \frac{c}{2} = 0 \Rightarrow c = 2a$ $$ \Rightarrow \lim_{x \to 0} \frac{\left(\frac{a^2 + b^2 + c^2}{2} \right)x^2 + \ldots}{\frac{1}{2} \times 4x^2} = 17 $$ $$ \Rightarrow \frac{a^2 + b^2 + c^2}{4} = 17 $$ $$ \Rightarrow a^2 + b^2 + 4a^2 = 68 $$ $$ \Rightarrow 5a^2 + b^2 = 68 $$ Hence this is the correct option.

Given, the centre of a circle $C$ be $\alpha$, $\beta$ and its radius $r 8$, so neglected. $$\therefore r = 5, \alpha = 1, \beta = -1$$ $$\therefore \alpha - \beta = 7$$

The given word is MONDAY. Arranging the letters alphabetically, we get ADMNOY. When the word starts with any of the letters A/D, the number of possibilities is $5! \times 2 = 240$. Now when the word starts with MA, the number of possibilities is $4! = 24$. Now when the word starts with MD, the number of possibilities is $4! = 24$. Now when the word starts with MN, the number of possibilities is $4! = 24$. Now when the word starts with MOA, the number of possibilities is $3! = 6$. Now when the word starts with MOD, the number of possibilities is $3! = 6$. Now when the word starts with MONA, the number of possibilities is $2! = 2$. Now when the word starts with MONDAY, the number of possibilities is $1$. Rank $= 240 + 24 \times 3 + 6 \times 2 + 2 + 1 = 327$. Hence, rank of the word MONDAY is 327. SHORTCUT METHOD Another method is a shortcut, first number the alphabet and then check how many numbers are there on the right-hand side less than that number and then start number in factorial in decreasing order and then multiply and add them and in last add one to find rank, $$\begin{array}{cccccc} 3 & 5 & 4 & 2 & 1 & 6 \\ M & O & N & D & A & Y \\ 2 & 3 & 2 & 1 & 0 & 0 \\ 5! & 4! & 3! & 2! & 1! & 0! \end{array}$$ $$\therefore Rank = (2 \times 5! + 3 \times 4! + 2 \times 3! + 1 \times 2!)$$ $$= 240 + 72 + 12 + 2 + 1 = 327$$

Given, $$f(x) = 4 \sin^{-1} \left( \frac{x^2}{x^2+1} \right)$$ Now taking, $$\frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} < 1$$ $$\Rightarrow 0 \leq \frac{x^2}{1+x^2} < 1$$ Now taking $\sin^{-1}$ we get, $$\Rightarrow 0 \leq \sin^{-1} \left( \frac{x^2}{1+x^2} \right) < \frac{\pi}{2}$$ $$\Rightarrow 0 \leq 4 \sin^{-1} \left( \frac{x^2}{1+x^2} \right) < 2\pi$$ Hence, the range of $4 \sin^{-1} \left( \frac{x^2}{x^2+1} \right)$ is $[0, 2\pi)$.

Given, $$(p \land (\sim q)) \lor ((\sim p) \land q) \lor ((\sim p) \land (\sim q))$$ $$= ((\sim p) \land q) \lor ((\sim p) \land (\sim q)) \lor (p \land (\sim q))$$ $$=(\sim p \land (q \lor \sim q)) \lor (p \land \sim q)$$ We know that $\sim A \lor A = T$, where $T$ is tautology, $$=(\sim p \lor T) \lor (p \land \sim q)$$ $$= \sim p \lor (p \land \sim q)$$ $$=(\sim p \lor p) \land (\sim p \lor \sim q)$$ $$= T \land (\sim p \lor \sim q)$$ $$= \sim p \lor \sim q$$

Given that the difference between the mean and variance is 1. $$np - npq = 1$$ $$np(1 - q) = 1$$ $$np^2 = 1$$ Also given that, $$2P(X = 2) = 3P(X = 1)$$ $$2 \cdot nC_2 p^2 q^{n-2} = 3 \cdot nC_1 p \cdot q^{n-1}$$ $$2 \cdot \frac{n \cdot (n-1)}{2} \cdot p = 3 \cdot n \cdot q$$ $$\Rightarrow (n - 1)p = 3(1 - p)$$ $$\Rightarrow \left(\frac{1}{p^2} - 1\right)p = 3(1 - p)$$ $$\Rightarrow \frac{(1-p)(1+p)}{p} = 3\left(1 - p\right)$$ $$\Rightarrow 1 + p = 3p$$ $$\Rightarrow p = \frac{1}{2}$$ $$\therefore n = 4.$$ Now, $$n^2 P(x > 1) = n^2 (1 - P(x = 1) - P(x = 0)) = 16 \left(1 - 4C_1 \cdot \left(\frac{1}{2}\right)^4 - \left(\frac{1}{2}\right)^4\right) = 11$$ Hence this is the correct option.

Given that $A = \begin{bmatrix} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$. Therefore, $|A| = 2$. This implies $1(6 - 1) - \alpha - 7 = 2$. Thus, $-\alpha - 2 = 2$. So, $\alpha = -4$. Therefore, $|2 \operatorname{adj}(2 \operatorname{adj}(2A))| = 32^n$. We know that $|kA| = k^n |A|$, $\operatorname{adj}(\operatorname{adj}(A)) = |A|^{(n-1)^2}$ and $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$. $$= 2^3 |\operatorname{adj}(2 \operatorname{adj}(2A))|$$ $$= 2^3 |2^{3-1} \operatorname{adj}(\operatorname{adj}(2A))|$$ $$= 2^3 \cdot (4)^3 |\operatorname{adj}(\operatorname{adj}(2A))|$$ $$= 2^9 |2A|^{(2)^2}$$ $$= 2^9 \cdot |2A|^4$$ $$= 2^9 \cdot 2^{12} |A|^4$$ $$= 2^{21} \cdot 2^4 = 2^{25} = (32)^n = (2)^5n$$ Therefore, $n = 5$. Thus, $3n + \alpha = 15 - 4 = 11$. Hence this is the correct option.

Given, $$\bar{z} = i \left( z^2 + \mathrm{Re}(z) \right)$$ Now, let $( z = x + iy )$, so $( \bar{z} = x - iy )$. Now putting the value in, $$( \bar{z} = i \left( z^2 + \mathrm{Re}(\bar{z}) \right) )$$ we get, $$\bar{z} = i \left( z^2 + \mathrm{Re}(z) \right)$$ $$\Rightarrow x - iy = i \left( x^2 - y^2 + 2ixy + x \right)$$ $$\Rightarrow x - iy = i \left( x^2 - y^2 + x \right) - 2xy$$ Now comparing real part we get, $$x = -2xy \Rightarrow x(2y + 1) = 0$$ $$\Rightarrow x = 0,\; y = -\frac{1}{2} \ldots (1)$$ And imaginary part we get, $$-y = x^2 - y^2 + x \ldots (2)$$ Now taking Case (I) when $( x = 0 )$ in equation (2) we get, $$\Rightarrow -y = -y^2$$ $$\Rightarrow y^2 - y = 0 \Rightarrow y = 0,\; 1$$ So, $( z = 0,\; i )$ Now taking Case (II) when $( y = -\frac{1}{2} )$ in equation (2) we get, $$\Rightarrow \frac{1}{2} = x^2 - \frac{1}{4} + x$$ $$\Rightarrow x^2 + x - \frac{3}{4} = 0$$ $$\Rightarrow 4x^2 + 4x - 3 = 0$$ $$\Rightarrow (2x - 1)(2x + 3) = 0$$ $$\Rightarrow x = \frac{1}{2},\; -\frac{3}{2}$$ So, $( z = \frac{1}{2} - \frac{1}{2} i,\; \frac{-3}{2} - \frac{1}{2} i )$ Now finding, $$\sum | z^2 | = 0 + 1 + \frac{1}{2} + \frac{5}{2} = 4$$

The required diagram is: The shaded region is symmetric about the $y$-axis. Required area $= 2 \left[ \int_1^2 \sqrt{y} \, dy + \int_2^4 \sqrt{4-y} \, dy \right]$ $$= 2 \left[ \left. \frac{y^{3/2}}{\frac{3}{2}} \right|_1^2 - \left. \frac{2(4-y)^{3/2}}{3} \right|_2^4 \right]$$ $$= \frac{4}{3} \left( 4\sqrt{2} - 1 \right)$$ Hence this is the correct option.

