JEE Main 25 January 2023 Shift 2 question paper with solutions

Given $$Y = \frac{Stress}{Strain} = \frac{F/A}{\Delta \ell / \ell} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$$ $$F = 6 \pi \eta r v \Rightarrow \eta = \frac{F}{6 \pi r v}$$ $$[\eta] = \frac{[MLT^{-2}]}{[L][LT^{-1}]} = [ML^{-1}T^{-1}]$$ $$E = h \nu \Rightarrow h = \frac{E}{\nu} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$$ Work function has same dimension as that of energy, so $[\phi] = [ML^2T^{-2}]$

Given $x = 4t^2$. The velocity $v$ is given by the derivative $\frac{dx}{dt} = 8t$. At $t = 5$ sec, $v = 8 \times 5 = 40 \, \mathrm{m/s}$.

Range $= \frac{u^2 \sin 2\theta}{g}$ Range for projection angle “$\alpha$” $$R_1 = \frac{u^2 \sin 2\alpha}{g}$$ Range for projection angle “$\beta$” $$R_2 = \frac{u^2 \sin 2\beta}{g}$$ $\alpha + \beta = 90^\circ$ (Given) $$\Rightarrow \beta = 90^\circ - \alpha$$ $$R_2 = \frac{u^2 \sin 2(90^\circ - \alpha)}{g}$$ $$R_2 = \frac{u^2 \sin (180^\circ - 2\alpha)}{g}$$ $$R_2 = \frac{u^2 \sin 2\alpha}{g}$$ $$\Rightarrow \frac{R_1}{R_2} = \frac{\left(\frac{u^2 \sin 2\alpha}{g}\right)}{\left(\frac{u^2 \sin 2\alpha}{g}\right)} = \frac{1}{1}$$

Given the forces acting on the block, we have: $$F_1 = mg \sin 45^\circ + f = mg \sin 45^\circ + \mu N$$ Substituting the values, we get: $$F_1 = \frac{mg}{\sqrt{2}} + \mu mg \cos 45^\circ$$ Simplifying further: $$F_1 = \frac{mg}{\sqrt{2}} (1 + \mu)$$ For the second block: $$F_2 = mg \sin 45^\circ - f = mg \sin 45^\circ - \mu N$$ This simplifies to: $$= \frac{mg}{\sqrt{2}} (1 - \mu)$$ Given that $F_1 = 2F_2$, we have: $$\frac{mg}{\sqrt{2}} (1 + \mu) = 2 \frac{mg}{\sqrt{2}} (1 - \mu)$$ Solving for $\mu$: $$1 + \mu = 2 - 2\mu$$ $$\mu = 1/3 = 0.33$$

Given $\frac{v_1}{v_2} = \frac{3}{2}$. Since $m_1 v_1 = m_2 v_2$, we have $\frac{m_1}{m_2} = \frac{2}{3}$. Since nuclear mass density is constant, $$\frac{m_1}{\frac{4}{3} \pi r_1^3} = \frac{m_2}{\frac{4}{3} \pi r_2^3}$$ This implies $$\left( \frac{r_1}{r_2} \right)^3 = \frac{m_1}{m_2}$$ Therefore, $$\frac{r_1}{r_2} = \left( \frac{2}{3} \right)^{\frac{1}{3}}$$ So, $x = 2$.

Given the initial conditions, we have: $$1 \times u_1 = -2 + 3v \implies u_1 = -2 + 3v (1)$$ $$1 = \frac{v + 2}{u_1} \implies v + 2 = u_1 (2)$$ Solving equations (1) and (2), we find: $$u_1 = 4 \, \mathrm{m/s}$$

The tangent will be $\frac{x}{7}$. The value of $x$ is given. For the solid sphere: Given $m_1 = 5 \, \mathrm{kg}$ and radius $= R$. $$I_1 = \frac{2}{5} m_1 R^2 + m_1 R^2$$ Simplifying, we have: $$I_1 = m_1 R^2 \left( \frac{7}{5} \right)$$ Thus, $$I_1 = 7 R^2$$ For the disc: Given $m_2 = 4 \, \mathrm{kg}$ and radius $= R$. $$I_2 = \frac{m_2 R^2}{4} + m_2 R^2$$ Simplifying, we have: $$I_2 = \frac{5}{4} m_2 R^2$$ Thus, $$I_2 = 5 R^2$$

The potential energy is given by $$U = \frac{-G M_e m}{r}$$ The initial potential energy is $$U_i = \frac{-G M_e m}{R_e}$$ The final potential energy is $$U_f = \frac{-G M_e m}{(R_e + h)} = \frac{-G M_e m}{R_e + 2R_e}$$ which simplifies to $$\frac{-G M_e m}{3R_e}$$ The increase in internal energy $\Delta U$ is given by $U_f - U_i$: $$\Delta U = \frac{2}{3} \frac{G M_e m}{R_e}$$ This can be rewritten as $$\frac{2}{3} \frac{G M_e}{R_e^2} m R_e$$ which simplifies to $$\frac{2}{3} mg R_e$$

Given $$F = \frac{G m_1 m_2}{r^2}$$ Therefore, $$F \propto \frac{1}{r^2}$$ Also, $$F \propto m_1 m_2$$ This force provides centripetal force and acts towards the sun. Thus, $$T^2 \propto a^3$$ (Kepler's third law)

Surface Tension = T R : Radius of bigger drop r : Radius of smaller drop Volume will remain same $$\frac{4}{3} \pi R^3 = 1000 \times \frac{4}{3} \pi r^3$$ R = 10 r $$u_i = T \cdot 4 \pi R^2$$ $$u_f = T \cdot 4 \pi r^2 \times 1000$$ $$\frac{u_f}{u_i} = \frac{1000 r^2}{R^2}$$ $$\frac{u_f}{u_i} = \frac{10}{1}$$ So, x = 1

Diatomic gas molecules have three translational degrees of freedom, two rotational degrees of freedom, and it is given that it has one vibrational mode. So there are two additional degrees of freedom corresponding to one vibrational mode, so the total degree of freedom is $7$. $$C_V = \frac{fR}{2} = \frac{7R}{2}$$

