JEE Main 8 April 2023 Shift 2 question paper with solutions

Given, the mean and variance of 12 observations be $\frac{9}{2}$ and 4 respectively. So, mean will be, $$\bar{x} = \frac{x_1 + x_2 + \ldots + 9 + 10 + \ldots + x_{12}}{12} = \frac{9}{2}$$ $$\Rightarrow x_1 + x_2 + \ldots + x_{12} + 19 = 54$$ Now removing the observation 9 and 10 and adding the observation 7 and 14, so, the new total sum of the observation will be, $$x_1 + x_2 + \ldots + x_{12} + 7 + 14 = 54 - 19 + 7 + 14$$ Now new mean will be, $$\frac{x_1 + x_2 + \ldots + x_{12} + 7 + 14}{12} = \frac{56}{12}$$ $$\Rightarrow \bar{x}_{new} = \frac{14}{3}$$ Now using the formula of the variance we get, $$\frac{x_1^2 + x_2^2 + \ldots + x_{12}^2 + 9^2 + 10^2}{12} - \left(\frac{9}{2}\right)^2 = 4$$ $$\Rightarrow x_1^2 + x_2^2 + \ldots + x_{12}^2 + 9^2 + 10^2 - 81 \times 3 = 4 \times 12$$ $$\Rightarrow x_1^2 + x_2^2 + \ldots + x_{12}^2 + 9^2 + 10^2 = 291$$ Now removing $9^2$ and $10^2$ and adding $7^2$ and $14^2$ we get, $$x_1^2 + x_2^2 + \ldots + x_{12}^2 + 7^2 + 14^2 = 355$$ New variance = $$\frac{\sum x_i^2}{N} - \left(\bar{x}\right)^2$$ $$= \frac{355}{12} - \left(\frac{14}{3}\right)^2$$ $$= \frac{281}{36}$$ Now on comparing with $\frac{m}{n} = \frac{281}{36}$ we get, $$\Rightarrow m = 281 and n = 36$$ Therefore, $m + n = 317$

The given series is $a_n$, then we can write as $a_n=5+8+14+23+\cdots+a_n$ $a_n=5+8+14+23+\cdots+a_{n-1}+a_n$ Subtracting above equations, we get $0=5+3+6+9+\cdots+(a_n-a_{n-1})-a_n$ $\Rightarrow\ a_n=5+\left[\frac{n-1}{2}\left(2\times3+(n-1-1)3\right)\right]$ $\Rightarrow\ a_n=5+\frac{n-1}{2}(3n)$ $\Rightarrow\ a_n=\frac{1}{2}(3n^2-3n+10)$ So, $\Rightarrow\ a_{40}=\frac{1}{2}(3\times40^2-3\times40+10)$ $\Rightarrow\ a_{40}=\frac{1}{2}(4800-120+10)$ $\Rightarrow\ a_{40}=2345$ $S_{30}=\sum_{n=1}^{30}a_n$ $\Rightarrow\ S_{30}=\frac{1}{2}\left(3\sum_{n=1}^{30}n^2-3\sum_{n=1}^{30}n+\sum_{n=1}^{30}10\right)$ $=\frac{1}{2}\left(\frac{3(30)(31)(61)}{6}-\frac{3(30)(31)}{2}+10\times30\right)$ $\Rightarrow\ S_{30}=13635$ $\Rightarrow\ S_{30}-a_{40}=13635-2345=11290$ Therefore, the required value is $11290$.

Given, $\n$ P be the plane passing through the line $\($ $\frac{x-1}{1}$ = $\frac{y-2}{-3}$ = $\frac{z+5}{7}$ $\)$ and the point $\($(2, 4, -3)$\)$, $\n$ So, equation of plane is given by, $\n$ $$ \begin{vmatrix} x-2 & y-4 & z+3 \\ 1 & -3 & 7 \\ 2-1 & 4-2 & -3-(-5) \end{vmatrix} = 0 $$ $\n$ $$ \Rightarrow \begin{vmatrix} x-2 & y-4 & z+3 \\ 1 & -3 & 7 \\ 1 & 2 & 2 \end{vmatrix} = 0 $$ $\n$ $$ \Rightarrow (x-2)(-20)-(y-4)(-5)+(z+3)(5)=0 $$ $\n$ $$ \Rightarrow 4x-y-z-7=0 $$ $\n$ Now finding the image $\($($\alpha$, $\beta$, $\gamma$)$\)$ of the point $\($(-1, 3, 4)$\)$ with respect to plane $\($4x-y-z-7=0$\)$ we get, $\n$ $$ \frac{x+1}{4} = \frac{y-3}{-1} = \frac{z-4}{-1} = -2 \left( \frac{-4-3-4-7}{16+1+1} \right) = 2 $$ $\n$ $$ \Rightarrow x = 7 = \alpha, \; y = 1 = \beta, \; z = 2 = \gamma $$ $\n$ Hence, $\($ $\alpha$ + $\beta$ + $\gamma$ = 10 $\)$

Let $$z = \frac{1 + 2i \sin \theta}{1 - i \sin \theta}$$ Multiply numerator and denominator by the conjugate of the denominator: $$\Rightarrow z = \frac{1 + 2i \sin \theta}{1 - i \sin \theta} \times \frac{1 + i \sin \theta}{1 + i \sin \theta}$$ Simplifying, we get: $$\Rightarrow z = \frac{1 - i \sin \theta + 2i \sin \theta - 2 \sin^2 \theta}{1 - 2 \sin^2 \theta + 3i \sin \theta}$$ Further simplification gives: $$\Rightarrow z = \frac{1 + \sin^2 \theta}{1 - 2 \sin^2 \theta} + i \left( \frac{3 \sin \theta}{1 + \sin^2 \theta} \right)$$ Since, $z$ is a purely imaginary number, so real part must be zero, hence $$\frac{1 - 2 \sin^2 \theta}{1 + \sin^2 \theta} = 0$$ This implies: $$\Rightarrow 1 - 2 \sin^2 \theta = 0$$ $$\Rightarrow \cos 2\theta = 0$$ Solving for $\theta$, we have: $$\Rightarrow \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, for \theta \in (0, 2\pi)$$ Therefore, the sum of all values is: $$\therefore Sum of all values = \frac{16\pi}{4} = 4\pi$$

The given expansion is $\left(2x^2 + \frac{1}{2x}\right)^{11}$. The general term in the binomial expansion of $(x + a)^n$ is given by $T_{r+1} = {}^{n}C_{r} x^{n-r} a^r$. $$\Rightarrow T_{r+1} = {}^{11}C_{r} (2x^2)^{11-r} \left(\frac{1}{2x}\right)^r$$ $$= {}^{11}C_{r} (2^{11-r-r}) x^{22-2r-r}$$ Now for coefficient of $x^7$ $$\Rightarrow 22 - 3r = 7$$ $$\Rightarrow r = 5$$ Coefficient of $x^7$ is ${}^{11}C_{5} (2^{11-5 \times 2})$ Now for coefficient of $x^{10}$ $$\Rightarrow 22 - 3r = 10$$ $$\Rightarrow r = 4$$ Coefficient of $x^{10}$ is ${}^{11}C_{4} (2^{11-2 \times 4})$ We need the absolute difference, we get by $\left| {}^{11}C_{5} (2^{11-5 \times 2}) - {}^{11}C_{4} (2^{11-4 \times 2}) \right|$. $$= |924 - 2640| = 1716$$ $$= 12^3 - 12$$ Hence the absolute difference is $12^3 - 12$.

The word 'MATHEMATICS' consists of 11 letters including two $M$'s, two $T$'s and two $A$'s. Now, first we will arrange letters $\_M\_A\_T\_H\_E\_M\_A\_T\_I\_$. Here, we have 10 gaps to fill $C$ and $S$. The number of words with or without meaning can be formed from the word MATHEMATICS when $C$, $S$ does not come together is $$= \frac{9!}{2! \cdot 2!} \times \binom{10}{2} \times 2!$$ $$= \frac{9 \times 8 \times 7 \times 6!}{2! \cdot 2!} \times \frac{10 \times 9}{2}$$ $$= 5670$$ Hence this is the correct option.

Given, $S$ be the set of all values of $\theta \in [-\pi, \pi]$ for which the system of linear equations $$x + y + \sqrt{3}z = 0$$ $$-x + \left( \tan \theta \right) y + \sqrt{7}z = 0$$ $$x + y + (\tan \theta) z = 0$$ has non-trivial solution, So, by condition of non-trivial solution we get, $$\begin{vmatrix} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{vmatrix} = 0$$ $$\Rightarrow \left( \tan^2 \theta - \sqrt{7} \right) - \left( - \tan \theta - \sqrt{7} \right) + \sqrt{3}(-1 - \tan \theta) = 0$$ $$\Rightarrow \tan^2 \theta + \left( 1 - \sqrt{3} \right) \tan \theta - \sqrt{3} = 0$$ $$\Rightarrow (\tan \theta + 1) \left( \tan \theta - \sqrt{3} \right) = 0$$ $$\Rightarrow \tan \theta = \sqrt{3}, -1$$ $$\Rightarrow \theta = \frac{\pi}{3}, -\frac{2\pi}{3}, -\frac{\pi}{4}, \frac{3\pi}{4}$$ Hence, $$\frac{120}{\pi} \sum \theta = 120 \left( \frac{1}{3} - \frac{2}{3} - \frac{1}{4} + \frac{3}{4} \right) = 20$$

