JEE Main 1 September 2021 Shift 2 question paper with solutions

Given the limit expression: $$\lim_{x \to \frac{\pi}{4}} \frac{\int_{2}^{\sec^2 x} f(x) dx}{x^2 - \frac{\pi^2}{16}}$$ We simplify it as follows: $$\lim_{x \to \frac{\pi}{4}} \cdot \frac{\left[f(\sec^2 x) \cdot 2 \sec x \cdot \sec x \tan x \right]}{2x}$$ This becomes: $$\lim_{x \to \frac{\pi}{4}} \frac{\pi}{4} f\left(\sec^2 x\right) \cdot \sec^3 x \cdot \frac{\sin x}{x}$$ Evaluating the limit, we have: $$\frac{\pi}{4} f(2) \cdot \left(\sqrt{2}\right)^3 \cdot \frac{1}{\sqrt{2}} \times \frac{4}{\pi}$$ This simplifies to: $$\Rightarrow 2f(2)$$

Given $$\cos^{-1}(\cos(-5)) + \sin^{-1}(\sin(6)) - \tan^{-1}(\tan(12))$$ This simplifies to $$(2\pi - 5) + (6 - 2\pi) - (12 - 4\pi)$$ Which further simplifies to $$4\pi - 11$$

Given $$\Delta = \begin{vmatrix} -1 & 1 & 2 \\ 3 & -a & 5 \\ 2 & -2 & -a \end{vmatrix}$$ $$= -1 \left(a^2 + 10\right) - 1(-3a - 10) + 2(-6 + 2a)$$ $$= -a^2 + 10 + 3a + 10 - 12 + 4a$$ $$\Delta = -a^2 + 7a - 12$$ $$\Delta = -\left[a^2 - 7a + 12\right]$$ $$\Delta = -[(a - 3)(a - 4)]$$ $$\Delta_1 = \begin{vmatrix} 0 & 1 & 2 \\ 1 & -a & 5 \\ 7 & -2 & -a \end{vmatrix}$$ $$= 0 - 1(-a - 35) + 2(-2 + 7a)$$ $$\Rightarrow a + 35 - 4 + 14a$$ $$15a + 31$$ Now $$\Delta_1 = 15a + 31$$ For inconsistent $$\Delta = 0$$. Therefore, $a = 3, a = 4$ and for $a = 3$ and $4$ $$\Delta_1 \neq 0$$ $$n(S_1) = 2$$ For infinite solution: $$\Delta = 0$$ and $$\Delta_1 = \Delta_2 = \Delta_3 = 0$$ Not possible Therefore, $$n(S_2) = 0$$

Given the planes: $$P_1: x - 2y - 2z + 1 = 0$$ $$P_2: 2x - 3y - 6z + 1 = 0$$ The equation of the bisector is given by: $$\frac{x - 2y - 2z + 1}{\sqrt{1 + 4 + 4}} = \frac{2x - 3y - 6z + 1}{\sqrt{2^2 + 3^2 + 6^2}}$$ Simplifying, we have: $$\frac{x - 2y - 2z + 1}{3} = \pm \frac{2x - 3y - 6z + 1}{7}$$ Since $a_1a_2 + b_1b_2 + c_1c_2 = 20 > 0$, the negative sign will give the acute bisector. Thus, we have: $$7x - 14y - 14z + 7 = -[6x - 9y - 18z + 3]$$ Simplifying further: $$13x - 23y - 32z + 10 = 0$$ The point $(-2, 0, -\frac{1}{2})$ satisfies it. Therefore, the answer is (2).

The truth tables are given for the expressions. The first table shows the values for $p$, $q$, $\sim p$, $\sim q$, $p \rightarrow q$, $\sim (p \rightarrow q)$, and $\sim (q \rightarrow p)$. The second table shows the values for $p \land \sim q$, $\sim p \rightarrow \sim q$, $p \rightarrow \sim q$, and $\sim (p \rightarrow \sim q)$. The equivalence $p \land \sim q \equiv \sim (p \rightarrow q)$ is shown, which corresponds to Option (4).

Total ways of choosing square = $\binom{64}{2}$ = $\frac{64 \times 63}{2 \times 1}$ = 32 $\times$ 63. Ways of choosing two squares having common side = 2(7 $\times$ 8) = 112. Required probability = $\frac{112}{32 \times 63}$ = $\frac{16}{32 \times 9}$ = $\frac{1}{18}$.

Given $x^2 dy + \left( y - \frac{1}{x} \right) dx = 0 : x > 0, y(1) = 1$. Rewriting, we have $x^2 dy + \frac{(xy - 1)}{x} dx = 0$. This simplifies to $x^2 dy = \frac{(xy - 1)}{x} dx$. Differentiating, $\frac{dy}{dx} = \frac{1 - xy}{x^3}$. This becomes $\frac{dy}{dx} = \frac{1}{x^3} - \frac{y}{x^2}$. Rearranging gives $\frac{dy}{dx} = \frac{1}{x^2} \cdot y = \frac{1}{x^3}$. If $e^{\int \frac{1}{x^2} dx} = e^{-\frac{1}{x}}$, then $ye^{-\frac{1}{x}} = \int \frac{1}{x^3} \cdot e^{-\frac{1}{x}}$. This results in $ye^{-\frac{1}{x}} = e^{-x} \left( 1 + \frac{1}{x} \right) + C$. Substituting $1 \cdot e^{-1} = e^{-1}(2) + C$, we find $C = -e^{-1} = -\frac{1}{e}$. Thus, $ye^{-\frac{1}{x}} = e^{-\frac{1}{x}} \left( 1 + \frac{1}{x} \right) - \frac{1}{e}$. Evaluating $y \left( \frac{1}{2} \right) = 3 - \frac{1}{e} \times e^2$, we conclude $y \left( \frac{1}{2} \right) = 3 - e$.

Given the equation: $$2 \cos x \left( 4 \sin \left( \frac{\pi}{4} + x \right) \sin \left( \frac{\pi}{4} - x \right) - 1 \right) = 1$$ Simplify using the identity $\sin(a + b) \sin(a - b) = \sin^2 a - \sin^2 b$: $$2 \cos x \left( 4 \left( \sin^2 \frac{\pi}{4} - \sin^2 x \right) - 1 \right) = 1$$ Since $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, we have $\sin^2 \frac{\pi}{4} = \frac{1}{2}$: $$2 \cos x \left( 4 \left( \frac{1}{2} - \sin^2 x \right) - 1 \right) = 1$$ Simplify further: $$2 \cos x (2 - 4 \sin^2 x - 1) = 1$$ $$2 \cos x (1 - 4 \sin^2 x) = 1$$ Using the identity $\sin^2 x = 1 - \cos^2 x$: $$2 \cos x (4 \cos^2 x - 3) = 1$$ Let $y = \cos x$, then: $$4 y^3 - 3 y = \frac{1}{2}$$ This simplifies to: $$\cos 3x = \frac{1}{2}$$ Given $x \in [0, \pi]$, therefore $3x \in [0, 3\pi]$.

