JEE Main 6 April 2023 Shift 2 question paper with solutions

Given that three dice are thrown. The total number of outcomes when three dice are thrown together is $6^3 = 6 \times 6 \times 6$. The number of outcomes such that all the outcomes are different is $6 \times 5 \times 4$. For example: If the outcome in dice 1 is "6" then the number of outcomes for dice 2 should be 5 (1, 2, 3, 4, 5) which excludes the outcome "6". Hence, the required probability is $$= \frac{6 \times 5 \times 4}{6 \times 6 \times 6}$$ $$= \frac{20}{36} = \frac{5}{9} = \frac{p}{q}$$ $$\Rightarrow q - p = 9 - 5 = 4$$ Therefore, the required answer is 4.

Given, (S1) : $(2023)^{2022}$ - $(1999)^{2022}$ is divisible by 8 Now we know that (x - y) divides $(x^n - y^n)$ $\forall$ n $\in$ $\mathbb{N}$ So, (2023 - 1999) divides $(2023)^{2022}$ - $(1999)^{2022}$ $\Rightarrow$ 24 divides $(2023)^{2022}$ - $(1999)^{2002}$ $\Rightarrow$ 8 will divide $(2023)^{2022}$ - $(1999)^{2002}$ As 8 divides 24 Hence, (S1) is correct Now solving, (S2) : $13(13)^n - 11n - 13$ is divisible by 144 for n $\in$ $\mathbb{N}$ So using binomial theorem in $(1+12)^{n}$ we get, $13(1+12)^{n}-11n-13$ $=13\left({}^{n}C_{0}+{}^{n}C_{1}12+{}^{n}C_{2}12^{2}+\cdots+{}^{n}C_{n}12^{n}\right)-11n-13$ $=12\times13n-11n+12\lambda$ $=145n+144\lambda$ which is not divisible by $144$. Hence, $(S2)$ is incorrect.

Let, $$L = \lim_{n \to \infty} \left\{ \left( 2^{\frac{1}{2}} - 2^{\frac{1}{3}} \right) \left( 2^{\frac{1}{2}} - 2^{\frac{1}{5}} \right) \cdots \left( 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} \right) \right\}$$ Now, $$2^{\frac{1}{2}} - 2^{\frac{1}{3}} < 1$$ $$2^{\frac{1}{2}} - 2^{\frac{1}{5}} < 1$$ ------------------ $$2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} < 1 \forall n \in \mathbb{N}$$ And $$\left( 2^{\frac{1}{2}} - 2^{\frac{1}{3}} \right)^n < \left( 2^{\frac{1}{2}} - 2^{\frac{1}{3}} \right) \left( 2^{\frac{1}{2}} - 2^{\frac{1}{5}} \right) \cdots \left( 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} \right) < \left( 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} \right)^n$$ $$\Rightarrow \lim_{n \to \infty} \left( 2^{\frac{1}{2}} - 2^{\frac{1}{3}} \right)^n < \lim_{n \to \infty} \left( \left( 2^{\frac{1}{2}} - 2^{\frac{1}{3}} \right) \left( 2^{\frac{1}{2}} - 2^{\frac{1}{5}} \right) \cdots \left( 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} \right) \right) < \lim_{n \to \infty} \left( 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} \right)^n$$ $$\Rightarrow \lim_{n \to \infty} \left( 2^{\frac{1}{2}} - 2^{\frac{1}{3}} \right)^n < L < \lim_{n \to \infty} \left( 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} \right)^n$$ And $$\lim_{n \to \infty} \left( 2^{\frac{1}{2}} - 2^{\frac{1}{3}} \right)^n = 0 \& \lim_{n \to \infty} \left( 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} \right)^n = 0$$ $$\left\{ as 2^{\frac{1}{2}} - 2^{\frac{1}{3}} < 1 \& 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} < 1 \right\}$$ Hence, $$\lim_{n \to \infty} \left\{ \left( 2^{\frac{1}{2}} - 2^{\frac{1}{3}} \right) \left( 2^{\frac{1}{2}} - 2^{\frac{1}{5}} \right) \cdots \left( 2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}} \right) \right\} = 0$$

Given, $$X = \{ z \in \mathbb{C} : \Re(az^2 + bz) = a and \Re(bz^2 + az) = b \}$$ Now, let $z = x + iy$. So, $\Re(az^2 + bz) = a$ $$\Rightarrow \Re\left(a(x + iy)^2 + b(x + iy)\right) = a$$ $$\Rightarrow \Re\left(a(x^2 - y^2 + 2ixy) + b(x + iy)\right) = a$$ $$\Rightarrow a(x^2 - y^2) + bx = a \ldots (i)$$ And, $\Re(bz^2 + az) = b$ $$\Rightarrow \Re\left(b(x + iy)^2 + a(x + iy)\right) = b$$ $$\Rightarrow \Re\left(b(x^2 - y^2 + 2ixy) + a(x + iy)\right) = b$$ $$\Rightarrow b(x^2 - y^2) + ax = b \ldots (ii)$$ From (i) and (ii), (i) - (ii) we get, $$(x^2 - y^2)(a - b) - x(a - b) = a - b$$ $$\Rightarrow x^2 - y^2 - x = 1 \ldots (iii)$$ From (i) and (ii), (i) + (ii) $$((x^2 - y^2) + x - 1)(a + b) = 0 (Note: here a + b \neq 0 is considered but it is not clear from the question)$$ $$\Rightarrow x^2 - y^2 + x - 1 = 0 \ldots (iv)$$ Now from (iii) and (iv) we get, $$x = 0, \; y^2 = -1 (No solution)$$

Given, $$f(x) = \frac{1}{\sqrt{[x] - x}}$$ We know that, $x = [x] + \{x\}$ where $\{x\}$ is the fractional part of $x$ whose value is $0 0$, but $-\{x\} \in (-1, 0)$ Hence, $\sqrt{-\{x\}}$ is not defined Hence, Domain and Range will be $= \emptyset$

The given differential equation is $\left(1 + \ln(x)\right) \frac{dx}{dy} - x \ln(x) = e^y$. Put $x \ln(x) = t$. $$\Rightarrow (1 + \ln(x)) dx = dt$$ $$\Rightarrow \frac{dt}{dy} - t = e^y$$ which is a linear differential equation. Integrating factor $= e^{-\int dy} = e^{-y}$. The solution of the differential equation is $$\Rightarrow t \times e^{-y} = \int e^y \times e^{-y} dy + c$$ $$\Rightarrow t \times e^{-y} = y + c$$ $$\Rightarrow x \ln(x) = y \, e^y + c \, e^y$$ Put $x = 1$, $y = 0$ $$\Rightarrow c = 0$$ Put $x = a$, $y = 2$ $$a \ln(a) = 2e^2$$ $$\therefore a^a = e^{2e^2}$$ Hence this is the correct option.

Let the given vectors be $\vec{A} = \hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{B} = 2\hat{i} - 3\hat{j} + 4\hat{k}$, $\vec{C} = (\alpha + 1)\hat{i} + 2\hat{k}$ and $\vec{D} = 9\hat{i} + (\alpha - 8)\hat{j} + 6\hat{k}$. We know that if the vectors are coplanar then $$\begin{vmatrix} \vec{AB} & \vec{AC} & \vec{AD} \end{vmatrix} = 0$$ $$\Rightarrow \vec{AB} = (2\hat{i} - 3\hat{j} + 4\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = \hat{i} - \hat{j} + \hat{k}$$ $$\Rightarrow \vec{AC} = ((\alpha + 1)\hat{i} + 2\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = \alpha \hat{i} + 2\hat{j} - \hat{k}$$ $$\Rightarrow \vec{AD} = (9\hat{i} + (\alpha - 8)\hat{j} + 6\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = 8\hat{i} + (\alpha - 6)\hat{j} + 3\hat{k}$$ Now, $$\begin{vmatrix} 1 & -1 & 1 \\ \alpha & 2 & -1 \\ 8 & \alpha - 6 & 3 \end{vmatrix} = 0$$ $$\Rightarrow 1(6 + \alpha - 6) + (3\alpha + 8) + (\alpha^2 - 6\alpha - 16) = 0$$ On simplifying we get, $$\Rightarrow \alpha^2 - 2\alpha - 8 = 0$$ Therefore, sum of the roots is $(-2) = 2$.

Given, the system of equations $$x + y + z = 6$$ $$x + 2y + \alpha z = 10$$ $$x + 3y + 5z = \beta$$ Now finding, $$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{vmatrix}$$ $$\Rightarrow \Delta = 1(10 - 3\alpha) - (5 - \alpha) + (3 - 2)$$ $$\Rightarrow \Delta = 6 - 2\alpha$$ Now for unique solution, $\Delta \neq 0 \Rightarrow \alpha \neq 3$ So, for unique solution $\Rightarrow \alpha \neq 3$ But in option $(D)$, system has unique solution for $\alpha = 3$ which is completely wrong.