The reading on scale minus the lower fixed point divided by the upper fixed point minus the lower fixed point is constant. $$\frac{t_P - 30}{180 - 30} = \frac{t_Q - 0}{100 - 0}$$ Simplifying gives: $$\frac{t_P - 30}{150} = \frac{t_Q}{100}$$

Given $\Delta U = n C_V \Delta T$. For isothermal process $T$ is constant. So $\Delta U = 0$. A $\longrightarrow$ II. Adiabatic process $\Delta Q = 0$. $\Delta Q = \Delta U + \Delta W$. $\Delta U = -\Delta W$. Work done by gas is positive. So $\Delta U$ is negative. B $\longrightarrow$ I. For Isochoric process $\Delta W = 0$. C $\longrightarrow$ IV. For Isobaric process $\Delta W = P \Delta V \neq 0$. $\Delta U = n C_V \Delta T \neq 0$. Heat absorbed goes partly to increase internal energy and partly do work.

Let time from 0 to A/2 is $t_1$ and from A/2 to A is $t_2$. Then $\omega t_1 = \pi/6$ and $\omega t_2 = \pi/3$. $$\frac{t_1}{t_2} = \frac{1}{2}$$ $$t_2 = 2t_1 = 2 \times 2 = 4 sec$$

NCERT fact based

The apparent frequency $f_{app}$ is given by the formula: $$f_{app} = f \left( \frac{v}{v - v_s} \right)$$ Substituting the given values: $$= 320 \left( \frac{330}{330 - 66} \right)$$ $$= 400 \, Hz$$

Given the equation for potential energy: $$E_P = \frac{K \times 10}{2^2} - \frac{K \times 40}{(x_0 - 2)^2} = 0$$ Simplifying the equation, we have: $$\frac{1}{2} = \frac{2}{x_0 - 2}$$ Solving for $x_0$, we get: $$x_0 - 2 = 4$$ Therefore, $$x_0 = 6 \, cm$$

Given $\frac{\varepsilon_0 A}{d} = 5 \, \mu \mathrm{F}$. For the new configuration, $$C_{new} = \frac{\varepsilon_0 A}{\frac{d}{2}} \left( \frac{1}{1.5} + \frac{1}{1} \right)$$ Simplifying, $$= \frac{\varepsilon_0 A}{\frac{d}{3} + \frac{d}{2}} = \frac{6 \varepsilon_0 A}{5d}$$ Therefore, $$= \frac{6}{5} \times 5 \, \mu \mathrm{F} = 6 \, \mu \mathrm{F}$$

Given $R_{initial} = \frac{\rho \ell}{A} = 5 \Omega$. The volume of wire is constant in stretching, so $V_i = V_f$. Therefore, $A_i \ell_i = A_f \ell_f$. We have $A \ell = A'(5 \ell)$. Thus, $A' = \frac{A}{5}$. The final resistance is $R_f = \frac{\rho \ell_f}{A_f} = \frac{\rho (5 \ell)}{\frac{A}{5}}$. This simplifies to $= 25 \left( \frac{\rho \ell}{A} \right) = 25 \times 5 = 125 \Omega$.

The equivalent voltage $E_{eq}$ is calculated as follows: $$E_{eq} = \frac{12}{3} - \frac{6}{6} \Bigg/ \frac{1}{3} + \frac{1}{6}$$ This simplifies to: $$E_{eq} = 6 \, \mathrm{V}$$ The equivalent resistance $r_{eq}$ is: $$r_{eq} = 2 \, \Omega$$ The resistance $R$ is: $$R = 4 \, \Omega$$ In the circuit, the current $i$ is given by: $$i = \frac{6}{2 + 4} = 1 \, \mathrm{A}$$

Given $\tau = K \theta$ and $\mathrm{NiAB} = K \theta$. $$A = \frac{K \theta}{\mathrm{NiB}} = \frac{4 \times 10^{-5} \times 0.05}{200 \times 10^{-3} \times 0.01}$$ On solving $A = 10^{-4} \, \mathrm{m^2} = 1 \, \mathrm{cm^2}$.

Magnetic fields due to both wires will be perpendicular to each other. $B_1 = \frac{\mu_0 i_1}{2 \pi d}$, $B_2 = \frac{\mu_0 i_2}{2 \pi d}$ $$B_{net} = \sqrt{B_1^2 + B_2^2} \Rightarrow \frac{\mu_0}{2 \pi d} \sqrt{i_1^2 + i_2^2}$$ $$\Rightarrow \frac{4 \pi \times 10^{-7}}{2 \pi \times (7/\sqrt{2}) \times 10^{-2}} \times \sqrt{8^2 + 15^2} (d = \frac{7}{\sqrt{2}} \, cm)$$ $$\Rightarrow 68 \times 10^{-6} \, T$$

Induced emf across the ends $= B v \ell$ $$= 2 \times 8 \times 1 = 16 \, \mathrm{V}$$

Given the formula for the power factor, we have: $$\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_C - X_L)^2}}$$ Substituting the given values: $$\cos \phi = \frac{80}{\sqrt{(80)^2 + (60)^2}}$$ Simplifying further: $$\cos \phi = \frac{80}{100} \Rightarrow \frac{8}{10}$$

Gauss's Law of electrostatics $$\phi = \oint \vec{E} \cdot d\vec{s} = \frac{q}{\varepsilon_0}$$ Faraday's law $$\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$$ Gauss's law of magnetism $$\oint \vec{B} \cdot d\vec{A} = 0$$ Ampere's Maxwell law $$\oint \vec{B} \cdot d\vec{l} = \mu_0 i_C + \mu_0 \varepsilon_0 \frac{d\phi_E}{dt}$$ Where $i_C$: Conduction current $\varepsilon_0 \frac{d\phi_E}{dt}$: Displacement current

Plane mirror forms erect, same sized, laterally inverted and virtual image of real object.

Given $f = 10 \, \mathrm{cm}$. $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{15} - \frac{1}{-u} = \frac{1}{10}$$ $$\Rightarrow \frac{1}{u} = \frac{1}{10} - \frac{1}{15}$$ On solving we get value of $u$ as $30 \, \mathrm{cm}$.

