TS EAMCET 2002 Chemistry Question Paper with Answer and Solution

(B) We need to evaluate the scalar triple product: $(a+b) \cdot ((b+c) \times (a+b+c))$.
First,expand the cross product term: $(b+c) \times (a+b+c) = (b \times a) + (b \times b) + (b \times c) + (c \times a) + (c \times b) + (c \times c)$.
Since the cross product of any vector with itself is zero ($b \times b = 0$ and $c \times c = 0$),and using the property $c \times b = -(b \times c)$,we get:
$(b+c) \times (a+b+c) = (b \times a) + (b \times c) + (c \times a) - (b \times c) = (b \times a) + (c \times a)$.
Now,take the dot product with $(a+b)$:
$(a+b) \cdot ((b \times a) + (c \times a)) = a \cdot (b \times a) + a \cdot (c \times a) + b \cdot (b \times a) + b \cdot (c \times a)$.
Using the properties of scalar triple products,$a \cdot (b \times a) = 0$,$a \cdot (c \times a) = 0$,and $b \cdot (b \times a) = 0$.
This leaves us with $b \cdot (c \times a) = [b c a]$.
Since $[b c a] = [a b c]$,the final result is $[a b c]$.

(A) Let $a = x\hat{i} + y\hat{j} + z\hat{k}$.
Given $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$.
From $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j})$,we have $a \cdot \hat{i} = a \cdot \hat{i} + a \cdot \hat{j}$,which implies $a \cdot \hat{j} = 0$. Thus,$y = 0$.
From $a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$,we have $a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j}) + a \cdot \hat{k}$,which implies $a \cdot \hat{k} = 0$. Thus,$z = 0$.
Since $a \cdot \hat{i} = x$,and the problem does not specify the magnitude,if we assume $a$ is a unit vector or simply look for the component form,$a = x\hat{i}$. Given the options,$a = \hat{i}$ satisfies the conditions.

(B) Let the vector $\vec{v} = \vec{PQ} = (0-0, 0-1, 1-0) = (0, -1, 1)$.
The normal vector to the plane $x+y+z=3$ is $\vec{n} = (1, 1, 1)$.
The length of the projection of a vector $\vec{v}$ on a plane with normal $\vec{n}$ is given by $|\vec{v}| \sin \theta$,where $\theta$ is the angle between the vector and the normal.
First,calculate the magnitude of $\vec{v}$: $|\vec{v}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
Next,calculate $\cos \theta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|} = \frac{|(0)(1) + (-1)(1) + (1)(1)|}{\sqrt{2} \cdot \sqrt{1^2+1^2+1^2}} = \frac{|0-1+1|}{\sqrt{2} \cdot \sqrt{3}} = 0$.
Since $\cos \theta = 0$,the vector $\vec{v}$ is parallel to the plane,so $\sin \theta = 1$.
The length of the projection is $|\vec{v}| \sin \theta = \sqrt{2} \cdot 1 = \sqrt{2}$.

(A) The given equation of the plane is $7x + 11y + 13z = 3003$.
Dividing the entire equation by $3003$,we get the intercept form of the plane:
$\frac{7x}{3003} + \frac{11y}{3003} + \frac{13z}{3003} = 1$
$\Rightarrow \frac{x}{429} + \frac{y}{273} + \frac{z}{231} = 1$.
The plane meets the coordinate axes at points $A, B,$ and $C$.
Setting $y=0, z=0$ gives $x=429$,so $A = (429, 0, 0)$.
Setting $x=0, z=0$ gives $y=273$,so $B = (0, 273, 0)$.
Setting $x=0, y=0$ gives $z=231$,so $C = (0, 0, 231)$.
The centroid of $\triangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Centroid $= \left(\frac{429+0+0}{3}, \frac{0+273+0}{3}, \frac{0+0+231}{3}\right)$
$= (143, 91, 77)$.

(C) First,we find the prime factors of $2001$.
$2001 = 3 \times 23 \times 29$.
Given that $a, b, c$ are the factors such that $a < b < c$,we have $a = 3$,$b = 23$,and $c = 29$.
The direction ratios of the normal to the plane are given as $b, c, a$,which are $23, 29, 3$.
The equation of the plane is $bx + cy + az + d = 0$,which becomes $23x + 29y + 3z + d = 0$.
Since the plane passes through the point $(1, 1, 1)$,we substitute these coordinates into the equation:
$23(1) + 29(1) + 3(1) + d = 0$
$23 + 29 + 3 + d = 0$
$55 + d = 0$
$d = -55$.
Thus,the equation of the plane is $23x + 29y + 3z - 55 = 0$,or $23x + 29y + 3z = 55$.