Given $f(x) = x^3 - 6x^2 + ax + b$. $f(2) = 8 - 24 + 2a + b = 0$ $2a + b = 16 \ldots (1)$ $f(4) = 64 - 96 + 4a + b = 0$ $4a + b = 32 \ldots (2)$ Solving (1) and (2) $a = 8, \ b = 0$ $f(x) = x^3 - 6x^2 + 8x$ $f(x) = x^3 - 6x^2 + 8x$ $f'(x) = 3x^2 - 12x + 8$ $f''(x) = 6x - 12$ $\Rightarrow f'(x)$ is $\uparrow$ for $x > 2$, and $f'(x)$ is $\downarrow$ for $x 0$ for $x \in (x_4, 4)$ $x_4 \in (3, 4)$ $f(x) = x^3 - 6x^2 + 8x$ $f(3) = 27 - 54 + 24 = -3$ $f(4) = 64 - 96 + 32 = 0$ For $x_4(3, 4)$ $f(x_4) -4$ $2f'(x_3) > -8$ So, $2f'(x_3) = \sqrt{3}f(x_4)$ Correct Ans. (1)

Given $$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ $$J_{6+i,3} - J_{i+3,3}; i \leq j$$ $$\Rightarrow \int_0^{1/2} \frac{x^{6+i} - x^{i+3}}{x^3-1} - \int_0^{1/2} \frac{x^{i+3}}{x^3-1}$$ $$\Rightarrow \int_0^{1/2} x^{i+3} (x^3-1)$$ $$\Rightarrow \frac{x^{3+i+1}}{3+i+1} = \left( \frac{x^{4+i+1}}{4+i} \right)^{1/2}$$ $$a_{ij} = J_{6+i,3} - J_{i+3,3,3} = \left( \frac{1}{2} \right)^{4+i} \frac{1}{4+i}$$ $$a_{11} = \left( \frac{1}{2} \right)^5 \frac{1}{5} = \frac{1}{5 \cdot 2^5}$$ $$a_{12} = \frac{1}{5 \cdot 2^5}$$ $$a_{13} = \frac{1}{5 \cdot 2^5}$$ $$a_{22} = \frac{1}{6 \cdot 2^6}$$ $$a_{23} = \frac{1}{6 \cdot 2^6}$$ $$a_{33} = \frac{1}{7 \cdot 2^7}$$ $$A = \begin{bmatrix} \frac{1}{5 \cdot 2^5} & \frac{1}{5 \cdot 2^5} & \frac{1}{5 \cdot 2^5} \\ 0 & \frac{1}{6 \cdot 2^6} & \frac{1}{6 \cdot 2^6} \\ 0 & 0 & \frac{1}{7 \cdot 2^7} \end{bmatrix}$$ $$|A| = \frac{1}{5 \cdot 2^5} \cdot \frac{1}{6 \cdot 2^6} \cdot \frac{1}{7 \cdot 2^7}$$ $$|A| = \frac{1}{210 \cdot 2^{18}}$$ $$|adjA^{-1}| = |A^{-1}|^{n-1} = |A^{-1}|^2 = \frac{1}{(|A|)^2}$$ $$\Rightarrow (210 \cdot 2^{18})^2$$ $$(105)^2 \times 2^{38}$$

Given $$A = \int_0^{\pi/2} ((\sin x + \cos x) - |\cos x - \sin x|) \, dx$$ We have $$A = \int_0^{\pi/2} ((\sin x + \cos x) - (\cos x - \sin x)) \, dx$$ plus $$\int_{\pi/4}^{\pi/2} ((\sin x + \cos x) - (\sin x - \cos x)) \, dx$$ This simplifies to $$A = 2 \int_0^{\pi/2} \sin x \, dx + 2 \int_{\pi/4}^{\pi/2} \cos x \, dx$$ Evaluating the integrals, we get $$A = -2 \left( \frac{1}{\sqrt{2}} - 1 \right) + 2 \left( 1 - \frac{1}{\sqrt{2}} \right)$$ Simplifying further, $$A = 4 - 2\sqrt{2} = 2\sqrt{2}(\sqrt{2} - 1)$$ Option (1)

Given the equations $3y - 2z - 1 = 0$ and $3x - z + 4 = 0$. The direction ratios (D.R's) are $(0, 3, -2)$ and $(3, -1, 0)$. Let the D.R's of the given line be $a, b, c$. Now $3b - 2c = 0$ and $3a - c = 0$. Therefore, $6a = 3b = 2c$. Thus, $a : b : c = 3 : 6 : 9$. Any point on the line is $3K - 1, 6K + 1, 9K + 1$. Now $3(3K - 1) + 6(6K + 1) + 9(9K + 1) = 0$. This implies $K = \frac{1}{3}$. The point on the line is $(0, 3, 4)$. Given point is $(2, -1, 6)$. Therefore, the distance is $\sqrt{4 + 16 + 4} = 2\sqrt{6}$.

Given: $y - \dfrac{3}{4} = \left(x - \dfrac{1}{2}\right)^2 \quad \cdots (1)$ **Step 1: Find point P** For $x = -\dfrac{1}{2}$: $$y - \frac{3}{4} = 1 \implies y = \frac{7}{4} \implies P\left(-\frac{1}{2}, \frac{7}{4}\right)$$ **Step 2: Find slope of normal** $$y' = 2\left(x - \frac{1}{2}\right)$$ At $x = -\dfrac{1}{2}$: $$m_T = -2, \qquad m_N = \frac{1}{2}$$ **Step 3: Equation of Normal** $$y - \frac{7}{4} = \frac{1}{2}\left(x + \frac{1}{2}\right)$$ $$y = \frac{x}{2} + 2$$ **Step 4: Substitute into (1) to find Q** $$\frac{x}{2} + 2 - \frac{3}{4} = \left(x - \frac{1}{2}\right)^2$$ $$\implies x = 2 \ \text{or} \ x = -\frac{1}{2}$$ $$\implies Q(2, 3)$$ **Step 5: Find $(PQ)^2$** $$(PQ)^2 = \left(2+\frac{1}{2}\right)^2 + \left(3-\frac{7}{4}\right)^2 = \frac{25}{4} + \frac{25}{16} = \frac{125}{16}$$ **Answer: Option (2)**