Stopping potential $V_S = \frac{\mathrm{KE_{max}}}{e}$ $$V_S = \frac{\frac{hC}{\lambda} - \phi}{e}$$ Stopping potential does not depend on intensity or power of light used, it only depends on frequency or wavelength of incident light. So both statements I and II are correct.

The wavelength $\lambda$ is given by the formula $\lambda = \frac{hc}{\Delta E}$. For $\Delta E_A = 2.2 \, \mathrm{eV}$, $\Delta E_B = 5.2 \, \mathrm{eV}$, $\Delta E_C = 3 \, \mathrm{eV}$, and $\Delta E_D = 10 \, \mathrm{eV}$, we calculate the wavelengths as follows: For $\lambda_A$: $$\lambda_A = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{2.2 \times 1.6 \times 10^{-19}}$$ $$= \frac{12.41 \times 10^{-7}}{2.2} \, \mathrm{m}$$ $$= \frac{1241}{2.2} \, \mathrm{nm} = 564 \, \mathrm{nm}$$ For $\lambda_B$: $$\lambda_B = \frac{1241}{5.2} \, \mathrm{nm} = 238.65 \, \mathrm{nm}$$ For $\lambda_C$: $$\lambda_C = \frac{1241}{3} \, \mathrm{nm} = 413.66 \, \mathrm{nm}$$ For $\lambda_D$: $$\lambda_D = \frac{1241}{10} = 124.1 \, \mathrm{nm}$$

Statement I is correct. When P-N junction is formed, an electric field is generated from N-side to P-side due to which barrier potential arises and majority charge carrier cannot flow through the junction due to barrier potential, so current is zero unless we apply forward bias voltage.

Element C has a percentage of 85.8. The moles are calculated as $\frac{85.8}{12} = 7.15$ with a mole ratio of 1. Element H has a percentage of 14.2. The moles are calculated as $\frac{14.2}{1} = 14.2$ with a mole ratio of 2. The empirical formula is $\mathrm{(CH_2)}$. Solving for $n$: $$14 \times n = 84$$ $$n = 6$$ Therefore, the molecular formula is $\mathrm{C_6H_{12}}$.

$p_x$, $p_y$, $p_z$, $d_{z^2}$, and $d_{x^2-y^2}$ are axial orbitals.

Statement II: The crossed arrow symbolises the direction of the shift of electron density in the molecule.

Let the volume of $C_2H_4$ be $x$ litre. $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$ Initial: $x$ Final: $2x$ $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ Initial: $(16.8-x)$ Final: $(16.8-x)$ Total volume of $CO_2 = 2x + (16.8-x)$ $\Rightarrow 28 = 16.8 + x$ $x = 11.2\,L$ $n_{CH_4} = \dfrac{PV}{RT} = \dfrac{1\times5.6}{0.082\times298} = 0.229\ \text{mole}$ $n_{C_2H_4} = \dfrac{PV}{RT} = \dfrac{1\times11.2}{0.082\times298} = 0.458\ \text{mole}$ $\therefore$ Heat evolved $= 0.229\times900 + 0.458\times1400$ $= 206.1 + 641.2$ $= 847.3\,kJ$

Given $\Delta[\mathrm{H}^+] = 1000$. $\Delta \mathrm{pH} = -\log \Delta[\mathrm{H}^+] = -\log 10^3$. Therefore, $\Delta \mathrm{pH} = -3$.

Cobalt catalyst leads to Methanol production. Syn gas leads to Coal gasification: $$\left( \mathrm{C_{(Red\, hot\, coke)}} + \mathrm{H_2O(g)} \rightarrow \mathrm{CO} + \mathrm{H_2} \right)$$ Nickel catalyst leads to Water gas production. Brine solution leads to Production: $$\left( \mathrm{(aq.\, NaCl)} \right)$$ $$\left( \begin{array}{c} \mathrm{H_2} \rightarrow Cathode \\ \mathrm{Cl_2} \rightarrow anode \end{array} \right)$$

The alkali metals and their salts impart characteristic colour to oxidizing flame.

Sodium have lowest oxidation potential in alkali metals. Hence it is weakest reducing agent among alkali metals.

Metallic character increases down the group and decreases along the period.

The oxide which form acid on dissolving in water is acidic oxide.

A. Propanamine and N-Methylethanamine are functional isomers. B. Hexan-2-one and Hexan-3-one are metamers. C. Ethanamide and Hydroxyethanimine are tautomers. D. o-Nitrophenol and p-nitrophenol are positional isomers.

Assume: Mass of solvent $\approx$ Mass of solution Case I: $$0.25 = \frac{W_1}{62} \times \frac{1000}{500}$$ Case II: $$0.25 = \frac{W_2}{62} \times \frac{1000}{250}$$ $$W_1 = 2$$

Given $\pi = iCRT$. Therefore, $\pi \propto iC$. A, B, D, and E have the same value of osmotic pressure.

Overall reaction: $$\mathrm{H_2}_{(g)} + \mathrm{M^{3+}}_{(aq)} \rightarrow \mathrm{M^+}_{(aq)} + 2\mathrm{H^+}_{(aq)}$$ The cell potential is given by: $$E_{Cell} = E^\circ_{Cathode} - E^\circ_{anode} - \frac{0.059}{2} \log \frac{[\mathrm{M^+}] \times 1^2}{[\mathrm{M^{+3}}] \times 1}$$ Substituting the values: $$0.1115 = 0.2 - \frac{0.059}{2} \log \frac{[\mathrm{M^+}]}{[\mathrm{M^{+3}}]}$$ Solving for the concentration ratio: $$3 = \log \frac{[\mathrm{M^+}]}{[\mathrm{M^{+3}}]}$$ Therefore, $a = 3$.