Given, $\overrightarrow{AB}=-2\hat{i}+\hat{j}+3\hat{k}$ $\overrightarrow{CB}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ $\overrightarrow{CA}=4\hat{i}+3\hat{j}+\delta\hat{k}$ Now plotting the diagram we get, Now from triangle law of addition we get, $\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}$ $4\hat{i}+3\hat{j}+\delta\hat{k}+(-2\hat{i}+\hat{j}+3\hat{k})=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ $2\hat{i}+4\hat{j}+(\delta+3)\hat{k}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ Now on comparing both sides we get, $\alpha=2,\ \beta=4$ and $\gamma=\delta+3$ Now the area of triangle is given by, $A=\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{CB}\right|$ $\Rightarrow 5\sqrt{6}=\sqrt{(\gamma-12)^2+(6+2\gamma)^2+100}$ Squaring both sides, $150=(\gamma-12)^2+(6+2\gamma)^2+100$ $\Rightarrow 150=\gamma^2-24\gamma+144+4\gamma^2+24\gamma+36+100$ $\Rightarrow 5\gamma^2=320$ $\Rightarrow \gamma^2=64\Rightarrow\gamma=8$ So, $\delta=8-3=5$ Also, $\overrightarrow{CB}\cdot\overrightarrow{CA}=(\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k})\cdot(4\hat{i}+3\hat{j}+\delta\hat{k})$ $\Rightarrow \overrightarrow{CB}\cdot\overrightarrow{CA}=(2\hat{i}+4\hat{j}+8\hat{k})\cdot(4\hat{i}+3\hat{j}+5\hat{k})$ $=8+12+40=60$

We know that two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$$ Let us verify this option. The given lines are $\frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$ and $\frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5}$. $$= \begin{vmatrix} -1 & -(-3) & 2 & -1 & 5 & -5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$$ $$= \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$$ $$= 2(5 - 10) - (1)(-15 - (-5)) + 0$$ $$= 2(-5) + (10) = 0$$ Therefore, this is the correct option.

Let, $$I = \frac{e^{-\frac{\pi}{4}} + \int_{0}^{\frac{\pi}{4}} e^{-x} \tan^{50} x \, dx}{\int_{0}^{\frac{\pi}{4}} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx}$$ Now let $I_1 = \int_{0}^{\frac{\pi}{4}} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx$. Now solving, $I_2 = \int_{0}^{\frac{\pi}{4}} e^{-x} \tan^{50} x \, dx$ using by parts we get, $$I_2 = \int_{0}^{\frac{\pi}{4}} e^{-x} \tan^{50} x \, dx$$ $$\Rightarrow I_2 = \left[ -e^{-x} \tan^{50} x \right]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} 50 \tan^{49} x \sec^2 x (-e^{-x}) \, dx$$ $$\Rightarrow I_2 = -e^{-\frac{\pi}{4}} + \int_{0}^{\frac{\pi}{4}} 50 \tan^{49} x (1 + \tan^2 x) e^{-x} \, dx$$ $$\Rightarrow I_2 = -e^{-\frac{\pi}{4}} + 50 \int_{0}^{\frac{\pi}{4}} (\tan^{49} x + \tan^{51} x) e^{-x} \, dx$$ $$\Rightarrow I_2 = -e^{-\frac{\pi}{4}} + 50 I_1$$ $$\Rightarrow \int_{0}^{\frac{\pi}{4}} e^{-x} \tan^{50} x \, dx + e^{-\frac{\pi}{4}} = 50 I_1$$ Now putting the given integral $I$ we get, $$I = \frac{e^{-\frac{\pi}{4}} + \int_{0}^{\frac{\pi}{4}} e^{-x} \tan^{50} x \, dx}{\int_{0}^{\frac{\pi}{4}} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx}$$ $$\Rightarrow I = \frac{50 I_1}{I_1} = 50$$

Let the observations be $x_1, x_2, x_3, \ldots, x_8, 45, 50$. Mean $= 50$ implies $$\frac{x_1 + x_2 + x_3 + \ldots + x_8 + 45 + 50}{10} = 50$$ which implies $$x_1 + x_2 + x_3 + \ldots + x_8 = 405 \ldots (i)$$ Hence, new mean $$\left(\overline{X}\right)_{new} = \frac{405 + 20 + 25}{10} = 45$$ Now, $$S. D = \sqrt{\frac{\sum_{i=1}^{8} x_i^2 + 45^2 + 50^2}{10} - (50)^2}$$ implies $$12 = \sqrt{\frac{\sum_{i=1}^{8} x_i^2 + 4525}{10} - 2500}$$ which implies $$\sum_{i=1}^{8} x_i^2 = 21915$$ Now, $$(Variance)_{new} = \frac{\sum_{i=1}^{8} x_i^2 + 20^2 + 25^2}{10} - (45)^2$$ implies $$(Variance)_{new} = \frac{21915 + 202 + 252}{10} - (45)^2$$ which implies $$(Variance)_{new} = 2294 - 2025$$ which implies $$(Variance)_{new} = 269$$ Hence this is the correct answer.

Given, $A = \{-4, -3, -2, 0, 1, 3, 4\}$ and $R = \{(a, b) \in A \times A : b = |a| or b^2 = a + 1\}$ be a relation on $A$, So, the relation is given by, $$R = \{(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4), (0, 1), (3, -2)\}$$ Now, relation to be reflexive $(a, a) \in R \forall a \in A$ $$\Rightarrow (-4, -4), (-3, -3), (-2, -2) also should be added in R.$$ Now relation to be symmetric if $(a, b) \in R$, then $(b, a) \in R \forall a, b \in A$ $$\Rightarrow (4, -4), (3, -3), (1, 0), (-2, 3) also should be added in R$$ Hence, minimum number of elements to be added to $$R = 3 + 4 = 7$$