Consider the equation $x^2 + ax + b = 0$. If has two roots (not necessarily real $\alpha$ and $\beta$). Either $\alpha = \beta$ or $\alpha \neq \beta$. Case(1) If $\alpha = \beta$, then it is repeated root. Given that $\alpha^2 - 2$ is also a root. So, $\alpha^2 - 2 \Rightarrow (\alpha + 1)(\alpha - 2) = 0$ $$\Rightarrow \alpha = -1 or \alpha = 2$$ When $\alpha = -1$ then $(a, b) = (2, 1)$ $\alpha = 2$ then $(a, b) = (-4, 4)$ Case(2) If $\alpha \neq \beta$ Then (I) $\alpha = \alpha^2 - 2$ and $\beta = \beta^2 - 2$ Here $(\alpha, \beta) = (2, -1) or (-1, 2)$ Hence $(a, b) = (-(\alpha + \beta), \alpha \beta)$ $$= (1, -2)$$ (II) $\alpha = \beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$ Since $\alpha \neq \beta$ we get $\alpha + \beta = \beta^2 + \alpha^2 - 4$ $\alpha + \beta = (\alpha + \beta)^2 - 2\alpha \beta - 4$ Thus $-1 = 1 - 2\alpha \beta - 4$ which implies $\alpha \beta = -1$ Therefore $(a, b) = (-(\alpha + \beta), \alpha \beta)$ $$= (1, -1)$$ (III) $\alpha = \alpha^2 - 2 = \beta^2 - 2 and \alpha \neq \beta$ $$\Rightarrow \alpha = -\beta$$ Thus $\alpha = 2, \beta = -2$ $\alpha = -1, \beta = 1$ Therefore $(a, b) = (0, -4) \& (0, -1)$ (IV) $\beta = \alpha^2 - 2 = \beta^2 - 2 and \alpha \neq \beta$ is same as (III) Therefore we get 6 pairs of $(a, b)$ Which are $(2, 1), (-4, 4), (-1, -2), (1, -2), (1, -1)(0, -4)$ Option (1)

Let $T_r = r(n - r)$. $T_r = nr - r^2$. Therefore, $$S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} (nr - r^2)$$ $$S_n = \frac{n \cdot (n)(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$$ $$S_n = \frac{n(n-1)(n+1)}{6}$$ Now $$\sum_{r=4}^{\infty} \left( \frac{2 \cdot S_n}{n!} - \frac{1}{(n-2)!} \right)$$ $$= \sum_{r=4}^{\infty} \left( 2 \cdot \frac{n(n-1)(n+1)}{6 \cdot n(n-1)(n-2)!} - \frac{1}{(n-2)!} \right)$$ $$= \sum_{r=4}^{\infty} \left( \frac{1}{3} \cdot \frac{n-2+3}{(n-2)!} - \frac{1}{(n-2)!} \right)$$ $$= \sum_{r=4}^{\infty} \frac{1}{3} \cdot \frac{1}{(n-3)!} = \frac{1}{3} (e - 1)$$ Option (1)

Total Number of Triangles = $$^{15}C_3$$ $$i + j + k = 15$$ (Given) Number of Possible triangles using the vertices $P_i, P_j, P_k$ such that $i + j + k \neq 15$ is equal to $$^{15}C_3 - 12 = 443$$ Option (3)

Given $$f(x) = \log_{\sqrt{5}} \left( 3 + \cos \left( \frac{3\pi}{4} + x \right) + \cos \left( \frac{\pi}{4} + x \right) + \cos \left( \frac{\pi}{4} - x \right) - \cos \left( \frac{3\pi}{4} - x \right) \right)$$ We have $$f(x) = \log_{\sqrt{5}} \left[ 3 + 2 \cos \left( \frac{\pi}{4} \right) \cos(x) - 2 \sin \left( \frac{3\pi}{4} \right) \sin(x) \right]$$ This simplifies to $$f(x) = \log_{\sqrt{5}} [3 + \sqrt{2}(\cos x - \sin x)]$$ Since $$-\sqrt{2} \leq \cos x - \sin x \leq \sqrt{2}$$ It follows that $$\Rightarrow \log_{\sqrt{5}} \left[ 3 + \sqrt{2}(-\sqrt{2}) \right] \leq f(x) \leq \log_{\sqrt{5}} [3 + \sqrt{2}(\sqrt{2})]$$ Thus, $$\Rightarrow \log_{\sqrt{5}}(1) \leq f(x) \leq \log_{\sqrt{5}}(5)$$ So the range of $f(x)$ is $[0, 2]$. Option (4)

Given $$\sum_{n=1}^{20} \frac{1}{a_n a_{n+1}} = \sum_{n=1}^{20} \frac{1}{a_n (a_n + d)}$$ $$= \frac{1}{d} \sum_{n=1}^{20} \left( \frac{1}{a_n} - \frac{1}{a_n + d} \right)$$ $$\Rightarrow \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_{21}} \right) = \frac{4}{9} (Given)$$ $$\Rightarrow \frac{1}{d} \left( \frac{a_{21} - a_1}{a_1 a_{21}} \right) = \frac{4}{9}$$ $$\Rightarrow \frac{1}{d} \left( \frac{a_1 + 20 \, d - a_1}{a_1 a_2} \right) = \frac{4}{9} \Rightarrow a_1 a_2 = 45 \ldots (1)$$ Now sum of first 21 terms $$= \frac{21}{2} (2a_1 + 20 \, d) = 189$$ $$\Rightarrow a_1 + 10d = 9 \ldots (2)$$ For equation (1) $\&$ (2) we get $$a_1 = 3 \& d = \frac{3}{5}$$ OR $$a_1 = 15 \& d = -\frac{3}{5}$$ So, $$a_6 \cdot a_{16} = (a_1 + 5d) \,(a_1 + 15d)$$ $$\Rightarrow a_6 a_{16} = 72$$ Option (2)

Given $$f(x) = x + \int_0^{\pi/2} \sin x \cos y f(y) \, dy$$ $$f(x) = x + \sin x \int_0^{\pi/2} \cos y f(y) \, dy$$ Let $$K = \int_0^{\pi/2} \cos y f(y) \, dy$$ Thus, $$f(x) = x + K \sin x$$ Similarly, $$f(y) = y + K \sin y$$ Now, $$K = \int_0^{\pi/2} \cos y (y + K \sin y) \, dy$$ $$K = \int_0^{\pi/2} y \cos y \, dy + \int_0^{\pi/2} \cos y \sin y \, dy$$ $$K = (y \sin y)\bigg|_0^{\pi/2} - \int_0^{\pi/2} \sin y \, dy + K \int_0^1 t \, dt$$ $$\Rightarrow K = \frac{\pi}{2} - 1 + K \left( \frac{1}{2} \right)$$ $$\Rightarrow K = \pi - 2$$ So, $$f(x) = x + (\pi - 2) \sin x$$ Option (4)

