TS EAMCET 2014 Chemistry Question Paper with Answer and Solution

(D) In Clark's process,temporary hardness of water is removed by adding a calculated amount of lime $(Ca(OH)_2)$. It reacts with soluble bicarbonates of calcium and magnesium to form insoluble carbonates and hydroxides which can be filtered out.
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$
$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2H_2O$

(D) Given,$f(x) = \frac{x}{1+x}$.
We are given $g(x) = f(f(x))$.
Substituting $f(x)$ into itself:
$g(x) = f\left(\frac{x}{1+x}\right) = \frac{\frac{x}{1+x}}{1 + \frac{x}{1+x}}$.
Multiplying the numerator and denominator by $(1+x)$:
$g(x) = \frac{x}{1+x+x} = \frac{x}{2x+1}$.
Now,we differentiate $g(x)$ with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:
$g^{\prime}(x) = \frac{(1)(2x+1) - (x)(2)}{(2x+1)^2}$.
$g^{\prime}(x) = \frac{2x+1 - 2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}$.

(C) The dissociation of $NaOH$ is: $NaOH \rightarrow Na^+ + OH^-$. Since $NaOH$ is a strong electrolyte,$[OH^-] = 1.0 \ M$.
Let the solubility of $Ni(OH)_2$ be $s \ M$. The dissociation is: $Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-$.
The concentration of $Ni^{2+}$ is $s$ and the total concentration of $OH^-$ is $(2s + 1.0) \ M$.
Given $K_{sp} = [Ni^{2+}][OH^-]^2 = 1.9 \times 10^{-15}$.
Substituting the values: $s(2s + 1.0)^2 = 1.9 \times 10^{-15}$.
Since $s$ is very small,we can neglect $2s$ compared to $1.0$.
Thus,$s(1.0)^2 = 1.9 \times 10^{-15}$.
$s = 1.9 \times 10^{-15} \ M$.

(D) For an ideal gas,the change in internal energy $\Delta U$ is given by the formula $\Delta U = n C_v \Delta T$.
Given that $n = 1$ mole,we have $\Delta U = C_v \Delta T$.
From the ideal gas equation $pV = nRT$,at constant pressure $p$,we have $p \Delta V = nR \Delta T$.
Since $n = 1$,$p \Delta V = R \Delta T$,which implies $\Delta T = \frac{p \Delta V}{R}$.
Substituting this into the internal energy formula: $\Delta U = C_v \left( \frac{p \Delta V}{R} \right)$.
We know that $C_v = \frac{R}{\gamma - 1}$.
Therefore,$\Delta U = \left( \frac{R}{\gamma - 1} \right) \left( \frac{p \Delta V}{R} \right) = \frac{p \Delta V}{\gamma - 1}$.
The change in volume is $\Delta V = 2V - V = V$.
Thus,$\Delta U = \frac{p V}{\gamma - 1}$.

(B) Let $T$ be the tension in the string and $F$ be the horizontal force applied to hold the mass at an angle $\theta = 60^{\circ}$ with the vertical.
At equilibrium,the forces acting on the mass are balanced:
$1$. Vertical direction: $T \cos \theta = M g$ (Equation $i$)
$2$. Horizontal direction: $T \sin \theta = F$ (Equation $ii$)
Dividing Equation $ii$ by Equation $i$:
$\frac{F}{M g} = \frac{T \sin \theta}{T \cos \theta} = \tan \theta$
$F = M g \tan \theta$
Given $\theta = 60^{\circ}$,we have:
$F = M g \tan 60^{\circ}$
Since $\tan 60^{\circ} = \sqrt{3}$,the required force is:
$F = \sqrt{3} M g$

(D) Given,the error in diameter $\Delta d = \pm 0.04 \text{ cm}$.
Since the radius $r = \frac{d}{2}$,the error in radius is $\Delta r = \frac{\Delta d}{2} = \pm \frac{0.04}{2} = \pm 0.02 \text{ cm}$.
The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$,we get $\frac{dV}{dr} = 4 \pi r^2$.
The approximate error in volume is $\Delta V = \frac{dV}{dr} \Delta r = 4 \pi r^2 \Delta r$.
The percentage error in volume is given by $\frac{\Delta V}{V} \times 100$.
Substituting the values,we get $\frac{4 \pi r^2 \Delta r}{\frac{4}{3} \pi r^3} \times 100 = \frac{3 \Delta r}{r} \times 100$.
Substituting $r = 10 \text{ cm}$ and $\Delta r = \pm 0.02 \text{ cm}$,we get $\frac{3 \times (\pm 0.02)}{10} \times 100 = \frac{\pm 0.06}{10} \times 100 = \pm 0.6 \%$.
Thus,the approximate percentage error is $\pm 0.6 \%$.