Given: $$y = f(x) = |x - 1| + |x - 2|$$ $$\Rightarrow f(x) = \begin{cases} 3 - 2x; & x < 1 \\ 1; & 1 \leq x < 2 \\ 2x - 3; & x \geq 2 \end{cases}$$ Let us draw diagram of $y = f(x)$ and $y = 3$ we get, Required area is the area of trapezium which is Area $= \frac{1}{2} [1 + 3] \times 2$ $= 4$ sq. units

Given, $P$ be a square matrix such that $P^2 = I - P$. Now solving $P^2 = I - P$ we get, $$P^4 = (I - P)(I - P)$$ $$P^4 = I + P^2 - 2P$$ $$P^4 = I + I - P - 2P = 2I - 3P$$ Now solving, $P^4 \cdot P^2 = P^6 = 2I - 5P + 3P^2$ $$P^6 = 2I - 5P + 3(I - P) = 5I - 8P \ldots (i)$$ Similarly $P^8 = 13I - 21P \ldots (ii)$ Now adding (ii) + (i) we get, $$P^8 + P^6 = 18I - 29P$$ Now subtracting (ii) - (i) we get, $$P^8 - P^6 = 8I - 13P$$ Now comparing with $P^\alpha - P^\beta = \delta I$ and $P^\alpha + P^\beta = \gamma I - 29P$ we get, $\alpha = 8$, $\beta = 6$, $\gamma = 18$, $\delta = 8$ Hence, $\alpha + \beta + \gamma - \delta = 8 + 6 + 18 - 8 = 24$

The given word is PUBLIC. Arranging the letters alphabetically, we get BCILPU. When the word starts with any of the letters B/C/I/L, the number of possibilities is $5! \times 4 = 480$. Now when the word starts with PB, then the number of possibilities is $4! = 24$. Now when the word starts with PC, then the number of possibilities is $4! = 24$. Now when the word starts with PI, then the number of possibilities is $4! = 24$. Now when the word starts with PL, then the number of possibilities is $4! = 24$. Now when the word starts with PUBC, then the number of possibilities is $2! = 2$. Now when the word starts with PUBI, then the number of possibilities is $2! = 2$. Now when the word starts with PUBLC, then the number of possibilities is $1$. Now when the word starts with PUBLIC, then the number of possibilities is $1$. Rank $= 480 + 24 \times 4 + 2 \times 2 + 1 \times 2 = 582$. Hence, rank of the word PUBLIC is 582.

Given, the line $L$ pass through the point $(0, 1, 2)$, intersect the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and be parallel to the plane $2x + y - 3z = 4$. Now plotting the diagram of above data we have, let the direction of line which passes through $(0, 1, 2)$ be $(a, b, c)$. Now the line $L$ and $L_1$ are coplanar, so coplanar condition we get, $$\begin{vmatrix} a & b & c \\ 2 & 3 & 4 \\ 2 & 3 & 4 \end{vmatrix} = 0$$ $$\Rightarrow \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \end{vmatrix} = 0$$ $$\Rightarrow a - 2b + c = 0 \ldots (1)$$ And the line $L$ is perpendicular to the normal vector of the plane, so by perpendicular condition we get, $$2a + b - 3c = 0 \ldots (2)$$ So, from equation (1) and (2) we get, $$a = b = c$$ So, equation of line $L$ will be, $$L = \frac{x-1}{a} = \frac{y-1}{a} = \frac{z-2}{a} = \lambda$$ So any point on $L$ can be taken as $A(\lambda, 1 + \lambda, 2 + \lambda)$. Now $\overrightarrow{AP}$ will be perpendicular to the direction ratio of the line $L$, so, again by perpendicular condition we get, $$\overrightarrow{AP} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$$ $$\Rightarrow \lambda - 1 + \lambda + 10 + \lambda = 0$$ $$\Rightarrow 3\lambda = 0$$ $$\Rightarrow \lambda = -3$$ Hence, $A(-3, -2, -1)$ and $P(1, -9, 2)$ So, by distance formula we get, $AP = \sqrt{74}$

Given, a plane $P$ contains the line of intersection of the plane $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6$ and $\vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5$. Now let, $P_1 : x + y + z - 6 = 0$ and $P_2 : 2x + 3y + 4z + 5 = 0$. So, by family of planes concept we get, $$P : (x + y + z - 6) + \lambda (2x + 3y + 4z + 5) = 0$$ Also given plane $P$ passes through $(0, 2, -2)$. So, $(0 + 2 - 2 - 6) + \lambda (4 + 6 - 8 + 5) = 0$ $$\Rightarrow -6 + \lambda (3) = 0$$ $$\Rightarrow \lambda = 2$$ Hence, equation of plane will be, $$P : 5x + 7y + 9z + 4 = 0$$ Now distance of point $(12, 12, 18)$ from plane will be, $$\left( \frac{5(12) + 7(12) + 9(18) + 4}{\sqrt{25 + 49 + 81}} \right)$$ And square of distance will be, $$= \left( \frac{310}{155} \right)^2$$ $$= 620$$

Let $$I = \int_0^\pi f(x) \sin x \, dx \cdots (1)$$ Now using the property $$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$$ we get, $$\Rightarrow I = \int_0^\pi f(\pi - x) \sin (\pi - x) \, dx$$ $$\Rightarrow I = \int_0^\pi f(\pi - x) \sin x \, dx \cdots (2)$$ Adding (1) and (2), we get $$2I = \int_0^\pi [f(x) + f(\pi - x)] \sin x \, dx$$ $$\Rightarrow 2I = \pi^2 \int_0^\pi \sin x \, dx \{as given f(x) + f(\pi - x) = \pi^2\}$$ $$\Rightarrow 2I = \pi^2 [-\cos x]_0^\pi$$ $$\Rightarrow 2I = 2\pi^2$$ $$\Rightarrow I = \pi^2$$ Hence this is the correct option.

The coefficient of $x^7$ in $\left(ax^2 + \frac{1}{2bx}\right)^{11}$ and $\left(ax - \frac{1}{3bx^2}\right)^{11}$ are equal, then $$T_{r+1} = {}^{11}C_{r} (ax^2)^{11-r} \left(\frac{1}{2bx}\right)^r$$ $$= {}^{11}C_{r} a^{11-r} \left(\frac{1}{2b}\right)^r x^{22-3r}$$ Now solving $22 - 3r = 7 \Rightarrow r = 5$ Coefficient of $x^{-7}$ in $\left(ax - \frac{1}{3bx^2}\right)^{11}$ $$T_{r+1} = {}^{11}C_{r} (ax)^{11-r} \left(-\frac{1}{3bx^2}\right)^r$$ $$= {}^{11}C_{r} a^{11-r} \left(-\frac{1}{3b}\right)^r x^{11-3r}$$ Now solving $11 - 3r = -7 \Rightarrow r = 6$ Now equating coefficient of $x^7$ and $x^{-7}$ we get, $$\therefore {}^{11}C_{5} (a)^6 \left(\frac{1}{2b}\right)^5 = {}^{11}C_{6} a^5 \left(-\frac{1}{3b}\right)^6$$ $$\Rightarrow 3^6 ab = 32$$ $$\Rightarrow 729ab = 32$$ Hence this is the correct option.

Plotting the Truth table of the given expression we get, $$ \begin{array}{|c|c|c|c|c|c|c|} \hline p & q & \neg p & \neg p \land q & p \Rightarrow q & q \Rightarrow p & (p \Rightarrow q) \lor (\neg p \land q) & (q \Rightarrow p) \lor (\neg p \land q) \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\ \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \hline \end{array} $$ Hence, $(q \Rightarrow p) \lor (\neg p \land q)$ is a tautology, and $(p \Rightarrow q) \lor (\neg p \land q)$ is neither tautology and contradiction. So both statements are false.

