TS EAMCET 2006 Chemistry Question Paper with Answer and Solution

(B) Isotones are species having the same number of neutrons.
In $_{20}Ca^{40}$,the number of neutrons is $40 - 20 = 20$.
In $_{18}Ar^{40}$,the number of neutrons is $40 - 18 = 22$.
Since the number of neutrons is different,$_{20}Ca^{40}$ and $_{18}Ar^{40}$ are not isotones.
Therefore,option $B$ is incorrect.

(A) Given:
$E = 3 \times 10^{-12} \ erg$
$h = 6.62 \times 10^{-27} \ erg \cdot s$
$c = 3 \times 10^{10} \ cm/s$
Using the formula $E = \frac{hc}{\lambda}$,we have:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{3 \times 10^{-12}} \ cm$
$\lambda = 6.62 \times 10^{-5} \ cm$
Since $1 \ cm = 10^7 \ nm$,
$\lambda = 6.62 \times 10^{-5} \times 10^7 \ nm$
$\lambda = 6.62 \times 10^2 \ nm = 662 \ nm$.

(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot m \cdot \Delta v = \frac{h}{4 \pi}$.
For particle $A$: $\Delta x_A \cdot m_A \cdot \Delta v_A = \frac{h}{4 \pi}$.
For particle $B$: $\Delta x_B \cdot m_B \cdot \Delta v_B = \frac{h}{4 \pi}$.
Given: $\Delta v_A = 0.05 \ m/s$,$\Delta v_B = 0.02 \ m/s$,and $m_B = 5 m_A$.
Equating the two expressions: $\Delta x_A \cdot m_A \cdot 0.05 = \Delta x_B \cdot (5 m_A) \cdot 0.02$.
$\Delta x_A \cdot m_A \cdot 0.05 = \Delta x_B \cdot m_A \cdot 0.10$.
$\frac{\Delta x_A}{\Delta x_B} = \frac{0.10}{0.05} = 2$.

(C) colloidal solution of gold is formed when the dispersed phase is a $solid$ and the dispersion medium is a $liquid$.
Metals like gold cannot form a colloidal state simply by mixing with water,requiring special preparation methods.
Therefore,they are classified as $lyophobic$ (solvent-fearing) colloids.

(D) Given the thermo emf equation: $V = 10 \times 10^{-6} t - \frac{1}{40} \times 10^{-6} t^2$.
The neutral temperature $(t_n)$ occurs when the thermo emf is maximum,which implies $\frac{dV}{dt} = 0$.
Differentiating $V$ with respect to $t$: $\frac{dV}{dt} = 10 \times 10^{-6} - \frac{2}{40} \times 10^{-6} t = 10 \times 10^{-6} - \frac{1}{20} \times 10^{-6} t$.
Setting $\frac{dV}{dt} = 0$: $10 \times 10^{-6} = \frac{1}{20} \times 10^{-6} t_n \implies t_n = 200^{\circ} C$.
Now,substitute $t_n = 200^{\circ} C$ into the equation for $V$ to find $V_{\max}$:
$V_{\max} = 10 \times 10^{-6} (200) - \frac{1}{40} \times 10^{-6} (200)^2$.
$V_{\max} = 2000 \times 10^{-6} - \frac{40000}{40} \times 10^{-6} = 2 \times 10^{-3} - 1 \times 10^{-3} = 1 \times 10^{-3} \text{ V} = 1 \text{ mV}$.

(B) According to the Stefan-Boltzmann law,the rate of heat loss $dQ/dt$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$.
For a sphere,$A = 4 \pi r^2$ and $dQ = mc(dT) = (\rho V c) dT = \rho (4/3 \pi r^3) c dT$.
Thus,the rate of change of temperature is given by: $\rho (4/3 \pi r^3) c (-dT/dt) = \sigma (4 \pi r^2) (T^4 - T_0^4)$.
Simplifying this,we get: $(-dT/dt) = [3 \sigma / (\rho c r)] (T^4 - T_0^4)$.
Since the material is the same,$\rho$ and $c$ are constant. For the same temperature difference,$(-dT/dt) \propto 1/r$.
Therefore,the ratio of the rate of change of temperature of $A$ and $B$ is: $(dT/dt)_A / (dT/dt)_B = r_B / r_A$.

(B) For an isothermal process,the temperature of the gas remains constant,so the gas obeys Boyle's law: $P_1 V_1 = P_2 V_2$.
Given that the pressure is doubled,$P_2 = 2 P_1$.
Thus,$P_1 V_1 = (2 P_1) V_2$,which gives $\frac{V_1}{V_2} = 2$.
Now,the gas expands adiabatically until its original volume is restored,so $V_3 = V_1$.
The adiabatic process follows $P V^\gamma = \text{constant}$,so $P_2 V_2^\gamma = P_3 V_3^\gamma$.
Given $P_3 = 0.75 P_1$ and $P_2 = 2 P_1$,we have:
$(2 P_1) V_2^\gamma = (0.75 P_1) V_1^\gamma$.
Dividing both sides by $P_1$,we get $2 V_2^\gamma = 0.75 V_1^\gamma$.
$\left(\frac{V_1}{V_2}\right)^\gamma = \frac{2}{0.75} = \frac{2}{3/4} = \frac{8}{3}$.
Since $\frac{V_1}{V_2} = 2$,we have $2^\gamma = \frac{8}{3} = 2.667$.
Taking the logarithm on both sides: $\gamma \log 2 = \log 2.667$.
$\gamma = \frac{\log 2.667}{\log 2} \approx \frac{0.426}{0.301} \approx 1.41$.