Given $$t_{10\%} = \frac{1}{K} \ln \left( \frac{a}{a-x} \right) = \frac{1}{K} \ln \left( \frac{100}{90} \right)$$ $$t_{10\%} = \frac{2.303}{K} (\log 10 - \log 9)$$ $$t_{10\%} = \frac{2.093}{K} \times (0.04)$$ Similarly $$t_{90\%} = \frac{1}{K} \ln \left( \frac{100}{10} \right)$$ $$t_{90\%} = \frac{2.303}{K}$$ $$\frac{t_{90\%}}{t_{10\%}} = \frac{1}{0.04} = 25$$ $$e^{kt} = \frac{a}{a-x}$$ $$\frac{a-x}{a} = e^{-kt}$$ $$1 - \frac{x}{a} = e^{-kt}$$ $$x = a(1 - e^{-kt})$$ $$\alpha = \frac{x}{a} = (1 - e^{-kt})$$

B and D options are correct

'A' water vapours are absorbed by calcium chloride. C. As the adsorption proceeds, $\Delta H$ becomes less and less negative.

In froth floatation method a rotating paddle draws in air and stirs the pulp.

The reaction of $\mathrm{Cl^-}$ with $\mathrm{AgNO_3}$ produces $\mathrm{AgCl}$, which is a curdy white precipitate. $$\mathrm{Cl^- + AgNO_3 \rightarrow AgCl}$$ This is labeled as $[A]$. Next, $\mathrm{AgCl}$ reacts with $\mathrm{NH_4OH}$ to form a soluble complex $\mathrm{[Ag(NH_3)_2]Cl}$. $$\mathrm{AgCl + NH_4OH \rightarrow [Ag(NH_3)_2]Cl}$$ This is labeled as $[B]$ (Soluble Complex).

The reaction is given by: $$14\mathrm{H}^+ + 6\mathrm{e}^- + \mathrm{Cr_2O_7}^{-2} \rightarrow 2\mathrm{Cr}^{+3} + 7\mathrm{H_2O}$$

Given the relationship $E = \frac{hc}{\lambda}$, we have $E \propto \frac{1}{\lambda}$. Therefore, $\Delta(CFSE) \propto \frac{1}{\lambda_{absorb}} \propto$ strength of ligand.

[$\mathrm{Co(NH_3)_4Cl_2}$]$\mathrm{Cl}$ $\Rightarrow$ Gives 1 mole AgCl [$\mathrm{Ni(H_2O)_6}$]$\mathrm{Cl_2}$ $\Rightarrow$ Gives 2 moles AgCl [$\mathrm{Pt(NH_3)_2Cl_2}$] $\Rightarrow$ Gives No AgCl [$\mathrm{Pd(NH_3)_4}$]$\mathrm{Cl_2}$ $\Rightarrow$ Gives 2 moles AgCl Total number of moles of AgCl = 5 mole

Butylated hydroxyl anisole is an antioxidant

Basic strength is proportional to $\frac{1}{pK_b}$. Order for $pK_b$: A > B > D > C.

Fact based

B. $\mathrm{Cl} + \mathrm{O_3} \longrightarrow \mathrm{O_2} + \mathrm{ClO}$ D. 'Blue baby' syndrome occurs due to the presence of excess of nitrate ions in water.

Given $x^2 + 60^{\frac{1}{4}} x + a = 0$ with roots $\alpha$ and $\beta$. We have $\alpha + \beta = -60^{\frac{1}{4}}$ and $\alpha \beta = a$. Given $\alpha^4 + \beta^4 = -30$. This implies $$\left(\alpha^2 + \beta^2\right)^2 - 2\alpha^2 \beta^2 = -30$$ which further implies $$\left((\alpha + \beta)^2 - 2\alpha \beta\right)^2 - 2a^2 = -30$$ Substituting the known values, we get $$\left(60^{\frac{1}{2}} - 2a\right)^2 - 2a^2 = -30$$ Expanding, $$60 + 4a^2 - 4a \times 60^{\frac{1}{2}} - 2a^2 = -30$$ Simplifying, $$2a^2 - 4 \cdot 60^{\frac{1}{2}} a + 90 = 0$$ The product is $$\frac{90}{2} = 45$$

Given $(z - 2i)(\overline{z} + 2i) = 4(z + i)(\overline{z} - i)$. Expanding both sides, we have: $$z \overline{z} + 4 + 2i(z - \overline{z}) = 4(z \overline{z} + 1 + i(\overline{z} - z))$$ Simplifying, we get: $$3z \overline{z} - 6i(z - \overline{z}) = 0$$ Separating real and imaginary parts, we have: $$x^2 + y^2 - 2i(2iy) = 0$$ This simplifies to: $$x^2 + y^2 + 4y = 0$$

Numbers between 5000 and 10000. Using digits 1, 3, 5, 7, 9. The first digit can be 5, 7, or 9, giving 3 options. The second digit can be any of the 4 remaining digits. The third digit can be any of the 3 remaining digits. The fourth digit can be any of the 2 remaining digits. Total numbers are calculated as follows: $$3 \times 4 \times 3 \times 2 = 72$$

7 Red apple (RA), 5 white apple (WA), 8 oranges (O) 5 fruits to be selected (Note: fruits taken different) Possible selections: (2O, 1RA, 2WA) or (2O, 2RA, 1WA) or (3O, 1RA, 1WA) $$\Rightarrow \binom{8}{2} \binom{7}{1} \binom{5}{2} + \binom{8}{2} \binom{7}{2} \binom{5}{1} + \binom{8}{3} \binom{7}{1} \binom{5}{1}$$ $$\Rightarrow 1960 + 2940 + 1960$$ $$\Rightarrow 6860$$

Given $f(x) = 2x^n + \lambda$. $f(4) = 133$ $f(5) = 255$ $133 = 2 \times 4^n + \lambda$ (1) $255 = 2 \times 5^n + \lambda$ (2) Subtracting (1) from (2): $$122 = 2(5^n - 4^n)$$ $$\Rightarrow 5^n - 4^n = 61$$ Therefore, $n = 3$ and $\lambda = 5$. Now, $f(3) - f(2) = 2(3^3 - 2^3) = 38$. The number of divisors is $1, 2, 19, 38$; their sum is $60$.