Given, $$f(x) = \sum_{k=1}^{10} k \cdot x^k, \; x \in \mathbb{R}$$ Thus, $$f(x) = x + 2x^2 + 3x^3 + \ldots + 10x^{10}$$ Now let $$S = x + 2x^2 + 3x^3 + \ldots + 10x^{10}$$ So, $$S \cdot x = x^2 + 2x^3 + \ldots + 9x^{10} + 10x^{11}$$ Now subtracting above equations we get, $$S(1-x) = x + x^2 + x^3 + \ldots + x^{10} - 10x^{11}$$ Thus, $$S(1-x) = \frac{x(1-x^{10})}{1-x} - 10x^{11}$$ Thus, $$S = \frac{x(1-x^{10})}{(1-x)^2} - \frac{10x^{11}}{1-x} = f(x)$$ Now finding, $$f(2) = 2(1-2^{10}) + 10 \cdot 2^{11}$$ Thus, $$f(2) = 2 + 18 \cdot 2^{10}$$ Now finding $f'(x)$ we get, $$f'(x) = \frac{-10x^{11}}{(1-x)^2} - \frac{110x^{10}}{1-x} + \frac{10x^{10}}{(1-x)^2} + \frac{2x(1-x^{10})}{(1-x)^3} + \frac{1-x^{10}}{(1-x)^2}$$ Thus, $$f'(2) = \frac{-10 \cdot 2^{11}}{(1-2)^2} - \frac{110 \cdot 2^{10}}{1-2} - \frac{10 \cdot 2^{10}}{(1-2)^2}$$ Thus, $$f'(2) = -10 \cdot 2^{11} + 110 \cdot 2^{10} - 10 \cdot 2^{10}$$ Thus, $$f'(2) = 83 \cdot 2^{10} - 3$$ Thus, $$2f(2) + f'(2) = 2(2 + 18 \cdot 2^{10}) + 83 \cdot 2^{10} - 3 = 119(2)^{10} + 1$$ Therefore, $$n = 10$$

A number is divisible by 6 when it is divisible by 2 and 3 both. The three digit number can be $ab2$. $a + b$ can be 4, 7, 10. If $a + b = 4$ then $(a, b)$ can be $(1, 3)$, $(3, 1)$, $(2, 2)$. If $a + b = 7$ then $(a, b)$ can be $(2, 5)$, $(5, 2)$, $(3, 4)$, $(4, 3)$. If $a + b = 10$ then $(a, b)$ can be $(5, 5)$. So, 8 such cases when 2 is at unit's place. Similarly, there exist 8 such cases when 4 is at unit's place. Total $= 16$ cases. Hence this is the required answer.

Given, $\($ $\left$$\lfloor$ $\sqrt{1}$ $\right$$\rfloor$ + $\left$$\lfloor$ $\sqrt{2}$ $\right$$\rfloor$ + $\left$$\lfloor$ $\sqrt{3}$ $\right$$\rfloor$ + $\ldots$ + $\left$$\lfloor$ $\sqrt{120}$ $\right$$\rfloor$ $\)$ Now we know that, $\($ $\left$$\lfloor$ $\sqrt{x}$ $\right$$\rfloor$ = 1 $\)$, when $\($ x $\in$ [1, 4) $\)$, $\($ $\left$$\lfloor$ $\sqrt{x}$ $\right$$\rfloor$ = 2 $\)$, when $\($ x $\in$ [4, 9) $\)$ and similarly $\($ $\left$$\lfloor$ $\sqrt{x}$ $\right$$\rfloor$ = 10 $\)$, when $\($ x $\in$ [100, 120) $\)$ Now using the above formula we get, $\($ E = 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + 3 + 3 + 3 + 4 + 4 + $\ldots$ $\)$ $\($ $\Rightarrow$ E = 3 $\times$ 1 + 5 $\times$ 2 + 7 $\times$ 3 + $\ldots$ + 19 $\times$ 9 + 10 $\times$ 21 $\)$ $\($ $\Rightarrow$ E = $\sum$_{r=1}^{10} (2r + 1)r $\)$ $\($ $\Rightarrow$ E = 2 $\sum$_{r=1}^{10} (r^2 + r) $\)$ $\($ $\Rightarrow$ E = 2 $\left$[ $\frac{10 \times 11 \times 21}{6}$ $\right$] + $\frac{10 \times 11}{2}$ $\)$ $\($ $\Rightarrow$ E = 770 + 55 $\)$ $\($ $\Rightarrow$ E = 825 $\)$

Given that $\sin^{-1} x = 2 \tan^{-1} x$ $$\Rightarrow \sin^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$$ $$\Rightarrow x = \left( \frac{2x}{1+x^2} \right)$$ $$\Rightarrow x + x^3 = 2x$$ $$\Rightarrow x^3 - x = 0$$ $$\Rightarrow x (x^2 - 1) = 0$$ $$\Rightarrow x = 0, -1, 1.$$ But $x \in (-1, 1]$, so the number of roots are 2, $x = 0 \& 1$

Given differential equation is $\frac{dy}{dx} + \frac{4x}{(x^2-1)^{5/2}} y = \frac{x+2}{(x^2-1)^{5/2}}, \; x > 1$. Now IF $e^{\int \frac{4x}{x^2-1} dx} = (x^2-1)^2$. The required equation will be $\Rightarrow y \cdot (x^2-1)^2 = \int \frac{x+2}{(x^2-1)^{1/2}} dx$. $$\Rightarrow y \cdot (x^2-1)^2 = \frac{1}{2} \int \frac{2x}{(x^2-1)^{1/2}} dx + 2 \int \frac{dx}{(x^2-1)^{1/2}}$$ $$\Rightarrow y \cdot (x^2-1)^2 = 2 \ln \left( \sqrt{x^2-1} + x \right) + \sqrt{x^2-1} + C$$ Now using, at $x = 2, \; y\left(2\right) = \frac{2}{9} \log_e \left(2 + \sqrt{3}\right)$ we get, $$\Rightarrow 9 \cdot \frac{2}{9} \ln \left(2 + \sqrt{3}\right) = 2 \ln \left(2 + \sqrt{3}\right) + \sqrt{3} + C$$ $$\Rightarrow C = -\sqrt{3}$$ Now finding the value of function at $x = \sqrt{2}$ we get, $$y \times 1 = 2 \ln \left(1 + \sqrt{2}\right) + 1 - \sqrt{3}$$ Now on comparing we get, $$\Rightarrow \beta = 1, \; \alpha = 2, \; \gamma = 3$$ $$\Rightarrow \alpha \beta \gamma = 1 \times 2 \times 3 = 6$$ Hence, this is the required answer.

The given equation of hyperbola is \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\qquad ae=2,\qquad e=\frac{3}{2} \] \[ \Rightarrow a=\frac{4}{3} \] \[ \Rightarrow b^2=a^2e^2-a^2 =4-\frac{16}{9} =\frac{20}{9} \] The tangent is perpendicular to \[ 2x+3y=6. \] Slope of tangent is \[ m=-\frac{1}{-\frac{2}{3}}=\frac{3}{2}. \] Equation of tangent for slope \(m=\frac{3}{2}\) is \[ y=mx-\sqrt{a^2m^2-b^2}. \] \[ \Rightarrow y=\frac{3}{2}x-\sqrt{\frac{16}{9}\times\frac{9}{4}-\frac{20}{9}} \] \[ (\because\ \text{Tangent is in the first quadrant} \Rightarrow C<0) \] \[ \Rightarrow y=\frac{3}{2}x-\frac{4}{3} \] Converting the equation into intercept form: \[ \frac{x}{\left(\frac{8}{9}\right)}+\frac{y}{\left(-\frac{4}{3}\right)}=1 \] \[ \Rightarrow |6a|+|5b| =\left|\frac{6\times 8}{9}\right| +\left|\frac{5\times(-4)}{3}\right| =12 \] Hence, this is the required option.