The point of intersection of the curves $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and $x^{2}+y^{2}=3$ in the first quadrant is $\left(\frac{3}{2},\frac{\sqrt{3}}{2}\right)$ Now slope of tangent to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ at $\left(\frac{3}{2},\frac{\sqrt{3}}{2}\right)$ is $m_{1}=-\frac{1}{3\sqrt{3}}$ and slope of tangent to the circle at $\left(\frac{3}{2},\frac{\sqrt{3}}{2}\right)$ is $m_{2}=-\sqrt{3}$ So, if angle between both curves is $\theta$ then $\tan\theta$ $=\left|\dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right|$ $=\left|\dfrac{-\frac{1}{3\sqrt{3}}+\sqrt{3}}{1+\left(-\frac{1}{3\sqrt{3}}\right)(-\sqrt{3})}\right|$ $=\frac{2}{\sqrt{3}}$ Option (2)

\begin{tabular}{|c|c|c|c|c|c|} \hline x & -2 & 1 & 3 & 4 & 6 \\ \hline P(X=x) & $\dfrac{1}{5}$ & a & $\dfrac{1}{3}$ & $\dfrac{1}{5}$ & b \\ \hline \end{tabular} Given $\bar{X} = 2.3$. $-a + 6b = \frac{9}{10}$ ......(1) $\sum P_i = \frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1$ $a + b = \frac{4}{15}$ ......(2) From equation (1) and (2) $a = \frac{1}{10}, \; b = \frac{1}{6}$ $\sigma^2 = \sum p_i x_i^2 - (\bar{X})^2$ $$\frac{1}{5}(4) + a(1) + \frac{1}{3}(9) + \frac{1}{5}(16) + b(36) - (2.3)^2$$ $$= \frac{4}{5} + a + 3 + \frac{16}{5} + 36 \, b - (2.3)^2$$ $$= 4 + a + 3 + 36 \, b - (2.3)^2$$ $$= 7 + a + 36 \, b - (2.3)^2$$ $$= 7 + \frac{1}{10} + 6 - (2.3)^2$$ $$= 13 + \frac{1}{10} - \left(\frac{23}{10}\right)^2$$ $$= \frac{131}{10} - \left(\frac{23}{10}\right)^2$$ $$= \frac{1310}{100} - (23)^2$$ $$= \frac{1310 - 529}{100}$$ $$= \frac{781}{100}$$ $\sigma^2 = \frac{781}{100}$ $100\sigma^2 = 781$

Given $f(n) = x^6 + 2x^4 + x^3 + 2x + 3$. $$\lim_{x \to 1} \frac{x^n f(1) - f(x)}{x-1} = 44$$ $$\lim_{x \to 1} \frac{9x^n - (x^6 + 2x^4 + x^3 + 2x + 3)}{x-1} = 44$$ $$\lim_{x \to 1} \frac{9nx^{n-1} - (6x^5 + 8x^3 + 3x^2 + 2)}{1} = 44$$ Thus, $9n - (19) = 44$. Therefore, $9n = 63$. Hence, $n = 7$.

Given $|z - 2 - 2i| \leq 1$. $|x + iy - 2 - 2i| \leq 1$ $|(x - 2) + i(y - 2)| \leq 1$ $(x - 2)^2 + (y - 2)^2 \leq 1$ $|3iz + 6|_{\max}$ at $a + ib$ $|3i| |z + \frac{6}{3i}|$ $3|z - 2i|_{\max}$ From the figure, the maximum distance is at $3 + 2i$. $a + ib = 3 + 2i = a + b = 3 + 2 = 5$ Ans.

The intersection points of the given lines are $(1, 2)$, $(7, 5)$, $(2, 3)$. $$\Delta = \frac{1}{2} \begin{vmatrix} 1 & 2 & 1 \\ 7 & 5 & 1 \\ 2 & 3 & 1 \end{vmatrix}$$ $$= \frac{1}{2} [1(5 - 3) - 2(7 - 2) + 1(21 - 10)]$$ $$= \frac{1}{2} [2 - 10 + 11]$$ $$\Delta DEF = \frac{1}{2} (3) = \frac{3}{2}$$ $$\Delta ABC = 4 \Delta DEF = 4 \left( \frac{3}{2} \right) = 6$$

Given $kf(k) + 2 = \lambda (x - 2)(x - 3)(x - 4)(x - 5) \ldots (1)$. Put $x = 0$, we get $\lambda = \frac{1}{60}$. Now put $\lambda$ in equation (1): $$kf(k) + 2 = \frac{1}{60} (x - 2)(x - 3)(x - 4)(x - 5).$$ Put $x = 10$: $$10f(10) + 2 = \frac{1}{60} (8)(7)(6)(5).$$ Therefore, $$52 - 10f(10) = 52 - 26 = 26.$$

Given the word FARMER with letters A, E, F, M, R, R. The arrangement is as follows: A E F A E F A M F A R E F A R M E R The calculations are: $$\left\lfloor \frac{5}{2} \right\rfloor - 4 = 60 - 24 = 36$$ $$\left\lfloor \frac{3}{2} \right\rfloor - 2 = 3 - 2 = 1$$ $$= 1$$ $$= 2$$ $$= 1$$ The final result is 77.

Given $(x+y)^n \Rightarrow 2^n = 4096$. Therefore, $2^n = 2^{12}$. Thus, $n = 12$. We have $2^{10} = 1024 \times 2$, $2^{11} = 2048$, and $2^{12} = 4096$. Now, $^{12}C_6 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 11 \times 3 \times 4 \times 7 = 924$.