Given, the probability that the random variable $X$ takes values $x$ is given by $P(X = x) = k(x + 1)3^{-x}$, $x = 0, 1, 2, 3, \ldots$, where $k$ is a constant. Now we know that, $$P(X = 0) + P(x = 1) + P(x = 2) + \ldots = 1$$ $$\Rightarrow \frac{k}{3^0} + \frac{2k}{3^1} + \frac{3k}{3^2} + \ldots = 1$$ $$\Rightarrow k \left(1 + \frac{2}{3} + \frac{3}{3^2} + \ldots \right) = 1$$ Now finding, $$S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \ldots (1)$$ $$\Rightarrow \frac{S}{3} = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \ldots \ldots (2)$$ Now on subtracting above two equation we get, $$\Rightarrow \frac{2S}{3} = 1 + \frac{1}{3} + \frac{1}{3^2} + \ldots$$ $$\Rightarrow S = \frac{9}{4}$$ Hence, $$k \left(1 + \frac{2}{3} + \frac{3}{3^2} + \ldots \right) = 1$$ $$\Rightarrow k = \frac{4}{9}$$ Now finding, $$P(X \geq 2) = P(2) + P(3) + \ldots$$ $$\Rightarrow P(X \geq 2) = 1 - P(0) - P(1)$$ $$\Rightarrow P(X \geq 2) = 1 - \left(\frac{k}{1} + \frac{2k}{3} \right) = 1 - \frac{20}{27} = \frac{7}{27}$$

We need to find the value of $$36 (4 \cos^2 9^\circ - 1)(4 \cos^2 27^\circ - 1)(4 \cos^2 81^\circ - 1)(4 \cos^2 243^\circ - 1)$$ $$= 36 \left[ (4 \cos^2 9^\circ - 1)(4 \sin^2 9^\circ - 1)(4 \cos^2 27^\circ - 1)(4 \sin^2 27^\circ - 1) \right]$$ $$= 36 \left[ \left( 4(2 \cos 9^\circ \sin 9^\circ)^2 - 4 \cos^2 9^\circ - 4 \sin^2 9^\circ + 1 \right) \left( 4(2 \sin 27^\circ \cos 9^\circ)^2 - 4 \cos^2 27^\circ - 4 \sin^2 27^\circ + 1 \right) \right]$$ $$= 36 \left[ (4 \sin^2 18^\circ - 4 + 1)(4 \sin^2 54^\circ - 4 + 1) \right]$$ $$= 36 \left[ (4 \sin^2 18^\circ - 3)(4 \sin^2 54^\circ - 3) \right]$$ $$= 36 \left\{ \left( \frac{\sqrt{5} - 1}{4} \right)^2 - 3 \right\} \left\{ \left( \frac{\sqrt{5} + 1}{4} \right)^2 - 3 \right\}$$ $$= 36 \left\{ \left( \frac{6 - 2\sqrt{5} - 12}{4} \right) \left( \frac{6 + 2\sqrt{5} - 12}{4} \right) \right\}$$ $$= 36 \left\{ \left( \frac{2\sqrt{5}}{16} \right)^2 - 36 \right\} = 36$$ Hence this is the correct option.

Given, $$I = \int \left( \left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x \right) \log_2 x \, dx$$ Now, let $$\left( \frac{x}{2} \right)^x = t$$ $$\Rightarrow x \log_2 \left( \frac{x}{2} \right) = \log_2 t$$ $$\Rightarrow x \log_e \left( \frac{x}{2} \right) \cdot \log_2 e = \log_e t \cdot \log_2 e$$ On differentiating both sides we get: $$\log_e \left( \frac{x}{2} \right) + x \cdot \frac{2}{x} \cdot \frac{1}{2} = \frac{1}{t} \frac{dt}{dx}$$ $$\Rightarrow \log_e \left( \frac{x}{2} \right) + 1 = \frac{1}{t} \frac{dt}{dx}$$ Then solution is not possible as there is no proper substitution. Note: This question was bonus in Jee Mains 2023 April session.

Given, the quadrilateral $ABCD$ with vertices $A(2, 1, 1)$, $B(1, 2, 5)$, $C(-2, -3, 5)$ and $D(1, -6, -7)$. Now finding, $\overrightarrow{AB} = -\hat{i} + \hat{j} + 4\hat{k}$ and $\overrightarrow{AD} = -\hat{i} - 7\hat{j} - 8\hat{k}$. And plotting the diagram we get, Now we know that, the area of $\triangle ABD$ is given by, $$= \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 4 \\ -1 & -7 & 8 \end{vmatrix}$$ $$= \frac{1}{2} \begin{vmatrix} 10\hat{i} - 6\hat{j} + 4\hat{k} \end{vmatrix} = 2\sqrt{38}$$ Now finding, $\overrightarrow{CB} = 3\hat{i} + 5\hat{j}$ and $\overrightarrow{CD} = 3\hat{i} - 3\hat{j} - 12\hat{k}$. So, the area of $\triangle CBD$ is $$= \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 0 \\ 3 & -3 & -12 \end{vmatrix}$$ $$= \frac{1}{2} \begin{vmatrix} 6 \left( 5\hat{i} - 3\hat{j} - 2\hat{k} \right) \end{vmatrix}$$ $$= 6\sqrt{38}$$ Therefore, the area of quadrilateral $ABCD = 8\sqrt{38}$ square units.

Given, for $a, b \in \mathbb{Z}$ and $|a - b| \leq 10$, let the angle between the plane $P: ax + y - z = b$ and the line $L: x - 1 = a - y = z + 1$ be $\cos^{-1}\left(\frac{1}{3}\right)$. Now rewriting the equation of line we get, $$L: \frac{x - 1}{1} = \frac{y - a}{-1} = \frac{z + 1}{1}$$ Now let, $\theta = \cos^{-1}\left(\frac{1}{3}\right) = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Now we know that, angle between plane and line is given by, $$\cos(90^\circ - \theta) = \frac{a + 1)(-1) + (-1)1}{\sqrt{a^2 + 2} \cdot \sqrt{3}}$$ $$\Rightarrow \frac{a - 2}{\sqrt{a^2 + 2}} = \frac{2\sqrt{2}}{\sqrt{3}}$$ $$\Rightarrow 3(a - 2)^2 = 8(a^2 + 2)$$ $$\Rightarrow 3(a^2 + 4 - 4a) = 8(a^2 + 2)$$ $$\Rightarrow 5a^2 + 12a + 4 = 0$$ So, $a = -\frac{2}{5}, a = -2$. Hence, $a = -2$ (since $a \in \mathbb{Z}$). So, the equation of plane will be, $$P: -2x + y - z - b = 0$$ Now given, distance of point $(6, -6, 4)$ from plane is $= 3\sqrt{6}$. $$\Rightarrow \frac{|12 - 6 - 4 - b|}{\sqrt{2^2 + 1^2 + 1^2}} = 3\sqrt{6}$$ $$\Rightarrow |22 + b| = 18$$ $$\Rightarrow b = -4, -40$$ So, $b = -4$ (since $|a - b| \leq 10$). Hence, $a^4 + b^2 = 16 + 16 = 32$

Given, $$25^{190} - 19^{190} - 8^{190} + 2^{190}$$ Now $\left(25^{190} - 19^{190}\right)$, $8^{190}$, $2^{190}$ are divisible by 2, as $25^{190}$ will give remainder as 1 and $19^{190}$ will give remainder as 1 when divided by 2, so combined $25^{190} - 19^{190}$ will give remainder 0. Now, $\left(25^{190} - 8^{190}\right)$ $\&$ $\left(-19^{190} + 2^{190}\right)$ are divisible by 17, $a^n - b^n$ is divisible by $a - b$ if $n$ is even. So given number is divisible by 34 as it is divisible by 2 $\&$ 17 both. Now, $25^{190} = (7 \times 3 + 4)^{190}$, dividing by 7 we get, $$(7 \times 3 + 4)^{190} = 4^{190} = 16^{95} = 2^{380} = 4 \cdot 8^{126}$$ Now when $4 \cdot (7 + 1)^{126}$ divided by 7 will give remainder as $4 \cdot 1 = 4$. And $19^{190} = (7 \times 3 - 2)^{190}$, when divided by 7 we get, $$(7 \times 3 - 2)^{190} = (-2)^{190} = 2^{190} = 2 \times (7 + 1)^{63}$$ So, when $2 \times (7 + 1)^{63}$ divided by 7 will give remainder as 2. Similarly $8^{190} = (7 + 1)^{190}$ will give remainder 1. And $2^{190} = 2 \times (7 + 1)^{63}$ will give remainder 2. So combining all we get, $$25^{190} - 19^{190} - 8^{190} + 2^{190} = 4 - 2 - 1 + 2 = 3$$ $\Rightarrow$ Not divisible by 7, so it will not be divisible by 14.

Given: $\vec{u}_1 = \hat{i} + \hat{j} + a\hat{k}$, $\vec{u}_2 = \hat{i} + b\hat{j} + \hat{k}$ and $\vec{u}_3 = c\hat{i} + \hat{j} + \hat{k}$ are coplanar. Also given that, $\vec{v}_1 = (a + b)\hat{i} + c\hat{j} + c\hat{k}$, $\vec{v}_2 = a\hat{i} + (b + c)\hat{j} + a\hat{k}$ and $\vec{v}_3 = b\hat{i} + b\hat{j} + (c + a)\hat{k}$ are coplanar. Now, using the condition of coplanar we get, $$\begin{vmatrix} 1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1 \end{vmatrix} = 0$$ Expanding the determinant along $R_1$. $$\Rightarrow (b - 1) - (1 - c) + a(1 - bc) = 0$$ $$\Rightarrow a + b + c = 2 + abc \cdots (i)$$ Again using the coplanar condition we get, $$\begin{vmatrix} a + b & c & c \\ a & b + c & a \\ b & b & c + a \end{vmatrix} = 0$$ Apply row transformations, $(R_3 \rightarrow R_3 - (R_1 + R_2))$ $$\Rightarrow \begin{vmatrix} a + b & c & c \\ a & b + c & a \\ -2a & -2c & 0 \end{vmatrix} = 0$$ Expand the determinant along $R_1$. $$\Rightarrow (a + b)(0 + 2ac) - c(0 + 2a^2) + c(-2ac + 2a(b + c)) = 0$$ $$\Rightarrow 2a^2c + 2abc - 2ac^2 - 2ac^2 + 2abc + 2ac^2 = 0$$ $$\Rightarrow abc = 0$$ Therefore, $a + b + c = 2$ (From eq $(i)$) Therefore, $6(a + b + c) = 12$ Hence this is the correct option.