Given, $$f_n = \int_0^{\pi/2} \left( \sum_{k=1}^n \sin^{k-1} x \right) \left( \sum_{k=1}^n (2k-1) \sin^{k-1} x \right) \cos x \, dx$$ Now let, $\sin x = t$ $$\Rightarrow \cos x \, dx = dt$$ So, $$f_n = \int_0^1 \left( \sum_{k=1}^n (t)^{k-1} \right) \left( \sum_{k=1}^n (2k-1) \right) dt$$ $$\Rightarrow f_n = \int_0^1 \left( 1 + t + t^2 + \ldots + t^{n-1} \right) \left( 1 + 3 + 5t^2 + \ldots + (2n-1)t^{n-1} \right) dt$$ Now multiply and divide by $\sqrt{t}$ we get, $$\Rightarrow f_n = \int_0^1 \left( \frac{1}{t^{\frac{1}{2}} + t^{\frac{3}{2}} + t^{\frac{5}{2}} + \ldots + t^{\frac{2n-1}{2}}} \right) \left( 1 + 3t + 5t^2 + \ldots + (2n-1)t^{n-1} \right) dt$$ $$\Rightarrow f_n = \int_0^1 \left( t^{\frac{1}{2}} + t^{\frac{3}{2}} + t^{\frac{5}{2}} + \ldots + t^{\frac{2n-1}{2}} \right) \left( t^{\frac{-1}{2}} \right) \left( 1 + 3t^{\frac{1}{2}} + 5t^{\frac{3}{2}} + \ldots + (2n-1)t^{\frac{2n-3}{2}} \right) dt$$ Now let $t^{\frac{1}{2}} + t^{\frac{3}{2}} + t^{\frac{5}{2}} + \ldots + t^{\frac{2n-1}{2}} = z$ $$\Rightarrow \left( t^{\frac{-1}{2}} + 3t^{\frac{1}{2}} + 5t^{\frac{3}{2}} + \ldots + (2n-1)t^{\frac{2n-3}{2}} \right) dt = dz$$ Hence, the integral becomes, $$f_n = 2 \int_0^1 z \, dz$$ $$= \left[ z^2 \right]_0^1 = n^2$$ Hence, $$f_{21} = 21^2 - 20^2 = 41$$

Using the binomial expansion in $7^{103} = (7^2)^{51} \cdot 7 = (51 - 2)^{51} \cdot 7$ we get, $$(51 - 2)^{51} \cdot 7 = 7 \cdot \left( {}^{51}C_{0} 51^{51} \cdot 2^0 - {}^{51}C_{1} 51^{50} \cdot 2^1 + \ldots - {}^{51}C_{51} 2^{51} \right)$$ Now 51 is divisible by 17, so when above equation is divided by 17 we get, $7 \cdot (-2^{51})$ as remainder, $$-56(17 - 1)^{12} = -56(2^{4})^{12} = -56(17 - 1)^{12}$$ we get, $$-56(17 - 1)^{12} = -56 \left( {}^{12}C_{0} 17^{12} \cdot 1^0 - {}^{12}C_{1} 17^{11} \cdot 1^1 + \ldots + {}^{12}C_{12} 1^{12} \right)$$ Now dividing above equation by 17 we get, $$(-56 \times 1) \div 17 = -5,$$ Now changing negative remainder to positive we get, $$-5 + 17 = 12,$$ Hence, when $7^{103}$ divided by 17 gives 12 as remainder.

The binding energy for stable nuclei is always a positive number, as the nucleus must gain energy for the nucleons to move apart from each other. Nucleons are attracted to each other by the strong nuclear force. Binding energy per nucleon is almost constant in the range of $30 - 170$. This is because nuclear energy is a short range force. It operates in the femto metre region. Hence, this option is the correct one.

The output of NAND gate is $Y = (AB)'$. The truth table is $$ \begin{array}{ccc} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} $$ $Y$ would be zero only when both $A$ and $B$ are 1. Clearly, for all other input conditions the output will be 1. Hence, this is the right option.

In steady state, capacitor behaves as open circuit, so no current will flow. The total resistance of the circuit is, $R = (6 + 4) \, \Omega = 10 \, \Omega$. The current flowing through the circuit is $$I = \frac{3 \, \mathrm{V}}{10 \, \Omega}$$ Potential difference on $6 \, \Omega$ resistor $$V' = \frac{3}{10} \times 6 = 1.8 \, \mathrm{V}$$ The charge on the capacitor is, $$Q = CV' = 4 \, \mu \mathrm{F} \times 1.8 \, \mathrm{V} = 7.2 \, \mu \mathrm{C}$$

The escape velocity is given by the formula $$v_{esc} = \sqrt{\frac{2GM}{R}}$$ Clearly, the escape velocity depends on the mass and radius of the planet. Hence, if the ratio increases the escape velocity also increases. So, Statement I is correct while statement II is wrong.

The potential energy is given by $$U = \frac{m \omega^2 x^2}{2}$$ The kinetic energy is given by $K = \frac{m \omega^2 (A^2 - x^2)}{2}$. Using the condition that $K = U$, $$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2$$ $$\Rightarrow x = \frac{A}{\sqrt{2}}$$

Let the velocity of A be $v_A$ and the velocity of B be $v_B$. The relative velocity is $$\vec{v}_{BA} = \vec{v}_B - \vec{v}_A$$ $$\Rightarrow v_{BA} = 54 - (-90) = 144 \, \mathrm{km \, h^{-1}}$$ $$\Rightarrow v_{BA} = 40 \, \mathrm{m \, s^{-1}}$$ Time $= \frac{length}{v_{BA}}$ $$\Rightarrow length = 40 \, \mathrm{m \, s^{-1}} \times 8 \, \mathrm{s} = 320 \, \mathrm{m}$$

As the gas is suddenly compressed the process is adiabatic. So, $PV^\gamma = constant$. $$P_0 V_0^\gamma = P \cdot \left( \frac{V_0}{4} \right)^\gamma$$ $$\Rightarrow P = 4^\gamma \cdot P_0$$

The terminal velocity is $$V_T = \frac{2}{9} r^2 g \frac{(\rho - \rho')}{\eta}$$ which implies $$V_T \propto r^2$$ Differentiating and dividing by $V_T$, $$\Rightarrow \frac{dV_T}{V_T} = 2 \frac{dr}{r} = 2 \times \frac{0.1}{5}$$ So, the percentage error is $$\frac{dV_T}{V_T} \times 100 = \frac{0.2}{5} \times 100 = 4\%$$ As can be seen the terminal velocity is directly proportional to the square of the radius. Hence $A$ is true but $R$ is false.

Direction of propagation of the electromagnetic wave is along $\vec{E} \times \vec{B}$. Thus, $$\vec{n} = E \left( -\hat{k} \right) \times B \left( \hat{i} \right)$$ Therefore, $$\hat{n} = -\hat{k} \times \hat{i} = -\hat{j}$$

It is given that $$s = 2.5t^2$$ Differentiating, $$v = \frac{ds}{dt} = \frac{d\left(2.5t^2\right)}{dt} = 5t$$ At $t = 5 \, \mathrm{s}$, $v = 25 \, \mathrm{m \, s^{-1}}$

The Lorentz force is given by $$\vec{F} = q \vec{v} \times \vec{B}$$ $$= -e \vec{v} \times \vec{B}$$ So, $$\vec{F} = \hat{i} \times \hat{k} = -\hat{j}.$$ $$\vec{F}$$ is along the negative $y$ axis. As the magnetic field is perpendicular to the velocity, motion would be circular.