Given $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{c} = 3\hat{i} + 2\hat{j} - \hat{k}$, $\vec{v} = x \vec{a} + y \vec{b}$. $\vec{v} (3\hat{i} + 2\hat{j} - \hat{k}) = 0$. $\vec{v} \cdot \hat{a} = 19$. $\vec{v} = \lambda \vec{c} \times (\vec{a} \times \vec{b})$. $\vec{v} = \lambda [(\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b}]$. $= \lambda [(3 + 4 + 1)(2\hat{i} - \hat{j} + 2\hat{k}) - \left(\frac{6 - 2 - 2}{2}\right)(\hat{i} + 2\hat{j} + \hat{k})]$. $= \lambda [16\hat{i} - 8\hat{j} + 16\hat{k} - 2\hat{i} - 4\hat{j} + 2\hat{k}]$. $\vec{v} = \lambda [14\hat{i} - 12\hat{j} + 18\hat{k}]$. $\lambda [14\hat{i} - 12\hat{j} + 18\hat{k}] \cdot \left(\frac{2\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{4+1+4}}\right) = 19$. $\lambda \left(\frac{28 + 12 + 36}{3}\right) = 19$. $\lambda \left(\frac{76}{3}\right) = 19$. $4\lambda = 3 \Rightarrow \lambda = \frac{3}{4}$. $2|\vec{v}| = |2 \times \frac{3}{4} [14\hat{i} - 12\hat{j} + 18\hat{k}]|^2$. $\frac{9}{4} \times 4(7^2 - 6^2 + 9^2) = 9(49 + 36 + 81)$. $= 9(166)$. $= 1494$.

Given $f(x) = \lfloor x \rfloor |x^2 - 1| + \sin \frac{\pi}{\lfloor x + 3 \rfloor} - \lfloor x + 1 \rfloor$. $$f(x) = \begin{cases} 3 - 2x^2, & -2 < x \leq -1 \\ x^2, & -1 < x < 0 \\ \frac{\sqrt{3}}{2} + 1, & 0 \leq x < 1 \\ x^2 + 1 + \frac{1}{\sqrt{2}}, & 1 \leq x < 2 \end{cases}$$ The function is discontinuous at $x = 0, 1$.

50 $\left$( PR^2 + RQ^2 $\right$) 50(20 + 5) 50(25) = 1250

As electric field is in y-direction so electric flux is only due to top and bottom surface. Bottom surface $y = 0$ implies $E = 0 \Rightarrow \phi = 0$. Top surface $y = 0.5 \, \mathrm{m}$ implies $E = 150(0.5)^2 = \frac{150}{4}$. Now flux $\phi = EA = \frac{150}{4} (0.5)^2 = \frac{150}{16}$. By Gauss's law $\phi = \frac{Q_{in}}{\varepsilon_0}$. Therefore, $$\frac{150}{16} = \frac{Q_{in}}{\varepsilon_0}$$ $$Q_{in} = \frac{150}{16} \times 8.85 \times 10^{-12} = 8.3 \times 10^{-11} \, \mathrm{C}$$ Option (2)

The equivalent circuit is shown. The current $i$ is given by $$i = \frac{V_0 B \ell}{4 + 1}$$ which implies $$V_0 = \frac{5(2 \, \mathrm{mA})}{5 \times .2} = 10^{-2} \, \mathrm{m/s} = 1 \, \mathrm{cm/s}.$$

De-Broglie wavelength $$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mE}}$$ Where $E$ is kinetic energy $$E = \frac{3kT}{2} for gas$$ $$\lambda = \frac{h}{\sqrt{3mkT}} = \frac{6.6 \times 10^{-34}}{\sqrt{3 \times 9 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}$$ $$\lambda = 6.26 \times 10^{-9} \, \mathrm{m} = 6.26 \, \mathrm{nm}$$ Option (1)

Work done = Change in kinetic energy $$W_{mg} + W_{air-friction} = \frac{1}{2} m(0.8 \sqrt{gh})^2 - \frac{1}{2} m(0)^2$$ $$W_{air-friction} = \frac{0.64}{2} mgh - mgh = -0.68 mgh$$ Option (1)

Range \[ R=\frac{u^2\sin2\theta}{g} \] and same for \[ \theta \] and \[ 90^\circ-\theta. \] So same for \[ 42^\circ \] and \[ 48^\circ. \] Maximum height \[ H=\frac{u^2\sin^2\theta}{2g} \] \(H\) is higher for higher \(\theta\). So \[ H \text{ for } 48^\circ \text{ is higher than } H \text{ for } 42^\circ. \] Option (2)

Let acceleration of wedge is $a_1$ and acceleration of block w.r.t. wedge is $a_2$. $$N \cos 60^\circ = Ma_1 = 16a_1$$ $$\Rightarrow N = 32a_1$$ F.B.D. of block w.r.t. wedge Perpendicular to incline $$N = 8g \cos 30^\circ - 8a_1 \sin 30^\circ \Rightarrow 32a_1 = 4\sqrt{3}g - 4a_1$$ $$\Rightarrow a_1 = \frac{\sqrt{3}}{9}g$$ Along incline $$8g \sin 30^\circ + 8a_1 \cos 30^\circ = ma_2 = 8a_2$$ $$a_2 = g \times \frac{1}{2} + \frac{\sqrt{3}}{9}g \cdot \frac{\sqrt{3}}{2} = \frac{2}{3}g$$ Option (4)

Mass of ice $m = \rho A \ell = 10^3 \times 10^{-4} \times 1 = 10^{-1} \, \mathrm{kg}$. Energy required to melt the ice $$Q = ms \wedge T + mL$$ $$= 10^{-1} \left(2 \times 10^3 \times 10 + 3.33 \times 10^5\right) = 3.53 \times 10^4 \, \mathrm{J}$$ $$Q = i^2 RT \Rightarrow 3.53 \times 10^4 = \left(\frac{1}{2}\right)^2 \left(4 \times 10^3\right) (t)$$ Time $= 35.3 \, \mathrm{sec}$ Option $(2)$

Given $$y = \frac{MgL^3}{4bd^3 \delta}$$ The relative error is given by $$\frac{\Delta y}{y} = \frac{\Delta M}{M} + \frac{3 \Delta L}{L} + \frac{3 \Delta d}{d} + \frac{\Delta \delta}{\delta}$$ Substituting the values, $$\frac{\Delta y}{y} = \frac{10^{-3}}{2} + \frac{3 \times 10^{-3}}{1} + \frac{10^{-2}}{4} + \frac{3 \times 10^{-2}}{4} + \frac{10^{-2}}{5}$$ Simplifying, $$= 10^{-3} [0.5 + 3 + 2.5 + 7.5 + 2] = 0.0155$$

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$ $\dfrac{1}{R_{eq}}=\dfrac14+\dfrac14$ $\Rightarrow R_{eq}=2\Omega$ Also, $\dfrac{\Delta R_{eq}}{R_{eq}^2}$ $=\dfrac{\Delta R_1}{R_1^2}$ $+\dfrac{\Delta R_2}{R_2^2}$ $\dfrac{\Delta R_{eq}}{4}$ $=\dfrac{0.8}{16}$ $+\dfrac{0.4}{16}$ $=\dfrac{1.2}{16}$ $\Delta R_{eq}=0.3\Omega$ $\therefore\ R_{eq}=(2\pm0.3)\Omega$