(D) Given the function $f(x) = a x^3 + b x^2 + c x + d$.
To find the extreme values,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 3 a x^2 + 2 b x + c$.
For the function to have no extreme values,the derivative $f'(x)$ must not change its sign,which means the quadratic equation $3 a x^2 + 2 b x + c = 0$ must have no real roots or a repeated root such that the sign does not change.
This occurs when the discriminant $D$ of the quadratic equation $3 a x^2 + 2 b x + c = 0$ is less than or equal to $0$.
The discriminant $D = (2 b)^2 - 4(3 a)(c) = 4 b^2 - 12 a c$.
Setting $D < 0$ for no real roots:
$4 b^2 - 12 a c < 0$
$b^2 - 3 a c < 0$
$b^2 < 3 a c$.
Thus,the condition for no extreme value is $b^2 < 3 a c$.

(B) Given,$I = \int \frac{d x}{\sqrt{\sin ^3 x \cos x}}$.
We can rewrite the integrand as:
$I = \int \frac{d x}{\sqrt{\sin ^4 x \cdot \frac{\cos x}{\sin x}}} = \int \frac{d x}{\sin ^2 x \sqrt{\cot x}}$.
Since $\frac{1}{\sin ^2 x} = \operatorname{cosec}^2 x$,we have:
$I = \int \frac{\operatorname{cosec}^2 x}{\sqrt{\cot x}} d x$.
Let $t = \cot x$. Then $dt = -\operatorname{cosec}^2 x d x$,which implies $\operatorname{cosec}^2 x d x = -dt$.
Substituting these into the integral:
$I = \int \frac{-dt}{\sqrt{t}} = -\int t^{-1/2} dt$.
Integrating with respect to $t$:
$I = -\frac{t^{1/2}}{1/2} + c = -2\sqrt{t} + c$.
Substituting $t = \cot x$ back:
$I = -2\sqrt{\cot x} + c = -\frac{2}{\sqrt{\tan x}} + c$.
Comparing this with $g(x) + c$,we get $g(x) = -\frac{2}{\sqrt{\tan x}}$.

(D) Given differential equation is $x \frac{dy}{dx} = y + x e^{y/x}$.
Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} + e^{y/x}$.
This is a homogeneous differential equation.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting this into the equation: $v + x \frac{dv}{dx} = v + e^v$.
This simplifies to $x \frac{dv}{dx} = e^v$.
Separating variables: $e^{-v} dv = \frac{1}{x} dx$.
Integrating both sides: $\int e^{-v} dv = \int \frac{1}{x} dx$,which gives $-e^{-v} = \log x + c$.
Substituting $v = y/x$: $-e^{-y/x} = \log x + c$.
Using the condition $y(1) = 0$: $-e^{-0/1} = \log 1 + c \Rightarrow -1 = 0 + c \Rightarrow c = -1$.
Thus,$-e^{-y/x} = \log x - 1$,which rearranges to $e^{-y/x} + \log x = 1$.

(A) Let the coordinates of point $B$ be $(x, y, z)$.
Using the section formula for internal division,the point $P$ dividing the line segment $AB$ in the ratio $m:n = 1:3$ is given by:
$P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$
Substituting the given values:
$(-2, 3, 5) = \left( \frac{1 \times x + 3 \times 1}{1+3}, \frac{1 \times y + 3 \times 3}{1+3}, \frac{1 \times z + 3 \times 4}{1+3} \right)$
$(-2, 3, 5) = \left( \frac{x+3}{4}, \frac{y+9}{4}, \frac{z+12}{4} \right)$
Equating the coordinates:
$1) \frac{x+3}{4} = -2 \Rightarrow x+3 = -8 \Rightarrow x = -11$
$2) \frac{y+9}{4} = 3 \Rightarrow y+9 = 12 \Rightarrow y = 3$
$3) \frac{z+12}{4} = 5 \Rightarrow z+12 = 20 \Rightarrow z = 8$
Thus,the coordinates of $B$ are $(-11, 3, 8)$.

(C) Given,mean of binomial variable,$np = 8$ and variance of binomial variable,$npq = 4$.
Since $npq = 4$ and $np = 8$,we have $8q = 4$,which implies $q = \frac{1}{2}$.
Then $p = 1 - q = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 8$,we get $n(\frac{1}{2}) = 8$,so $n = 16$.
We need to find $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$.
Using the binomial probability formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:
$P(X = 0) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} = 1 \times \frac{1}{2^{16}} = \frac{1}{2^{16}}$.
$P(X = 1) = {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} = 16 \times \frac{1}{2^{16}} = \frac{16}{2^{16}}$.
$P(X = 2) = {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14} = \frac{16 \times 15}{2} \times \frac{1}{2^{16}} = 120 \times \frac{1}{2^{16}} = \frac{120}{2^{16}}$.
Therefore,$P(X < 3) = \frac{1 + 16 + 120}{2^{16}} = \frac{137}{2^{16}}$.