The given information can be represented in the form of the diagram below. Since angle in a semicircle is a right angle. Hence, the other circle will be passing through the centre of the first circle. The two ends of diameter of the circle is $O(0, 0)$ and $C(3, -2)$. Hence, the required equation of circle is $$(x - 0)(x - 3) + (y - 0)(y - (-2)) = 0$$ $$\Rightarrow x^2 + y^2 - 3x + 2y = 0$$ Put $\left( \alpha, \frac{1}{2} \right)$ in the above equation $$\Rightarrow \alpha^2 + \frac{1}{4} - 3\alpha + 1 = 0$$ $$\Rightarrow \alpha^2 - 3\alpha + \frac{5}{4} = 0$$ $$\Rightarrow 4\alpha^2 - 12\alpha + 5 = 0$$ $$\Rightarrow 4\alpha^2 - 10\alpha - 2\alpha + 5 = 0$$ $$\Rightarrow 2\alpha(2\alpha - 5) - 1(2\alpha - 5) = 0 \Rightarrow \alpha = \frac{1}{2}, \frac{5}{2}$$ Therefore this is the correct option.

The given statement is $(p \land (\sim q)) \lor (\sim p)$. Let us take $(A \cap B^c) \cup (A^c)$. Now let us apply distributive law to the above statement. $$\equiv (A \cup A^c) \cap (B^c \cup A^c)$$ $$\equiv (U) \cap (B^c \cup A^c)$$ $$\equiv (B^c \cup A^c)$$ Now the given statement $(p \land (\sim q)) \lor (\sim p)$ will be equivalent to $\sim q \lor \sim p$ or $\sim p \lor \sim q$. Hence the given statement is equivalent to $\sim p \lor \sim q$. Now negation of given statement by demorgan's law will be, $$\sim (\sim p \lor \sim q) \equiv p \land q$$

Since $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + 1 = 0$, therefore $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ would be the roots of $x^2 + bx + a = 0$. Let $$L = \lim_{x \to \frac{1}{\alpha}} \left[ \frac{1 - \cos \left( x^2 + bx + a \right)}{2 \left( 1 - \alpha x \right)^2} \right]^{\frac{1}{2}} \to \frac{0}{0}$$ $$\Rightarrow L = \lim_{x \to \frac{1}{\alpha}} \left[ \frac{2 \sin^2 \left( \frac{x^2 + bx + a}{2} \right)}{2 \left( 1 - \alpha x \right)^2} \right]^{\frac{1}{2}}$$ $$\Rightarrow L = \lim_{x \to \frac{1}{\alpha}} \left[ \frac{\sin^2 \left( \frac{1}{2} \left( x - \frac{1}{\alpha} \right) \left( x - \frac{1}{\beta} \right) \right)}{a^2 \left( x - \frac{1}{\alpha} \right)^2} \right]^{\frac{1}{2}}$$ $$\Rightarrow L = \lim_{x \to \frac{1}{\alpha}} \left[ \frac{\left( x - \frac{1}{\beta} \right)^2}{4a^2} \times \frac{\sin \left( \frac{1}{2} \left( x - \frac{1}{\alpha} \right) \right)}{\left( \frac{1}{2} \right)^2 \left( x - \frac{1}{\alpha} \right)^2} \right]^{\frac{1}{2}}$$ $$\Rightarrow L = \frac{\left( \frac{1}{\alpha} - \frac{1}{\beta} \right)}{2a}$$ On comparing this value with the value given in the question we get, $k = 2\alpha$. Hence this is the correct option.

Given, $$A = \begin{bmatrix} 1 & 5 \\ \lambda & 10 \end{bmatrix}, \ A^{-1} = \alpha A + \beta I \ and \ \alpha + \beta = -2,$$ Now solving by using characteristic equation, we get $$\left| A - k \right| = 0 \Rightarrow \left| \begin{matrix} 1 - k & 5 \\ \lambda & 10 - k \end{matrix} \right| = 0$$ $$\Rightarrow k^2 - 11k + 10 - 5\lambda = 0$$ $$\Rightarrow A^2 - 11A + (10 - 5\lambda)I = 0 \ \{by putting \ k = A\}$$ $$\Rightarrow A^{-1} = \frac{1}{10 - 5\lambda} \left( -A + 11I \right)$$ Now on comparing with $A^{-1} = \alpha A + \beta I$ we get, $$\Rightarrow \alpha = -\frac{1}{10 - 5\lambda}, \ \beta = \frac{11}{10 - 5\lambda}$$ And given, $$\alpha + \beta = -2 \Rightarrow \lambda = 3, \ \alpha = \frac{1}{5}, \ \beta = \frac{-11}{5}$$ Hence, the value of $6\alpha^2 + \beta^2 + \lambda^2 = 14$

Given, $A(0, 1)$, $B(1, 1)$ and $C(1, 0)$ be the mid-points of the sides of a triangle with incentre at the point $D$. Now finding the vertices of the triangle by using midpoint formula and plotting the diagram we get, Now finding the incentre of the above triangle using the formula, $$I \equiv \left( \frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c} \right)$$ We get, Incentre $D = \left( \frac{4}{4+2\sqrt{2}}, \frac{4}{4+2\sqrt{2}} \right)$ Now given parabola $y^2 = 4ax$ passes through the incentre $D$ we get, $$\left( \frac{4}{4+2\sqrt{2}} \right)^2 = 4a \left( \frac{4}{4+2\sqrt{2}} \right)$$ $$\Rightarrow a = \frac{1}{4+2\sqrt{2}}$$ Now we know that focus of parabola is given by, focus $(a, 0)$ Hence, focus will be, $\left( \frac{1}{4+2\sqrt{2}}, 0 \right) = \left( \frac{4-2\sqrt{2}}{8}, 0 \right)$ Now comparing with $\left( \alpha + \beta \sqrt{2}, 0 \right)$ we get, $$\alpha = \frac{4}{8}, \beta = -\frac{1}{4}$$ Hence, $$\frac{\alpha}{\beta^2} = \frac{\frac{1}{2}}{\left( -\frac{1}{4} \right)^2} = \frac{16}{2} = 8$$

We have, $A = \{1, 2, 3, 4, 5, 6, 7\}$ Reflexive: A relation $R$ on a set $A$ is said to be reflexive if every element of $A$ is related to itself. Thus, $R$ is reflexive $\iff (a, a) \in R$ for all $a \in A$ $$\therefore (1, 1), (2, 2), (3, 3), \ldots, (7, 7)$$ does not satisfy $x + y = 7$ Hence $R$ is not reflexive. Symmetric: A relation $R$ is symmetric on a set $A$ iff $$(a, b) \in R \Rightarrow (b, a) \in R for all a, b \in A$$ $$\Rightarrow x + y = 7$$ Now on interchanging $y$ and $x$ we get, $y + x = 7$ which is always true for given set, Hence $R$ is symmetric. Transitive: A relation $R$ on $A$ is said to be transitive relation iff $$(a, b) \in R and (b, c) \in R$$ $$\Rightarrow (a, c) \in R for all a, b, c \in A$$ Now taking $(a, b) \equiv (3, 4)$ and $(b, c) \equiv (4, 3)$ so $(a, c) \equiv (3, 3)$ does not satisfy $x + y = 7$, Hence, $R$ is not transitive and not equivalence. Therefore, $R$ is only Symmetric.

Given that $$\int_0^{2.4} [x^2] \, dx = \alpha + \beta \sqrt{2} + \gamma \sqrt{3} + \delta \sqrt{5}$$, then $$\alpha + \beta + \gamma + \delta$$. We know that for $$1 \leq x < 2$$, $$[x] = 1$$. Therefore, $$\int_0^{2.4} [x^2] \, dx = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{\sqrt{5}} 3 \, dx + \int_{\sqrt{5}}^2 4 \, dx + \int_2^{2.4} 5 \, dx$$. $$= 0 + \left( \sqrt{2} - 1 \right) + 2 \left( \sqrt{3} - \sqrt{2} \right) + 3 \left( 2 - \sqrt{3} \right) + 4 \left( \sqrt{5} - 2 \right) + 5 \left( 2.4 - \sqrt{5} \right)$$ $$= 9 - \sqrt{2} - \sqrt{3} - \sqrt{5}$$. This is in the form of $$\alpha + \beta \sqrt{2} + \gamma \sqrt{3} + \delta \sqrt{5}$$. $$\alpha = 9$$, $$\beta = -1$$, $$\gamma = -1$$, $$\delta = -1$$. Therefore, $$\alpha + \beta + \gamma + \delta = 9 - 1 - 1 - 1 = 6$$. Hence this is the required answer.