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous species.
For $\Delta H \neq \Delta E$,the condition $\Delta n_g \neq 0$ must be satisfied.
$(A)$ $\Delta n_g = 1 - 1 = 0$.
$(B)$ $\Delta n_g = 2 - (1 + 1) = 0$.
$(C)$ $\Delta n_g = 2 - (1 + 1) = 0$.
$(D)$ $\Delta n_g = 1 - (1 + 0.5) = -0.5$.
Since $\Delta n_g \neq 0$ for option $(D)$,$\Delta H \neq \Delta E$ for this reaction.

(D) The lattice energy $(U)$ of an ionic solid is defined as the energy released when gaseous ions combine to form one mole of the crystalline solid.
According to the Born-Landé equation,the lattice energy is inversely proportional to the inter-ionic distance $(r_0)$.
For an ionic solid $AB$,the inter-ionic distance $r_0$ is the sum of the ionic radii of the cation $(r_c)$ and the anion $(r_a)$,i.e.,$r_0 = r_c + r_a$.
Therefore,the lattice energy $U$ is proportional to $\frac{1}{(r_c+r_a)}$.

(D) The dimensional formulas are:
$[C] = [M^{-1} L^{-2} T^4 A^2]$
$[R] = [M L^2 T^{-3} A^{-2}]$
$[L] = [M L^2 T^{-2} A^{-2}]$
$[I] = [A]$
$(1)$ $[CR] = [M^{-1} L^{-2} T^4 A^2] \times [M L^2 T^{-3} A^{-2}] = [T^1]$
$(2)$ $[L/R] = [M L^2 T^{-2} A^{-2}] / [M L^2 T^{-3} A^{-2}] = [T^1]$
$(3)$ $[\sqrt{LC}] = ([M L^2 T^{-2} A^{-2}] \times [M^{-1} L^{-2} T^4 A^2])^{1/2} = [T^2]^{1/2} = [T^1]$
$(4)$ $[LI^2] = [M L^2 T^{-2} A^{-2}] \times [A^2] = [M L^2 T^{-2}]$ (This is energy,not time).
Thus,quantities $(1)$,$(2)$,and $(3)$ have the dimensions of time. Hence,option $(d)$ is correct.

(C) The intensity of light $I$ is directly proportional to the width of the slit $w$,i.e.,$I \propto w$.
Given that the width of the first slit $w_1 = 4 w_2$,the intensities are $I_1 = 4 I_2$.
Let $I_2 = I$,then $I_1 = 4 I$.
The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{4I} + \sqrt{I})^2}{(\sqrt{4I} - \sqrt{I})^2} = \frac{(2\sqrt{I} + \sqrt{I})^2}{(2\sqrt{I} - \sqrt{I})^2} = \frac{(3\sqrt{I})^2}{(\sqrt{I})^2} = \frac{9I}{I} = \frac{9}{1}$.
Thus,the ratio is $9: 1$.

(B) The frequency of the sound heard directly by the observer from the moving source is given by the Doppler effect formula: $f_1 = f_0 \left( \frac{v}{v + v_s} \right)$,where $f_0 = 1000 ~Hz$,$v = 330 ~m/s$,and $f_1 = 970 ~Hz$.
Substituting the values: $970 = 1000 \left( \frac{330}{330 + v_s} \right)$.
Solving for $v_s$: $330 + v_s = \frac{330000}{970} \approx 340.2 ~m/s$,so $v_s \approx 10.2 ~m/s$.
The sound reflected from the hill acts as if it is coming from a stationary source (the hill) towards the observer,but the source (van) is moving towards the hill. The hill receives the sound at a frequency $f' = f_0 \left( \frac{v}{v - v_s} \right)$.
Since the hill is stationary,it reflects this frequency back to the observer: $f_2 = f' = 1000 \left( \frac{330}{330 - 10.2} \right)$.
$f_2 = \frac{330000}{319.8} \approx 1031.89 ~Hz$.
Rounding to the nearest integer,the frequency is approximately $1032 ~Hz$.

(C) Given: $T_A = T_B$,$f_A = f_B$,$L_A = 80 \ cm$,$L_B = x \ cm$,$d_A / d_B = 0.81$,and $D_B = D_A / 2$.
Linear mass density $\mu$ is given by $\mu = \text{Area} \times \text{density} = \pi (D/2)^2 \times d = \frac{\pi D^2 d}{4}$.
Therefore,the ratio of linear densities is $\frac{\mu_A}{\mu_B} = \left(\frac{D_A}{D_B}\right)^2 \times \frac{d_A}{d_B} = (2)^2 \times 0.81 = 4 \times 0.81 = 3.24$.
The fundamental frequency of a string is $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $f_A = f_B$ and $T_A = T_B$,we have $\frac{1}{L_A \sqrt{\mu_A}} = \frac{1}{L_B \sqrt{\mu_B}}$,which implies $\frac{L_B}{L_A} = \sqrt{\frac{\mu_B}{\mu_A}}$.
Substituting the values: $\frac{x}{80} = \sqrt{\frac{1}{3.24}} = \frac{1}{1.8}$.
$x = \frac{80}{1.8} = \frac{800}{18} \approx 44.44 \ cm$. Wait,re-evaluating the ratio: $\frac{L_B}{L_A} = \sqrt{\frac{\mu_B}{\mu_A}} \implies \frac{x}{80} = \sqrt{\frac{1}{3.24}} = \frac{1}{1.8}$.
Actually,$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \implies L \propto \frac{1}{\sqrt{\mu}}$. So $\frac{L_B}{L_A} = \sqrt{\frac{\mu_A}{\mu_B}} = \sqrt{3.24} = 1.8$.
$x = 80 \times 1.8 = 144 \ cm$.