Given $a, b, \frac{1}{18} \to GP$. $$\frac{a}{18} = b^2 ..... (i)$$ Given $\frac{1}{a}, 10, \frac{1}{b} \to AP$. $$\frac{1}{a} + \frac{1}{b} = 20$$ This implies $a + b = 20ab$, from equation (i); we get $$18b^2 + b = 360b^3$$ $$\Rightarrow 360b^2 - 18b - 1 = 0 \{\because b \neq 0\}$$ $$\Rightarrow b = \frac{18 \pm \sqrt{324 + 1440}}{720}$$ $$\Rightarrow b = \frac{18 + \sqrt{1764}}{720} \{\because b > 0\}$$ $$\Rightarrow b = \frac{1}{12}$$ $$\Rightarrow a = 18 \times \frac{1}{144} = \frac{1}{8}$$ Now, $16a + 12b = 16 \times \frac{1}{8} + 12 \times \frac{1}{12} = 3$

$\sum_{k=0}^{6} {}^{51-k}C_{3}$ $={}^{51}C_{3}+{}^{50}C_{3}+{}^{49}C_{3}+\cdots+{}^{45}C_{3}$ $={}^{45}C_{3}+{}^{46}C_{3}+\cdots+{}^{51}C_{3}$ $={}^{45}C_{4}+{}^{45}C_{3}+{}^{46}C_{3}+\cdots+{}^{51}C_{3}-{}^{45}C_{4}$ $\left({}^{n}C_{r}+{}^{n}C_{r-1}={}^{n+1}C_{r}\right)$ $={}^{52}C_{4}-{}^{45}C_{4}$

$(2023)^{2023}$ $=(2030-7)^{2023}$ $=(35K-7)^{2023}$ $={}^{2023}C_{0}(35K)^{2023}(-7)^{0} +{}^{2023}C_{1}(35K)^{2022}(-7) +\cdots +{}^{2023}C_{2023}(-7)^{2023}$ $=35N-7^{2023}$ Now, $-7^{2023}$ $=-7\times7^{2022}$ $=-7(7^{2})^{1011}$ $=-7(50-1)^{1011}$ $=-7\left( {}^{1011}C_{0}50^{1011} -{}^{1011}C_{1}50^{1010} +\cdots +{}^{1011}C_{1011} \right)$ $=-7(5\lambda-1)$ $=-35\lambda+7$ $\therefore$ when $(2023)^{2023}$ is divided by $35,$ the remainder is $7.$

Given $\cos 2\theta \cdot \cos \frac{-}{2} = \cos 3\theta \cdot \cos \frac{-}{2}$. This implies $2 \cos 2\theta \cdot \cos \frac{\theta}{2} = 2 \cos \frac{90}{2} \cdot \cos 3\theta$. Therefore, $\cos \frac{50}{2} + \cos \frac{30}{2} = \cos \frac{150}{2} + \cos \frac{30}{2}$. This implies $\cos \frac{150}{2} = \cos \frac{50}{2}$. Thus, $\frac{150\theta}{2} = 2k\pi \pm \frac{50}{2}$. This gives $50\theta = 2k\pi$ or $100 = 2k\pi$. Therefore, $\theta = \frac{2k\pi}{5}$ or $\theta = \frac{k\pi}{5}$. Thus, $\therefore \theta = \left\{ -\pi, \frac{-4\pi}{5}, \frac{-3\pi}{5}, \frac{-2\pi}{5}, \frac{-\pi}{5}, 0, \frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}, \pi \right\}$. Given $m = 5$, $n = 5$. Therefore, $m \cdot n = 25$.

Given the equation $2\alpha + 24 - 12\alpha + 3 = 0$. Simplifying, we have $9\alpha + 27 = 0$. Solving for $\alpha$, we get $\alpha = -3$, $\beta = 5$. So $BC = \sqrt{122}$ and $(BC)^2 = 122$. If $x = 1$, $y = \frac{57}{4} = 14.25$. For the points $(1, 1)$, $(1, 2)$ to $(1, 14)$, this gives $14$ points. If $x = 2$, $y = \frac{27}{2} = 13.5$. For the points $(2, 2)$, $(2, 4)$ to $(2, 12)$, this gives $6$ points. If $x = 3$, $y = \frac{51}{4} = 12.75$. For the points $(3, 3)$, $(3, 6)$ to $(3, 12)$, this gives $4$ points. If $x = 4$, $y = 12$. For the points $(4, 4)$, $(4, 8)$, this gives $2$ points. If $x = 5$, $y = \frac{45}{4} = 11.25$. For the points $(5, 5)$, $(5, 10)$, this gives $2$ points. If $x = 6$, $y = \frac{21}{2} = 10.5$.If $x = 7$, $y = \frac{39}{4} = 9.75$

Given $\theta_2 = \frac{\pi}{2}$. $\newline$ $m_{PQ} \cdot m_{QR} = -1$ $\newline$ $$\frac{10 - 2}{9 + 3} \times \frac{10 - 4}{9 - \alpha} = -1 \implies \alpha = 13$$ $\newline$ $m_{QP} \cdot m_{QS} = -1 \implies m_{QS} = -\frac{4}{7}$ $\newline$ Equation of $QS$: $\newline$ $$y - 10 = -\frac{4}{7}(x - 9)$$ $\newline$ $$\implies 4x + 7y = 106 \ldots (1)$$ $\newline$ $m_{OR} \cdot m_{RS} = -1 \implies m_{RS} = -8$ $\newline$ Equation of $RS$: $\newline$ $$y - 4 = -8(x - 13)$$ $\newline$ $$\implies 8x + y = 108 \ldots (2)$$ $\newline$ Solving eq. (1) $\&$ (2) $\newline$ $$x_1 = \frac{25}{8}, \; y = 8$$ $\newline$ $$\implies 8k = 24$$

Given $y^2 = 6x$ and $y^2 = 4ax$. Therefore, $4a = 6$ implies $a = \frac{3}{2}$. The line equation is $y = mx + \frac{3}{2m}$; $(m \neq 0)$. The coordinates are $h = \frac{6m - 3}{4m^2}$, $k = \frac{6m + 3}{4m}$. Now eliminating $m$, we get $$3h = 2(-2k^2 + 9k - 9)$$ $$4y^2 - 18y + 3x + 18 = 0$$ $$21\alpha + 12(2 - \alpha) + 3 = 0$$