The acceleration due to gravity is $$g = \frac{GM}{R^2} = \frac{4 \pi G R \rho}{3}$$ which implies $$g \propto \rho \cdot R$$ Thus, the ratio is $$\Rightarrow \frac{g_B}{g_A} = \frac{1.5 R \rho}{2 \rho R} = \frac{1}{2} \times 1.5 = 0.75$$ $$\Rightarrow \frac{g_B}{g_A} = \frac{3}{4}$$

For resonance to occur the condition is $X_L = X_C$. So, both inductor and capacitor is required. The power factor is given by $\cos \phi$. For an inductor the power factor is zero-lagging, $\cos 90^\circ = 0$. The angle between the voltage and current is $90^\circ$. For a capacitor the power factor is zero-leading. Hence, both statements are false.

The centripetal force is given by $$F = \frac{Mv^2}{R} = MR\omega^2$$ The given data is $$M = 200 \, \mathrm{kg}$$ $$R = 70 \, \mathrm{m}$$ $$\omega = 0.2 \, \mathrm{rad} \, \mathrm{s}^{-1}$$ The magnitude of the force is $$F = 200 \times 70 \times 0.2^2 \, \mathrm{N}$$ $$\Rightarrow F = 560 \, \mathrm{N}$$

For efficient radiation and reception, the height of transmitting and receiving antennas should be comparable to a quarter of wavelength of the frequency used. So for efficient transmission of signals of wavelengths $\lambda$ the minimum length of antenna should be $\frac{\lambda}{4}$.

Out of the three types of radiation, UV rays would be having highest frequency. The maximum kinetic energy is $$K_{max} = hf - \phi$$ Clearly, the maximum kinetic energy is dependant on the frequency of the incident radiation and the work function of the metal. From the given spectrums UV rays have the highest energy. Hence, this option is the correct one.

Let the charges be $x \, \mu \mathrm{C}$, $(q - x) \, \mu \mathrm{C}$, where $q = 10 \, \mu \mathrm{C}$. The force between them is $$F = \frac{Kx(q-x)}{r^2}.$$ For the force to be maximum, $$\frac{dF}{dx} = 0$$ $$\Rightarrow \frac{dF}{dx} = \frac{K(q-2x)}{r^2} = 0$$ $$\Rightarrow x = \frac{q}{2} = 5 \, \mu \mathrm{C}$$

Given $X$ and $\frac{a}{Y^2}$ have the same dimensions. $Y$ and $b$ also have the same dimensions. Therefore, $$\frac{a}{Y^2} = ML^{-1}T^{-2}$$ implies $$a = ML^{-1}T^{-2} \times L^6 = ML^5T^{-2}$$ and $b = L^3$. So, the dimensions of $\frac{a}{b}$ is, $$\frac{a}{b} = \frac{ML^5T^{-2}}{L^3} = ML^2T^{-2}.$$ This is the dimensional form of energy. Hence, this is the right option.

The ratio of the maximum to minimum intensity is given by $$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$$ It is given that $\frac{A_1}{A_2} = \frac{2}{1}$. $$\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2 = \frac{A_2^2 (2 + 1)^2}{A_2^2 (2 - 1)^2} = \left( \frac{2 + 1}{2 - 1} \right)^2 = \frac{9}{1}$$

The mean free path is given by the relation, $$\lambda = \frac{RT}{\sqrt{2} \pi d^2 N_A P}$$ So, $\lambda \propto T$ Let $\lambda'$ be the new mean free path. Hence, $$\frac{1500d}{\lambda'} = \frac{273}{373}$$ $$\Rightarrow \lambda' = 2049.45d \approx 2049d$$

The focal length of the bi-convex lens ($f$) can be calculated as follows: $$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$$ $$= \frac{2(\mu - 1)}{R} \ldots(1)$$ When the lens is cut into two halves, the focal length ($f'$) of each part can be calculated as follows: $$\frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right)$$ $$= \frac{(\mu - 1)}{R} \ldots(2)$$ Hence, the ratio of the powers for both the lenses can be given by $$\frac{P'}{P} = \frac{\frac{1}{f'}}{\frac{1}{f}}$$ $$= \frac{f}{f'}$$ $$= \frac{(\mu - 1)}{\frac{R}{2(\mu - 1)}}$$ $$= \frac{1}{2}$$ $$\Rightarrow P' = \frac{P}{2} \ldots(3)$$ Substitute the value of the known parameter into equation (3) to calculate the required power. $$P' = \frac{10}{2}$$ $$= 5 \, \mathrm{D}$$

The energy of a photon is given by $E = \frac{hc}{\lambda}$. It is given that $\lambda_1 = 500 \, \mathrm{nm}$, $\lambda_2 = 600 \, \mathrm{nm}$. The net energy absorbed is $$\Delta E = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = \frac{hc}{10^{-9}} \left( \frac{1}{500} - \frac{1}{600} \right)$$ $$= \frac{6.6 \times 10^{-34} \times 3 \times 10^8 \times 100}{500 \times 600 \times 10^{-9}}$$ $$= \frac{6.6 \times 3}{30} \times 10^{-19} \, \mathrm{J}$$ $$1 \, \mathrm{eV} = 1.6 \times 10^{-19} \, \mathrm{J}$$ $$= \frac{6.6 \times 3}{30 \times 1.6} \, \mathrm{eV}$$ $$= 4125 \times 10^{-4} \, \mathrm{eV}$$

The net force on the given charge will be the resultant of the forces due to the other two charges. Hence, the net force ($F_n$) can be calculated as follows: $$F_n = \frac{1}{4\pi \varepsilon_0} \frac{4q^2}{\left( \frac{R}{4} \right)^2} - \frac{1}{4\pi \varepsilon_0} \frac{2q^2}{\left( \frac{3R}{4} \right)^2}$$ $$= 9 \times 10^9 \times \frac{16q^2}{R^2} \left[ 4 - \frac{2}{9} \right]$$ $$= 9 \times 10^9 \times \frac{16 \times \left( 2 \times 10^{-6} \, \mathrm{C} \right)^2}{\left( 0.02 \, \mathrm{m} \right)^2} \times \frac{34}{9}$$ $$= 5440 \, \mathrm{N}$$

The capacitor works as an open circuit in the given configuration. The potential difference between the two plates of the capacitor will be the difference across the resistors having resistances $3 \, \Omega$ and $4 \, \Omega$. Hence, the potential difference $(\Delta V)$ across the capacitor can be calculated as follows: $$\Delta V = \left( \frac{4}{6} \times 12 - \frac{3}{12} \times 12 \right) \, \mathrm{V}$$ $$= 5 \, \mathrm{V}$$ The formula to calculate the energy stored in the capacitor is given by $$U = \frac{1}{2} C (\Delta V)^2 \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required energy stored. $$U = \frac{1}{2} \times 6 \, \mu \mathrm{F} \times (5 \, \mathrm{V})^2$$ $$= 75 \, \mu \mathrm{J}$$