(C) Let the density of water be $\rho$,the cross-sectional area of the pipe be $A$,and the velocity of water be $v$.
The mass of water flowing out per second is given by $m = A v \rho$.
The kinetic energy imparted to the water per second is the power $P$ required by the motor:
$P = \frac{1}{2} m v^2 = \frac{1}{2} (A v \rho) v^2 = \frac{1}{2} A \rho v^3$.
This implies that $P \propto v^3$.
If we want to deliver $n$-times the mass of water in the same time,the mass flow rate $m' = n m$.
Since $m = A v \rho$,we have $m' = A v' \rho = n (A v \rho)$,which implies $v' = n v$.
The new power $P'$ required is $P' = \frac{1}{2} A \rho (v')^3$.
Taking the ratio of the new power to the original power:
$\frac{P'}{P} = \frac{\frac{1}{2} A \rho (n v)^3}{\frac{1}{2} A \rho v^3} = n^3$.
Therefore,the power must be increased $n^3$-times.

(B) The work-energy theorem states that the work done by all forces equals the change in kinetic energy.
$W_{total} = \Delta K = K_f - K_i$
Here,the total work done is the sum of work done by gravity $(W_g)$ and work done against air resistance $(W_{air})$.
$W_g + W_{air} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$
Given:
Mass $m = 10 ~g = 0.01 ~kg$
Initial velocity $u = 1000 ~ms^{-1}$
Final velocity $v = 500 ~ms^{-1}$
Height $h = 50 ~m$
Gravity $g = 10 ~ms^{-2}$
Work done by gravity $W_g = mgh = 0.01 \times 10 \times 50 = 5 ~J$
Change in kinetic energy $\Delta K = \frac{1}{2} \times 0.01 \times (500^2 - 1000^2)$
$\Delta K = 0.005 \times (250000 - 1000000) = 0.005 \times (-750000) = -3750 ~J$
Now,$W_g + W_{air} = \Delta K$
$5 + W_{air} = -3750$
$W_{air} = -3750 - 5 = -3755 ~J$
The work done against air resistance is the magnitude of the work done by air resistance,which is $3755 ~J$.

(D) The molecular formula of benzene is $C_6H_6$.
In the structure of benzene,there are $6$ $C-C$ bonds and $6$ $C-H$ bonds.
Each $C-C$ bond consists of one $\sigma$ bond,and three of these bonds are double bonds,each containing one $\pi$ bond.
Thus,there are $6$ $\sigma$ bonds between carbon atoms and $6$ $\sigma$ bonds between carbon and hydrogen atoms,totaling $12$ $\sigma$ bonds.
There are $3$ $\pi$ bonds present in the ring.
Therefore,the number of $\sigma$ and $\pi$ bonds are $12$ and $3$ respectively.

(D) In ethene $(C_2H_4)$,each carbon atom is $sp^2$ hybridized.
Each carbon atom forms two $\sigma$ bonds with hydrogen atoms by the overlap of $sp^2$ and $s$ orbitals. Since there are two carbon atoms,there are $2 \times 2 = 4$ such $\sigma$ bonds.
The two carbon atoms are bonded to each other by a $\sigma$ bond formed by the overlap of $sp^2$ and $sp^2$ orbitals.
Additionally,the unhybridized $p_z$ orbitals of the two carbon atoms overlap laterally to form one $\pi$ bond.
Thus,the molecule $(X)$ is $C_2H_4$.

(C) catalyst increases the rate of both forward and backward reactions equally by lowering the activation energy.
It does not affect the position of equilibrium or the equilibrium constant $(K_c)$.
Therefore,the statement that the value of $K_c$ decreases in the presence of a catalyst is incorrect.

(B) Statement $I$ is correct because the modern periodic law states that physical and chemical properties of elements are periodic functions of their atomic numbers,which is equivalent to their electronic configuration.
Statement $II$ is incorrect because fluorine $(F)$ has the highest electronegativity $(4.0)$ in the periodic table,which is greater than that of chlorine ($Cl$,$3.0$).
Statement $III$ is incorrect because electropositive nature (metallic character) increases from top to bottom in a group as the ionization energy decreases.
Therefore,only statement $I$ is correct.

(A) Given the inequality: $\sqrt{9 x^2+6 x+1} < (2-x)$
Since $\sqrt{(3x+1)^2} = |3x+1|$,the inequality becomes $|3x+1| < 2-x$.
For the square root to be defined,we must have $9x^2+6x+1 \ge 0$,which is always true as $(3x+1)^2 \ge 0$.
Also,for the inequality to hold,the right side must be positive: $2-x > 0 \implies x < 2$.
Now,solve $|3x+1| < 2-x$:
This is equivalent to $-(2-x) < 3x+1 < 2-x$.
Case $1$: $3x+1 < 2-x \implies 4x < 1 \implies x < \frac{1}{4}$.
Case $2$: $3x+1 > -(2-x) \implies 3x+1 > -2+x \implies 2x > -3 \implies x > -\frac{3}{2}$.
Combining these,we get $-\frac{3}{2} < x < \frac{1}{4}$.
Thus,$x \in \left(-\frac{3}{2}, \frac{1}{4}\right)$.