Since, $f(x) = \begin{cases} 3x^2 + k\sqrt{x} + 1; & 0 1 \end{cases}$$ So, $$f'(1^-) = f'(1^+)$$ $$\Rightarrow 6 + \frac{k}{2\sqrt{2}} = 2m$$ $$\Rightarrow m = 3 + \frac{k}{4\sqrt{2}} \ldots (2)$$ Putting in (1), we get $$k^2 - k\sqrt{2} + 3 + \frac{k}{4\sqrt{2}} - 3 = 0$$ $$\Rightarrow k^2 - \left( \sqrt{2} - \frac{1}{4\sqrt{2}} \right) k = 0$$ $$\Rightarrow k^2 - \left( \frac{7\sqrt{2}}{8} \right) k = 0$$ $$\Rightarrow k \left( k - \frac{7\sqrt{2}}{8} \right) = 0$$ $$\Rightarrow k = \frac{7\sqrt{2}}{8}$$ So, $$m = 1 + \frac{-k}{12\sqrt{2}} = 1 + \frac{7\sqrt{2}}{96\sqrt{2}} = \frac{103}{96}$$ Hence, $$f'(x) = \begin{cases} 6x + \frac{7\sqrt{2}}{16\sqrt{x}}; & 0 1 \end{cases}$$ So, $$\frac{f'(8)}{f'(\frac{1}{8})} = \frac{103 \cdot 8}{16} \times \frac{1}{\left( 6 \cdot \frac{1}{8} + \frac{7\sqrt{2}}{16\sqrt{\frac{1}{8}}} \right)}$$ $$\Rightarrow \frac{f'(8)}{f'(\frac{1}{8})} = \frac{103}{2} \times \frac{3}{4}$$ $$\Rightarrow \frac{f'(8)}{f'(\frac{1}{8})} = 309$$

Given that $\frac{1}{x}$, $\frac{1}{y}$, $\frac{1}{z}$ are in AP and $x$, $\sqrt{2}y$, $z$ are in GP. As given, $\frac{2}{y}=\frac{1}{x}+\frac{1}{z} \tag{i}$ Also, $2y^2=xz \tag{ii}$ Also given that $xy+yz+zx=\frac{3}{\sqrt{2}}xyz$. $\Rightarrow\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{\sqrt{2}} \tag{iii}$ From (i) and (iii), we get $\frac{3}{y}=\frac{3}{\sqrt{2}}$ $\Rightarrow\ y=\sqrt{2} \tag{iv}$ Now from (ii), $xz=2y^2=2(\sqrt{2})^2=4 \tag{v}$ Now using (ii), (iv) and (v), $\Rightarrow\ x+z=4\sqrt{2}$ Hence, $3(x+y+z)^2=3\left(\sqrt{2}+4\sqrt{2}\right)^2$ $=3(5\sqrt{2})^2$ $=150$ Therefore, this is the required answer.

We need to find the domain of the function $\log_e \left( \frac{6x^2 + 5x + 1}{2x - 1} \right) + \cos^{-1} \left( \frac{2x^2 - 3x + 4}{3x - 5} \right)$. $$\Rightarrow \frac{6x^2 + 5x + 1}{2x - 1} > 0 \ldots (1)$$ and $$-1 \leq \frac{2x^2 - 3x + 4}{3x - 5} \leq 1 \ldots (2)$$ Now from (1), $$\Rightarrow \frac{6x^2 + 5x + 1}{2x - 1} > 0$$ $$(3x + 1)(2x + 1) > 0$$ Now we get the common region here as $$\Rightarrow x \in \left( -\frac{1}{2}, -\frac{1}{3} \right) \cup \left( \frac{1}{2}, \infty \right) \ldots (a)$$ From (2) we get that $$\Rightarrow -1 \leq \frac{2x^2 - 3x + 4}{3x - 5} \leq 1$$ $$\Rightarrow \frac{2x^2 - 3x + 4}{3x - 5} \geq -1 \ldots (3)$$ $$\Rightarrow \frac{2x^2 - 3x + 4}{3x - 5} \leq 1 \ldots (4)$$ From (3) we get $$\Rightarrow \frac{2x^2 - 3x + 4}{3x - 5} + 1 \geq 0$$ $$\Rightarrow x \in \left[ -\frac{1}{\sqrt{2}}, \frac{1}{3} \right) \cup \left( \frac{5}{3}, \infty \right) \ldots (b)$$ From (4) we get $$\Rightarrow \frac{2x^2 - 3x + 4}{3x - 5} - 1 \leq 0$$ $$\Rightarrow \frac{2x^2 - 6x + 9}{3x - 5} \leq 0$$ As $D = (-6)^2 - 4(2)(9) 0$, we observe that $2x^2 - 6x + 9 > 0 \forall x \in \mathbb{R}$ $$\Rightarrow \frac{1}{3x - 5} \leq 0 (\because 2x^2 - 6x + 9 > 0 \forall x \in \mathbb{R})$$ $$\Rightarrow x \in \left( -\infty, \frac{5}{3} \right) \ldots (c)$$ Intersection of (a), (b) and (c) gives us $$x \in \left( -\frac{1}{2}, -\frac{1}{3} \right) \cup \left( \frac{1}{2}, \frac{1}{\sqrt{2}} \right).$$ On comparing this with $(\alpha, \beta) \cup (\gamma, \delta)$, $$\Rightarrow \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = \frac{10}{9}$$ $$\Rightarrow 18(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) = 20.$$ Hence this is the required answer.

Given, $$x^2 - 12x + [x] + 31 = 0$$ $$\Rightarrow x^2 - 12x + \underbrace{31}_{\geq -5} = -[x]$$ Now from above equation we say that, it could have its solution in $[5, 6)$ but it does not exist as at $x = 5$ as LHS = 1, So no solution, hence $m = 0$ Now solving, $$x^2 - 5|x + 2| - 4 = 0$$ Taking Case 1 when $x \geq -2$ we get, $$x^2 - 5(x + 2) - 4 = 0$$ $$\Rightarrow x^2 - 5x - 14 = 0 \Rightarrow x = 7, -2$$ Now taking Case 2 when $x < -2$ we get, $$x^2 + 5x + 10 - 4 = 0$$ $$\Rightarrow x = -2, -3$$ So, total 3 solution i.e., $x = -3, -2, 7$ Hence, $n = 3$ So, the value of $m^2 + mn + n^2 = 0 + 0 + 3^2 = 9$

Given, Focus $(3, 0)$, Directrix, $x = -3$. So, the equation of parabola will be, $y^2 = 12x$. Now taking the parametric point $(3t^2, 6t)$ and plotting the diagram we get, Now we know that $R(\alpha, \beta)$ will be the arithmetic mean of ordinate and geometric mean of abscissa will be, Hence, $\beta = \frac{6\left(t_1 + t_2\right)}{2}$ and $\alpha = \sqrt{3t_1^2 \times 3t_2^2} = 3t_1 t_2$. So, putting the value in $\frac{\beta^2}{\alpha}$, $\Rightarrow \frac{\beta^2}{\alpha} = \frac{9\left(t_1 + t_2\right)^2}{3t_1 t_2} = \frac{9\left(4t_2\right)^2}{3 \times 3t_2^2} = 16 \left(\because \frac{6t_1}{6t_2} = 3, given\right)$

Given, $P_1$ be the plane $3x - y - 7z = 11$, $P_2$ be the plane passing through the points $(2, -1, 0)$, $(2, 0, -1)$, and $(5, 1, 1)$. Now finding the equation of plane $P_2$, we get $$\begin{vmatrix} x - 2 & y + 1 & z \\ 2 - 2 & 0 - (-1) & 1 - 0 \\ 5 - 2 & 1 - 0 & 1 - (-1) \end{vmatrix} = 0$$ $$\Rightarrow \begin{vmatrix} x - 2 & y + 1 & z \\ 0 & 1 & 1 \\ 3 & 1 & 2 \end{vmatrix} = 0$$ $$\Rightarrow 3(x - 2) + (z)(-3) - 3z = 0$$ $$\Rightarrow x - 2y - 1 - z = 0$$ $$\Rightarrow x - y - z = 3$$ Now direction ratio of line of intersection of planes $3x - y - 7z = 11$ and $x - y - z = 3$ is given by, $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & -7 \\ 1 & -1 & -1 \end{vmatrix}$$ $$= \hat{i}(1 \cdot (-7)) - \hat{j}(-3 + 7) + \hat{k}(-3 + 1)$$ $$= -6\hat{i} - 4\hat{j} - 2\hat{k}$$ Or direction ratio is given by $3\hat{i} + 2\hat{j} + \hat{k}$. Now finding the point on the line of intersection by taking $z = 0$ in both planes, We get, $3x - y = 11$ and $x - y = 3$. Now solving above two lines we get, $(4, 1, 0)$ is a point on line of intersection. Hence, the equation of line will be, $$\frac{x - 4}{3} = \frac{y - 1}{2} = \frac{z}{1} = \lambda$$ Now any point on the line is given by, $(3\lambda + 4, 2\lambda + 1, \lambda)$. Now finding the distance of the point $(7, 4, -1)$ from the line, $$(7, 4, -1)$$ $$(4 + 3\lambda, 1 + 2\lambda, 0 + \lambda)$$ Now from above diagram using perpendicular condition we get, $$3(3\lambda - 3) + 2(2\lambda - 3) + 1(\lambda + 1) = 0$$ $$\Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$$ Hence, the point is $(7, 3, 1)$ and comparing with $(\alpha, \beta, \gamma)$ we get, $\alpha + \beta + \gamma = 11$

Given, $R = \{a, b, c, d, e\}$ and $S = \{1, 2, 3, 4\}$. Now taking, $f(a) = 1$ we get, one of $f(b), f(c), f(d), f(e) = 1$ then total such cases $= 4 \cdot 3! = 24$. Now if, only $f(a) = 1$, then we have distribute $\{2, 3, 4\}$ amongst $\{b, c, d, e\}$, So, total cases $= 3^4 - \left(\binom{3}{1} \cdot 2^4\right) + \left(\binom{3}{2} \cdot 1\right)$ $$= 36$$ So, number of onto functions when $f(a) = 1$ is $24 + 36 = 60$ Now finding, total number of onto functions, $$= 4^5 - \left(\binom{4}{1} \cdot 3^5\right) + \left(\binom{4}{2} \cdot 2^5\right) - \left(\binom{4}{3} \cdot 1\right)$$ $$= 1024 - 973 + 192 - 4$$ $$= 240$$ Number of required functions when $f(a) \neq 1$ will be, $$= 240 - 60 = 180$$