Given, the tangents at the points $P$ and $Q$ on the circle $x^2 + y^2 - 2x + y = 5$ meet at the point $R\left(\frac{9}{4}, 2\right)$. Now plotting the diagram of the above data we have, Now we know that, length of tangent is given by, $PR = QR = \sqrt{S_1}$. So, $L = \sqrt{S_1} = \sqrt{\left(\frac{9}{4}\right)^2 + (2)^2 - 2 \times \frac{9}{4} + 2 - 5}$ $$\Rightarrow L = \frac{5}{4}$$ Now distance between $CR$ will be, $$CR = \sqrt{\left(\frac{9}{4} - 1\right)^2 + \left(2 + \frac{1}{2}\right)^2} = \sqrt{\frac{25}{16} + \frac{25}{4}} = \sqrt{\frac{125}{16}} = \frac{5\sqrt{5}}{4}$$ Now finding $PQ$ by equating area, we get $$\frac{1}{2} \times \frac{5}{2} \times L = \frac{1}{2} \times PQ \times \frac{CR}{2}$$ $$\Rightarrow \frac{1}{2} \times \frac{5}{4} = \frac{1}{2} \times PQ \times \frac{5\sqrt{5}}{4}$$ $$\Rightarrow PQ = \sqrt{5}$$ Now finding area of triangle $PQR$ by Heron's formula, $$s = \frac{2\sqrt{5} + 5}{4}$$ So, area will be $$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$ $$\Rightarrow \Delta = \sqrt{\frac{2\sqrt{5} + 5}{4} \left(\frac{2\sqrt{5} + 5}{4} - \frac{5}{4}\right) \left(\frac{2\sqrt{5} + 5}{4} - \sqrt{5}\right) \left(\frac{5}{4}\right)}$$ $$\Rightarrow \Delta = \sqrt{\frac{2\sqrt{5} + 5}{4} \left(\frac{2\sqrt{5}}{4}\right) \left(\frac{2\sqrt{5}}{4}\right) \left(\frac{5 - 2\sqrt{5}}{4}\right)}$$ $$\Rightarrow \Delta = \sqrt{\left(\frac{5}{4}\right) \left(\frac{5}{4}\right) \left(\frac{25 - 20}{16}\right)} = \frac{5}{8}$$

We know that, Volume of parallelepiped whose edges determined by vectors $\vec{a}, \vec{b}, \vec{c}$ is $\left[ \vec{a} \vec{b} \vec{c} \right]$. Now it is given that $V$ is volume of parallelepiped $\vec{a}, \vec{b}, \vec{c}$, so required volume of parallelepiped with new edges will be $$= \left[ \vec{a} \vec{b} + \vec{c} \vec{a} + 2\vec{b} + 3\vec{c} \right]$$ $$= \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} \begin{vmatrix} \vec{a} \vec{b} \vec{c} \end{vmatrix}$$ $$= \{(1)(3 - 2) - 0 + 0\} \begin{vmatrix} \vec{a} \vec{b} \vec{c} \end{vmatrix}$$ $$= \left[ \vec{a} \vec{b} \vec{c} \right]$$ $$= V$$ Hence this is the correct option.

Let $S=1^2-2^2+3^2-4^2+\cdots+(2021)^2-(2022)^2+(2023)^2$ $\Rightarrow S=(1-2)(1+2)+(3-4)(3+4)+\cdots+(2021-2022)(2021+2022)+(2023)^2$ $\Rightarrow S=-[3+7+11+15+\cdots+4043]+(2023)^2$ The number of terms in the bracket are $\frac{2022}{2}=1011$ $\Rightarrow S=-\frac{1011}{2}(6+1010\times4)+(2023)^2$ $\Rightarrow S=-1011\times2023+(2023)^2$ $\Rightarrow S=2023\times1012$ $\Rightarrow S=17^2\times7\times1012$ So, $m=17,\ n=7$ and $gcd(17,7)=1$ Hence, $m^2-n^2=17^2-7^2=240$ Hence this is the correct option.

Given, in a group of 100 persons 75 speak English and 40 speak Hindi, and each person speaks at least one of the two languages. And the number of persons who speak only English is $\alpha$ and the number of persons who speaks only Hindi is $\beta$. Now from above diagram we can see that, $$\beta = 100 - 75 = 25$$ So, $$\alpha = 75 - (40 - 25) = 60$$ Hence, number of people who speak English only is $\alpha = 60$ and Hindi is $\beta = 25$. Now putting the value in given equation of ellipse we get, $$25 \left[ \frac{x^2}{60^2} + \frac{y^2}{25^2} \right] = 1$$ $$\Rightarrow \frac{x^2}{36 \times 4} + \frac{y^2}{25} = 1$$ Now comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ we get, $a = 12$ and $b = 5$. Now we know that, eccentricity of ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$. $$\Rightarrow e = \sqrt{1 - \frac{25}{36 \times 4}} = \frac{\sqrt{119}}{12}$$

Given, $f(x) = \frac{x}{(1+x^n)^n}$, $x \in \mathbb{R} - \{-1\}$, $n \in \mathbb{N}$, $n > 2$, $f^n(x) = (f \circ f \circ \ldots up to n times ) (x)$. Now finding $f(f(x)) = \frac{f(x)}{\left[ 1 + (f(x))^n \right]^{\frac{1}{n}}}$. $$\Rightarrow f(f(x)) = \frac{\frac{x}{(1+x^n)^n}}{\left[ 1 + \left( \frac{x}{(1+x^n)^n} \right)^n \right]^{\frac{1}{n}}}$$ $$\Rightarrow f(f(x)) = \frac{x}{\left[ 1 + 2x^n \right]^{\frac{1}{n}}}$$ Similarly, $f^n(x) = \frac{x}{(1+n \cdot x)^n}$. Now solving the integral we get, $$I = \int_0^1 \frac{x^{n-1}}{(1+n x)^{\frac{1}{n}}} \, dx$$ Now let, $1 + n x = t^n$ $$\Rightarrow n^2 x^{n-1} \, dx = n t^{n-1} \, dt$$ $$I = \int_1^{(1+n)^n} \frac{1}{n} \cdot \frac{1}{t} \, dt$$ $$\Rightarrow I = \left[ \frac{1}{n} \ln t \right]_1^{(1+n)^n}$$ $$\Rightarrow I = \frac{1}{n} \left[ n \ln (1+n) - 0 \right]$$ $$\Rightarrow I = n (n-1) \left( (1+n)^{1-\frac{1}{n}} - 1 \right)$$ Now solving the limit we get, $$\lim_{n \to \infty} \frac{(1+n)^{1-\frac{1}{n}} - 1}{n(n-1)}$$ Now let $n = \frac{1}{h}$ so the limit changes to, $$\lim_{h \to 0} \frac{\left( 1 + \frac{1}{h} \right)^{h-1} - 1}{\frac{1}{h} \left( \frac{1}{h} - 1 \right)}$$ $$= \lim_{h \to 0} \frac{h \left( 1 + \frac{1}{h} \right)^{h-1} - h}{1-h}$$ $$= \lim_{h \to 0} \frac{h \left( 1 - h \right)}{1-h}$$ $$= \lim_{h \to 0} h (1-h)$$ $$= \lim_{h \to 0} h = 0$$

Given, $$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$$ $$= (\cot 81^\circ + \tan 81^\circ) - (\tan 27^\circ + \cot 27^\circ)$$ $$= (\tan 9^\circ + \cot 9^\circ) - (\tan 27^\circ + \cot 27^\circ)$$ $$= \left( \frac{\sin 9^\circ}{\cos 9^\circ} + \frac{\cos 9^\circ}{\sin 9^\circ} \right) - \left( \frac{\sin 27^\circ}{\cos 27^\circ} + \frac{\cos 27^\circ}{\sin 27^\circ} \right)$$ $$= \frac{2}{\sin 18^\circ} - \frac{2}{\sin 54^\circ}$$ $$= \frac{2 \times 4}{\sqrt{5} - 1} - \frac{2 \times 4}{\sqrt{5} + 1}$$ $$= 8 \left( \frac{1}{\sqrt{5} - 1} - \frac{1}{\sqrt{5} + 1} \right)$$ $$= 8 \left( \frac{\sqrt{5} + 1 - (\sqrt{5} - 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} \right)$$ $$= 8 \left( \frac{2}{5 - 1} \right) = 4$$

Let, $$L_1 = \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{\alpha} = \lambda$$ $$L_2 = \frac{x-4}{5} = \frac{y-1}{2} = \frac{z-0}{\beta} = \mu$$ For point of intersection $$2\lambda + 1 = 5\mu + 4 \ldots \ (i)$$ $$3\lambda + 2 = 2\mu + 1 \ldots \ (ii)$$ $$\alpha \lambda + 3 = \beta \mu + 0 \ldots \ (iii)$$ From (i) and (ii), we get $\lambda = \mu = -1$. Now, from (iii), we get $\alpha - \beta = 3$. Let $y = 8\alpha \beta$. $$\Rightarrow y = 8\alpha (\alpha - 3)$$ $$\Rightarrow y = 8 \left( \alpha^2 - 3\alpha + \frac{9}{4} \right) - 18$$ $$\Rightarrow y = 8 \left( \alpha - \frac{3}{2} \right)^2 - 18$$ So, Minimum value of $y = -18$ at $\alpha = \frac{3}{2}$.