(D) Given $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$.
Rationalizing the denominator inside the square root:
$x = \sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}} = \sqrt{\frac{(2+\sqrt{3})^2}{4-3}} = 2+\sqrt{3}$.
Now,we need to evaluate $x^2(x-4)^2 = [x(x-4)]^2$.
Substitute $x = 2+\sqrt{3}$:
$x(x-4) = (2+\sqrt{3})(2+\sqrt{3}-4) = (2+\sqrt{3})(\sqrt{3}-2)$.
Using the identity $(a+b)(a-b) = a^2-b^2$:
$x(x-4) = (\sqrt{3}+2)(\sqrt{3}-2) = (\sqrt{3})^2 - (2)^2 = 3 - 4 = -1$.
Therefore,$x^2(x-4)^2 = (-1)^2 = 1$.

(A) Given the equation $\left|\frac{z-1}{z+1}\right| = 1$.
Substitute $z = x + iy$:
$\left|\frac{(x-1) + iy}{(x+1) + iy}\right| = 1$
Taking the modulus on both sides:
$|(x-1) + iy| = |(x+1) + iy|$
Squaring both sides:
$(x-1)^2 + y^2 = (x+1)^2 + y^2$
Expanding the squares:
$x^2 - 2x + 1 + y^2 = x^2 + 2x + 1 + y^2$
Simplifying the equation:
$-2x = 2x$
$4x = 0$
$x = 0$
Thus,the locus is the imaginary axis,represented by $x = 0$.

(B) Given $\left|\frac{z-i}{z i}\right|=2$.
Since $z=x iy$,we have $\left|\frac{x i(y-1)}{x i(y 1)}\right|=2$.
Squaring both sides,we get $\frac{x^2 (y-1)^2}{x^2 (y 1)^2}=4$.
$x^2 (y-1)^2=4(x^2 (y 1)^2)$.
$x^2 y^2-2y 1=4(x^2 y^2 2y 1)$.
$x^2 y^2-2y 1=4x^2 4y^2 8y 4$.
Rearranging the terms,we get $3x^2 3y^2 10y 3=0$.

(A) Natural numbers less than $1000$ can be $1$-digit,$2$-digit,or $3$-digit numbers.
Case $1$: $1$-digit numbers. The digits can be ${1, 2, 3, 4, 5, 6, 7, 8, 9}$. Total $= 9$.
Case $2$: $2$-digit numbers. The first digit can be chosen in $9$ ways (excluding $0$) and the second digit in $9$ ways (including $0$ but excluding the first digit). Total $= 9 \times 9 = 81$.
Case $3$: $3$-digit numbers. The first digit can be chosen in $9$ ways (excluding $0$),the second in $9$ ways (including $0$ but excluding the first digit),and the third in $8$ ways (excluding the first two digits). Total $= 9 \times 9 \times 8 = 648$.
Total number of natural numbers $= 9 + 81 + 648 = 738$.

(C) The total number of words of length $4$ that can be formed using $8$ different letters with repetition allowed is $8 \times 8 \times 8 \times 8 = 8^{4}$.
The number of words of length $4$ that can be formed using $8$ different letters without any repetition is given by the permutation formula ${}^{n}P_{r} = {}^{8}P_{4}$.
The number of words with at least one letter repeated is the total number of words minus the number of words with no letters repeated.
Therefore,the required number of words is $8^{4} - {}^{8}P_{4}$.

(B) Let $S=1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\ldots$
Comparing this with the binomial expansion $(1-x)^n = 1 + n(-x) + \frac{n(n-1)}{2!}(-x)^2 + \ldots$
We have $nx = \frac{2}{4} = \frac{1}{2}$ $(i)$
And $\frac{n(n-1)}{2!}x^2 = \frac{2 \cdot 5}{4 \cdot 8} = \frac{10}{32} = \frac{5}{16}$ (ii)
Dividing (ii) by the square of $(i)$: $\frac{\frac{n(n-1)}{2}x^2}{n^2x^2} = \frac{5/16}{1/4} \Rightarrow \frac{n-1}{2n} = \frac{5}{4}$
$4n - 4 = 10n$ $\Rightarrow -6n = 4$ $\Rightarrow n = -2/3$
Substituting $n$ into $(i)$: $(-2/3)x = 1/2 \Rightarrow x = -3/4$
Thus,$S = (1 - x)^n = (1 - (-3/4))^{-2/3} = (7/4)^{-2/3}$ is incorrect. Re-evaluating: The series is $(1-x)^n$ where $nx = 1/2$ and $x = 1/2$ is not correct. Using $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \ldots$
Here $nx = 1/2$ and $\frac{n(n+1)}{2}x^2 = 5/16$. Solving gives $n=2/3$ and $x=3/4$.
$S = (1-3/4)^{-2/3} = (1/4)^{-2/3} = 4^{2/3} = \sqrt[3]{16}$.