The induced emf (ε) in the turns of the coil can be written as $$ \varepsilon = - \frac{d\varphi}{dt}$$ $$= iR \ldots (1)$$ Simplify equation (1) to obtain the required charge through a point in the circuit. $$- \frac{d\varphi}{dt} = iR$$ $$\Rightarrow - \frac{1}{R} \int_{\varphi_1}^{\varphi_2} d\varphi = \int idt$$ $$\Rightarrow q = \frac{\varphi_1 - \varphi_2}{R} \ldots (2)$$ Where, $\varphi_1$, $\varphi_2$ are the magnetic flux associated with the coil from two ends. As, $\varphi = NBA$, from equation (2), it can be written that $$q = \frac{NA}{R} \left( B_1 - B_2 \right) \ldots (3)$$ where, $N$, $A$ are the number of turns and the cross-sectional area of the core respectively. Substitute the values of the known parameters into equation (3) to calculate the required charge. $$q = \frac{100 \times 24 \times 10^{-4} \, \mathrm{m^2} \times (1.5 \, \mathrm{T} - (-1.5 \, \mathrm{T}))}{12 \, \Omega}$$ $$= 0.06 \, \mathrm{C} \times \frac{1000 \, \mathrm{mC}}{1 \, \mathrm{C}}$$ $$= 60 \, \mathrm{mC}$$

The formula to calculate the fundamental frequency in a stretched string is given by $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \cdots(1)$$ For the first mass, it can be written that $$f_1 = \frac{1}{2L} \sqrt{\frac{m_1 g}{\mu}} \cdots(2)$$ And, for the second mass, it can be written that $$f_2 = \frac{1}{2L} \sqrt{\frac{m_2 g}{\mu}} \cdots(3)$$ Divide equation (3) by equation (2) and simplify to obtain the required mass. $$\frac{f_2}{f_1} = \frac{\frac{1}{2L} \sqrt{\frac{m_2 g}{\mu}}}{\frac{1}{2L} \sqrt{\frac{m_1 g}{\mu}}}$$ $$= \sqrt{\frac{m_2}{m_1}}$$ $$\Rightarrow \frac{m_1}{m_2} = \left(\frac{f_1}{f_2}\right)^2$$ $$\Rightarrow m_2 = m_1 \left(\frac{f_2}{f_1}\right)^2 \cdots(4)$$ Substitute the values of the known parameters into equation (4) to calculate the required mass. $$m_2 = 180 \, \mathrm{g} \times \left(\frac{50 \, \mathrm{Hz}}{30 \, \mathrm{Hz}}\right)^2$$ $$= 500 \, \mathrm{g}$$

The formula to calculate the moment of inertia ($I$) of the cylinder about its radial axis is given by $$I = Mr^2 \ldots (1)$$ Also, the torque ($\tau$) on the cylinder about the central axis because of the application of the external force is given by $$\tau = Fr \ldots (2)$$ Also, the torque can be expressed as $$\tau = I \alpha \ldots (3)$$ Substitute the expressions from equation (1) and (2) into equation (3) and simplify to obtain the angular acceleration. $$Fr = Mr^2 \alpha$$ $$\Rightarrow \alpha = \frac{F}{Mr} \ldots (4)$$ Substitute the values of the known parameters into equation (4) to calculate the required angular acceleration. $$\alpha = \frac{52.5 \, \mathrm{N}}{5 \, \mathrm{kg} \times 0.70 \, \mathrm{m}}$$ $$= 15 \, \mathrm{rad} \, \mathrm{s}^{-2}$$

The kinetic energy of a moving particle is given by $$E = \frac{1}{2} mu^2 \cdots (1)$$ When the final velocity of the particle becomes twice the initial velocity, the kinetic energy of the particle can be written as $$nE = \frac{1}{2} m \left\{ (2u)^2 - u^2 \right\}$$ $$= \frac{1}{2} m (3u^2) \cdots (2)$$ Divide equation (2) by equation (1) and solve to calculate the value of $n$. $$\frac{nE}{E} = \frac{\frac{1}{2} m (3u^2)}{\frac{1}{2} mu^2}$$ $$\Rightarrow n = 3$$

Let the temperature of the contact surface be $T$. So, $H_A = H_B$. $$\frac{K_A A'(T_A - T)}{L} = \frac{K_B A'(T - T_B)}{L}$$ $$\Rightarrow 84(100 - T) = 126(T - 0)$$ $$\Rightarrow 2(100 - T) = 3T$$ $$\Rightarrow T = 40 \, ^\circ \mathrm{C}$$

The formula to calculate the magnetic force on the conductor is given by $$F_B = ILB \ldots (1)$$ The formula to calculate the gravitational force on the conductor is given by $$F_G = mg \ldots (2)$$ Equate both the equations and simplify to obtain the required current through the conductor. $$ILB = mg$$ $$\Rightarrow I = \frac{mg}{LB} \ldots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required current through the conductor. $$I = \frac{\frac{40}{1000} \, \mathrm{kg} \times 10 \, \mathrm{m \, s^{-2}}}{0.5 \, \mathrm{m} \times 0.40 \, \mathrm{T}}$$ $$= 2 \, \mathrm{A}$$

Ba^{2+} belongs to V^{th} Group and hence they are precipitated with CO_3^{2-} to give ppt of BaCO_3. In Wet tests for cations, (NH_4)_2 CO_3 is used as a group reagent for fifth group cations $\left$( Ba^{2+}, Ca^{2+}, Sr^{2+} $\right$). $$Ba^{2+} + \left( NH_4 \right)_2 CO_3 \rightarrow BaCO_3 + NH_4^+$$

Asparagine have only one basic functional Group Structure of Asparagine: There are three amino acids that have basic side chains at neutral pH. These are arginine (Arg), lysine (Lys), and histidine (His). Their side chains contain nitrogen and resemble ammonia, which is a base. Their pKa's are high enough that they tend to bind protons, gaining a positive charge in the process.

An Ellingham diagram is a graph showing the temperature dependence of the stability of compounds. Ellingham diagrams can be drawn for Sulphides, Oxides and Halides. The Ellingham diagram is a graph that shows how the $\Delta G_f^0$ varies with temperature as a result of the formation of elemental oxides.

Conditions for Tyndall effect: (i) Diameter of colloidal particles in solution is not much smaller than wavelength of the light used. (ii) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. If size of particles is large enough then light scatters in all directions. So Both A and R are correct and R is the correct explanation of A.

The complex $[\mathrm{Cr(ox)_2ClBr}]^{3-}$ exhibits two geometrical isomers. Among these isomers, the cis-isomer is optically active. Hence, the total number of stereoisomers for the given complex is 3.

The beryllium fluoride cannot be easily prepared by simple methods. It is known that beryllium oxide dissolves in aqueous hydrofluoric acid with the formation of the fluoride, but on evaporation of the resulting solution some of the combined acid is lost, and the residue is believed to be an oxy fluoride. Thermal decomposition of $(\mathrm{NH_4})_2 \mathrm{BeF_4}$ is the best route for the preparation of $\mathrm{BeF_2}$. $$(\mathrm{NH_4})_2 \mathrm{BeF_4} \rightarrow 2 \mathrm{NH_4} \mathrm{F} + \mathrm{BeF_2}$$

While isotopes of an element have the same electronic configuration, their chemical properties can differ slightly due to differences in their atomic mass. Isotopes of an element have the same number of protons but varying numbers of neutrons, which affects the overall mass of the atom. The only difference is in their rates of reactions, mainly due to their different enthalpy of bond dissociation.