Let, $$S = (20)^{19} + 2(21)(20)^{18} + 3(21)^{2}(20)^{17} + \ldots + 20(21)^{19}$$ Thus, $$S = 20^{19} + 2 \cdot (20)^{19} \cdot \left( \frac{21}{20} \right) + 3(20)^{19} \cdot \left( \frac{21}{20} \right)^{2} + \ldots + 20(20)^{19} \cdot \left( \frac{21}{20} \right)^{19}$$ $$\Rightarrow S = 20^{19} \left( 1 + 2 \cdot \frac{21}{20} + 3 \left( \frac{21}{20} \right)^{2} + \ldots + 20 \left( \frac{21}{20} \right)^{19} \right)$$ Now on comparing with $S = k \cdot (20)^{19}$ we get, $$\Rightarrow k = 1 + 2 \cdot \left( \frac{21}{20} \right) + 3 \left( \frac{21}{20} \right)^{2} + \ldots + 20 \left( \frac{21}{20} \right)^{19} \ldots (1)$$ $$\Rightarrow \frac{21}{20} k = \frac{21}{20} + 2 \left( \frac{21}{20} \right)^{2} + \ldots + 19 \left( \frac{21}{20} \right)^{19} + 20 \left( \frac{21}{20} \right)^{20} \ldots (2)$$ Now on subtracting above equations we get, $$\Rightarrow k - \frac{21}{20} k = 1 + \frac{21}{20} + \left( \frac{21}{20} \right)^{2} + \ldots + \left( \frac{21}{20} \right)^{19} - 20 \left( \frac{21}{20} \right)^{20}$$ $$\Rightarrow \frac{-k}{20} = \left( \frac{21}{20} \right)^{20 - 1} - 20 \left( \frac{21}{20} \right)^{20}$$ $$\Rightarrow \frac{-k}{20} = \left( \left( \frac{21}{20} \right)^{20} - 1 \right) \times 20 - 20 \left( \frac{21}{20} \right)^{20}$$ $$\Rightarrow \frac{-k}{20} = -20$$ $$\Rightarrow k = 400$$

Given, we have to make four letter word from UNIVERSE each consisting 2 vowels and 2 consonants. Now in UNIVERSE, we have, E, E, I, U, (Vowels) + N, R, S, V (Consonants). 2 different vowels + 2 consonants can be arrange in $^3C_2 \cdot ^4C_2 \cdot 4! = 3 \times 6 \times 24 = 432$ ways. $^3C_2$ is taken because of repetition of letter E.

Let, $f(x) = x^5 - 20x^3 + 50x + 2$. Now differentiating the function $f(x)$ with respect to $x$, we get, $$f'(x) = 5x^4 - 60x^2 + 50$$ $$\Rightarrow f'(x) = 5(x^4 - 12x^2 + 10)$$ Now equating $f'(x) = 0$ we get, $$x^4 - 12x^2 + 10 = 0$$ $$\Rightarrow x^2 = \frac{12 \pm \sqrt{144 - 40}}{2}$$ $$\Rightarrow x^2 = 6 \pm \sqrt{26}$$ $$\Rightarrow x^2 \approx 11.1, \ 0.9$$ $$\Rightarrow x \approx \pm 3.31, \ \pm 0.95$$ Now finding the nature of function at integer value near by above value we get, $f(0) = 2$, $f(1) > 0$, $f(2) 0$. Now plotting the graph by observing above values, we get, Hence, it will cross $x$-axis 5 times.

Given, $\alpha$, $\beta$, $z \in \mathbb{C}$ and $\lambda > 1$. And $\sqrt{\lambda - 1}$ is the radius of the circle $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$. Now we know that equation of circle is given by, $$|z - z_1|^2 + |z - z_2|^2 = |z_1 - z_2|^2,$$ where radius is given by $r = \frac{|z_1 - z_2|}{2}$, On comparing with $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$, we get $$\frac{|z_1 - z_2|}{2} = \frac{|\alpha - \beta|}{2}$$ and $|\alpha - \beta|^2 = 2\lambda$. $$\Rightarrow \frac{|\alpha - \beta|}{2} = \sqrt{\lambda - 1} (given)$$ $$\Rightarrow |\alpha - \beta| = 2\sqrt{\lambda - 1}$$ $$\Rightarrow |\alpha - \beta|^2 = 4(\lambda - 1)$$ $$\Rightarrow 2\lambda = 4(\lambda - 1)$$ $$\Rightarrow \lambda = 2$$ Hence, $|\alpha - \beta|^2 = 4 = |\alpha - \beta| = 2$

Given, a curve $y = f(x)$, $x \in (0, \infty)$ pass through the points $P \left( 1, \frac{3}{2} \right)$ and $Q \left( a, \frac{1}{2} \right)$, and the tangent at any point $R(b, f(b))$ to the given curve cuts the $y$-axis at the point $S(0, c)$ such that $bc = 3$. Now tangent to the curve is given by, $$Y - y = m(X - x), m = \frac{dy}{dx}$$ Now at $Y$-axis, put $X = 0$ we get, $$Y = y - mx or c = y - mx as point S(0, c) given,$$ Now given $bc = 3$ $$\Rightarrow x(y - mx) = 3 \{here b = x on the curve\}$$ $$\Rightarrow y - \frac{xdy}{dx} = \frac{3}{x}$$ $$\Rightarrow ydx - xdy = \frac{3dx}{x} \cdot \frac{1}{x^2}$$ $$\Rightarrow d \left( \frac{y}{x} \right) = 3d \left( \frac{x^{-2}}{-2} \right)$$ Now integrating both side we get, $$\Rightarrow \frac{y}{x} = \frac{3}{2x^2} + C,$$ Now given curve passes through $P \left( 1, \frac{3}{2} \right)$, so $C = 0$. So, equation of curve will be $2xy = 3$. Now curve also passes through the point $Q \left( a, \frac{1}{2} \right)$. So, $a = 3$. Hence, $P \left( 1, \frac{3}{2} \right), Q \left( 3, \frac{1}{2} \right)$. So, by distance formula we get, $(PQ)^2 = 4 + 1 = 5$.

Given, the eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is reciprocal to that of the hyperbola $2x^2 - 2y^2 = 1$. Now eccentricity of rectangular hyperbola $2x^2 - 2y^2 = 1$ is $e_H = \sqrt{2}$. So, $e_e = \frac{1}{\sqrt{2}}$. Now, focus of hyperbola $=(\pm 1, 0)$. Now given that both curve intersect orthogonally, so ellipse and hyperbola are confocal. So, for ellipse $ae_e = 1 \Rightarrow a = \sqrt{2}$. Now length of latus rectum L. R. $= \frac{2b^2}{a} = 2a(1 - e_e^2) = 2\sqrt{2} \cdot \frac{1}{2} = \sqrt{2}$.

Given, the mean and variance of the frequency distribution are 9 and 15.08 respectively. Now mean is given by, $$Mean = \frac{8 + 16 + 120 + 80 + 56 + 80 + 6\alpha + 12\beta}{40 + \alpha + \beta} = 9$$ $$\Rightarrow 360 + 9\alpha + 9\beta = 360 + 6\alpha + 12\beta$$ $$\Rightarrow 3\alpha - 3\beta = 0$$ $$\Rightarrow \alpha = \beta \ldots\ldots(1)$$ Now variance is given by, $$15.08 = \frac{16 + 64 + 36\alpha + 960 + 800 + 144\alpha + 784 + 1280}{40 + 2\alpha} - 9^2$$ $$\Rightarrow 96.08 = \frac{3904 + 180\alpha}{40 + 2\alpha}$$ $$\Rightarrow (40 + 2\alpha)(96.08) = 3904 + 180\alpha$$ $$\Rightarrow 3843.20 + (192.16)\alpha = 3904 + 180\alpha$$ $$\Rightarrow (12.16)\alpha = 60.80$$ $$\Rightarrow \alpha = 5 = \beta$$ Hence, $$\alpha^2 + \beta^2 - \alpha\beta = 25 + 25 - 25 = 25$$

The formula for a lens is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ $\ldots$ (i) $u = (100 \pm 0.2) \, \mathrm{cm} - (80 \pm 0.2) \, \mathrm{cm} = (20 \pm 0.4) \, \mathrm{cm}$ $v = (180 \pm 0.2) \, \mathrm{cm} - (100 \pm 0.2) \, \mathrm{cm} = (80 \pm 0.4) \, \mathrm{cm}$ Substituting the values in the equation of focal length, $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{80} - \left(-\frac{1}{20}\right) = \frac{5}{80}$$ $$\Rightarrow \; f = 16 \, \mathrm{cm}$$ Differentiating the equation (i) $$\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$$ (Only positive sign is considered for error calculation) $$= \frac{0.4}{6400} + \frac{0.4}{400} = 0.4 \times \frac{17}{6400}$$ $$\Rightarrow df = 0.4 \times \frac{17}{6400} \times 16^2 = 0.272$$ The percentage error is $$\frac{df}{f} \times 100 = \frac{0.272}{16} \times 100 = 1.70\%$$

The formula for maximum current is given by $I_{max} = \omega C V_m \ldots (i)$ The given data is $\omega = 120 \pi$ $V_m = 36 \, \mathrm{V}$ $C = 150 \, \mu \mathrm{F}$ Substituting the values in equation (i) $$I_{max} = (120 \pi) \times (150 \times 10^{-6}) \times 36$$ $$= 2.036 \, \mathrm{A}$$ $$\approx 2 \, \mathrm{A}$$

Pascal's law states that the pressure in a closed incompressible fluid is transmitted equally in all directions. The formula for pressure is given by $$P = \frac{F}{A}$$ When a toothpaste is squeezed let the pressure transmitted by $P$. At the mouth of the toothpaste the pressure will also be equal to $P$ due to Pascal's law. The pressure applied to the tube is transmitted equally throughout the toothpaste. When the pressure reaches the open end, it forces toothpaste out through the opening. Hence, both $A$ & $R$ is correct and $R$ is correct explanation for $A$.