(D) Chlorofluorocarbons $(CFCs)$ are responsible for the depletion of ozone in the stratosphere.
$CFCl_3$ decomposes in the presence of ultraviolet radiation to produce chlorine free radicals $(Cl^{\bullet})$.
These chlorine free radicals react with ozone $(O_3)$ molecules to form oxygen $(O_2)$ and chlorine monoxide $(ClO^{\bullet})$.
The reaction is: $Cl^{\bullet} + O_3 \rightarrow ClO^{\bullet} + O_2$.

(A) First,calculate the value of $x$:
$x = \tan 15^{\circ} = \tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \approx 2 - 1.732 = 0.268$.
Next,calculate the value of $y$:
$y = \operatorname{cosec} 75^{\circ} = \frac{1}{\sin(45^{\circ} + 30^{\circ})} = \frac{1}{\sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}} = \frac{1}{\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}} = \frac{2\sqrt{2}}{\sqrt{3} + 1} = \sqrt{2}(\sqrt{3} - 1) = \sqrt{6} - \sqrt{2} \approx 2.449 - 1.414 = 1.035$.
Finally,calculate the value of $z$:
$z = 4 \sin 18^{\circ} = 4 \left( \frac{\sqrt{5} - 1}{4} \right) = \sqrt{5} - 1 \approx 2.236 - 1 = 1.236$.
Comparing the values: $0.268 < 1.035 < 1.236$,which implies $x < y < z$.

(C) We have $\operatorname{cosec} 15^{\circ} + \sec 15^{\circ} = \frac{1}{\sin 15^{\circ}} + \frac{1}{\cos 15^{\circ}}$
$= \frac{\cos 15^{\circ} + \sin 15^{\circ}}{\sin 15^{\circ} \cos 15^{\circ}}$
Multiply numerator and denominator by $2$:
$= \frac{2(\cos 15^{\circ} + \sin 15^{\circ})}{2 \sin 15^{\circ} \cos 15^{\circ}} = \frac{2(\cos 15^{\circ} + \sin 15^{\circ})}{\sin 30^{\circ}}$
Since $\sin 30^{\circ} = \frac{1}{2}$,we have:
$= 4(\cos 15^{\circ} + \sin 15^{\circ})$
Using $\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$:
$= 4 \left( \frac{\sqrt{6} + \sqrt{2}}{4} + \frac{\sqrt{6} - \sqrt{2}}{4} \right)$
$= 4 \left( \frac{2 \sqrt{6}}{4} \right) = 2 \sqrt{6}$

(B) We use the values of trigonometric functions for allied angles:
$\sin 120^{\circ} = \sin(180^{\circ} - 60^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$
$\cos 150^{\circ} = \cos(180^{\circ} - 30^{\circ}) = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}$
$\cos 240^{\circ} = \cos(180^{\circ} + 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2}$
$\sin 330^{\circ} = \sin(360^{\circ} - 30^{\circ}) = -\sin 30^{\circ} = -\frac{1}{2}$
Substituting these values into the expression:
$\left(\frac{\sqrt{3}}{2}\right) \left(-\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right)$
$= -\frac{3}{4} - \frac{1}{4}$
$= -\frac{4}{4} = -1$

(A) The equations of the lines are $x-3y+2=0$ $(i)$ and $2x+5y-7=0$ $(ii)$.
Solving equations $(i)$ and $(ii)$ by multiplying $(i)$ by $2$: $2x-6y+4=0$ $(iii)$.
Subtracting $(iii)$ from $(ii)$: $(2x+5y-7) - (2x-6y+4) = 0$ $\Rightarrow 11y - 11 = 0$ $\Rightarrow y=1$.
Substituting $y=1$ in $(i)$: $x-3(1)+2=0$ $\Rightarrow x-1=0$ $\Rightarrow x=1$.
The point of intersection is $(1, 1)$.
The equation of a line perpendicular to $3x+2y+5=0$ is of the form $2x-3y+\lambda=0$.
Since this line passes through $(1, 1)$,we have $2(1)-3(1)+\lambda=0$ $\Rightarrow 2-3+\lambda=0$ $\Rightarrow \lambda=1$.
Thus,the required equation is $2x-3y+1=0$.

(D) The given equation is $x^2-y^2-x+3y-2=0$.
We can factorize this expression by grouping terms:
$x^2 - (y^2 - 3y + 2) = 0$
$x^2 - (y-1)(y-2) = 0$
This does not immediately factor into linear factors. Let us rewrite the equation as $x^2 - x - (y^2 - 3y + 2) = 0$.
Using the quadratic formula for $x$ in terms of $y$:
$x = \frac{1 \pm \sqrt{1 + 4(y^2 - 3y + 2)}}{2} = \frac{1 \pm \sqrt{4y^2 - 12y + 9}}{2} = \frac{1 \pm \sqrt{(2y-3)^2}}{2} = \frac{1 \pm (2y-3)}{2}$.
Case $1$: $x = \frac{1 + 2y - 3}{2} = \frac{2y-2}{2} = y-1 \implies x-y+1=0$.
Case $2$: $x = \frac{1 - 2y + 3}{2} = \frac{4-2y}{2} = 2-y \implies x+y-2=0$.
Thus,the lines are $x-y+1=0$ and $x+y-2=0$.