(A) The given equation of the plane is $by + cz + d = 0$.
Since the equation does not contain the variable $x$,the normal vector to the plane is $\vec{n} = 0\hat{i} + b\hat{j} + c\hat{k}$.
The normal vector lies in the $YOZ$-plane.
Therefore,the plane $by + cz + d = 0$ is perpendicular to the $YOZ$-plane.

(A) Let the points be $A(0,0,1)$,$B(0,1,2)$,and $C(1,0,3)$.
Two vectors lying on the plane are $\vec{AB} = (0-0)\hat{i} + (1-0)\hat{j} + (2-1)\hat{k} = \hat{j} + \hat{k}$ and $\vec{AC} = (1-0)\hat{i} + (0-0)\hat{j} + (3-1)\hat{k} = \hat{i} + 2\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{AB} \times \vec{AC}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{vmatrix} = \hat{i}(2-0) - \hat{j}(0-1) + \hat{k}(0-1) = 2\hat{i} + \hat{j} - \hat{k}$.
Thus,the direction ratios of the normal are $(2, 1, -1)$.

(D) Given that the probability of success is $p = \frac{1}{4}$.
Then,the probability of failure is $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.
The standard deviation $(SD)$ of a binomial distribution is given by $\sqrt{npq} = 3$.
Squaring both sides,we get $npq = 9$.
Substituting the values of $p$ and $q$:
$n \times \frac{1}{4} \times \frac{3}{4} = 9$
$n \times \frac{3}{16} = 9$
$n = 9 \times \frac{16}{3} = 3 \times 16 = 48$.
The mean of a binomial distribution is given by $np$.
Mean $= 48 \times \frac{1}{4} = 12$.

(B) Let the original length be $l$ and the original radius be $r$. The area of cross-section is $A = \pi r^2$.
The fractional change in area is given by $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Given $\frac{\Delta A}{A} = -2 \% = -0.02$,so $2 \frac{\Delta r}{r} = -0.02$,which implies $\frac{\Delta r}{r} = -0.01$.
The Poisson's ratio $\sigma$ is defined as $\sigma = - \frac{\Delta r / r}{\Delta l / l}$.
Given $\sigma = 0.4$,we have $0.4 = - \frac{-0.01}{\Delta l / l}$.
Therefore,$\frac{\Delta l}{l} = \frac{0.01}{0.4} = 0.025$.
Converting to percentage,the increase in length is $0.025 \times 100 = 2.5 \%$.

(B) The balanced chemical equation for the reaction of fluorine with hot concentrated $KOH$ is:
$2F_2 + 4KOH \longrightarrow 4KF + 2H_2O + O_2$
Dividing the entire equation by $2$ to match the stoichiometry for $1 \text{ mole}$ of $F_2$:
$F_2 + 2KOH \longrightarrow 2KF + H_2O + 0.5O_2$
From the balanced equation,the molar ratio of $KF : H_2O : O_2$ is $2 : 1 : 0.5$.

(C) Gypsum $(CaSO_4 \cdot 2H_2O)$ reacts with an aqueous solution of ammonium sulphate $((NH_4)_2SO_4)$ to form a double salt known as ammonium calcium sulphate,which is represented by the formula $CaSO_4 \cdot (NH_4)_2SO_4 \cdot 2H_2O$.

(B) Mass of $C = \frac{12}{44} \times 12.571 \ g = 3.428 \ g$
Mass of $H = \frac{2}{18} \times 5.143 \ g = 0.571 \ g$
Moles of $C = \frac{3.428}{12} = 0.2857$
Moles of $H = \frac{0.571}{1} = 0.571$
Ratio of $C:H = 0.2857 : 0.571 \approx 1 : 2$

Element	Mass $(g)$	Moles	Simple Ratio
$C$	$3.428$	$0.2857$	$1$
$H$	$0.571$	$0.571$	$2$

$\therefore$ Empirical formula $= CH_2$

(B) The balanced chemical equation for the oxidation of carbon monoxide is:
$CO(g) + \frac{1}{2} O_2(g) \longrightarrow CO_2(g)$
According to Avogadro's Law,at constant temperature and pressure,the volume of gases is directly proportional to their moles.
From the stoichiometry of the reaction,$1$ mole of $CO$ produces $1$ mole of $CO_2$.
Therefore,$1$ volume of $CO$ produces $1$ volume of $CO_2$ at $STP$.
Given that $11.207$ litres of $CO_2$ is formed,the volume of $CO$ required is also $11.207$ litres.
Thus,$X = 11.207$.