When light travels through a medium, its velocity depends on the density of the medium. The formula to calculate the speed of light can be written as $$v = \nu \lambda \ldots (1)$$ From equation (1), it can be concluded that higher the speed, higher is the wavelength. Hence, the wavelength of light depends on the refractive index of the medium. Because of changed wavelength the phase difference changes while the two waves travel the same distance. Thus, both A and R are correct and R is the correct explanation of A.

Let the potential be $x \, \mathrm{V}$ at the junction. By using $V = IR$ we get, $$30 - x = 10i_1$$ $$x - 12 = 20i_2$$ $$x - 2 = 30i_3$$ By using Kirchhoff's law, $$i_1 = i_2 + i_3$$ $$\Rightarrow \frac{30-x}{10} + \frac{12-x}{20} + \frac{2-x}{30} = 0$$ $$\Rightarrow \frac{90 - 3x + 18 - 1.5x + 2 - x}{30} = 0$$ $$\Rightarrow x = 20 \, \mathrm{V}$$ The current through the $20 \, \Omega$ resistor is $$i_2 = \left( \frac{30 - 12}{20} \right) \, \mathrm{A}$$ $$\Rightarrow i_2 = \frac{8}{20} \, \mathrm{A} = 0.4 \, \mathrm{A}$$

The given data is $v = \pi \, \mathrm{m \, s^{-1}}$ By using trigonometric ratios in the triangle shown above the displacement can be found out by $$\sin 60^\circ = \frac{AB}{2R}$$ $$\Rightarrow AB = R\sqrt{3}$$ The formula for distance is $$d = vt$$ $$\Rightarrow t = \frac{d}{v}$$ $$\Rightarrow t = \frac{120^\circ}{360^\circ} \times \frac{2\pi R}{\pi} = \frac{2R}{3}$$ So, the average velocity is $$v_{avg} = \frac{\sqrt{3}R}{\frac{2R}{3}} = \frac{3\sqrt{3}}{2}$$ $$= 1.5\sqrt{3} \, \mathrm{m \, s^{-1}}$$

The data given is $\phi_{Al} = 4.1 \, \mathrm{eV}$ and $\phi_{Au} = 5.1 \, \mathrm{eV}$. The energy is given by $$h\nu = \phi + KE$$ $$\Rightarrow h\nu = \phi + eV_0$$ where $V_0$ is the stopping potential and $K = eV_0$. Thus, the above equation can be written as $$eV_0 = h\nu - \phi$$ $$\Rightarrow V_0 = \frac{h\nu}{e} - \frac{\phi}{e}$$ The equation is of the form of $y = mx + c$. The slope is $$m = \frac{h}{e}.$$ Since the Planck constant $h$ and charge of an electron is a constant $e$, the slope does not depend on the work function. Hence the ratio of the slope of the plot of stopping potential and frequency for both the given metals is 1.

The formula to calculate the speed of sound in hydrogen gas is given by $$v_{H_2} = \sqrt{\frac{\gamma RT}{M_{H_2}}} \ldots (1)$$ The formula to calculate the speed of sound in oxygen is given by $$v_{O_2} = \sqrt{\frac{\gamma RT}{M_{O_2}}} \ldots (2)$$ Divide equation (1) by (2) and simplify to obtain the required ratio. $$\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} \ldots (3)$$ Substitute the values of the known parameters into equation (3) to calculate the required ratio. $$\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32}{2}}$$ $$= 4 : 1$$

The angular speed ($\omega$) of the child can be calculated as follows: $$\omega = \frac{2\pi}{3.14}$$ $$= 2 \, \mathrm{rad} \, \mathrm{s}^{-1}$$ The formula to calculate the centrifugal force on the child is given by $$F_r = m\omega^2 r \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required centrifugal force. $$F_r = 5 \, \mathrm{kg} \times (2 \, \mathrm{rad} \, \mathrm{s}^{-1})^2 \times 2 \, \mathrm{m}$$ $$= 40 \, \mathrm{N}$$

The given data is $v = 60 \, \mathrm{m \, s^{-1}}$ $u = 10 \, \mathrm{m \, s^{-1}}$ $a = 2 \, \mathrm{m \, s^{-2}}$ Using the equation $v = u + at$, we get $$60 = 10 + 2t$$ $$\Rightarrow t = \frac{(60 - 10)}{2}$$ $$\Rightarrow t = 25 \, \mathrm{s}$$

A planet revolves round the Sun by the action of the gravitational force between the sun and the planet. Depending upon the distance from the sun in the elliptic orbit, the speed of revolution of the planet changes. The formula to calculate the linear speed of a planet in an elliptical orbit can be written as $$v = \sqrt{GM \left( \frac{2}{r} - \frac{1}{a} \right)}$$ From the above expression, it can be concluded that the linear speed of the planet is not constant during its revolution.

The electric current due to the charge concentration gradient is known as diffusion current. When the charge carriers move in a semiconductor due to the applied electric field, the resulting current is known as drift current. If the positive terminal of the battery is connected to the p-type while the negative terminal of the battery is connected to the n-type, the movement of electrons is eased as they can travel to the positive terminal of the battery. This causes an increase in the diffusion current. Hence, the diffusion current is more than the drift current. So, A is correct. In an n-type semiconductor, the concentration of electrons is more than that of the holes. While in a p-type semiconductor, the concentration of holes is more than that of the electrons. During the formation of a p-n junction diffusion and drift are the two processes taking place. When a p-n junction is formed, holes diffuse from the p-side to the n-side while electrons diffuse from the n-side to the p-side. This result due to the concentration gradient across p and n sides, which gives rise to a diffusion current across the junction. The diffusion results in an electric field from p-side to n-side. Thus, R is incorrect as diffusion current in $p-n$ junction is from $p$ side to $n$-side.

The data given is $l =$ distance between the charges $q =$ charge $m =$ mass of the positive charge $2m =$ mass of negative charge The moment of inertia of the system is $$I = \left( \frac{m \times 2m}{m + 2m} \right) (l)^2 = \frac{2ml^2}{3}$$ The angular frequency can be written as $$\omega = \sqrt{\frac{pE}{I}} \cdots (i)$$ Substituting the value of moment of inertia in the equation (i) we get $$\omega = \sqrt{\frac{pE}{\frac{2ml^2}{3}}} = \sqrt{\frac{3pE}{2ml^2}}$$ The value of the dipole moment is $p = ql$ Hence, the angular frequency becomes $$\omega = \sqrt{\frac{3qE}{2ml}}$$ This question was given bonus by NTA as none of the options matched in the original paper.

When an electromagnetic wave propagates from the source, it transfers energy to the objects in its path. Electric and magnetic fields serve as energy reservoirs for electromagnetic waves. An electromagnetic wave contains the same amount of energy as are contained in the electric and magnetic fields together. In such situation, the energy density of the electromagnetic wave is equal to the sum of the energies of the electric and magnetic fields. The average electric and magnetic energy densities are given by $$U_E = \frac{1}{2} \varepsilon_0 E^2$$ $$U_B = \frac{1}{2} \frac{B^2}{\mu_0}$$

At 200 K, the rms speed is given by $$v_0 = \sqrt{\frac{3R \times 200}{M}} \cdots (1)$$ At 800 K, the rms speed is given by $$v' = \sqrt{\frac{3R \times 800}{M}} \cdots (2)$$ Divide equation (2) by equation (1) and solve to calculate the required speed. $$\frac{v'}{v_0} = \frac{\sqrt{\frac{3R \times 800}{M}}}{\sqrt{\frac{3R \times 200}{M}}}$$ $$= \sqrt{\frac{800}{200}}$$ $$= 2$$ $$\Rightarrow v' = 2v_0$$

To convert galvanometer into ammeter, low resistances should be added in parallel, and for voltmeter conversion, a very high resistance should be added in series.