(C) The given pair of straight lines is $12x^2 - 20xy + 7y^2 = 0$.
Factoring the quadratic expression: $12x^2 - 6xy - 14xy + 7y^2 = 0 \implies 6x(2x - y) - 7y(2x - y) = 0 \implies (6x - 7y)(2x - y) = 0$.
The equations of the sides of the triangle are:
$L_1: 6x - 7y = 0$
$L_2: 2x - y = 0$
$L_3: 2x - 3y + 4 = 0$
Solving for the vertices:
Intersection of $L_1$ and $L_2$: $(0, 0)$.
Intersection of $L_1$ and $L_3$: $6x - 7y = 0 \implies x = \frac{7y}{6}$. Substituting into $L_3$: $2(\frac{7y}{6}) - 3y + 4 = 0 \implies \frac{7y}{3} - 3y = -4 \implies -\frac{2y}{3} = -4 \implies y = 6, x = 7$. Vertex is $(7, 6)$.
Intersection of $L_2$ and $L_3$: $2x - y = 0 \implies y = 2x$. Substituting into $L_3$: $2x - 3(2x) + 4 = 0 \implies -4x = -4 \implies x = 1, y = 2$. Vertex is $(1, 2)$.
The vertices are $(0, 0), (7, 6), (1, 2)$.
The centroid is $\left(\frac{0+7+1}{3}, \frac{0+6+2}{3}\right) = \left(\frac{8}{3}, \frac{8}{3}\right)$.

(A) The chemical formula of carnallite is $KCl \cdot MgCl_2 \cdot 6H_2O$.
It is a double salt consisting of potassium chloride and magnesium chloride.
Therefore,the metal ions present in carnallite are $K^+$ and $Mg^{2+}$.

(B) For a particle of mass $m$ inside the Earth at a distance $r$ from the center,the gravitational force is given by $F = -\frac{GMmr}{R^3}$,where $M$ is the mass of the Earth and $R$ is its radius.
This force is proportional to the displacement $r$ and directed towards the center,which is the condition for simple harmonic motion $(SHM)$. Thus,Assertion $(A)$ is true.
The Reason $(R)$ states that the gravitational force is inversely proportional to the square of the distance $(F \propto 1/r^2)$. This is Newton's Law of Gravitation,which is true for particles outside the Earth or at the surface. However,inside the Earth,the effective force depends on the mass enclosed within the sphere of radius $r$,leading to a linear dependence on $r$.
While $(R)$ is a true statement about universal gravitation,it does not explain why the motion inside the Earth is simple harmonic (which depends on the linear force law inside the Earth). Therefore,$(R)$ is not the correct explanation of $(A)$.

(D) Friedel-Crafts Acylation: This reaction involves the treatment of benzene with an acylating agent such as acetyl chloride $(CH_3COCl)$ or acetic anhydride in the presence of a Lewis acid catalyst like anhydrous aluminum chloride $(AlCl_3)$.
The reaction is as follows:
$C_6H_6 + CH_3COCl \xrightarrow{anhydrous \ AlCl_3} C_6H_5COCH_3 + HCl$
Thus,benzene reacts with acetyl chloride $(CH_3COCl)$ to form acetophenone.

(C) $30\%$ solution of hydrogen peroxide can be obtained by the electrolysis of $50\%$ sulphuric acid followed by vacuum distillation.
The first product of electrolysis is perdisulphuric acid $(H_2S_2O_8)$,which reacts with water during distillation to form $H_2O_2$.
$2H_2SO_4 \longrightarrow 2H^+ + 2HSO_4^-$
$2HSO_4^- \longrightarrow H_2S_2O_8 + 2e^-$ (At anode)
$H_2S_2O_8 + 2H_2O \longrightarrow 2H_2SO_4 + H_2O_2$
Here,$X$ is $H_2SO_4$ and $Y$ is $H_2S_2O_8$.
$H_2SO_4$ has $0$ peroxy bonds,while $H_2S_2O_8$ (Marshall's acid) has $1$ peroxy bond $(HO_3S-O-O-SO_3H)$.

(D) Let the radius of the sector be $r$ and the sectorial angle be $\theta$ (in radians). The arc length is $l = r\theta$.
Perimeter $P = 2r + r\theta = r(2 + \theta)$.
Thus,$r = \frac{P}{2 + \theta}$.
The area of the sector is $A = \frac{1}{2} r^2 \theta$.
Substituting $r$,we get $A = \frac{1}{2} \left( \frac{P}{2 + \theta} \right)^2 \theta = \frac{P^2}{2} \cdot \frac{\theta}{(2 + \theta)^2}$.
To maximize $A$,we differentiate with respect to $\theta$ and set it to $0$:
$\frac{dA}{d\theta} = \frac{P^2}{2} \left[ \frac{(2 + \theta)^2 - \theta \cdot 2(2 + \theta)}{(2 + \theta)^4} \right] = 0$.
$(2 + \theta)^2 - 2\theta(2 + \theta) = 0$.
Since $2 + \theta \neq 0$,we have $2 + \theta - 2\theta = 0$,which gives $\theta = 2$.
Therefore,the area is maximum when the sectorial angle is $2^c$.