Given, the lines $x + y = 2$, $y = 0$, $x = 0$ and the curve $f(x) = \min \left\{ x^2 + \frac{3}{4}, 1 + \left\lfloor x \right\rfloor \right\}$ where $\left\lfloor x \right\rfloor$ denotes the greatest integer $\leq x$. Now plotting the diagram of the given function we get, now from the above diagram, the area enclosed is given by, $$A = \left[ \int_0^{\frac{1}{2}} \left( x^2 + \frac{3}{4} \right) dx \right] + \frac{1}{2} \left( \frac{1}{2} + \frac{3}{2} \right) \times 1 = \frac{5}{12} + 1$$ $$\Rightarrow 12A = 17$$

The electric potential for a point charge is given by $$V = \frac{Q}{4 \pi \varepsilon_0 r}$$ The data given is $$V = 50 \, \mathrm{V}$$ $$Q = 5 \times 10^{-9} \, \mathrm{C}$$ So, the distance of the point is $$r = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{V}$$ $$\Rightarrow r = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{50}$$ $$\Rightarrow r = 0.9 \, \mathrm{m} = 90 \, \mathrm{cm}$$

As can be seen from the diagram, the initial phase of the particle is given by $\varphi = 30^\circ = \frac{\pi}{6}$. The equation of motion of the particle at its initial position at time $t = 0$ can be written as $$x(t) = r \cos \theta \cdots (1)$$ where, $\theta$ is the angular displacement of the particle at time $t = t$. Since, the position of the particle at a later instant of time always lags behind its initial position at time $t = 0$, the equation of motion of the particle at any instant of time is given by $$x(t) = r \cos(\theta + \varphi) \cdots (2)$$ Substitute the values of the known parameters into equation (2) to obtain the required expression $$x\left(t\right) = r \cos\left(\omega t + \frac{\pi}{6}\right)$$

The dimensions of the quantities are as follows. $$F = M \, LT^{-2}$$ $$r = L$$ $$v = LT^{-1}$$ $$A = L^2$$ The dimensional form of stress is same as pressure. Torque $=$ Force $\times$ distance $=$ $ML^2 \, T^{-2}$ Stress $=$ $\frac{Force}{Area} = M \, L^{-1} \, T^{-2}$ Pressure Gradient $=$ $\frac{dp}{dx} = ML^{-2} \, T^{-2}$ $$\eta = \frac{F}{6 \pi rv} = \frac{MLT^{-2}}{L \, LT^{-1}} = ML^{-1} \, T^{-1}$$

Considering the transistor in saturation mode, $V_{CE} = 0$. Using Kirchhoff's voltage rule, $$-I_C R_C + V_{CC} = 0$$ implies $$I_C = \frac{V_{CC}}{R_C}$$ which gives $$I_C = \frac{1 \, \mathrm{V}}{10^3 \, \Omega} = 10^{-3} \, \mathrm{A}$$ So, $$\beta = \frac{I_C}{I_b} = \frac{10^{-3} \, \mathrm{A}}{I_b}$$ implies $$I_b = \frac{10^{-3}}{10^2} = 10^{-5} \, \mathrm{A} = 10 \, \mu \mathrm{A}$$

The trajectory of a projectile is parabolic in nature. At maximum height, it can be written that $$\frac{dy}{dx} = 0 \ldots (1)$$ Substitute the expression for trajectory into equation (1) and solve to calculate the horizontal distance, for which the projectile attains the maximum height. $$\frac{d}{dx} \left[ x - \frac{x^2}{20} \right] = 0$$ $$\Rightarrow 1 - \frac{x}{10} = 0$$ $$\Rightarrow x = 10 \, \mathrm{m}$$ Substitute the value of $x$ in the given expression for the trajectory to obtain the maximum height. $$y = \left( 10 - \frac{10^2}{20} \right) \, \mathrm{m}$$ $$= 5 \, \mathrm{m}$$

The initial and final amount of the radioactive substance are related by the formula $$\frac{N}{N_0} = \left( \frac{1}{2} \right)^n \ldots (1)$$ where, $n$ is the number of half lives. Substitute $\frac{1}{8}$ for $\frac{N}{N_0}$ into equation (1) and solve to calculate the value of $n$. $$\frac{1}{8} = \left( \frac{1}{2} \right)^n$$ $$\Rightarrow \left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^n$$ $$\Rightarrow n = 3$$ The half-life of the radioactive element can be calculated as follows: $$\frac{t}{t_{1/2}} = n \ldots (2)$$ Substitute the values of the known parameters into equation (2) and solve to calculate the half-life. $$\frac{3 \, days}{t_{1/2}} = 3$$ $$\Rightarrow t_{1/2} = 1 \, day$$ The initial and the final mass of the radioactive element are related by $$\frac{m}{m_0} = \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \ldots (3)$$ Substitute the values of the known parameters into equation (3) and solve to calculate the initial mass of the substance. $$\frac{8 \times 10^{-3} \, kg}{m_0} = \left( \frac{1}{2} \right)^{\frac{5}{1}}$$ $$\Rightarrow m_0 = 2^5 \times 8 \times 10^{-3} \, kg$$ $$= 256 \, g$$

The equivalent circuit of the given configuration can be drawn as follows: With reference to the above diagram, the equivalent resistance of the given configuration can be calculated as follows: $$\frac{1}{R} = \frac{1}{20 \, \Omega} + \frac{1}{20 \, \Omega} + \frac{1}{10 \, \Omega}$$ $$= \frac{1}{5 \, \Omega}$$ $$\Rightarrow R = 5 \, \Omega$$

The given data is $m = 5000 \, \mathrm{kg}$ $g = 10 \, \mathrm{m \, s^{-2}}$ $A = 250 \times 10^{-4} \, \mathrm{m^2}$ Using the formula of pressure, $$P = \frac{F}{A} = \frac{mg}{A}$$ $$= \frac{5000 \times 10}{250 \times 10^{-4}}$$ $$= 2 \times 10^6 \, \mathrm{N \, m^{-2}}$$ $$= 2 \times 10^6 \, \mathrm{Pa}$$

For a satellite revolving around the Earth in a circular orbit, it can be written that $$\frac{mv^2}{r} = \frac{GMm}{r^2} \ldots(1) \; (r = R + h \; is the distance from the centre of the Earth)$$ Simplify equation (1) to obtain the linear speed of the satellite. $$v = \sqrt{\frac{GM}{r}}$$ Thus, the angular momentum of the satellite is given by $$L = mvr$$ $$= mr \sqrt{\frac{GM}{r}}$$ $$= m\sqrt{GMr} \ldots(2)$$ Similarly, for distance $r'$ from the centre of the Earth, the angular momentum is given by $$L' = m\sqrt{GMr'} \ldots(3)$$ Divide equation (3) by equation (2) to obtain the required angular momentum. $$\frac{L'}{L} = \frac{m\sqrt{GMr'}}{m\sqrt{GMr}}$$ $$= \sqrt{\frac{r'}{r}} \ldots(4)$$ Now, $r' = r + 8r = 9r$ Substitute the known values of the parameters into equation (4) to obtain the required angular momentum. $$\frac{L'}{L} = \sqrt{9}$$ $$\Rightarrow L' = 3L$$

The kinetic energy is given by $KE = \frac{f}{2} kT$. Let the kinetic energy be $K_1$ at $27 \, ^\circ \mathrm{C}$ and the kinetic energy be $K'$ at $T'$ temperature. As given in the question, $K' = 2K_1$. Hence, $$\frac{K'}{K_1} = \frac{T'}{T}$$ $$\Rightarrow 2 = \frac{T'}{300}$$ $$\Rightarrow T' = 600 \, \mathrm{K}$$ In Celsius the temperature is $T' - 273 = 327 \, ^\circ \mathrm{C}$.

The acceleration due to gravity $g'$ at a height $h$ from the surface of Earth is given by $$g' = \frac{GM}{(R+h)^2} \cdots (1)$$ Use the method of binomial expansion to obtain the approximate expression for the acceleration due to gravity at a particular height from the surface of the Earth. $$g' = \frac{GM}{R^2 \left(1 + \frac{h}{R}\right)^2}$$ $$= \frac{GM}{R^2} \left(1 + \frac{h}{R}\right)^{-2}$$ $$\approx \frac{GM}{R^2} \left(1 - \frac{2h}{R}\right) \cdots (2)$$ The formula to calculate the acceleration due to gravity at the surface of the Earth is given by $$g = \frac{GM}{R^2} \cdots (3)$$ Divide equation (2) by equation (3) and simplify to obtain the required relation. $$\frac{g'}{g} = \frac{\frac{GM}{R^2} \left(1 - \frac{2h}{R}\right)}{\frac{GM}{R^2}}$$ $$= \left(1 - \frac{2h}{R}\right)$$ $$\Rightarrow g' = g \left(1 - \frac{2h}{R}\right)$$

The temperatures in Kelvin scales are $$T_2 = 27 \, ^\circ \mathrm{C} + 273 = 300 \, \mathrm{K}$$ $$T_1 = 127 \, ^\circ \mathrm{C} + 273 = 400 \, \mathrm{K}$$ The efficiency of the Carnot engine is $$\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{400} = \frac{1}{4}$$ Using the relation, $$\eta = \frac{W}{Q}$$ $$\Rightarrow \frac{2 \, \mathrm{kJ}}{Q} = \frac{1}{4} \Rightarrow Q = 8 \, \mathrm{kJ}$$

The definition of displacement is given by the product of velocity and time. $$\int v dt = \Delta x$$ = displacement of body in given time. Distance is a scalar quantity and displacement is a vector quantity defined by the shortest distance between the initial and the final position. Acceleration is defined by the rate of change of velocity. $$\int a dt = \Delta v$$ = change in velocity in given time. Thus, the area under the velocity time graph gives the displacement of the body and the area under the acceleration time graph gives the change in velocity of the body.