Newton's law of cooling states that $$\frac{(T_1 - T_2)}{t} = K \left( \frac{T_1 + T_2}{2} - T_0 \right) \ldots (i)$$ where $T_0$ is the temperature of the surrounding. From equation (i), $$\frac{60 - 40}{7 \times 60} = K \left( \frac{60 + 40}{2} - 10 \right)$$ $$\Rightarrow \frac{20}{7 \times 60} = 40K$$ $$\Rightarrow K = \frac{20}{7 \times 60 \times 40}$$ Let the temperature of the body after the given time be $x \, ^\circ \mathrm{C}$. By Newton's law of cooling, $$\frac{40 - x}{7 \times 60} = K \left( \frac{40 + x}{2} - 10 \right)$$ $$\Rightarrow \frac{40 - x}{7 \times 60} = \frac{1}{14 \times 60} \left( \frac{20 + x}{2} \right)$$ $$\Rightarrow 160 - 4x = 20 + x$$ $$\Rightarrow x = 28 \, ^\circ \mathrm{C}$$

The data given is $$U = \frac{1}{2} m \omega^2 r^2$$ From Bohr's quantization, $$mvr = \frac{nh}{2\pi}$$ $$\Rightarrow v^2 = \left( \frac{nh}{2\pi rm} \right)^2 \cdots (i)$$ In an orbit the value of the kinetic energy is half of the potential energy, $$K = \frac{m \omega^2 r^2}{4}$$ Substituting the value of equation (i) in the kinetic energy, $$K = \frac{1}{2} m \left( \frac{n^2 h^2}{4\pi^2 m^2 r^2} \right) = \frac{m \omega^2 r^2}{4}$$ $$\Rightarrow \frac{n^2 h^2}{2\pi^2 m^2 \omega^2} = r^4$$ $$\Rightarrow r \propto \sqrt{n}$$

Modulation index is the ratio of the difference in amplitudes to the sum of the amplitudes. The formula to calculate the modulation index of a wave is given by $$\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}} \ldots (1)$$ Substitute the values of the known parameters into equation (1) and solve to calculate the required value of maximum amplitude. $$0.6 = \frac{A_{\max} - 3}{A_{\max} + 3}$$ $$\Rightarrow 0.6 A_{\max} + 1.8 = A_{\max} - 3$$ $$\Rightarrow 0.4 A_{\max} = 4.8$$ $$\Rightarrow A_{\max} = \frac{4.8}{0.4}$$ $$= 12 \, \mathrm{V}$$

The acceleration due to gravity changes as one goes to a height from the surface of the earth. Let the new acceleration due to gravity be $g'$. The formula for acceleration due to gravity at a height is given by $g' = \frac{gR^2}{(R+h)^2}$. The weight of the body on surface is $W = mg = 100 \, \mathrm{N}$. The weight at the given height is $W = mg'$. $$\Rightarrow \, W' = 100 \times \frac{R^2}{\left(R + \frac{R}{4}\right)^2}$$ $$\Rightarrow \, W' = 100 \times \frac{R^2}{\frac{25R^2}{16}} = 64 \, \mathrm{N}$$

Let the resistance of the voltmeter be $R \, \Omega$. The voltmeter is in parallel with the $5 \, \Omega$. So the total resistance in the circuit is $$\left( \frac{5R}{5+R} + 2 \right) \, \Omega$$ Applying Ohm's law in the whole circuit, $V = IR$, $$I = \left( \frac{3}{2 + \frac{5R}{5+R}} \right) \, \mathrm{A}$$ It is given that the potential difference across $5 \, \Omega$ is $2 \, \mathrm{V}$. Thus, by Ohm's law, $$\left( \frac{3}{2 + \frac{5R}{5+R}} \right) \times \left( \frac{5R}{R+5} \right) = 2$$ $$\Rightarrow \frac{15 + 3R}{10 + 7R} \times \frac{5R}{5+R} = 2$$ $$\Rightarrow 15R = 20 + 14R$$ $$\Rightarrow R = 20 \, \Omega$$

The data given is $T = 7 \times 10^5 \, \mathrm{N \, m^{-2}}$ $r = 7 \times 10^{-3} \, \mathrm{m}$ Using the equation $$\frac{mg}{A} = T$$ $$\Rightarrow mg = TA$$ $$\Rightarrow mg = T \pi r^2$$ $$\Rightarrow mg = 7 \times 10^5 \times \frac{22}{7} \times 7^2 \times 10^{-6}$$ $$\Rightarrow mg = \frac{49 \times 22}{10}$$ $$\Rightarrow m = \frac{49 \times 22}{98} = 11 \, \mathrm{kg}$$

The given data is $A = 200 \times 10^{-4} \, \mathrm{m^2}$ d $d = 5 \times 10^{-3} \, \mathrm{m}$ The value of $a + b = 4 \times 10^{-3} \, \mathrm{m}$ The formula for a parallel plate capacitor is $$C = \frac{\varepsilon_0 A}{D}$$ Let the capacitor with gap $a$ be $C_1$ and capacitor with gap $b$ be $C_2$ $$C_1 = \frac{\varepsilon_0 A}{a}, C_2 = \frac{\varepsilon_0 A}{b}$$ The equivalent capacitance is $$\frac{1}{C_{eq}} = \frac{a}{\varepsilon_0 A} + \frac{b}{\varepsilon_0 A}$$ $$\Rightarrow C_{eq} = \frac{\varepsilon_0 A}{a+b}$$ It is given that $C_{eq} = x \varepsilon_0$ So, $$\frac{\varepsilon_0 A}{a+b} = x \varepsilon_0$$ $$\Rightarrow x = \frac{A}{a+b}$$ $$\Rightarrow x = \frac{200 \times 10^{-4}}{4 \times 10^{-3}} = 5$$

The moment of inertia of the ring about an axis perpendicular to the plane of ring is $I_r = mk^2$. The moment of inertia of the sphere is $I_s = \frac{2}{5} m R_s^2$. It is given that $$mk^2 = m R_r^2 = \frac{2}{5} m R_s^2 \cdots (i)$$ From equation (i) $$R_r = \sqrt{\frac{2}{5}} R_s$$ From the question, it can be written that $$\frac{R_r}{R_s} = \sqrt{\frac{2}{x}}$$ $$\Rightarrow \frac{R_r}{R_s} = \sqrt{\frac{2}{5}}$$ $$\Rightarrow x = 5$$

From the given figure we can see that $\sin \theta_0 = \frac{A}{l} = \frac{10}{100} = \frac{1}{10}$. From conservation of energy $\frac{mv^2}{2} = mgl \left(1 - \cos \theta_0 \right)$. Maximum tension occurs at mean position $$T = mg + \frac{mv^2}{l}$$ $$\Rightarrow T = mg + 2mg \left(1 - \cos \theta_0 \right)$$ Substituting the values, $$T = mg \left(1 + 2 \left(1 - \sqrt{1 - \sin^2 \theta_0} \right) \right)$$ $$\Rightarrow T = mg \left(3 - 2 \sqrt{1 - \frac{1}{100}} \right)$$ $$\Rightarrow T = 0.25 \times 9.8 \left(3 - 2 \left(1 - \frac{1}{200} \right) \right)$$ $$\Rightarrow T = 0.25 \times 9.8 \times 1.01 = 2.4745 \, \mathrm{N}$$ It is given that $$T = \frac{x}{40}$$ $$\Rightarrow x = 40 \times 2.4745 = 98.98 \approx 99$$

Let the magnetic field of the 200 turns be $B_1$. The formula of magnetic field is given by $B = \frac{\mu_0 i}{2r}$. Let $r = 1000 \, \mathrm{cm}$. Thus, $$B_1 = \frac{n \mu_0 \times I}{2 \times 10} = \frac{200 \mu_0 I}{2 \times 10}$$ The flux through the coil of radius $r_1 = 1 \, \mathrm{cm}$ is $\phi = 10 \times \left( \vec{B} \cdot \vec{A} \right) = 10 \times B_1 \pi r_1^2$. $$\Rightarrow \phi = 10 \times \frac{200 \mu_0 I}{2 \times 10} \times \pi (0.01)^2$$ $$L = 10 \times \frac{200 \mu_0}{2 \times 10} \times \pi (0.01)^2$$ $$= 10 \times \frac{200 \left(4 \pi \times 10^{-7}\right)}{2 \times 10} \times \pi (0.01)^2$$ $$= 4 \times 10^{-8} \, \mathrm{H}$$

The given data is $\lambda_1 = 7000 \, Å$, $\lambda_2 = 5500 \, Å$, $d = 2.5 \times 10^{-3} \, m$, $D = 1.5 \, m$. The path difference is given by $n \lambda_1 = m \lambda_2$. $7n = 5.5 \, m$ implies $14n = 11 \, m$, which implies $n = 11$ and $m = 14$. The formula for the distance of a bright fringe is $y = \frac{n \lambda_1 D}{d}$. Therefore, $$y = \frac{11 \times 7 \times 10^{-7} \times 1.5}{2.5 \times 10^{-3}}$$ $$= 46.2 \times 10^{-4} = 462 \times 10^{-5}$$ It is given that $n \times 10^{-5} = 462 \times 10^{-5}$, which implies $n = 462$.