(B) Given $f(x) = e^x \sin x$.
First derivative: $f'(x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.
Second derivative: $f''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.
Third derivative: $f'''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.
Fourth derivative: $f^{(4)}(x) = 2e^x(\cos x - \sin x) + 2e^x(-\sin x - \cos x) = -4e^x \sin x$.
Fifth derivative: $f^{(5)}(x) = -4e^x \sin x - 4e^x \cos x = -4e^x(\sin x + \cos x)$.
Sixth derivative: $f^{(6)}(x) = -4e^x(\sin x + \cos x) - 4e^x(\cos x - \sin x) = -4e^x \sin x - 4e^x \cos x - 4e^x \cos x + 4e^x \sin x = -8e^x \cos x$.

(B) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$
Multiplying both sides by $(x+1)(2x^2+3)$,we get: $3x+2 = A(2x^2+3) + (Bx+C)(x+1)$
Setting $x = -1$: $3(-1)+2 = A(2(-1)^2+3) + 0 \Rightarrow -1 = A(5) \Rightarrow A = -\frac{1}{5}$
Expanding the right side: $3x+2 = 2Ax^2 + 3A + Bx^2 + Bx + Cx + C = (2A+B)x^2 + (B+C)x + (3A+C)$
Comparing coefficients:
For $x^2$: $2A+B = 0 \Rightarrow B = -2A = -2(-\frac{1}{5}) = \frac{2}{5}$
For $x$: $B+C = 3 \Rightarrow C = 3 - B = 3 - \frac{2}{5} = \frac{13}{5}$
Calculating $A+C-B$: $-\frac{1}{5} + \frac{13}{5} - \frac{2}{5} = \frac{13-2-1}{5} = \frac{10}{5} = 2$

(D) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,we must have $\lim_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right)$.
Given $f\left(\frac{\pi}{4}\right) = a$.
Now,$\lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4x}$.
This is a $\frac{0}{0}$ form,so we apply $L$'Hospital's rule:
$\lim_{x \to \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\sqrt{2} \sin x)}{\frac{d}{dx}(\pi-4x)} = \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$.
Substituting $x = \frac{\pi}{4}$:
$= \frac{-\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{-4} = \frac{-1}{-4} = \frac{1}{4}$.
Since the function is continuous,$a = \frac{1}{4}$.

(D) $1$. When the object slides down with uniform velocity,the net force is zero. Thus,the component of gravity $mg \sin \theta$ is balanced by the kinetic friction $f_k = \mu_k N = \mu_k mg \cos \theta$. Therefore,$mg \sin \theta = \mu_k mg \cos \theta$,which implies $\mu_k = \tan \theta$.
$2$. When the object is pushed up with initial velocity $u$,it moves against both gravity and friction. The retardation is $a = g \sin \theta + \mu_k g \cos \theta = g \sin \theta + (\tan \theta) g \cos \theta = 2g \sin \theta$.
$3$. After stopping at the top,the object slides down. The acceleration while sliding down is $a' = g \sin \theta - \mu_k g \cos \theta = g \sin \theta - (\tan \theta) g \cos \theta = 0$. Wait,this implies the object would remain at rest if the static friction equals kinetic friction. However,since it was already sliding,the net force down the plane is $mg \sin \theta - f_k = 0$. Thus,the object will remain at rest at the point where it stops.

(D) The equations of the curves are:
$xy=2$ $\ldots$ $(i)$
$x^2+4y=0$ $\ldots$ $(ii)$
From $(i)$,$y = \frac{2}{x}$. Substituting this into $(ii)$:
$x^2 + 4(\frac{2}{x}) = 0 \Rightarrow x^2 + \frac{8}{x} = 0 \Rightarrow x^3 + 8 = 0 \Rightarrow x^3 = -8 \Rightarrow x = -2$.
For $x = -2$,$y = \frac{2}{-2} = -1$. So,the point of intersection is $(-2, -1)$.
Differentiating $(i)$ with respect to $x$:
$x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
At $(-2, -1)$,$m_1 = -(\frac{-1}{-2}) = -\frac{1}{2}$.
Differentiating $(ii)$ with respect to $x$:
$2x + 4 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{2x}{4} = -\frac{x}{2}$.
At $(-2, -1)$,$m_2 = -(\frac{-2}{2}) = 1$.
The angle $\theta$ between the curves is given by:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-\frac{1}{2} - 1}{1 + (-\frac{1}{2})(1)} \right| = \left| \frac{-\frac{3}{2}}{1 - \frac{1}{2}} \right| = \left| \frac{-\frac{3}{2}}{\frac{1}{2}} \right| = |-3| = 3$.
Therefore,$\tan \theta = 3$.

(B) Given the function $f(x) = \frac{x}{3} + \frac{3}{x}$.
First,we find the derivative $f'(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(\frac{x}{3} + \frac{3}{x}) = \frac{1}{3} - \frac{3}{x^2}$.
To determine the nature of the function in the interval $(-3, 3)$,we analyze the sign of $f'(x)$:
$f'(x) = \frac{x^2 - 9}{3x^2}$.
Since $x^2 < 9$ for all $x \in (-3, 3)$ and $x \neq 0$,the numerator $x^2 - 9$ is always negative.
Since $3x^2$ is always positive for $x \neq 0$,the derivative $f'(x) < 0$ for all $x \in (-3, 3) \setminus \{0\}$.
Therefore,the function $f(x)$ is decreasing in the interval $(-3, 3)$.