(A) The elastic potential energy per unit volume $(u)$ of a stretched wire is given by the formula:
$u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} \times \frac{F}{A} \times \frac{F}{AY} = \frac{F^2}{2AY}$
Where $F$ is the force,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since $A = \pi r^2$,we have $u \propto \frac{1}{A^2} \propto \frac{1}{r^4}$ (as $F$ and $Y$ are constant).
Given the ratio of diameters $d_1:d_2 = 1:2$,the ratio of radii $r_1:r_2 = 1:2$.
Therefore,the ratio of elastic potential energy per unit volume is:
$\frac{u_1}{u_2} = \left(\frac{r_2}{r_1}\right)^4 = \left(\frac{2}{1}\right)^4 = \frac{16}{1}$.
Thus,the ratio is $16:1$.

(D) ,$Si$,and $Ge$ do not react with water or steam.
$Sn$ reacts with steam at high temperatures to form tin$(IV)$ oxide and hydrogen gas.
The chemical equation is:
$Sn_{(s)} + 2H_2O_{(g)} \xrightarrow{\Delta} SnO_{2(s)} + 2H_{2(g)}$

(B) Moles of $KI = 0.1 \times 0.250 = 0.025 \ mol$.
Moles of $KMnO_4 = 0.02 \times 0.250 = 0.005 \ mol$.
The balanced chemical equation is: $2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^-$.
According to the stoichiometry,$2 \ moles$ of $KMnO_4$ react with $6 \ moles$ of $KI$ to produce $3 \ moles$ of $I_2$.
Here,$KMnO_4$ is the limiting reagent because $0.005 \ mol$ of $KMnO_4$ requires $0.015 \ mol$ of $KI$ (which is less than the $0.025 \ mol$ available).
Moles of $I_2$ formed $= \frac{3}{2} \times \text{moles of } KMnO_4 = \frac{3}{2} \times 0.005 = 0.0075 \ mol$.

(D) $KO_2$ is a superoxide compound consisting of $K^{+}$ and $O_2^{-}$ ions.
The $K^{+}$ ion has a noble gas configuration and is diamagnetic.
The superoxide ion $O_2^{-}$ has $17$ electrons.
According to molecular orbital theory,the electronic configuration of $O_2^{-}$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$.
Due to the presence of one unpaired electron in the $\pi^* 2p$ orbital,$O_2^{-}$ is paramagnetic.

(A) Given that the oxide contains $40 \%$ oxygen,the metal content is $100 \% - 40 \% = 60 \%$.
Since $40 \ g$ of oxygen combines with $60 \ g$ of metal,
$8 \ g$ of oxygen (equivalent mass of oxygen) will combine with:
$\text{Equivalent weight of metal} = \frac{60 \times 8}{40} = 12 \ g$.
Atomic weight is calculated as:
$\text{Atomic weight} = \text{Equivalent weight} \times \text{Valency}$.
$\text{Atomic weight} = 12 \times 2 = 24$.

(C) According to Graham's law of diffusion:
$\frac{r_X}{r_Y} = \frac{1}{5}$ $(i)$
$\frac{r_Y}{r_Z} = \frac{1}{6}$ $(ii)$
Multiplying equation $(i)$ and $(ii)$:
$\frac{r_X}{r_Y} \times \frac{r_Y}{r_Z} = \frac{1}{5} \times \frac{1}{6}$
$\frac{r_X}{r_Z} = \frac{1}{30}$
Therefore,the ratio of rates of diffusion of $Z$ and $X$ is:
$\frac{r_Z}{r_X} = \frac{30}{1}$
Thus,$r_Z : r_X = 30:1$.

(A) For $4d$ orbital,the principal quantum number $n = 4$ and the azimuthal quantum number $l = 2$.
Number of angular nodes $= l = 2$.
Number of radial nodes $= n - l - 1 = 4 - 2 - 1 = 1$.
Therefore,the number of angular and radial nodes are $2$ and $1$ respectively.

(D) The order of increasing energy is determined by the $(n+l)$ rule.
If two orbitals have the same value of $(n+l)$,the orbital with the lower value of $n$ has lower energy.
$(i)$ For $n=4, l=1$,$(n+l) = 4+1 = 5$.
(ii) For $n=4, l=0$,$(n+l) = 4+0 = 4$.
(iii) For $n=3, l=2$,$(n+l) = 3+2 = 5$.
(iv) For $n=3, l=1$,$(n+l) = 3+1 = 4$.
Comparing the values: (iv) and (ii) have $(n+l) = 4$. Since (iv) has lower $n$,$(iv) < (ii)$.
$(i)$ and (iii) have $(n+l) = 5$. Since (iii) has lower $n$,$(iii) < (i)$.
Thus,the increasing order of energy is $(iv) < (ii) < (iii) < (i)$.

(B) For a reaction,the Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T \Delta S$.
Given that $\Delta G = 0$,we have $0 = \Delta H - T \Delta S$,which implies $T = \frac{\Delta H}{\Delta S}$.
Substituting the given values: $\Delta H = -20.5 \ kJ \ mol^{-1} = -20500 \ J \ mol^{-1}$ and $\Delta S = -50.0 \ J \ K^{-1} \ mol^{-1}$.
$T = \frac{-20500 \ J \ mol^{-1}}{-50.0 \ J \ K^{-1} \ mol^{-1}} = 410 \ K$.