Given $(p \rightarrow q) \Delta (p \lor q)$. Option 1: $\Delta = \land$, $\nabla = \lor$ $\[$ \begin{tabular}{|l|l|l|l|} \hline p & q & (p $\rightarrow$ q) & (p $\lor$ q) & (p $\rightarrow$ q) $\land$ (p $\lor$ q) \\ \hline T & T & T & T & T \\ \hline T & F & F & T & F \\ \hline F & T & T & T & T \\ \hline F & F & T & F & F\\ \hline \end{tabular} Option 2: $\Delta = \lor$, $\nabla = \land$ $\[$ \begin{tabular}{|l|l|l|l|} \hline p & q & (p $\rightarrow$ q) & (p $\lor$ q) & (p $\rightarrow$ q) $\land$ (p $\lor$ q) \\ \hline T & T & T & T & T \\ \hline T & F & F & T & F \\ \hline F & T & T & T & T \\ \hline F & F & T & F & T \\ \hline \end{tabular} Option 3: $\Delta = \lor$, $\nabla = \lor$ $\[$ \begin{tabular}{|l|l|l|l|} \hline p & q & (p $\rightarrow$ q) & (p $\lor$ q) & (p $\rightarrow$ q) $\land$ (p $\lor$ q) \\ \hline T & T & T & T & T \\ \hline T & F & F & T & T \\ \hline F & T & T & T & T \\ \hline F & F & T & F & T \\ \hline \end{tabular} Hence, it is tautology. Option 4: $\Delta = \land$, $\nabla = \land$ $\[$ \begin{tabular}{|l|l|l|l|} \hline p & q & (p $\rightarrow$ q) & (p $\lor$ q) & (p $\rightarrow$ q) $\land$ (p $\lor$ q) \\ \hline T & T & T & T & T \\ \hline T & F & F & F & F \\ \hline F & T & T & F & F \\ \hline F & F & T & F & F \\ \hline \end{tabular}

Given, $A^T = A$, $B^T = -B$, $C^T = -C$. Let $M = A^{13} B^{26} - B^{26} A^{13}$. Then, $M^T = (A^{13} B^{26} - B^{26} A^{13})^T$ $$= (A^{13} B^{26})^T - (B^{26} A^{13})^T$$ $$= (B^T)^{26} (A^T)^{13} - (A^T)^{13} (B^T)^{26}$$ $$= (B)^{26} (A)^{13} - (A)^{13} (B)^{26}$$ $$= B^{26} A^{13} - A^{13} B^{26} = -M$$ Hence, $M$ is skew symmetric. Let, $N = A^{26} C^{13} - C^{13} A^{26}$ then, $N^T = (A^{26} C^{13})^T - (C^{13} A^{26})^T$ $$= -(C)^{13} (A)^{26} + A^{26} C^{13} = N$$ Hence, $N$ is symmetric. Therefore, only $S2$ is true.

Given $$AA^T = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ $$B^2 = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2i \\ 0 & 1 \end{bmatrix}$$ $$B^3 = \begin{bmatrix} 1 & -3i \\ 0 & 1 \end{bmatrix}$$ $$\vdots$$ $$B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix}$$ Let $M = A^TBA$. Then $$M^2 = M.M = A^TBA \cdot A^TBA = A^TB^2A$$ $$M^3 = M^2.M = A^TB^2AA^TBA = A^TB^3A$$ $$\vdots$$ $$M^{2023} = \ldots = A^TB^{2023}A$$ Thus, $$AM^{2023}A^T = AA^TB^{2023}AA^T = B^{2023}$$ Therefore, $$= \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix}$$

Since, $-\sqrt{2} \leq \sin x - \cos x \leq \sqrt{2}$. Therefore, $-2 \leq \sqrt{2} (\sin x - \cos x) \leq 2$. Assume $\sqrt{2} (\sin x - \cos x) = k$. Then $-2 \leq k \leq 2$ $\ldots$ (i). $f(x) = \log_{\sqrt{m}} (k + m - 2)$. Given, $0 \leq f(x) \leq 2$. So, $0 \leq \log_{\sqrt{m}} (k + m - 2) \leq 2$. This implies $1 \leq k + m - 2 \leq m$. Therefore, $-m + 3 \leq k \leq 2$ $\ldots$ (ii). From eq. (i) $\&$ (ii), we get $-m + 3 = -2$. Thus, $m = 5$.

Given the function $f: \{1, 2, 3, 4\} \to \{a \in \mathbb{Z} : |a| \leq 8\}$. The equation is $f(n) + \frac{1}{n} f(n+1) = 1$, for all $n \in \{1, 2, 3\}$. The value $f(n+1)$ must be divisible by $n$. For $f(4)$, the possible values are $-6, -3, 0, 3, 6$. For $f(3)$, the possible values are $-8, -6, -4, -2, 0, 2, 4, 6, 8$. For $f(2)$, the possible values are $-8, \ldots, 8$. For $f(1)$, the possible values are $-8, \ldots, 8$. The value $\frac{f(4)}{3}$ must be odd since $f(3)$ should be even. Therefore, 2 solutions are possible. $\begin{array}{cccc}$ f(4) & f(3) & f(2) & f(1) $\\$ -3 & 2 & 0 & 1 $\\$ 3 & 0 & 1 & 0 $\\$ $\end{array}$

\[ \begin{aligned} &\lim_{x\to \frac{\pi}{2}^{+}} e^{\frac{\cot 6x}{\cot 4x}} =\lim_{x\to \frac{\pi}{2}^{+}} e^{\frac{\sin 4x\cos 6x}{\sin 6x\cos 4x}} =e^{\frac{2}{3}}.\\[6pt] &\Rightarrow \lim_{x\to \frac{\pi}{2}^{-}} (1+|\cos x|)^{\frac{\lambda}{|\cos x|}} =e^{\lambda}.\\[6pt] &\Rightarrow f\!\left(\frac{\pi}{2}\right)=\mu.\\[6pt] &\text{By continuity, } e^{\frac{2}{3}}=e^{\lambda}=\mu.\\[6pt] &\Rightarrow \lambda=\frac{2}{3},\qquad \mu=e^{\frac{2}{3}}.\\[6pt] &9\lambda+6\log_e\mu+\mu^6-e^{6\lambda}=10. \end{aligned} \]

Given $f(x) = 2x^3 + (2p - 7)x^2 + 3(2p - 9)x - 6$. The derivative is $f'(x) = 6x^2 + 2(2p - 7)x + 3(2p - 9)$. We have $f'(0) < 0$. Therefore, $3(2p - 9) < 0$. Solving gives $p < \frac{9}{2}$. Thus, $p \in \left(-\infty, \frac{9}{2}\right)$.

