TS EAMCET 2011 Chemistry Question Paper with Answer and Solution

(C) Given,$I_n = \int_0^{\pi / 4} \tan^n \theta \, d\theta$.
We need to find $I_{n-1} + I_{n+1}$.
$I_{n-1} + I_{n+1} = \int_0^{\pi / 4} \tan^{n-1} \theta \, d\theta + \int_0^{\pi / 4} \tan^{n+1} \theta \, d\theta$.
Factor out $\tan^{n-1} \theta$:
$I_{n-1} + I_{n+1} = \int_0^{\pi / 4} \tan^{n-1} \theta (1 + \tan^2 \theta) \, d\theta$.
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have:
$I_{n-1} + I_{n+1} = \int_0^{\pi / 4} \tan^{n-1} \theta \sec^2 \theta \, d\theta$.
Let $u = \tan \theta$,then $du = \sec^2 \theta \, d\theta$.
When $\theta = 0$,$u = 0$. When $\theta = \pi / 4$,$u = 1$.
Substituting these into the integral:
$I_{n-1} + I_{n+1} = \int_0^1 u^{n-1} \, du = \left[ \frac{u^n}{n} \right]_0^1 = \frac{1}{n} - 0 = \frac{1}{n}$.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(y/x)}{\phi'(y/x)}$.
Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting this into the equation gives: $v + x \frac{dv}{dx} = v + \frac{\phi(v)}{\phi'(v)}$.
Subtracting $v$ from both sides,we get: $x \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)}$.
Separating the variables,we have: $\frac{\phi'(v)}{\phi(v)} dv = \frac{dx}{x}$.
Integrating both sides: $\int \frac{\phi'(v)}{\phi(v)} dv = \int \frac{dx}{x}$.
This yields: $\ln|\phi(v)| = \ln|x| + \ln|k|$.
Using logarithmic properties,$\phi(v) = kx$.
Substituting $v = \frac{y}{x}$ back,we get: $\phi\left(\frac{y}{x}\right) = kx$.

(B) Let the initial vector be $\vec{A} = 4\hat{i} + 10\hat{j}$.
The magnitude of the vector is $|\vec{A}| = \sqrt{(4)^2 + (10)^2} = \sqrt{16 + 100} = \sqrt{116} ~m$.
When the vector is rotated,its magnitude remains constant.
Let the new vector be $\vec{A}' = 8\hat{i} + n\hat{j}$,where the $x$-component is doubled $(4 \times 2 = 8)$.
Since the magnitude is conserved,$|\vec{A}'| = |\vec{A}|$.
Therefore,$\sqrt{(8)^2 + n^2} = \sqrt{116}$.
Squaring both sides,we get $64 + n^2 = 116$.
$n^2 = 116 - 64 = 52$.
$n = \sqrt{52} \approx 7.21 ~m$.
Thus,the new $y$-component is approximately $7.2 ~m$.

(D) The equation of a plane passing through the point $(x_1, y_1, z_1)$ with normal vector having direction ratios $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Given the point $(2, 3, -1)$,the equation is $a(x-2) + b(y-3) + c(z+1) = 0 \dots (i)$.
Since the plane is perpendicular to a line with direction ratios $(3, -4, 7)$,the normal vector of the plane is parallel to this line. Thus,we can take $a=3, b=-4, c=7$.
Substituting these into equation $(i)$,we get:
$3(x-2) - 4(y-3) + 7(z+1) = 0$
$3x - 6 - 4y + 12 + 7z + 7 = 0$
$3x - 4y + 7z + 13 = 0$.
The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A=3, B=-4, C=7, D=13$.
$d = \frac{|13|}{\sqrt{3^2 + (-4)^2 + 7^2}} = \frac{13}{\sqrt{9 + 16 + 49}} = \frac{13}{\sqrt{74}}$.

(B) Total number of ways to arrange $7$ white balls and $3$ black balls in a row is $\frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
To ensure no two black balls are adjacent,we place the $7$ white balls first,creating $8$ possible gaps (including ends) where the $3$ black balls can be placed.
The number of ways to choose $3$ gaps out of $8$ is $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
The required probability is $\frac{\binom{8}{3}}{\binom{10}{3}} = \frac{56}{120} = \frac{7}{15}$.