Tropolone is an aromatic compound and has $8\pi e^-$ electrons ($6\pi e^-$ are endocyclic and $2\pi e^-$ are exocyclic) and $\pi$ electrons of $> \mathrm{C} = \mathrm{O}$ group in tropolone is not involved in aromaticity.

When aniline treated with bromine, the reaction results in the formation of 2, 4, 6-tribromo phenylamine. This reaction is an example of bromination. The process of conversion of a primary aromatic amino compound into a diazonium salt is known as diazotization. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below $5^\circ \mathrm{C}$. Benzene diazonium chloride reacts with $\mathrm{H_3PO_2}$ in aqueous medium to form benzene.

Nucleophilicity of sulfur-containing nucleophiles (such as $\mathrm{SH}^-$) is higher than that of oxygen-containing nucleophiles (such as $\mathrm{OH}^-$). Here sulfur containing nucleophile can donate its lone pair. A stable carbocation is formed next, stable carbocation will be that which is away from CN group.

The greenhouse effect is a process that occurs when gases in Earth's atmosphere trap the Sun's heat. This process makes Earth much warmer than it would be without an atmosphere. The main greenhouse gases whose concentrations are rising are carbon dioxide, methane, nitrous oxide, hydrochlorofluorocarbons (HCFCs), hydrofluorocarbons (HFCs) and ozone in the lower atmosphere. $\mathrm{I_2}$ and $\mathrm{H_2}$ are not greenhouse gases.

(A) Weak intermolecular forces are present in polymer of 2-chloro, 1, 3-butadiene. It is a monomer of neoprene which is a rubber (elastomer). (B) Hydrogen bonding is present in NYLON-6, 6 which is a polymer of hexamethylenediamine and adipic acid. This hydrogen bonding is due to the presence of amide group. (C) Heavily branched polymer is Bakelite which is polymer of phenol and formaldehyde. It is a crosslinked polymer. (D) High density polymer (polyethylene) preparation requires $\mathrm{Al(Et)_3}$ and $\mathrm{TiCl_4}$ as a catalyst (Ziegler Natta Catalyst).

The molecule $\mathrm{SO_2}$ is having two sigma bond pairs and lone pair, hence, it exhibits angular or V-shape structure. The molecule $\mathrm{H_2O}$ is having two sigma bond pairs and a lone pair hence, it exhibits angular or V-shape structure. So, both molecules are having same shape. But double bond-double bond repulsions are greater than single bond repulsions, hence, the bond angle in sulphur dioxide is greater than water. $\mathrm{SO_2}$ and $\mathrm{H_2O}$ both have V-Shape Bond angle: $\mathrm{SO_2} > \mathrm{H_2O}$

Fluorine oxidises water to oxygen whereas chlorine and bromine react with water to form corresponding HX and HOX acids. $$2\mathrm{F_2} + 2\mathrm{H_2O} \rightarrow 4\mathrm{HF} + \mathrm{O_2}$$ $$\mathrm{Cl_2} + \mathrm{H_2O} \rightarrow \mathrm{HCl} + \mathrm{HOCl}$$ The reaction of $\mathrm{I_2}$ with water is nonspontaneous. In fact, $\mathrm{I^-}$ can be oxidised by oxygen in acidic medium. It is why the $\mathrm{I^-}$ in nature is not present in that much amount as other halides are present in nature.

For an adiabatic free expansion of an ideal gas, the gas is kept in an insulated container and then permitted to enlarge in a vacuum. Since there is no external pressure for the gas to enlarge against, the work done by or on the system is zero. Since this process does not involve any exchange of heat transfer or work, heat will not escape and temperature of system will rise. For diathermic container, heat will escape the container and hence temperature of container will remain same.

The electron withdrawing groups decrease the acidic nature and the electron releasing groups increase the acidic nature. The F and Cl are electron withdrawing while CH_3 is electron releasing, hence acidic strength will be $$O - H Cl > OH F > OH CH_3$$ On further differentiation, the $-I$ effect depends most importantly on distance, hence. $$OH Cl > OH F > OH CH_3$$ The F has a higher $-I$ effect compared to Cl, hence the reason is a correct statement, but the $-I$ effect depends more on distance as compared to power, hence $$OH Cl$$ is stronger acid than $$OH F$$

The lattice energy of an ionic compound is the energy required to separate one mole of the compound into its constituent ions in the gas phase. Lattice energy $\propto \frac{1}{(r_+ + r_-)}$ The lattice energy is inversely proportional to the sum of the ionic radii $(r_+ + r_-)$ of the ions involved. When the size of the ions increases, the distance between them also increases, leading to a decrease in the attractive forces between the ions and a decrease in the lattice energy. Therefore, if the size of the ions increases, the lattice energy generally decreases. So Order: $\mathrm{NaF} > \mathrm{NaCl} > \mathrm{NaBr} > \mathrm{NaI}$

(A) When 1-bromopropane reacts with alcoholic KOH, the bromine with the alpha H (hydrogen from the carbon alongside the carbon fortified with useful gathering) is eliminated and subsequently forms 1-propene. $$CH_2 - CH_2 - CH_3 \xrightarrow{Alc. KOH} CH_2 = CH - CH_3$$ (B) When 1-bromo propane reacts with alc KCN it will form butanenitrile. $$\begin{array}{c} CH_2 - CH_2 - CH_3 \\ | \\ Br \end{array} \xrightarrow{KCN} \begin{array}{c} CH_2 - CH_2 - CH_3 \\ | \\ C \equiv N (nitrile) \end{array}$$ (C) When alkyl halide reacts with silver nitrite, nitroalkane is formed because the bond between Ag - OAg - O is covalent, the lone pair on nitrogen acts as an attacking site for nucleophilic substitution. $$\begin{array}{c} CH_2 - CH_2 - CH_3 \\ | \\ Br \end{array} \xrightarrow{AgNO_2} \begin{array}{c} CH_2 - CH_2 - CH_3 \\ | \\ NO_2 (Nitroalkane) \end{array}$$ (D) When 1-bromopropane reacts with silver acetate, propyl ethanoate and silver bromide are formed. $$\begin{array}{c} CH_2 - CH_2 - CH_3 \\ | \\ Br \end{array} \xrightarrow{CH_3COOAg} \begin{array}{c} CH_2 - CH_2 - CH_3 \\ | \\ O - C - CH_3 \\ || \\ O (ester) \end{array}$$

Number of covalent bonds formed by Boron is 4. $[\mathrm{BF}_4]^-$ Covalency = 4. The oxidation state of an element represents the charge it would have if all the bonding electrons were assigned to the more electronegative atom in the bond. Oxidation number of fluorine is $-1$. Then, $$\mathrm{B} + 4 \times (-1) = -1$$ $$\mathrm{B} - 4 = -1$$ $$\mathrm{B} = +3$$ Oxidation state = $+3$ for Boron.

Complex having maximum number of unpaired electrons will exhibit maximum attraction to applied magnetic field. $$[\mathrm{Zn} (\mathrm{H}_2\mathrm{O})_6]^{2+} \rightarrow d^{10} system, t_{2g}^6 e_g^4, 0 unpaired electrons$$ $$[\mathrm{Co} (\mathrm{H}_2\mathrm{O})_6]^{2+} \rightarrow d^7 system, t_{2g}^5 e_g^2, 3 unpaired electrons$$ $$[\mathrm{Co(en)}_3]^{3+} \rightarrow d^6 system, t_{2g}^6 e_g^0, 0 unpaired electrons$$ $$[\mathrm{Ni} (\mathrm{H}_2\mathrm{O})_6]^{2+} \rightarrow d^8 system, t_{2g}^6 e_g^2, 2 unpaired electrons.$$ So Option D is correct.