Given $$I = 16 \int_1^2 \frac{dx}{x^3 (x^2 + 2)^2}$$ This can be rewritten as $$= 16 \int_1^2 \frac{dx}{x^3 x^4 \left(1 + \frac{2}{x^2}\right)^2}$$ Let, $$1 + \frac{2}{x^2} = t \implies -\frac{4}{x^3} dx = dt$$ Then, $$I = 4 \int_{\frac{3}{2}}^{\frac{7}{3}} \frac{dt}{\left(\frac{2}{t-1}\right) t^2}$$ This simplifies to $$I = -4 \int_{\frac{3}{2}}^{\frac{7}{3}} \left(\frac{t-1}{2}\right)^2 \frac{dt}{t^2}$$ Further simplifying, $$I = \frac{4}{4} \int_{\frac{3}{2}}^{\frac{7}{3}} \left(1 - \frac{2}{t} + \frac{1}{t^2}\right) dt$$ This results in $$I = -1 \left[t - 2 \ln|t| - \frac{1}{t}\right]_{\frac{3}{2}}^{\frac{7}{3}}$$ Evaluating the integral, $$I = -1 \left[\left(\frac{3}{2} - 2 \ln \frac{3}{2} - \frac{3}{2}\right) - \left(3 - 2 \ln 3 - \frac{1}{3}\right)\right]$$ This simplifies to $$I = -1 \left[2 \ln 2 - \frac{11}{6}\right]$$ Finally, $$I = \frac{11}{6} - \ln 4$$

Given $\int_{1/3}^{3} \ln x \, dx = \int_{1/3}^{1} (-\ln x) \, dx + \int_{1}^{3} (\ln x) \, dx$. $$= -\left[ x \ln x - x \right]_{1/3}^{1} + \left[ x \ln x - x \right]_{1}^{3}$$ $$= -\left[ -1 \left( \frac{1}{3} \ln 1 - \frac{1}{3} \right) \right] + \left[ 3 \ln 3 - 3 - (-1) \right]$$ $$= \left[ -\frac{2}{3} - \frac{1}{3} \ln 1 \right] + \left[ 3 \ln 3 - 2 \right]$$ $$= -\frac{4}{3} + \frac{8}{3} \ln 3$$ $$= \frac{4}{3} (2 \ln 3 - 1)$$ $$= \frac{4}{3} \left( \ln \frac{9}{e} \right)$$ Therefore, $m = 4$, $n = 3$. Now, $m^2 + n^2 - 5 = 16 + 9 - 5 = 20$

Given the equation $16(x^2 + 4x) - (y^2 - 4y) + 44 = 0$. Simplifying, we have: $$16(x + 2)^2 - 64 - (y - 2)^2 + 4 + 44 = 0$$ $$16(x + 2)^2 - (y - 2)^2 = 16$$ This can be rewritten as: $$\frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{16} = 1$$ The area $A$ is given by the integral: $$A = \int_{2}^{\sqrt{6}} \left(2 - (x^2 - 4)\right) \, dx$$ Simplifying the integral: $$A = \int_{2}^{\sqrt{6}} (6 - x^2) \, dx = \left(6x - \frac{x^3}{3}\right) \bigg|_{2}^{\sqrt{6}}$$ Evaluating the integral: $$A = \left(6\sqrt{6} - \frac{6\sqrt{6}}{3}\right) - \left(-12 + \frac{8}{3}\right)$$ Simplifying further: $$A = \frac{12\sqrt{6}}{3} + \frac{28}{3}$$ Thus, the area is: $$A = 4\sqrt{6} + \frac{28}{3}$$

Given $\($ $\frac{dy}{dt}$ + $\alpha$ y = $\gamma$ e^{-$\beta$ t} $\)$. The integrating factor (I.F.) is $\($ e^{$\int$ $\alpha$ $\,$ dt} = e^{$\alpha$ t} $\)$. Solution implies $\($ y $\cdot$ e^{$\alpha$ t} = $\int$ $\gamma$ e^{-$\beta$ t} $\cdot$ e^{$\alpha$ t} $\,$ dt $\)$. $\($ $\Rightarrow$ ye^{$\alpha$ t} = $\gamma$ $\frac{e^{(\alpha - \beta)t}}{(\alpha - \beta)}$ + c $\)$. $\($ $\Rightarrow$ y = $\frac{\gamma}{e^{\beta t}(\alpha - \beta)}$ + $\frac{c}{e^{\alpha t}}$ $\)$. So, $\($ $\lim$_{t $\to$ $\infty$} y(t) = $\frac{\gamma}{\infty}$ + $\frac{c}{\infty}$ = 0 $\)$.