(D) Given,$p = 0.001$ and $n = 2000$.
Since $n$ is large and $p$ is very small,we use the Poisson distribution.
The mean $\lambda = np = 2000 \times 0.001 = 2$.
The probability mass function for the Poisson distribution is given by $P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$.
For $x = 3$,we have:
$P(X = 3) = \frac{2^3 e^{-2}}{3!} = \frac{8 \times e^{-2}}{6} = \frac{4}{3e^2}$.
Thus,the correct option is $D$.

(A) The mean $E(X)$ is calculated as:
$E(X) = \sum x_i p_i = 0 \times \frac{1}{10} + 1 \times \frac{2}{10} + 2 \times \frac{3}{10} + 3 \times \frac{4}{10}$
$E(X) = 0 + \frac{2}{10} + \frac{6}{10} + \frac{12}{10} = \frac{20}{10} = 2$
The variance $Var(X)$ is calculated as:
$Var(X) = E(X^2) - [E(X)]^2$
First,calculate $E(X^2) = \sum x_i^2 p_i$:
$E(X^2) = 0^2 \times \frac{1}{10} + 1^2 \times \frac{2}{10} + 2^2 \times \frac{3}{10} + 3^2 \times \frac{4}{10}$
$E(X^2) = 0 + \frac{2}{10} + \frac{12}{10} + \frac{36}{10} = \frac{50}{10} = 5$
Now,$Var(X) = 5 - (2)^2 = 5 - 4 = 1$
Thus,the variance is $1$.

(A) Let $V$ be the volume of the body and $\rho_L$ be the density of the liquid.
Since the body floats,the weight of the body equals the weight of the displaced liquid: $mg = V_{imm} \rho_L g$.
Thus,$V_{imm} \rho_L = \text{constant}$.
At $t_1 = 20^{\circ}C$,$V_{imm,1} = \frac{2}{3}V$,so $\frac{2}{3}V \rho_1 = m$.
At $t_2 = 100^{\circ}C$,$V_{imm,2} = \frac{3}{4}V$,so $\frac{3}{4}V \rho_2 = m$.
Equating the two: $\frac{2}{3} \rho_1 = \frac{3}{4} \rho_2$,which gives $\rho_2 = \frac{8}{9} \rho_1$.
Using the relation $\rho_2 = \frac{\rho_1}{1 + \gamma_R \Delta t}$,we have $1 + \gamma_R \Delta t = \frac{\rho_1}{\rho_2} = \frac{9}{8}$.
$\gamma_R \Delta t = \frac{9}{8} - 1 = \frac{1}{8}$.
Given $\Delta t = 100 - 20 = 80^{\circ}C$,so $\gamma_R = \frac{1}{8 \times 80} = \frac{1}{640} = 0.0015625 = 15.6 \times 10^{-4} \, ^{\circ}C^{-1}$.

(C) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4T}{r}$,where $T$ is the surface tension of the soap solution.
Since the pressure is inversely proportional to the radius $(P \propto \frac{1}{r})$,the smaller bubble has a higher internal pressure than the larger bubble.
When connected by a tube,air flows from the region of higher pressure to the region of lower pressure.
Therefore,air flows from the smaller bubble to the larger bubble.

(B) Boron reacts with concentrated $HNO_3$ to form orthoboric acid $(H_3BO_3)$ and nitrogen dioxide $(NO_2)$.
The balanced chemical equation is:
$B + 3HNO_3 \longrightarrow H_3BO_3 + 3NO_2$
Therefore,the statement that boron reacts with concentrated $HNO_3$ to give $N_2O$ is incorrect.

(B) The structure of $ClO_2$ is angular (bent).
In $ClO_2$,the central chlorine atom has one unpaired electron and is $sp^3$-hybridised.
Due to the presence of the unpaired electron and lone pairs,the bond angle is observed to be $118^{\circ}$,and the $Cl-O$ bond length is $1.47 \mathring{A}$.