The given ratio of the velocities is $\frac{v_1}{v_2} = 4$. The percentage loss in kinetic energy is $$\frac{\frac{mv_1^2}{2} - \frac{mv_2^2}{2}}{\frac{mv_1^2}{2}} \times 100 = \frac{x}{4}$$ $$\Rightarrow \frac{v_1^2 - v_2^2}{v_1^2} = \frac{x}{400}$$ $$\Rightarrow 1 - \frac{v_2^2}{v_1^2} = \frac{x}{400}$$ $$\Rightarrow 1 - \frac{1}{16} = \frac{x}{400}$$ $$\Rightarrow x = 375$$

The given data is $q = 1.6 \times 10^{-19} \, \mathrm{C}$ and $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \mathrm{N \, m^2 \, C^{-2}}$. The formula for the value of energy in a hydrogen atom is given by $$E = \frac{q^2}{8 \pi \varepsilon_0 r}$$ which implies $$E = \frac{q^2}{2(4 \pi \varepsilon_0)r}$$ Therefore, $$r = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{12.8 \times 2}$$ which simplifies to $$r = \frac{9}{16} \times 10^{-10} \, \mathrm{m}$$ It is given that $$r = \frac{9}{x} \times 10^{-10} \, \mathrm{m}$$ Therefore, $$\frac{9}{16} \times 10^{-10} = \frac{9}{x} \times 10^{-10}$$ Solving for $x$, we find $$x = 16$$

The given data is $\mathrm{K.E} = 2 \, \mathrm{eV}$ $B = \frac{\pi}{2} \times 10^{-3} \, \mathrm{T}$ $\theta = 60^\circ$ The pitch of the proton is given by $$P = \frac{2\pi m}{qB} \times v \cos \theta \cdots (i)$$ By using $$K = \frac{mv^2}{2}$$ $$\Rightarrow v = \sqrt{\frac{2K}{m}}$$ Substituting the values in equation (i) $$\frac{2\pi \times \sqrt{2mK}E^{\frac{1}{2}} \times 2}{1.6 \times 10^{-19} \times \pi \times 10^{-3}}$$ $$= \frac{2 \times \sqrt{2 \times 1.6 \times 10^{-27} \times 2 \times 1.6 \times 10^{-19} \times 10^3}}{1.6 \times 10^{-19}}$$ $$= 2 \times 2 \times 10^{-1} = 0.4 \, \mathrm{m}$$

Smaller the size of the cation, more the hydration of ion, hence the more values of the hydration enthalpies. Now, as the down the group the size increases so the hydration enthalpies decreases, hence as the increasing order of the size of the ions is as follows: $$\mathrm{Be^{2+} } Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}$$

Ligands that include a numerical prefix in the name use the prefixes bis for 2, tris for 3, or tetrakis for 4 to indicate their number. If the complex ion is an anion, we drop the ending of the metal name and add –ate. In the given coordination compound, Co is the cobalt metal, $\mathrm{C_2O_4^{2-}}$ are the ligands named as oxalato. The charge on $[\mathrm{Co(C_2O_4)_3}]$ is $-3$ and oxalate is having charge of $-2$. $$X + 3(-2) = -3$$ $$X = +3$$ Hence, the IUPAC name of the compound $\mathrm{K_3[Co(C_2O_4)_3]}$ is Potassium trioxalatocobaltate(III).

The single letter amino acid codes is fairly easy to remember. In eleven cases out of twenty, it is just the first letter.

Nessler's reagent is an aqueous solution of potassium iodide, mercuric chloride, and potassium hydroxide. Its chemical formula is $\mathrm{K_2HgI_4}$. It is used to detect the presence of ammonia. It turns into pale brown colour in presence of ammonia. So, N is not present in Nessler's reagent.

BeCl$_2$ is dimeric in vapour phase. The coordination number of beryllium is 3. BeCl$_2$ is monomeric at high temperature. The coordination number of beryllium is 2. BeCl$_2$ is polymeric in solid state. The coordination number of beryllium is 4.

In the case of substituted phenols, the acidity of phenols increases in the presence of the electron-withdrawing group. This is due to the stability of the phenoxide ion generated. In case of meta substituted phenols, the acidic nature can be explained on the basis of inductive effect. Among the given groups $-\mathrm{NO_2}$ and $-\mathrm{Cl}$ groups are electron withdrawing groups and methyl group is electron releasing group. Acidic strength order is: $$OH (with \mathrm{NO_2} ) > OH (with \mathrm{Cl} ) > OH (with \mathrm{CH_3} )$$

The basicity of an oxide is closely related to the electropositive character of the element. The more electropositive the element, the more basic its oxide will be. This is because the oxide ion can react with a cation to form a basic oxide. As we move down Group 13, the electropositive character of the elements increases, and so does the basicity of their oxides. $$\mathrm{B_2O_3 < Al_2O_3 < Ga_2O_3 < In_2O_3 < Tl_2O_3}$$ So most basic oxide among the given options is $\mathrm{Tl_2O_3}$.

The curve contains three activated complexes as shown below in the curve. In the curve, P and Q are the intermediates. The step with the highest activation energy is the rate determining step. Hence, in the given reaction, two intermediates, three activated complexes are involved and step-II is the rate determining step.

The ligand carbon monoxide has no charge, hence metals have zero oxidation state in both $\mathrm{Ni(CO)_4}$ and $\mathrm{Fe(CO)_5}$. Low oxidation states are found when a complex has ligands capable of $\pi$-acceptor character in addition to the $\sigma$-bonding. In the complexes, $\mathrm{Ni(CO)_4}$ and $\mathrm{Fe(CO)_5}$, the ligand CO is $\pi$-acceptor character in addition to the $\sigma$-bonding.

In neutral or weakly alkaline solution oxidation state of Mn changes by 3 unit. The reaction between permanganate and thiosulfate is a redox reaction, in which permanganate acts as an oxidizing agent and thiosulfate acts as a reducing agent. $$\mathrm{MnO_4^- + S_2O_3^{2-} \rightarrow MnO_2 + SO_4^{2-}}$$ This ionic mechanism is favoured in neutral aqueous medium.

α, β-unsaturated carbonyl compounds undergo 1,4-addition reaction with Grignard reagent in the presence of cuprous iodide. Now the carbanion formed undergo nucleophilic substitution reaction with propyl iodide as shown below.

On adding $\mathrm{H_2SO_4}$ to $\mathrm{Pb^{2+}}$, white precipitate of lead sulphate, $\mathrm{PbSO_4}$ is formed. By adding potassium iodide solution to $\mathrm{Pb^{2+}}$ - Yellow precipitate of lead iodide is formed. The ppt dissolves in boiling water and on cooling recrystallises. By adding potassium chromate solution to $\mathrm{Pb^{2+}}$ - Yellow precipitate of lead chromate is formed. But lead nitrate is a water soluble compound.

Column chromatography separates compounds based on their polarity. The more polar a compound is, the more it interacts with the stationary phase and the slower it moves through the column. The less polar a compound is, the faster it moves through the column. As the chromatogram, degree of polarity $$\rightarrow a > b > c$$ Therefore, statements A, B are incorrect as b comes out before 'C' the statement C is also incorrect. As a is most polar, it spends most time. Hence, A, B and C are incorrect statements.

Opioids (narcotic analgesics) are a class of medicines that are used to provide relief from moderate-to-severe acute or chronic pain. They may also be called opiates, opioid analgesics, or narcotics. Analgesic is another name for a medicine that relieves pain. Morphine is a narcotic analgesic. It helps in relieving pain with producing sleep.

The balanced chemical equation for the reaction between HBr and Ba(OH)_2 is: $$2 \mathrm{HBr} + \mathrm{Ba(OH)_2} \rightarrow \mathrm{BaBr_2} + 2 \mathrm{H_2O}$$ From this equation, we can see that 2 moles of HBr react with 1 mole of Ba(OH)_2. Equal equivalents will react. Number of equivalents = Normality $\times$ Volume So, $$N_1 V_1 = N_2 V_2$$ $$0.02 \times V_1 = 0.02 \times 10$$ $$V_1 = 10 \, \mathrm{ml}$$

The aqueous solution of sodium chloride is known as brine solution. On electrolysis of brine solution, the following reactions are observed. Oxidation reaction takes place at anode and reduction reaction takes place at cathode. The reaction at the anode is: $$2 OH^- \rightarrow \frac{1}{2} O_2 + H_2O + 2e^-$$ At the cathode: $$H_2O + e^- \rightarrow \frac{1}{2} H_2 + OH^-$$ Hydrogen and oxygen are electrode products. The remaining solution contains NaOH. Hence, HCl is not obtained.