The bombarding electrons can eject electrons from the inner shells of the atoms of the metal target. Those vacancies will be quickly filled by electrons dropping from higher levels, emitting x-rays with sharply defined frequencies associated with the difference between the atomic energy levels of the target atoms. Hence, X-rays are emitted when a metal target is bombarded with high energy electrons.

For a light of wavelength $\lambda$, the fringe width in double slit experiment is given by $$\beta = \frac{\lambda D}{d} \cdots (1)$$ For a light of wavelength $\lambda'$, the fringe width is given by $$\beta' = \frac{\lambda' D}{d} \cdots (2)$$ Divide equation (2) by equation (1) and simplify to obtain the fringe width for the later case. $$\frac{\beta'}{\beta} = \frac{\frac{\lambda' D}{d}}{\frac{\lambda D}{d}}$$ $$= \frac{\lambda'}{\lambda}$$ $$\Rightarrow \beta' = \frac{\lambda'}{\lambda} \beta \cdots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required fringe width. $$\beta' = \frac{600 \, \mathrm{nm}}{400 \, \mathrm{nm}} \times 2 \, \mathrm{mm}$$ $$= 3 \, \mathrm{mm}$$

A coil surrounding a magnetic core having high permeability and low retentivity, through which an electric current is fed to magnetise the core is what makes up an electromagnet. Wherever controllable magnets are needed, as in devices where the magnetic flux needs to be adjusted, reversed, or turned on and off, an electromagnet is used. Hence, both A and R are correct and R is the correct explanation of A.

With proper energetic radiation, the photoelectric current increases with an increase in accelerating potential until it reaches a point where it reaches its maximum and does not increase with an additional increase in accelerating potential. For a given potential difference, the photocurrent is directly proportional to the intensity of the incident radiation. The maximum kinetic energy of the emitted photoelectrons can be written as $$K_{max} = h\nu - \varphi$$ From above equation, it can be seen that the maximum kinetic energy of the photoelectron depends on the frequency of the light used. Hence, the statements A and C are correct.

The formula to calculate the motional emf induced in a moving conductor in a magnetic field is given by $\varepsilon = Blv (1)$ Substitute the values of the known parameters into equation (1) and solve to calculate the required velocity of the rod. $$0.08 \, \mathrm{V} = 0.4 \, \mathrm{T} \times \left(10 \, \mathrm{cm} \times \frac{1 \, \mathrm{m}}{100 \, \mathrm{cm}}\right) \times v$$ $$\Rightarrow v = \frac{0.08 \, \mathrm{V}}{0.4 \, \mathrm{T} \times 10 \, \mathrm{cm} \times \frac{1 \, \mathrm{m}}{100 \, \mathrm{cm}}}$$ $$= 2 \, \mathrm{m} \, \mathrm{s}^{-1}$$

Let the mass of the bullet be $m$ and mass of the block be $M$. It is given that $V = 400 \, \mathrm{m \, s^{-1}}$. By conservation of momentum, $$mV = (m + M)v$$ $$\Rightarrow \, 0.1 \times 400 = (0.1 + 3.9)v$$ $$\Rightarrow \, v = 10 \, \mathrm{m \, s^{-1}}$$ The force acting on the block $$f = \mu m'g$$ $$\Rightarrow \, a = \mu g$$ Using $v^2 - u^2 = -2aS$ $$\Rightarrow \, 0^2 - 10^2 = 2(-\mu g)(S)$$ $$\Rightarrow \, \mu = \frac{100}{2 \times 10 \times 20} = 0.25$$

Linear antennas are considered to be those antennas that imply the use of electrically thin conductors having diameter much less than the wavelength of the electromagnetic wave passing through them. The formula to calculate the effective power radiated by a linear antenna is given by $$P_{eff} = k \frac{l^2}{\lambda^2}$$ where, $k$ is a constant.

The formula to calculate the quality factor of an LCR circuit is given by $$Q = \frac{1}{R} \sqrt{\frac{L}{C}} \cdots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required quality factor of the circuit. $$Q = \frac{1}{100} \sqrt{\frac{1}{6.25 \times 10^{-6}}}$$ $$= \frac{1}{100} \times \frac{1000}{2.5}$$ $$= 4$$

The formula to calculate the fundamental frequency produced in a vibrating string is given by $$f_0 = \frac{v}{2L} \ldots (1)$$ For the first case, it can be written that $$120 = \frac{v}{2(90)} \ldots (2)$$ Similarly, for the second case, it can be written that $$180 = \frac{v}{2(L)} \ldots (3)$$ Divide equation (3) by equation (2) and solve to obtain the required length of the string. $$\frac{180}{120} = \frac{\frac{v}{2(L)}}{\frac{v}{2(90)}}$$ $$= \frac{90}{L}$$ $$\Rightarrow L = 90 \times \frac{120}{180}$$ $$= 60$$

The formula to calculate the wavelength of $H_\alpha$-line in Balmer series is given by $$\frac{hc}{\lambda_\alpha} = 13.6 \left[ \frac{1}{4} - \frac{1}{9} \right] \cdots (1)$$ The formula to calculate the wavelength of $H_\beta$-line in Balmer series is given by $$\frac{hc}{\lambda_\beta} = 13.6 \left[ \frac{1}{4} - \frac{1}{16} \right] \cdots (2)$$ Divide equation (2) by equation (1) $$\Rightarrow \frac{\lambda_\alpha}{\lambda_\beta} = \frac{\frac{1}{4} - \frac{1}{16}}{\frac{1}{4} - \frac{1}{9}}$$ $$= \frac{36}{64} \times \frac{12}{5}$$ $$= \frac{27}{20}$$

The formula to calculate the current through any conductor is given by $$I = n A e v_d \ldots (1)$$ Substitute the values of the given parameters into equation (1) and solve to calculate the required drift velocity of the electrons. $$3.2 = 8 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19} \times v_d$$ $$\Rightarrow v_d = \frac{3.2}{16 \times 1.6 \times 1000} \, \mathrm{m \, s^{-1}}$$ $$= 125 \times 10^{-6} \, \mathrm{m \, s^{-1}}$$

The formula to calculate the Young's modulus of the material of the wire is given by $$Y = \frac{FL}{A \Delta L} \ldots (1)$$ The increase in length of the wire due to increase in temperature is given by $$\Delta L = L \alpha (T_2 - T_1) \ldots (2)$$ Substitute the expression for the extension in length from equation (2) into equation (1) and simplify to obtain the required compressive tension. $$Y = \frac{\frac{FL}{A}}{L \alpha (T_2 - T_1)}$$ $$\Rightarrow F = A \alpha Y (T_2 - T_1) \ldots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required compressive tension in the wire. $$F = 10^{-4} \, \mathrm{m}^2 \times 10^{-5} \, \mathrm{K}^{-1} \times 2 \times 10^{11} \, \mathrm{Nm}^{-2} \times (200^\circ \, \mathrm{C} - 0^\circ \, \mathrm{C})$$ $$= 4 \times 10^4 \, \mathrm{N}$$

The total kinetic energy of the ball at the bottom of the curve can be calculated as follows: $$K = \frac{1}{2} M v^2 + \frac{1}{2} \cdot \frac{2}{3} M R^2 \omega^2$$ $$= \frac{1}{2} M v^2 + \frac{1}{3} M (R \omega)^2$$ $$= \frac{1}{2} M v^2 + \frac{1}{3} M v^2 [\because \ v = \omega R]$$ $$= \frac{5}{6} M v^2 \cdots (1)$$ The formula to calculate the potential energy acquired by the ball when it attains the maximum height is given by $$U = M g h \cdots (2)$$ Equate equation (1) and (2) and simplify to obtain the required height. $$M g h = \frac{5}{6} M v^2$$ $$\Rightarrow h = \frac{5 v^2}{6 g} \cdots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required height attained by the ball. $$h = \frac{5 \times (3 \, \mathrm{m/s})^2}{6 \times 10 \, \mathrm{m/s^2}}$$ $$= 0.75 \, \mathrm{m} \times \frac{100 \, \mathrm{cm}}{1 \, \mathrm{m}}$$ $$= 75 \, \mathrm{cm}$$

The formula to calculate the change in momentum of the object is given by $$p_f - p_i = \int F dt \ldots (1)$$ Substitute the values of the known parameters into equation (1) and solve to calculate the final momentum of the object. $$p_f - 10 \, \mathrm{kg \, m \, s^{-1}} = 2 \, \mathrm{N} \times 5 \, \mathrm{s}$$ $$\Rightarrow p_f = (10 + 10) \, \mathrm{kg \, m \, s^{-1}}$$ $$= 20 \, \mathrm{kg \, m \, s^{-1}}$$ The formula to calculate increase in kinetic energy ($\Delta K$) of the particle is given by $$\Delta K = \frac{1}{2m} \left( p_f^2 - p_i^2 \right) \ldots (2)$$ Substitute the values of the known parameters into equation (2) to calculate the required increase in kinetic energy. $$\Delta K = \frac{1}{2 \times 5 \, \mathrm{kg}} \times (20^2 - 10^2) \, \mathrm{kg^2 \, m^2 \, s^{-2}}$$ $$= 30 \, \mathrm{J}$$

The formula to calculate the initial energy stored in the first capacitor is given by $$U_i = \frac{1}{2} CV^2 \ldots (1)$$ The formula to calculate the final energy stored when two capacitors are connected is given by $$U_f = \frac{1}{2} C \left( \frac{V}{2} \right)^2 \times 2$$ $$= \frac{1}{4} CV^2 \ldots (2)$$ Subtract equation (2) from equation (1) to obtain the energy loss ($\Delta U$). $$\Delta U = \frac{1}{2} CV^2 - \frac{1}{4} CV^2$$ $$= \frac{CV^2}{4} \ldots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required loss. $$\Delta U = \frac{600 \times 10^{-12} \times (200)^2}{4} \mathrm{J}$$ $$= 6 \, \mu \mathrm{J}$$

Let's consider the following diagram. The relation between the object distance, the image distance and the radius of curvature of the spherical surface can be written as $$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \ldots (1)$$ Substitute the values of the known parameters into equation (1) and solve to calculate the required image distance. $$\frac{1.5}{v} - \frac{1}{-15} = \frac{1.5 - 1.0}{30}$$ $$\Rightarrow \frac{1.5}{v} = \frac{1}{60} - \frac{1}{15}$$ $$= -\frac{1}{20}$$ $$\Rightarrow v = -30$$ Hence, the image is formed at a distance of 30 cm from the pole on the same direction as the object is placed.