Also initially $N_x = N_y = N_0$. Activity $A = \lambda N$. As $\lambda_x < \lambda_y \Rightarrow A_x < A_y$. Therefore, $y$ will decay faster than $x$. Option (2)

Conceptual question Option (1)

Height of water observed by observer $$= \frac{H}{\mu_w} = \frac{H}{(4/3)} = \frac{3H}{4}$$ Height of air observed by observer $= 17.5 - H$ According to question, both height observed by observer is same. $$\frac{3H}{4} = 17.5 - H \Rightarrow H = 10 \, cm$$ Option (2)

As per diagram, Diode $D_1$ and $D_2$ are in forward bias i.e. $R = 30 \Omega$ whereas diode $D_3$ is in reverse bias i.e. $R = infinite$. Equivalent circuit will be Applying KVL starting from point A $$-\left(\frac{I_1}{2}\right) \times 30 - \left(\frac{I_1}{2}\right) \times 130 - I_1 \times 20 + 200 = 0$$ $$\Rightarrow -100 I_1 + 200 = 0$$ $$I_1 = 2$$ Option (3)

In steady state, inductor behaves as a conducting wire. So, equivalent circuit becomes $$\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$$ $$\Rightarrow R_{eq} = 1\, \Omega$$ Circuit becomes $$i = \frac{30}{3} = 10\, A$$

During upward motion $F = 2 \, \mathrm{N} = (+\mathrm{ve})$ constant During downward motion $F = 2 \, \mathrm{N} = (-\mathrm{ve})$ constant Therefore, the best possible answer is option (2).

From potential energy curve $$U_{max} = \frac{1}{2} k A^2 \Rightarrow 10 = \frac{1}{2} k (2)^2$$ $$\Rightarrow k = 5$$ Now $T_{spring} = T_{pendulum}$ $$2\pi \sqrt{\frac{5}{5}} = 2\pi \sqrt{\frac{4}{g}}$$ $$\Rightarrow 1 = \sqrt{\frac{4}{g}} \Rightarrow g = 4 on planet$$ Option (3)

Given $V = V_0 \left(1 - e^{-t/RC}\right)$. $$2 = 20 \left(1 - e^{-1/RC}\right)$$ $$\frac{1}{10} = 1 - e^{-t/RC}$$ $$e^{-t/RC} = \frac{9}{10}$$ $$e^{t/RC} = \frac{10}{9}$$ $$\frac{t}{RC} = \ln \left(\frac{10}{9}\right) \Rightarrow C = \frac{t}{R \ln \left(\frac{10}{9}\right)}$$ $$C = \frac{10^{-6}}{10 \times .105} = .95 \, \mu F$$

The net force $F_{net}$ is given by $$F_{net} = \frac{MV^2}{R}.$$ The equation becomes $$\sqrt{2}F + F_1 = \frac{MV^2}{R}.$$ Substituting the gravitational forces, we have $$\sqrt{2} \frac{GMM}{(\sqrt{2}R)^2} + \frac{GMM}{(2R)^2} = \frac{MV^2}{R}.$$ Simplifying, $$\frac{GM}{R} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = V^2.$$ Further simplification gives $$\frac{GM}{R} \left( \frac{4 + \sqrt{2}}{4\sqrt{2}} \right) = V^2.$$ Solving for $V$, we find $$V = \frac{1}{2} \sqrt{\frac{GM(2\sqrt{2}+1)}{R}}.$$

Speed of wave $= \dfrac{2 \times 10^{10}}{200} = 10^8\,\mathrm{m/s}$ Refractive index $= \dfrac{3 \times 10^8}{10^8} = 3$ Now refractive index $= \sqrt{\varepsilon_r \mu_r}$ $$3 = \sqrt{\varepsilon_r(1)}$$ $\Rightarrow \varepsilon_r = 9$

The magnetic field due to wire (1) is given by $$B_{due to wire (1)} = \frac{\mu_0 I}{4 \pi y} [\sin 90 + \sin \theta_1]$$ which simplifies to $$= \frac{\mu_0 I}{4 \pi y} \left( 1 + \frac{x}{\sqrt{x^2 + y^2}} \right) \cdots (1)$$ The magnetic field due to wire (2) is given by $$B_{due to wire (2)} = \frac{\mu_0 I}{4 \pi x} (\sin 90^\circ + \sin \theta_2)$$ which simplifies to $$= \frac{\mu_0 I}{4 \pi x} \left( 1 + \frac{y}{\sqrt{x^2 + y^2}} \right) \cdots (2)$$ The total magnetic field is $$B = B_1 + B_2 = \frac{\mu_0 I}{4 \pi} \left[ \frac{1}{y} + \frac{x}{y \sqrt{x^2 + y^2}} + \frac{1}{x} + \frac{y}{x \sqrt{x^2 + y^2}} \right]$$ which simplifies to $$B = \frac{\mu_0 I}{4 \pi} \left[ \frac{x+y}{xy} + \frac{\sqrt{x^2 + y^2}}{xy} \right] = \frac{\mu_0 I}{4 \pi xy} \left[ \sqrt{x^2 + y^2} + (x + y) \right]$$ Option (1)

Given amplitude $\propto$ slit width. Also intensity $\propto$ (Amplitude)$^2 \propto$ (Slit width)$^2$. $$\frac{I_1}{I_2} = \left(\frac{3}{1}\right)^2 = 9 \Rightarrow I_1 = 9I_2$$ $$\frac{I_{\min}}{I_{\max}} = \left(\frac{\sqrt{I_1} - \sqrt{I_2}}{\sqrt{I_1} + \sqrt{I_2}}\right)^2 = \left(\frac{3-1}{3+1}\right)^2 = \frac{1}{4} = \frac{x}{4}$$ $\($$\Rightarrow$ x = 1.00$\)$

Given $T_1 = 1$ hour. Therefore, $\omega_1 = 2\pi rad/hour$. $T_2 = 8$ hours. Therefore, $\omega_2 = \frac{\pi}{4} rad/hour$. $R_1 = 2 \times 10^3 km$. As $T^2 \propto R^3$, $$\left( \frac{R_2}{R_1} \right)^3 = \left( \frac{T_2}{T_1} \right)^2$$ $$\Rightarrow \frac{R_2}{R_1} = \left( \frac{8}{1} \right)^{2/3} = 4 \Rightarrow R_2 = 8 \times 10^3 km$$ $V_1 = \omega_1 R_1 = 4\pi \times 10^3 km/h$. $V_2 = \omega_2 R_2 = 2\pi \times 10^3 km/h$. Relative $\omega = \frac{V_1 - V_2}{R_2 - R_1} = \frac{2\pi \times 10^3}{6 \times 10^3} = \frac{\pi}{3} rad/hour$. $x = 3$.