Let \[ A:(3,-4,2),\qquad B:(1,2,-1),\qquad C:(-2,-1,3) \] and \[ D:(5,-2\alpha,4). \] As \[ A,\ B,\ C,\ D \] are coplanar points, therefore \[ \begin{vmatrix} 1-3 & 2+4 & -1-2\\ -2-3 & -1+4 & 3-2\\ 5-3 & -2\alpha+4 & 4-2 \end{vmatrix} =0 \] \[ \Rightarrow \begin{vmatrix} -2 & 6 & -3\\ -5 & 3 & 1\\ 2 & 4-2\alpha & 2 \end{vmatrix} =0 \] \[ \Rightarrow \alpha=-\frac{73}{17}. \]

Given $\vec{a} \times \vec{b} = (\hat{i} - \hat{j})$. Taking cross product with $\vec{a}$, we have: $$\vec{a} \times (\vec{a} \times \vec{b}) = \vec{a} \times (\hat{i} - \hat{j})$$ $$\Rightarrow (\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b} = \hat{i} + \hat{j} + 2\hat{k}$$ $$\Rightarrow \vec{a} - 3\vec{b} = \hat{i} + \hat{j} + 2\hat{k}$$ $$\Rightarrow 2\vec{a} - 6\vec{b} = 2\hat{i} + 2\hat{j} + 4\hat{k}$$ $$\Rightarrow \vec{a} - 6\vec{b} = 3\hat{i} + 3\hat{j} + 3\hat{k}$$

Given $\($ $\frac{x+1}{1}$ = $\frac{y}{\frac{1}{2}}$ = $\frac{z}{-\frac{1}{12}}$ $\)$ and $\($ $\frac{x}{1}$ = $\frac{y+2}{1}$ = $\frac{z-1}{\frac{1}{6}}$ $\)$. Therefore, the shortest distance is given by: $$ \frac{(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q})}{|\mathbf{p} \times \mathbf{q}|} $$ The scalar distance (S.D.) is: $$ S.D. = \frac{(-\hat{i} + 2\hat{j} - \hat{k}) \cdot (\mathbf{p} \times \mathbf{q})}{|\mathbf{p} \times \mathbf{q}|} $$ Calculating $\($ $\mathbf{p}$ $\times$ $\mathbf{q}$ $\)$: $$ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \frac{1}{2} & -\frac{1}{12} \\ 1 & 1 & \frac{1}{6} \end{vmatrix} = \frac{1}{6} \hat{i} - \frac{1}{4} \hat{j} + \frac{1}{2} \hat{k} or 2\hat{i} - 3\hat{j} + 6\hat{k} $$ Thus, the scalar distance is: $$ S.D. = \frac{(-\hat{i} + 2\hat{j} - \hat{k}) \cdot (2\hat{i} - 3\hat{j} + 6\hat{k})}{\sqrt{2^2 + 3^2 + 6^2}} = \frac{| -14 |}{7} = 2 $$

Given $L: \frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1} = \lambda$ (let). Let foot of perpendicular is $P(2\lambda - 1, 5\lambda + 1, -\lambda - 1)$. $$\overrightarrow{PA} = (3 - 2\lambda) \hat{i} - (5\lambda + 1) \hat{j} + (6 + \lambda) \hat{k}$$ Direction ratio of line $\Rightarrow \overrightarrow{b} = 2\hat{i} + 5\hat{j} - \hat{k}$. Now, $\overrightarrow{PA} \cdot \overrightarrow{b} = 0$. $$\Rightarrow 2(3 - 2\lambda) - 5(5\lambda + 1) - (6 + \lambda) = 0$$ $$\Rightarrow \lambda = -\frac{1}{6}$$ $$P(2\lambda - 1, 5\lambda + 1, -\lambda - 1) \equiv P(\alpha, \beta, \gamma)$$ $$\Rightarrow \alpha = 2 \left( -\frac{1}{6} \right) - 1 = -\frac{4}{3} \Rightarrow \alpha = -\frac{4}{3}$$ $$\Rightarrow \beta = 5 \left( -\frac{1}{6} \right) + 1 = \frac{1}{6} \Rightarrow \beta = \frac{1}{6}$$ $$\Rightarrow \gamma = -\lambda - 1 = \frac{1}{6} - 1 = -\frac{5}{6} \Rightarrow \gamma = -\frac{5}{6}$$

Given $\mathbf{r} = (i + 2j + 3k) + \lambda (i + j + k)$ and $\mathbf{r} = \mathbf{a} + \lambda \mathbf{p}$. Also, $\mathbf{r} = (+i - j + 2k) + \mu (2i - j)$ and $\mathbf{r} = \mathbf{b} + \mu \mathbf{q}$. The cross product $\mathbf{p} \times \mathbf{q}$ is given by: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} = \hat{i} + 2\hat{j} - 3\hat{k}$$ The distance $d$ is: $$d = \frac{|(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q})|}{|\mathbf{p} \times \mathbf{q}|}$$ Substituting the values: $$d = \frac{|(-3\hat{j} - \hat{k}) \cdot (\hat{i} + 2\hat{j} - 3\hat{k})|}{\sqrt{14}}$$ Calculating the dot product: $$= \frac{|-6 + 3|}{\sqrt{14}} = \frac{3}{\sqrt{14}}$$ Thus, $\alpha = \frac{3}{\sqrt{14}}$. Now, $28\alpha^2 = \frac{9}{14} \times 28 = 18$.

Given: $N - 2$, $\sqrt{3N}$, $N + 2$ are in G.P. $$3N = (N - 2)(N + 2)$$ $$3N = N^2 - 4$$ $$\Rightarrow N^2 - 3N - 4 = 0$$ $$(N - 4)(N + 1) = 0 \Rightarrow N = 4 or N = -1 rejected$$ (Sum = 4) $\equiv \{(1, 3), (3, 1), (2, 2)\}$ $n(A) = 3$ $$P(A) = \frac{3}{36} = \frac{1}{12} = \frac{4}{48} \Rightarrow k = 4$$

Let $E_1$ be smokers. The probability $P(E_1) = \frac{1}{4}$. Let $E_2$ be non-smokers. The probability $P(E_2) = \frac{3}{4}$. Let $E$ be diagnosed with lung cancer. The probability $P(E/E_1) = \frac{27}{28}$ and $P(E/E_2) = \frac{1}{28}$. The probability $P(E_1/E)$ is given by: $$P(E_1/E) = \frac{P(E_1)P(E/E_1)}{P(E)}$$ Substituting the values, we have: $$= \frac{\frac{1}{4} \times \frac{27}{28}}{\frac{1}{4} \times \frac{27}{28} + \frac{3}{4} \times \frac{1}{28}} = \frac{\frac{27}{112}}{\frac{30}{112}} = \frac{9}{10}$$ Thus, $K = 9$.