(C) The molarity formula is given by $Y = \frac{X \times 1000}{m \times V}$.
Given $m = 106$,$V = 100 \ mL$,and $Y$ is the molarity.
Substituting the values: $Y = \frac{X \times 1000}{106 \times 100} = \frac{10X}{106}$.
Rearranging gives $106Y = 10X$.
For option $C$: $X = 1.06$ and $Y = 0.1$.
$106(0.1) = 10.6$ and $10(1.06) = 10.6$.
Since both sides are equal,option $C$ is correct.

(B) Given: $W = 4 \ g$,$V = 5.6035 \ L$,$T = 546 \ K$,$P = 2 \ atm$.
Using the ideal gas equation: $PV = nRT$,where $n = \frac{W}{M}$.
$PV = \frac{W}{M} RT$
$M = \frac{WRT}{PV}$
Substituting the values: $M = \frac{4 \times 0.0821 \times 546}{2 \times 5.6035}$
$M = \frac{179.3304}{11.207} \approx 16 \ g/mol$.

(B) Let the weight of each gas be $64 \ g$.
The number of moles $(n)$ for each gas is calculated as $n = \frac{\text{weight}}{\text{molar mass}}$.
For $He$: $n_1 = \frac{64}{4} = 16 \ mol$.
For $CH_4$: $n_2 = \frac{64}{16} = 4 \ mol$.
For $SO_2$: $n_3 = \frac{64}{64} = 1 \ mol$.
Total moles = $16 + 4 + 1 = 21 \ mol$.
Partial pressure $p_i = \chi_i \times P_{total}$,where $\chi_i$ is the mole fraction.
Since $p_i \propto n_i$,the gas with the highest number of moles will have the highest partial pressure.
Comparing the moles: $16 (He) > 4 (CH_4) > 1 (SO_2)$.
Therefore,$p_1 > p_2 > p_3$.

(C) Iso-electronic species are those that have the same number of electrons.
$1$. For $Mg^{2+}$ ($Z=12$,$12-2=10$ electrons) and $C^{4-}$ ($Z=6$,$6+4=10$ electrons),both have $10$ electrons. Thus,it is an iso-electronic pair.
$2$. For $N^{3-}$ ($Z=7$,$7+3=10$ electrons) and $O^{2-}$ ($Z=8$,$8+2=10$ electrons),both have $10$ electrons. Thus,it is an iso-electronic pair.
$3$. For $N^{2-}$ ($Z=7$,$7+2=9$ electrons) and $O^{2-}$ ($Z=8$,$8+2=10$ electrons),they have different numbers of electrons ($9$ and $10$). Thus,it is not an iso-electronic pair.
$4$. For $F^{-}$ ($Z=9$,$9+1=10$ electrons) and $Al^{3+}$ ($Z=13$,$13-3=10$ electrons),both have $10$ electrons. Thus,it is an iso-electronic pair.
Therefore,the correct option is $C$.

(D) The energy of radiation is given by the formula: $E = h \nu = \frac{hc}{\lambda} = hc \bar{\nu}$,where $\bar{\nu}$ is the wave number.
Rearranging for the wave number: $\bar{\nu} = \frac{E}{hc}$.
Substituting the given values: $\bar{\nu} = \frac{19.875 \times 10^{-13} \ erg}{6.625 \times 10^{-27} \ erg \ sec \times 3 \times 10^{10} \ cm \ sec^{-1}}$.
$\bar{\nu} = \frac{19.875 \times 10^{-13}}{19.875 \times 10^{-17}} = 10^4 \ cm^{-1}$.

(D) . Rydberg's constant $(R_H)$ has units of $cm^{-1}$ or $m^{-1}$,and wave number $(\bar{\nu})$ also has units of $cm^{-1}$ or $m^{-1}$. Thus,this statement is correct.
$B$. Lyman series corresponds to transitions to $n=1$,which falls in the ultraviolet region. This statement is correct.
$C$. According to Bohr's postulate,angular momentum $mvr = \frac{nh}{2\pi}$. For the ground state $(n=1)$,angular momentum is $\frac{h}{2\pi}$. This statement is correct.
$D$. The radius of the first Bohr orbit of the hydrogen atom is $0.529 \times 10^{-8} \ cm$ (or $0.529 \ \mathring{A}$). The value given in the option $(2116 \times 10^{-8} \ cm)$ is incorrect.

(A) The combustion reaction for methane is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H = ?$
To obtain this,we manipulate the given equations:
$1$. Reverse equation $(i)$: $CH_{4(g)} \rightarrow C_{\text{(graphite)}} + 2H_{2(g)} \quad \Delta H = +74.8 \ kJ$
$2$. Keep equation (ii) as is: $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
$3$. Multiply equation (iii) by $2$: $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)} \quad \Delta H = 2 \times (-286.2) = -572.4 \ kJ$
Adding these equations gives the target reaction:
$\Delta H = 74.8 + (-393.5) + (-572.4) = -891.1 \ kJ$