On smooth incline $a = g \sin 30^\circ$ by $S = ut + \frac{1}{2} at^2$ $$S = \frac{1}{2} \frac{g}{2} T^2 = \frac{g}{4} T^2 \ldots$$ On rough incline $$a = g \sin 30^\circ - \mu g \cos 30^\circ$$ by $S = ut + \frac{1}{2} at^2$ $$S = \frac{1}{4} g (1 - \sqrt{3} \mu) (\alpha T)^2$$ By (i) and (ii) $$\frac{1}{4} g T^2 = \frac{1}{4} g (1 - \sqrt{3} \mu) \alpha^2 T^2$$ $$\Rightarrow 1 - \sqrt{3} g = \frac{1}{\alpha^2} \Rightarrow g = \left( \frac{\alpha^2 - 1}{\alpha^2} \right) \cdot \frac{1}{\sqrt{3}}$$ $$\Rightarrow x = 3.00$$

Given Translation K.E. of $\mathrm{N_2} = K.E. of electron$ $$\frac{3}{2} kT = eV$$ $$\frac{3}{2} \times 1.38 \times 10^{-23} T = 1.6 \times 10^{-19} \times 0.1$$ $$\Rightarrow T = 773 k$$ $$T = 773 - 273 = 500^\circ C$$

First case $P_1 = \frac{V^2}{R} = \frac{(240)^2}{36}$. Second case Resistance of each half $= 18\, \Omega$. $$P_2 = \frac{(240)^2}{18} + \frac{(240)^2}{18} = \frac{(240)^2}{9}$$ $$\frac{P_1}{P_2} = \frac{1}{4}$$ $x = 4.00$

Thermal force $F = Ay \propto \Delta T$ $$F = (10 \times 10^4) (2 \times 10^{11}) (10^{-5}) (400)$$ $$F = 8 \times 10^5 \, \mathrm{N}$$ $$\Rightarrow x = 8$$

By energy conservation, $mg\ell = \frac{1}{2} I \omega^2 = \frac{1}{2} m \frac{\ell^2 \omega^2}{3}$. This implies $\omega = \sqrt{\frac{6g}{\ell}}$. Speed $v = \omega r = \omega \ell = \sqrt{6g\ell}$. $v = \sqrt{6 \times 10 \times 0.6} = 6 \, \mathrm{m/s}$.

Given $$A_{\max} = A_C + A_m = 250 + 150 = 400$$ $$A_{\min} = A_C - A_m = 250 - 150 = 100$$ $$\frac{A_{\min}}{A_{\max}} = \frac{100}{400} = \frac{1}{4} = \frac{50}{200}$$ $$x = 200$$

Work = $\Delta$ K, E $W_{friction} + W_{spring} = 0 - \frac{1}{2} mv^2$ $-\frac{90}{100} \left( \frac{1}{2} mv^2 \right) + W_{Spring} = -\frac{1}{2} mv^2$ $W_{spring} = -\frac{10}{100} \times \frac{1}{2} mv^2$ $-\frac{1}{2} kx^2 = -\frac{1}{20} mv^2$ $$\Rightarrow \; k = \frac{40000 \times (20)^2}{10 \times (1)^2} = 16 \times 10^5$$

Clean water could have BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more.

Calamine is $\mathrm{ZnCO_3}$. Malachite is $\mathrm{Cu(OH)_2 \cdot CuCO_3}$.

Solution $\mathrm{NaH_2}$ is not reducing agent.

By observation, we obtain this plot in the measurable temperature range. Hence, the correct answer is the $3^{\text{rd}}$ option.

Aniline is oxidized using $\mathrm{K_2Cr_2O_7}$ in the presence of an oxidizing agent $[O]$ to form a compound $[A]$. The structure of $[A]$ is shown as a benzene ring with two ketone groups attached at the para positions.

The given compound is $CH_3-CH_2-CH_2-C(=O)-O-CH_2-CH_3$ with molecular formula $C_6H_{12}O_2$. Upon reduction with $LiAlH_4$ followed by hydrolysis with $H_3O^+$, it forms $CH_3-CH_2-CH_2-CH_2-OH$ and $CH_3-CH_2-OH$. The primary alcohol $CH_3-CH_2-CH_2-CH_2-OH$ is then oxidized to form $CH_3-CH_2-CH_2-C(=O)-OH$, which is a $C_4$ carboxylic acid.

According to type of reactions for preparation, colloids have been classified.

Terephthalic acid reacts with ethylene glycol to form Dacron, a type of polyester. The reaction involves the formation of ester linkages between the acid and the alcohol groups.

Option (1) Acenaphthene has $10 \, \pi e^-$ in cyclic conjugation, which makes it aromatic. Option (2) has $4 \, \pi e^-$ in ring conjugation, which makes it anti-aromatic. Option (3) has $4 \, \pi e^-$ in ring conjugation, which makes it antiaromatic. Option (4) Cyclopentadienyl anion has $6 \, \pi e^-$ in ring conjugation, which makes it aromatic.

Colour of $\mathrm{Fe^{2+}}$ is observed green and $\mathrm{Fe^{3+}}$ is yellow.

Trans-2-butene reacts with $\mathrm{Br_2/CCL_4}$ in an anti addition to form a meso product. The structure of the product is shown with $\mathrm{CH_3}$ groups and $\mathrm{Br}$ atoms added across the double bond.

The balanced chemical equation is: $$\mathrm{I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + O_2}$$

The reaction starts with $\mathrm{CH_3-CH=CH_2}$, which is propene ($\mathrm{C_3H_6}$). In the presence of $\mathrm{H^+/H_2O}$, it undergoes hydration to form $\mathrm{CH_3-CH(OH)-CH_3}$, which is compound (A). Compound (A) is then subjected to the iodoform reaction using $\mathrm{KOI/dil.\ KOH}$. This results in the formation of $\mathrm{CHI_3}$ (iodoform) and $\mathrm{CH_3-C(=O)OK}$, which is compound (C).