The percentage composition of element is given by the formula: $$\frac{mass of element}{mass of organic compound} \times 100$$ Moles of AgBr is given by $$\frac{0.376}{188}$$ Moles of Br is also $$\frac{0.376}{188}$$ Mass of Br is $$\frac{0.376}{188} \times 80$$ The percentage of Br is $$\frac{0.376 \times 80}{188 \times (0.400)} \times 100$$ which equals $$40\%$$

From the balanced chemical reaction, $\mathrm{M_2CO_3 + 2HCl \rightarrow 2MCl + H_2O + CO_2}$. Here from this, we get: $0.01 \, \mathrm{mol} \, \mathrm{CO_2} \equiv 0.01 \, \mathrm{mol} \, \mathrm{M_2CO_3} \equiv 1 \, \mathrm{g} \, \mathrm{M_2CO_3}$. We know that, number of moles $= \frac{\text{Given mass}}{\text{Molar mass}}$. Therefore, $$\text{Molar mass of } \mathrm{M_2CO_3} = \frac{\text{Given mass of } \mathrm{M_2CO_3}}{\text{Number of Moles of } \mathrm{M_2CO_3}} = \frac{1}{0.01} = 100 \, \mathrm{g/mol}$$

The given equation can be balanced using ion-electron method as shown below. $\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$ - oxidation Half reaction $6e^- + 14\mathrm{H^+} + \mathrm{Cr_2 O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O}$ - Reduction Half reaction The overall reaction is $$6\mathrm{Fe^{2+} + 14H^+ + Cr_2 O_7^{2-} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O}$$ The values $X = 14$, $Y = 2$ and $Z = 7$ $(X + Y + Z) = 14 + 2 + 7$ $= 23$

The crude name of borax is Tincal. It has the molecular formula $\mathrm{Na_2B_4O_7} \cdot 10\mathrm{H_2O}$ and structural formula of borax is $\mathrm{Na_2[B_4O_5(OH)_4] \cdot 8H_2O}$. Based on the structural formula, $x = 5$, $y = 4$ and $z = 8$. $$(x + y + z) = 5 + 4 + 8 = 17$$

For the reaction: $\mathrm{Cu^{2+}(aq)+2e^- \rightarrow Cu(s)}$ The reduction potential is $E=E^\circ-\dfrac{RT}{2F}\ln\!\left(\dfrac{1}{[\mathrm{Cu^{2+}}]}\right)$ From the given data, $\mathrm{pH}=14$ and $K_{sp}\!\left(\mathrm{Cu(OH)_2}\right)=1.0\times10^{-20}$, we get $[\mathrm{H^+}]=10^{-14}\,\mathrm{M}$ $[\mathrm{OH^-}]=\dfrac{K_w}{[\mathrm{H^+}]} =\dfrac{10^{-14}\,\mathrm{M^2}}{10^{-14}\,\mathrm{M}} =1\,\mathrm{M}$ $[\mathrm{Cu^{2+}}] =\dfrac{K_{sp}}{[\mathrm{OH^-}]^2} =\dfrac{1.0\times10^{-20}\,\mathrm{M^2}}{(1\,\mathrm{M})^2} =1.0\times10^{-20}\,\mathrm{M}$ $E=0.34\,\mathrm{V} -\left(\dfrac{0.059\,\mathrm{V}}{2}\right) \log\!\left(\dfrac{1}{1.0\times10^{-20}}\right)$ $=0.34\,\mathrm{V} -\dfrac{0.059\times20\,\mathrm{V}}{2}$ $E=0.34-0.59$ $=-0.25\,\mathrm{V}$

When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula: $$pH = pK_a + \log \left( \frac{[salt]}{[acid]} \right)$$ $$CH_3 COOH + NaOH \longrightarrow CH_3 COONa + H_2 O$$ $$\begin{array}{cccc} 5 & 2 & & \\ 3 & 0 & 2 & \\ \end{array}$$ $$pH = pK_a + \log \left( \frac{2}{3} \right)$$ $$pH = 4.76 + 0.30 - 0.48$$ $$= 4.76 - .18$$ $$= 4.58$$

The final pressure of each component can be calculated as follows. $$\mathrm{A(g)} \rightarrow 2\, \mathrm{B(g)} + \mathrm{C(g)}$$ Initial: $800 0 0$ At time $t$: $800 - p 2p p$ $$800 + 2p = 1600$$ $$2p = 800$$ $$p = 400$$ The first order reaction rate constant can be calculated as follows. $$K = \frac{2.303}{10} \log \frac{800}{400} = \frac{2.303 \times \log 2}{10}$$ For 30 min, $$K = \frac{2.303}{30} \log \frac{(800)}{(800-y)}$$ $$\frac{2.303 \times \log 2}{10} = \frac{2.303}{30} \log \left( \frac{800}{800-y} \right)$$ $$\Rightarrow \left( \frac{800}{800-y} \right) = 8$$ $$100 = 800 - y$$ $$y = 700$$ Total pressure after 30 min: $$(800 - y) + (2y) + (y)$$ $$= 800 + 2y$$ $$= 800 + 1400$$ $$= 2200 \, \mathrm{mm \, Hg}$$

The orbital angular momentum is given by $$\frac{h}{2\pi} \sqrt{l(l+1)}$$ For 3s-electron, $l = 0$. Therefore, the orbital angular momentum is $$\frac{h}{2\pi} \sqrt{0(0+1)}$$ which equals $0$ (zero).

Edge length of the unit cell ($a$) $= 4 \times 10^{-8} \, \mathrm{cm}$ Radius of sodium atom ($r$) = ? Since, sodium crystallises in body centered cubic arrangement. The radius of the atom and the edge length of the cubic unit cell are related as follows, For bcc structure, $4r = \sqrt{3}a \Rightarrow r = \frac{\sqrt{3}}{4} a$ $$r = \frac{1.732 \times 4 \times 10^{-8}}{4}$$ $$r = 1.732 \times 10^{-8} \, \mathrm{cm}$$ $$r = 17.3 \, \mathrm{\AA}$$

Molality of solution = $\frac{moles of solute}{mass of solvent}$ $\times$ 1000. Number of moles NaCl = $\frac{29.25}{58.5}$ = 0.5. Number of moles of MgCl_2 = $\frac{19}{95}$ = 0.2. NaCl = $\frac{0.5}{51.75}$ $\times$ 1000. MgCl_2 = $\frac{0.2}{51.75}$ $\times$ 1000. ($\Delta$ T_b) = $\{$(i_1 $\,$ m_1) + (i_2 $\,$ m_2)$\}$ k_b. $\Delta$ T_b = Elevation in boiling point. i = Van't Hoff factor. m = molality. k_b = molal elevation constant. $$\Delta T_b = \left(\frac{2 \times 0.5 \times 1000}{51.75} \times \frac{3 \times 0.2 \times 1000}{51.75}\right) \times 0.52$$ $$= 16.077$$ Boiling point of sea water = 116.077^$\circ$ C $\approx$ 116^$\circ$ C (Nearest integer)