The formula to calculate the magnetic field at any axial point situated at a distance $x$ from the centre of a circular current carrying conductor of radius $r$ can be written as $$B = \frac{\mu_0 I r^2}{2 \left( r^2 + x^2 \right)^{3/2}} \ldots (1)$$ Substitute $0$ for $x$ in equation (1) to calculate the magnetic field $(B_c)$ at the centre of the conductor. $$B_c = \frac{\mu_0 I r^2}{2 \left( r^2 + 0^2 \right)^{3/2}}$$ $$= \frac{\mu_0 I r^2}{2 r^3}$$ $$= \frac{\mu_0 I}{2 r} \ldots (2)$$ Substitute $r$ for $x$ in equation (1) to calculate the magnetic field $(B_r)$ at the given distance from the centre of the conductor. $$B_r = \frac{\mu_0 I r^2}{2 \left( r^2 + r^2 \right)^{3/2}}$$ $$= \frac{\mu_0 I r^2}{4 \sqrt{2} r^3}$$ $$= \frac{\mu_0 I}{4 \sqrt{2} r} \ldots (3)$$ Divide equation (2) by equation (3) to calculate the required ratio of the magnetic fields. $$\frac{B_c}{B_r} = \frac{\frac{\mu_0 I}{2 r}}{\frac{\mu_0 I}{4 \sqrt{2} r}}$$ $$= 2 \sqrt{2}$$ $$= \sqrt{8}$$

All nonzero digits are significant. All zeros that are found between nonzero digits are significant. Leading zeros (to the left of the first nonzero digit) are not significant. Trailing zeros for a whole number that ends with a decimal point are significant. The number of significant figures of the given numbers are as follows: $\therefore (A), (C)$ and $(D)$ have the same number of significant figures.

Chlorofluorocarbon compounds (CFCs) are non-reactive, non-flammable molecules that are used in refrigerators, air conditioners, plastics, and electronic industries. Once released, CFCs mix with atmospheric gases and reach the stratosphere, where they are decomposed by UV radiations. $CF_2Cl_2(g)\xrightarrow{UV}Cl(g)+CF_2Cl(g)$ The chlorine radical released will react with ozone. $Cl(g)+O_3(g)\rightarrow O_2(g)+ClO(g)$ The ClO will generate chlorine radicals again. $ClO(g)+O(g)\rightarrow O_2(g)+Cl$

IUPAC name of the following compound is 2-Methyl-5-oxo-hexanoic acid 1. Identify the longest continuous carbon chain that contains the carboxyl group (COOH). In this case, it is a six-carbon chain. 2. Number the carbon chain starting from the end closest to the carboxyl group. The carboxyl carbon is assigned the number 1. 3. Identify and name any substituents attached to the carbon chain. In this case, there is a methyl group attached to the second carbon atom of the chain. 4. Combine the substituent name with the parent chain name, with a hyphen in between. The resulting name is "2-methylhexanoic acid". 5. Finally, add the prefix "oxo" to indicate the presence of a ketone group (C = O) on the fifth carbon of the chain.

The van der Waals' constant '$a$' is directly proportional to the intermolecular force of attraction; hence, the gas with greater value of '$a$' can be most easily liquefied. The values of Vander Waal's constant '$a$' for the given gases are: $$ \begin{array}{cc} \text{Gases} & \text{Vander Waal's Constant 'a'} \\ A. \, \mathrm{Ar} & 1.35 \, \mathrm{L^2 \, atm \, mol^{-2}} \\ B. \, \mathrm{CH_4} & 2.25 \, \mathrm{L^2 \, atm \, mol^{-2}} \\ C. \, \mathrm{H_2O} & 5.536 \, \mathrm{L^2 \, atm \, mol^{-2}} \\ D. \, \mathrm{C_6H_6} & 18.57 \, \mathrm{L^2 \, atm \, mol^{-2}} \end{array} $$ Therefore, increasing order of '$a$' for the given gases is: $A < B < C < D$

Methyl orange is a weak acid. Additionally, in acidic conditions, methyl orange exists in the quinonoid form, which is red in color. In alkaline conditions, it exists in the benzenoid form, which is yellow in color. Since red is more deeply coloured than yellow, Statement-2 is wrong.

In redox titrations, the indicator changes color when the titration reaches the equivalence point, which corresponds to the point at which the reducing agent has been completely oxidized or the oxidizing agent has been completely reduced. So these are used to change in oxidation potential and not change in pH. In acid base titration, the indicators are sensitive to change in pH and not change in oxidation potential. Therefore, both the statements are incorrect.

The first step of the sequence is involved with electrophilic substitution reaction. Nitro group is deactivating group and methyl group is activating group, hence, the substitution reaction takes place with respect to methyl group. In the second step nitro group is reduced to amine group with hydrogen in the presence of palladium metal. The aniline undergo diazotisation in the third step. The diazonium salt loses nitrogen gas in the presence of $\mathrm{H_3PO_2}$.

Most of the chemical reactions have energy barrier. A positive catalyst in a chemical reaction increases the rate of reaction by lowering the energy barrier. It does not alter the energy of reactant as well as product. In the above graph, the curve in the presence of catalyst is having less activation energy. The enthalpy change is constant after addition of the positive catalyst.

$H_2O_2$ decomposes on exposure to sunlight or on heating. It also decomposes on standing but at a slow rate. These reactions occur explosively when $H_2O_2$ is in contact with dust. Urea acts as a negative catalyst to this reaction and slows down the decomposition of hydrogen peroxide.

Drugs that bind to the receptor site and inhibit its natural function are called antagonists. Antagonists are molecules that bind to receptors but do not activate them. Instead, they block the action of other molecules, including natural ligands and agonists, that normally bind to the receptor and activate it. Antagonists inhibit the natural communication of the body by preventing the binding of the natural ligands, which can result in physiological effects such as decreased heart rate, blood pressure, and inflammation.

For $[\mathrm{Cr(CN)_6}]^{3-}$, the oxidation state of Cr is $+3$ and its electronic configuration is $3d^3$. There are three unpaired electrons in the $3d$ orbitals. (A) $[\mathrm{Cr(CN)_6}]^{3-}$ Number of unpaired electrons $= 3$ For $[\mathrm{Fe(H_2O)_6}]^{2+}$, the oxidation state of Fe is $+2$ and its electronic configuration is $3d^6$. There are four unpaired electrons in the $3d$ orbitals. Hence, the hybridisation is $sp^3d^2$. (B) $[\mathrm{Fe(H_2O)_6}]^{2+}$ Number of unpaired electrons $= 4$ For $[\mathrm{Co(NH_3)_6}]^{3+}$, the oxidation state of Co is $+3$ and it has no unpaired electrons. (C) $[\mathrm{Co(NH_3)_6}]^{3+}$ Number of unpaired electrons $= 0$ For $[\mathrm{Ni(NH_3)_6}]^{2+}$, the oxidation state of Ni is $+2$ and it has two unpaired electrons. (D) $[\mathrm{Ni(NH_3)_6}]^{2+}$ Number of unpaired electrons $= 2$ Therefore, the correct option is (A).

In the first step bromine attacks the carbon-carbon double bond by electrophilic addition reaction mechanism and forms the bromonium cyclic intermediate. In the presence of sodium bicarbonate carboxylic acid group present in the reactant converts to carboxylate ion. This carboxylate ion opens the cyclic bromonium ion ring and forms the desired product.

The process of extracting Al metal by electrolysis of a mixture of purified $\mathrm{Al_2O_3}$ and $\mathrm{Na_3AlF_6}$ is known as Hall and Heroult process. Graphite is used as anode which is useful in the reduction of alumina oxide to the Al metal. The graphite electrode reacts with the oxygen released at cathode to form carbon dioxide and carbon monoxide. Therefore, the role of graphite as an anode in the extraction of Al is to check the formation of oxygen. In other words, we can say its role is to prevent the oxidation of Al back to $\mathrm{Al_2O_3}$ by oxygen.

Sodium is indeed about 30 times more abundant than potassium in seawater. This is because sodium is more soluble in water than potassium, and is therefore more readily dissolved from rocks and minerals into seawater. Potassium is bigger than sodium in size as atomic number of potassium is more than sodium and Potassium has one additional electron shell compared to sodium, which means that its atomic radius is larger. So Both A and R are true but R is NOT the correct explanation of A.

Amino acids are organic compounds that contain amine and carboxyl functional groups, as well as a side chain specific to each amino acid. There are 20 standard amino acids that are used to create proteins. One letter codes of the given natural amino acids are Natural amino acid One Letter Code (A) Glutamic acid (III) E (B) Glutamine (I) Q (C) Tyrosine (IV) Y (D) Tryptophan (II) W Therefore, the correct match is (A)-III, B-I, (C)-IV, (D)-II

Moseley's observation on characteristic X-ray spectra of elements relates frequency ($\nu$) of X-rays emitted to the atomic number by the following relation: $$\sqrt{\nu} = a(Z - b)$$ The plot of $\sqrt{\nu}$ vs $Z$ is linear. Based on this observation, the periodic law defined as the physical and chemical properties of the elements are periodic functions of their atomic numbers.