Statement I: Correct

The reactions shown involve the formation of diazonium salts using $\mathrm{NaNO_2 + HCl}$. 1. The first reaction converts an aliphatic amine to a diazonium salt: $$\mathrm{NH_2} \xrightarrow{\mathrm{NaNO_2 + HCl}} \mathrm{N^+ \equiv N}$$ 2. The second reaction involves an aromatic amine, which forms a stable diazonium salt due to the $+H$ effect: $$\mathrm{H_3C} on benzene ring \xrightarrow{\mathrm{NaNO_2 + HCl}} \mathrm{N^+ \equiv N} (Most stable)$$ 3. The third reaction shows the formation of a diazonium salt from a secondary aliphatic amine: $$\mathrm{CH_3-CH-CH_2-NH_2} \xrightarrow{\mathrm{NaNO_2 + HCl}} \mathrm{CH_3-CH-CH_2-N^+ \equiv N}$$ 4. The fourth reaction involves an aromatic amine with a methyl group, which does not form a diazonium salt but instead forms an alkyl nitroso amine: $$\mathrm{C_6H_5-NH-CH_3} \xrightarrow{\mathrm{NaNO_2 + HCl}} Diazonium salt not form$$ $$\mathrm{C_6H_5-N(CH_3)-N=O} (N, alkyl nitroso amine)$$

The reaction is given by: $$\mathrm{FeCl_3} + \mathrm{K_4[Fe(CN)_6]} \rightarrow \mathrm{Fe_4[Fe(CN)_6]_3}$$ This forms Prussian blue.

$\mathrm{Li_2O} \Rightarrow 2\mathrm{Li}^{+} \qquad \mathrm{O}^{2-}$ $\mathrm{CaO} \Rightarrow \mathrm{Ca}^{2+} \qquad \mathrm{O}^{2-}$ $\mathrm{Na_2O_2} \Rightarrow 2\mathrm{Na}^{+} \qquad \mathrm{O_2}^{2-}$ $\mathrm{KO_2} \Rightarrow \mathrm{K}^{+} \qquad \mathrm{O_2}^{-}$ $\mathrm{MgO} \Rightarrow \mathrm{Mg}^{2+} \qquad \mathrm{O}^{2-}$ $\mathrm{K_2O} \Rightarrow 2\mathrm{K}^{+} \qquad \mathrm{O}^{2-}$ $\mathrm{O_2}^{2-} \Rightarrow$ Complete octet, diamagnetic $\mathrm{O_2}^{2-} \Rightarrow \sigma_{1s}^{2}\,\sigma_{1s}^{*2}\,\sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\sigma_{2p_x}^{2}\,\pi_{2p_y}^{2}\,\pi_{2p_z}^{2}\,\pi_{2p_y}^{*2}\,\pi_{2p_z}^{*2}$ (dia) $\mathrm{O_2}^{-} \Rightarrow \sigma_{1s}^{2}\,\sigma_{1s}^{*2}\,\sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\sigma_{2p_x}^{2}\,\pi_{2p_y}^{2}\,\pi_{2p_z}^{2}\,\pi_{2p_y}^{*2}\,\pi_{2p_z}^{*1}$ (para)

The electronic configuration of $\mathrm{Fe^{3+}}$ is $[\mathrm{Ar}]3d^5$.

Initial mass of gas = 29 - 14.8 = 14.2 Kg Mass of gas used = 29 - 23 = 6 Kg Gas left = 14.2 - 6 = 8.2 Kg (1) $3.47$ $\times$ V = $\left$( $\frac{14.2 \times 10^3}{M}$ $\right$) $\times$ R $\times$ T (2) $p$ $\times$ V = $\left$( $\frac{8.2 \times 10^3}{M}$ $\right$) $\times$ R $\times$ T Divide : $\frac{(1)}{(2)}$ $\Rightarrow$ $\frac{3.47}{P}$ = $\frac{14.2}{8.2}$ $P$ = $2.003$

The reaction is given by: $$\mathrm{Zn(OH)_2 (s) \rightleftharpoons Zn^{+2} (aq) + 2OH^- (aq)}$$ Assume $S(0.1 + 2s) \simeq 0.1$. The solubility product is: $$K_{sp} = S(0.1)^2$$ Given: $$2 \times 10^{-20} = s \times 10^{-2} \Rightarrow s = 2 \times 10^{-18}$$ Thus: $$= x \times 10^{-18}$$ Solving for $x$ gives: $$x = 2$$

Given the equation $\Delta G = \Delta H - T \Delta S$. $$\Delta G = 57.8 - \frac{298(-176)}{1000}$$ Calculating gives: $$\Delta G = -5.352 \, \mathrm{kJ/mole}$$ The nearest integer value is 5.

The complex $[\mathrm{Ag(NH_3)_2}]^+$ and $[\mathrm{Ag(CN)_2}]^-$ both have a charge of $+1$.

Number of atoms = $\($ $\frac{8}{23}$ $\times$ 6.02 $\times$ 10^{23} = 2.09 $\times$ 10^{23} $\)$ $\($ $\simeq$ 2 $\times$ 10^{23} $\)$ $\($ = x $\times$ 10^{23} $\)$ $\($ x = 2 $\)$

Moles of $\mathrm{CuSO_4 \cdot 5H_2O} = \frac{80}{249.54}$. Molarity $= \frac{\frac{80}{249.54}}{5} = 64.117 \times 10^{-3}$. Nearest integer, $x = 64$.

Total energy per sec. = 50 J $$50 = \frac{n \times 6.63 \times 10^{-34} \times 3 \times 10^8}{795 \times 10^{-9}}$$ $$n = 1998.49 \times 10^{17} \ [n = no. of photons per second]$$ $$= 1.998 \times 10^{20}$$ $$\simeq 2 \times 10^{20}$$ $$= x \times 10^{20}$$ $$x = 2$$

For $\mathrm{B_2^+}$, the electronic configuration is $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2py}^1 \simeq \pi_{2pz}^0$. This implies $9e^-$. The magnetic moment $\mu$ is given by $$\mu = \sqrt{1(1+2)} = \sqrt{3} \, \mathrm{BM}$$ which equals $1.73 \, \mathrm{BM}$. Therefore, $$= 1.73 \times 10^{-2} \, \mathrm{BM}$$

$$k = 1.07 \times 10^6 \ \text{Sm}^{-1}, \quad R = 0.243 \ \Omega$$ $$G = \frac{1}{R} = \frac{1}{0.243} \ \Omega^{-1}$$ $$k = G \times G^*$$ $$G^* = \frac{k}{G} = \frac{1.07 \times 10^6}{\dfrac{1}{0.243}} \simeq 26 \times 10^4 \ \text{m}^{-1}$$

Total (3) peptide linkages are present. 3 peptide linkage Ans. (3)