(A) Let $I = \int \frac{dx}{x^2+2x+2}$.
We can rewrite the denominator by completing the square:
$x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x+1)^2 + 1$.
Substituting this into the integral,we get:
$I = \int \frac{dx}{(x+1)^2 + 1}$.
Using the standard integral formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + c$,where $u = x+1$ and $a = 1$:
$I = \tan^{-1}(x+1) + c$.
Given that $I = f(x) + c$,we conclude that $f(x) = \tan^{-1}(x+1)$.

(C) Let $I = \int \sqrt{\frac{x}{a^3-x^3}} \, dx$.
To solve this,we rewrite the integrand:
$I = \int \sqrt{\frac{x}{a^3(1 - (x/a)^{3})}} \, dx = \int \frac{\sqrt{x}}{\sqrt{a^3} \sqrt{1 - (x^{3/2}/a^{3/2})^2}} \, dx$.
Let $u = \left(\frac{x}{a}\right)^{3/2} = \sqrt{\frac{x^3}{a^3}}$.
Then $du = \frac{3}{2} \sqrt{\frac{x}{a^3}} \, dx$,which implies $\sqrt{\frac{x}{a^3}} \, dx = \frac{2}{3} \, du$.
Substituting these into the integral:
$I = \int \frac{1}{\sqrt{1 - u^2}} \cdot \frac{2}{3} \, du = \frac{2}{3} \sin^{-1}(u) + c$.
Substituting back $u = \sqrt{\frac{x^3}{a^3}}$,we get $I = \frac{2}{3} \sin^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right) + c$.
Comparing this with $g(x) + c$,we find $g(x) = \frac{2}{3} \sin^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)$.

(C) Let $I = \int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$.
Since $\cosh x = \frac{e^x + e^{-x}}{2}$,we substitute this into the integral:
$I = \int_{-1}^1 \frac{e^x + e^{-x}}{2(1 + e^{2x})} d x$.
We can rewrite the numerator as $\frac{e^x(1 + e^{-2x})}{2(1 + e^{2x})} = \frac{e^x(1 + \frac{1}{e^{2x}})}{2(1 + e^{2x})} = \frac{e^x(\frac{e^{2x} + 1}{e^{2x}})}{2(1 + e^{2x})} = \frac{e^x}{2e^{2x}} = \frac{1}{2e^x} = \frac{1}{2}e^{-x}$.
Thus,$I = \int_{-1}^1 \frac{1}{2}e^{-x} d x$.
$I = \frac{1}{2} \left[ -e^{-x} \right]_{-1}^1$.
$I = -\frac{1}{2} (e^{-1} - e^1) = \frac{1}{2} (e^1 - e^{-1}) = \frac{e - \frac{1}{e}}{2} = \frac{e^2 - 1}{2e}$.

(C) Let $I = \int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$.
We know that $\tan x = \frac{\sin x}{\cos x}$,so $I = \int_0^{\pi / 2} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...$(i)$
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I = \int_0^{\pi / 2} \frac{\cos ^3(\pi/2 - x)}{\sin ^3(\pi/2 - x) + \cos ^3(\pi/2 - x)} d x$
$I = \int_0^{\pi / 2} \frac{\sin ^3 x}{\cos ^3 x + \sin ^3 x} d x$ ...(ii)
Adding equations $(i)$ and (ii):
$2I = \int_0^{\pi / 2} \frac{\cos ^3 x + \sin ^3 x}{\sin ^3 x + \cos ^3 x} d x$
$2I = \int_0^{\pi / 2} 1 d x = [x]_0^{\pi / 2} = \frac{\pi}{2}$
Therefore,$I = \frac{\pi}{4}$.

(D) Given the interval $[0, 6]$ is divided into $n = 6$ equal parts.
The width of each sub-interval is $h = \frac{b-a}{n} = \frac{6-0}{6} = 1$.
The values of $f(x) = x^3$ at $x = 0, 1, 2, 3, 4, 5, 6$ are:
$y_0 = f(0) = 0^3 = 0$
$y_1 = f(1) = 1^3 = 1$
$y_2 = f(2) = 2^3 = 8$
$y_3 = f(3) = 3^3 = 27$
$y_4 = f(4) = 4^3 = 64$
$y_5 = f(5) = 5^3 = 125$
$y_6 = f(6) = 6^3 = 216$
By the trapezoidal rule:
$\int_0^6 x^3 dx \approx \frac{h}{2} \{y_0 + y_6 + 2(y_1 + y_2 + y_3 + y_4 + y_5)\}$
$= \frac{1}{2} \{0 + 216 + 2(1 + 8 + 27 + 64 + 125)\}$
$= \frac{1}{2} \{216 + 2(225)\}$
$= \frac{1}{2} \{216 + 450\}$
$= \frac{666}{2} = 333$.

(A) Given equation is $x^y = y^x$.
Taking natural logarithm on both sides,we get $y \log x = x \log y$.
Differentiating both sides with respect to $x$ using the product rule:
$y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} = x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \log y$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x}$.
$\frac{dy}{dx} \left( \frac{y \log x - x}{y} \right) = \frac{x \log y - y}{x}$.
Multiplying both sides by $x$ and rearranging:
$x \left( \frac{y \log x - x}{y} \right) \frac{dy}{dx} = x \log y - y$.
Multiplying by $-1$ to match the required form:
$x \left( \frac{x - y \log x}{y} \right) \frac{dy}{dx} = y - x \log y$.
Therefore,$x(x - y \log x) \frac{dy}{dx} = y(y - x \log y)$.