The correct order of acidic strength of the given carboxylic acids will be decided by the stability of their conjugate bases. The $+I$ effect of $\mathrm{CH_3}$ group destabilises the carboxylate anion. But the $-I$ effect of halogen substituted methyl group stabilises the carboxylate anion. Higher the $-I$ effect of the halogen substituted methyl group, higher the acidic strength of the corresponding acid. Therefore, the correct order of acidic strength is $$\mathrm{F_3C - COOH > FCH_2 - COOH > ClCH_2 - COOH > BrCH_2 - COOH > CH_3 - COOH}$$

The intermediate formed in aromatic nucleophilic substitution reaction is carbanion. Carbanion stabilised by electron withdrawing groups. Aryl halides are less reactive towards nucleophilic aromatic substitution with aqueous NaOH. Introduction of electron releasing group like OCH$_3$ group further decreases its reactivity towards nucleophilic substitution. But the introduction of electron withdrawing groups like NO$_2$ group particularly at the ortho and para position increases its reactivity towards nucleophilic substitution. Higher the number of electron withdrawing groups, higher the reactivity. Therefore, the correct reactivity order is $$NO_2-Cl-NO_2-NO_2 > Cl > Cl-OMe$$

Cement is made by strongly heating a mixture of lime stone and clay in a rotatory kiln. Lime stone and clay are finely powdered, and a little water is added. For a good quality cement, the ratio of silica ($\mathrm{SiO_2}$) to alumina ($\mathrm{Al_2 O_3}$) should be between 2.5 and 4 and the ratio of lime ($\mathrm{CaO}$) to the total of the oxides of silicon ($\mathrm{SiO_2}$), aluminium ($\mathrm{Al_2 O_3}$) and iron ($\mathrm{Fe_2 O_3}$) should be as close as possible to 2. If lime is in excess, the cement cracks during setting. On the other hand, if lime is less than the required, the cement is weak in strength. Therefore, a proper composition of cement must be maintained to get cement of good quality.

Carbolic acid or phenol reacts with phthalic anhydride in presence of conc. $\mathrm{H_2SO_4}$ to form phenolphthalein which is used as an acid base indicator. The type of reaction involved is electrophilic substitution reaction. Phenol is ortho-para directing group. Hence, substitution occurs at para position. The phenolphthalein is colourless below pH 8.5 and attains a pink to deep red hue above pH 9.0.

For the insoluble salt barium sulphate, sulphate ion added is the common ion. Due to the common ion effect the solubility of barium sulphate decreases. Let us assume solubility of barium sulphate is $S$. $$\mathrm{BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)}$$ $$\mathrm{K_2SO_4 \rightarrow K^+(aq) + SO_4^{2-}(aq)}$$ $$K_{\mathrm{SP}} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}]$$ $$1 \times 10^{-10} = S(S + 0.1)$$ $$\simeq 0.1 \, S$$ $$S = 10^{-9} \, \mathrm{M} = 233 \times 10^{-9} \, \mathrm{gL^{-1}}$$

The coagulating power of an electrolyte is a measure of its ability to cause the coagulation or precipitation of a colloidal solution. Coagulation power is proportional to $\($ $\frac{1}{Coagulation value}$ $\)$. Coagulating value of $\($ AlCl_3 $\)$ for $\($ As_2 S_3 = 0.09 $\)$. Coagulating value of $\($ NaCl $\)$ for $\($ As_2 S_3 = 50.04 $\)$. Then, $$ \frac{(C.V)_{AlCl_3}}{(C.V)_{NaCl}} = \frac{(C.P)_{NaCl}}{(C.P)_{AlCl_3}} $$ $$ \frac{0.09}{50.04} = \frac{(C.P)_{NaCl}}{(C.P)_{AlCl_3}} $$ $$ \left( C.P \right)_{AlCl_3} = 556 \left( C.P \right)_{NaCl} $$ So the answer is 556.

The number of radial nodes for an orbital with principal quantum number $n$ and azimuthal quantum number $l$ is given by the expression: Number of radial nodes $= n - l - 1$. Using this expression, we can determine the number of radial nodes for each of the given atomic orbitals: $7s$: $n = 7$, $l = 0$ $\Rightarrow 7 - 0 - 1 = 6$ $7p$: $n = 7$, $l = 1$ $\Rightarrow 7 - 1 - 1 = 5$ $6s$: $n = 6$, $l = 0$ $\Rightarrow 6 - 0 - 1 = 5$ $8p$: $n = 8$, $l = 1$ $\Rightarrow 8 - 1 - 1 = 6$ $8d$: $n = 8$, $l = 2$ $\Rightarrow 8 - 2 - 1 = 5$ Therefore, the atomic orbitals with 5 radial nodes are $7p$, $6s$, and $8d$.

The reaction is given by: $$\mathrm{C_2H_4\,(g) + 3O_2\,(g) \rightarrow 2CO_2\,(g) + 2H_2O(l)}$$ The change in internal energy is: $$\Delta U = -1406 \, \mathrm{kJ \, mol^{-1}}$$ Temperature is: $$T = 300 \, \mathrm{K}$$ The enthalpy change is given by: $$\Delta H = \Delta U + \Delta n_g \, RT$$ $\Delta U$ is the change in the internal energy of the system, which is the sum of the kinetic and potential energies of all the molecules in the system. $R$ is the gas constant $(8.314 \, \mathrm{J/(mol \, K)})$ and $T$ is the temperature in Kelvin. Calculating $\Delta H$: $$\Delta H = -1406 + \left(-2\right) \times 8.3 \times 300 = -1406 - 4.98$$ $$= -1410.98 \, \mathrm{KJ \, mol^{-1}} \approx -1411$$ The minimum value of $T \Delta S$ at equilibrium $\Delta H$ is: $$\Delta H = T \Delta S = -1411 \, \mathrm{KJ \, mol^{-1}}$$

The central Xe atom in $\mathrm{XeO_3}$ has three bonding domains and one lone pair of electrons. Hence, the electron geometry is tetrahedral and molecular geometry is pyramidal. $\mathrm{XeO_2F_2}$ is trigonal bipyramidal and the shape is a see-saw. $\mathrm{XeOF_2}$ is in square pyramidal shape. So, $\mathrm{XeF_5^+}$, $\mathrm{XeO_3}$, $\mathrm{XeO_2F_2}$ and $\mathrm{XeOF_4}$ have single lone pair on central atom.

The boiling point elevation constant ($K_b$) of a volatile liquid is given by the following expression $$K_b = \frac{R T_b^2 M}{\Delta H_v}$$ where $T_b$ is the boiling point of the liquid, $M$ is the molar mass of the liquid, and $\Delta H_v$ is the enthalpy of vaporisation. The boiling points of the two solvents are in the ratio $2:1$, which means that the boiling point elevation constants are also in the ratio $2:1$. Let's call the boiling point elevation constant of solvent $X$ as $(T_b)_x$, and the boiling point elevation constant of solvent $Y$ as $(T_b)_y$. Given: $M_x = M_y$ $$(T_b)_x = 2(T_b)_y$$ $$2(\Delta H_v)_x = (\Delta H_v)_y$$ $$\therefore \frac{(K_b)_x}{(K_b)_y} = \left(\frac{2}{1}\right)^2 \times \frac{2}{1} = \frac{8}{1} = m$$ Therefore, the value of $m$ is $8$ (nearest integer).

In the complex $\mathrm{Fe(CO)_5}$, take the oxidation number of Fe be $x$. Carbonyl is a neutral ligand which doesn't have any charge. So the oxidation number of Fe would be $$x - 0 \times 5 = 0 = 0$$ Hence, the oxidation number of iron in this complex is 0. Let oxidation state of vanadium in $\mathrm{VO_2^{2+}} = x$. So, $$x + (-2) = +2$$ $$\Rightarrow x = +2 + 2 = +4$$ Tungsten trioxide consists of one tungsten atom and three oxygen atoms. Tungsten is a d-block metal from group 6 and has an oxidation state $+6$ in the compound. Then, Sum of oxidation state of metals in the given compounds $= 10$

The complex $[\mathrm{Mn(NCS)}_6]^{x-}$ has a magnetic moment of $6.06 \, \mathrm{BM}$. This indicates that the complex has unpaired electrons, which contribute to its magnetic moment. Number of unpaired electrons $= 5 \left(\mathrm{Mn^{2+} : 3d^5}\right)$. So Mn must be in $+2$ oxidation state $\mathrm{Mn^{2+}}$. $$\Rightarrow 2 + (-6) = -x$$ $$\Rightarrow -4 = -x$$ $$\Rightarrow x = 4$$ The numerical value of $x$ is $4$, so the complex is $[\mathrm{Mn(NCS)}_6]^{4-}$.

E^{$\circ$}_{cell} is dependent on the equilibrium constant, K, which is not dependent on the pressure of any gas involved in the cell. Relation between Electrochemical Cell Potential and Gibbs Free Energy change is given by: $$\Delta G = -nFE$$ where Delta G is Gibbs Free energy change, n is the number of electrons involved in the reaction and F is Faraday and E is the Cell potential. We know that, $$dG = VdP - SdT$$ At constant pressure $$dP = 0$$ $$\Rightarrow dG = -SdT$$ $$S = -\frac{dG}{dT}$$ $$\Delta S = -\left(\frac{\Delta (dG)}{dT}\right)_P$$ $$\Rightarrow \frac{dE^{\circ}_{cell}}{dT} = \frac{\Delta S}{nF}$$ A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.

To determine the ratio of sigma and pi bonds in pyrophosphoric acid, we first need to understand the molecular structure of the compound. Pyrophosphoric acid has the chemical formula $\mathrm{H_4P_2O_7}$. The molecule contains two phosphorus atoms, seven oxygen atoms, and four hydrogen atoms. The structure of pyrophosphoric acid is It contains total 12 sigma bonds and 2 pi bonds. So $$\frac{No. of \sigma bonds}{No. of \pi bonds} = \frac{12}{2} = 6 : 1$$