(B) Phosphorous acid $(H_3PO_3)$ is a dibasic acid,meaning it provides $2$ moles of $H^+$ ions per mole of acid.
The normality of the $H_3PO_3$ solution is calculated as: $\text{Normality} = \text{Molarity} \times \text{Basicity} = 0.3 \ M \times 2 = 0.6 \ N$.
Using the neutralization principle $N_1 V_1 = N_2 V_2$:
For $NaOH$,$N_1 = 0.1 \ N$ and for $H_3PO_3$,$N_2 = 0.6 \ N$.
$0.1 \times V_1 = 0.6 \times 100$.
$V_1 = \frac{0.6 \times 100}{0.1} = 600 \ mL$.

(B) Let us analyze each reaction:
$1$. Reaction of fused $NaOH$ with $C$: $2NaOH + C \rightarrow Na_2CO_3 + H_2$. Hydrogen is liberated.
$2$. Reaction of $NaOH$ with sulphur: $6NaOH + 3S \rightarrow 2Na_2S + Na_2SO_3 + 3H_2O$. No hydrogen gas is liberated.
$3$. Heating concentrated $NaOH$ with $Si$: $Si + 2NaOH + H_2O \rightarrow Na_2SiO_3 + 2H_2$. Hydrogen is liberated.
$4$. Reaction of zinc with $NaOH$: $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$. Hydrogen is liberated.
Therefore,the correct option is $B$.

(D) The reaction for the heating of $NaHCO_3$ is: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$.
$Na_2CO_3$ does not decompose on heating.
Let the mass of $NaHCO_3$ be $x \ g$ and $Na_2CO_3$ be $y \ g$.
$x + y = 19$ (Equation $1$).
Moles of $CO_2$ produced $= \frac{1.12 \ L}{22.4 \ L/mol} = 0.05 \ mol$.
From the stoichiometry,$2 \ mol$ of $NaHCO_3$ produces $1 \ mol$ of $CO_2$.
So,moles of $NaHCO_3 = 2 \times 0.05 = 0.1 \ mol$.
Mass of $NaHCO_3 = 0.1 \ mol \times 84 \ g/mol = 8.4 \ g$.
Substituting in Equation $1$: $8.4 + y = 19 \Rightarrow y = 10.6 \ g$.

(C) Real gases deviate from ideal behaviour due to intermolecular forces and the finite volume of gas molecules.
At high temperature,the kinetic energy of gas molecules is high,making intermolecular forces negligible.
At low pressure,the volume occupied by the gas molecules becomes negligible compared to the total volume of the container.
Therefore,real gases approach ideal gas behaviour under conditions of high temperature and low pressure.

(A) According to Bohr's postulate,the angular momentum $(L)$ of an electron in a hydrogen atom is quantized and is given by the formula: $L = n \frac{h}{2 \pi}$,where $n$ is an integer $(n = 1, 2, 3, \dots)$.
This can be rewritten as $L = \frac{n}{2} \cdot \frac{h}{\pi}$.
For $n = 1$,$L = 0.5 \frac{h}{\pi}$.
For $n = 2$,$L = 1.0 \frac{h}{\pi}$.
For $n = 3$,$L = 1.5 \frac{h}{\pi}$.
For $n = 4$,$L = 2.0 \frac{h}{\pi}$.
Comparing these values with the given options,$1.25 \frac{h}{\pi}$ does not correspond to any integer value of $n$ (as $n = 2.5$ is not an integer). Therefore,it is not a permitted value.

(D) The relationship between frequency $(v)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $v = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^8 \ m/s$ and $\lambda = 600 \ nm = 600 \times 10^{-9} \ m$.
Substituting the values:
$v = \frac{3 \times 10^8 \ m/s}{600 \times 10^{-9} \ m} = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} \ Hz = 5.0 \times 10^{14} \ Hz$.

(D) The entropy change during a phase transition is given by the formula $\Delta S = \frac{q_{rev}}{T} = \frac{m \times L_f}{T}$.
Given:
Mass of ice $(m)$ $= 27.3 \ g$
Latent heat of fusion $(L_f)$ $= 330 \ J g^{-1}$
Temperature $(T)$ $= 0^{\circ} C = 273 \ K$
Substituting the values:
$\Delta S = \frac{27.3 \ g \times 330 \ J g^{-1}}{273 \ K}$
$\Delta S = \frac{9009 \ J}{273 \ K} = 33 \ J K^{-1}$.