DDT (dichlorodiphenyltrichloroethane) and Aldrin are two synthetic pesticides that were widely used for insect control. Pesticides are groups of chemicals used for destruction of insects, weeds, fungi, bacteria etc. They are generally called insecticides, fungicides, bactericides, herbicides and rodenticides.

The given reaction is an example of dehydration of alcohol. In this reaction the first step is protonation of alcohol which leads to the formation of carbocation. The carbocation formed undergoes rearrangement to get stability. The tertiary carbocation formed in this reaction is involved in ring closure, which loses a proton to form alkene. The steps are shown below.

Gallium (Ga) remains as liquid inside boiling water, and the other elements mentioned in options can react with water. Bromine is a halogen that reacts readily with water to form hydrobromic acid and hypobromous acid, which can further react to form bromic acid. Lithium and Caesium are alkali metals that react violently with water, producing hydrogen gas and the corresponding metal hydroxide.

According to Bohr's angular momentum equation, $$mvr = \frac{nh}{2\pi}$$ The angular momentum equation for third orbit is $$mvr = \frac{3h}{2\pi}$$ $$\frac{2\pi r}{3} = \frac{h}{mv}$$ According de Broglie's equation $$\lambda = \frac{h}{mv}$$ $$\therefore \lambda = \frac{2\pi 9 a_0}{3} = 6 \pi a_0$$

Hydrogen bonding occurs between hydrogen and strongly electronegative atoms like F, N, O. In water or ice hydrogen bonding occurs as there is hydrogen as well as oxygen both are present. Water has 2 hydrogen atoms and 1 oxygen atom. The oxygen of one water molecule has two lone pairs of electrons, each of which can form hydrogen bonds with hydrogen on other two water molecules. Each water molecule is H-bonded to 4 neighbouring molecules.

Given the reaction $\mathrm{PCl_3(g)} + \mathrm{Cl_2(g)} \rightleftharpoons \mathrm{PCl_5(g)}$ with initial concentrations $0.2$, $0.1$, and $0.4$ respectively. The equilibrium constant $K_c$ is given by $$K_c = \frac{0.4}{0.2 \times 0.1} = 20.$$ If $0.2$ moles of $\mathrm{Cl_2}$ is added, the reaction takes place in the forward direction. We have $$20 = K_c = \frac{0.4 + X}{(0.3 - X)(0.2 - X)}.$$ This implies $$0.4 + X = 20(0.3 - X)(0.2 - X).$$ Therefore, $$0.4 + X = 20(0.06 + X^2 - 0.5X).$$ Simplifying, $$0.4 + X = 1.2 + 20X^2 - 10X.$$ Rearranging gives $$20X^2 - 11X + 0.8 = 0.$$ Solving for $X$, $$X = \frac{11 \pm \sqrt{121 - 64}}{40} = \frac{11 - 7.55}{40} \approx 0.08625.$$ Thus, the concentration of $\mathrm{PCl_5}$ is $$0.48625 \approx 48.625 \times 10^{-2} or 49 \times 10^{-2}.$$

A solution having the same solute and solvent concentration as a cell across the semipermeable membrane is referred to as isotonic solution. The effective concentration of each solution can be calculated as follows, Effective concentration = $i(Von't Hoff factor) \times C$ 1M NaCl has effective concentration = $2 \times 1 = 2\, M$ and hence it is isotonic with two molar urea solution. 1M CaCl$_2$ has effective concentration = $3\, M$ and 1.5M KCl has also $3\, M$, hence, both are isotonic. 1.5M AlCl$_3 \cong 6\, M$ and 2M Na$_2$SO$_4 \cong 6\, M$, Hence both are isotonic. 2.5M KCl $\cong 5\, M$ and 1M Al$_2$(SO$_4$)$_3 \cong 5\, M$, Hence both are isotonic.

For feasibility, check, $E^\circ_{cell} = E^\circ_{cathode (reduction)} - E^\circ_{anode (oxidation)} > 0$. The $E^\circ$ values when metal acts as anode and $\mathrm{NO_3^-}$ reaction is cathodic reaction. For Vanadium metal, $E^\circ_{cell} = 0.97 + 1.19 = 2.16\, \mathrm{V}$. For Iron metal, $E^\circ_{cell} = 0.97 + 0.04 = 1.01\, \mathrm{V}$. For silver metal, $E^\circ_{cell} = 0.97 - 0.80 = 0.17\, \mathrm{V}$. For Gold metal, $E^\circ_{cell} = 0.97 - 1.140 = -0.17\, \mathrm{V}$. For electrodes having oxidation potential greater than $-0.97\, \mathrm{V}$, $E^\circ_{cell} > 0$. Therefore, $\mathrm{Ag}$, $\mathrm{Fe}$ and $\mathrm{V}$ can be oxidised.

The phase that is scattered or present in the form of colloidal particles is called dispersed phase and the medium in which the colloidal particles are dispersed is called the dispersion medium. Coloured gemstones are solid sols, colloids in which solid particles are dispersed in solid medium. In paints, small amount of solid pigments are dispersed in a large amount of liquid solvent. So, it is an example of a solid in liquid mixture. Smoke is a solid aerosol-type colloid consisting of the solid dispersed phase in the gaseous dispersed medium. Cheese is an example of gel in the colloidal systems, in which the dispersed phase is a liquid while the dispersion medium is a solid. Milk is a type of 'liquid in liquid' colloid, also known as emulsion. Hair cream is an example of a colloidal system where dispersed Phase is Liquid and dispersion Medium is also Liquid. Insecticide sprays is an example of a colloidal solution in which liquid particles are dispersed in the gaseous phase. Froth is a colloidal solution of gas in liquid. Soap lather is an example of colloidal system foam in which the dispersed phase is gas and the dispersion medium is liquid. Paints, milk, froth, soap lather and hair cream have liquid as dispersion medium.

XeF_4 is $sp^3d^2$ hybridised with two lone pairs of electrons, hence, it is square planar. SF_4 is $sp^3d$ hybridised with one lone pair of electrons, hence, it is See saw. SiF_4 is $sp^3$ hybridised with zero lone pairs of electrons, hence, it is Tetrahedral. BF_4^- is $sp^3$ hybridised with zero lone pairs of electrons, hence, it is Tetrahedral. $[\mathrm{Cu(NH_3)_4}]^{2+}$ is $dsp^2$ hybridised and is Square planar. $[\mathrm{FeCl_4}]^{2-}$ is $sp^3$ hybridised and is Tetrahedral. $[\mathrm{PtCl_4}]^{2-}$ is $dsp^2$ hybridised and is Square planar. BrF_4^- is $sp^3d^2$ with Square planar geometry. So, 4 square planar shape compounds are present.

The reaction is $\mathrm{C_2H_5OH(l)} \rightarrow 2 \mathrm{CO_2(g)} + 3 \mathrm{H_2O(l)} + 3/2 \mathrm{O_2(g)}$. Heat of combustion = Heat of formation of products - Heat of formation of reactants. Heat of combustion of hydrogen is equal to heat formation of water, and heat of combustion of carbon is equal to heat of formation of carbon dioxide. $$\Delta H_C = [2 \Delta H_f^\circ (\mathrm{CO_2}) + 3 \Delta H_f^\circ (\mathrm{H_2O})] - [\Delta H_f^\circ (\mathrm{C_2H_5OH(l)})]$$ $$-1234.7 = [2 \times (-393.5) + 3 \times (-241.8)] - [\Delta H_f^\circ (\mathrm{C_2H_5OH(l)})]$$ $$\Delta H_f^\circ \mathrm{C_2H_5OH} = -277.7 \, \mathrm{kJ/mol}$$ $$\simeq -278 \, \mathrm{kJ/mol}$$

Iodoform test is used to check the presence of carbonyl compounds with the structure \[ \mathrm{R-CO-CH_3} \] or alcohols with the structure \[ \mathrm{R-CH(OH)-CH_3} \] in a given unknown substance. The reaction of iodine, a base and a methyl ketone gives a yellow precipitate along with an “antiseptic” smell. Hence, (c), (d), (e) and (f) give iodoform reaction.

$C_8H_{11}N$ Degree of unsaturation = 4. The degree of unsaturation indicates the total number of pi bonds and rings within a molecule which makes it easier for one to figure out the molecular structure. Aniline derivatives cannot be prepared by Gabriel phthalimide synthesis. So, number of aromatic amines = 5

Bravais Lattice refers to the 14 different 3-dimensional configurations into which atoms can be arranged in crystals. The smallest group of symmetrically aligned atoms which can be repeated in an array to make up the entire crystal is called a unit cell. Crystal systems where body centred unit cell can be found are Cubic, orthorhombic and tetragonal. Hence, correct answer is 3.