TS EAMCET 2010 Chemistry Question Paper with Answer and Solution

(D) The probability that event $A_i$ does not occur is given by $P(\bar{A}_i) = 1 - P(A_i)$.
Given $P(A_i) = \frac{1}{1+i}$,we have $P(\bar{A}_i) = 1 - \frac{1}{1+i} = \frac{1+i-1}{1+i} = \frac{i}{1+i}$.
Since $A_i$ are independent events,the probability that none of $A_i$ occurs is the product of the probabilities of their complements:
$P(\text{none occurs}) = P(\bar{A}_1) \times P(\bar{A}_2) \times \ldots \times P(\bar{A}_n)$.
Substituting the values:
$P(\text{none occurs}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{n}{n+1}$.
Observing the product,the terms cancel out in a telescoping manner:
$P(\text{none occurs}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{n}{n+1} = \frac{1}{n+1}$.

(B) According to the equation of continuity,the product of cross-sectional area and velocity remains constant for an incompressible fluid: $A_1 v_1 = A_2 v_2$.
Given $A_1 = A$,$v_1 = 4 ~m/s$,and $A_2 = \frac{2}{3} A$.
Substituting these values: $A \times 4 = \frac{2}{3} A \times v_2$,which gives $v_2 = 6 ~m/s$.
Using the equation of motion $v_2^2 = v_1^2 + 2gh$ for a freely falling body under gravity:
$(6)^2 = (4)^2 + 2(10)h$.
$36 = 16 + 20h$.
$20 = 20h$.
Therefore,$h = 1 ~m$.

(B) The excess pressure inside a soap bubble is given by $p = \frac{4T}{R}$.
This pressure is balanced by the pressure exerted by the oil column,which is $p = h \rho g$.
Equating the two,we get $h \rho g = \frac{4T}{R}$.
Rearranging for surface tension $T$,we have $T = \frac{R h \rho g}{4}$.
Given values: $R = 1 ~cm = 10^{-2} ~m$,$h = 2 ~mm = 2 \times 10^{-3} ~m$,$\rho = 0.8 \times 10^3 ~kg/m^3$,and $g = 9.8 ~m/s^2$.
Substituting these values into the formula:
$T = \frac{10^{-2} \times 2 \times 10^{-3} \times 0.8 \times 10^3 \times 9.8}{4}$
$T = \frac{2 \times 0.8 \times 9.8 \times 10^{-2}}{4}$
$T = 0.4 \times 0.8 \times 9.8 \times 10^{-2} = 0.392 \times 10^{-1} = 0.0392 ~N/m$.

(B) Diborane $(B_2H_6)$ reacts with ammonia $(NH_3)$ under different conditions to yield various products:
$1$. At low temperature,it forms an ionic addition product: $B_2H_6 + 2NH_3 \rightarrow [BH_2(NH_3)_2]^+ [BH_4]^-$ (often written as $B_2H_6 \cdot 2NH_3$).
$2$. At high temperatures,it forms borazine $(B_3N_3H_6)$,which is known as inorganic benzene: $3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$.
$3$. At very high temperatures,it forms boron nitride $((BN)_n)$,which is isoelectronic with graphite: $B_2H_6 + 2NH_3 \rightarrow 2BN + 6H_2$.
Therefore,$B_{12}H_{12}$ is not formed in the reaction between diborane and ammonia.

(D) mixture of helium and oxygen is used in the treatment of asthma.
Because helium has a low density,this mixture flows easily through restricted respiratory passages,making it easier for patients with asthma to breathe.

(D) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the angle of incidence $i$ is equal to the angle of emergence $e$,and each is equal to $3/4$ of the angle of the prism:
$i = e = \frac{3}{4} \times A$
$i = e = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
The relation between the angle of incidence,angle of emergence,angle of the prism,and the angle of deviation $\delta$ is given by:
$i + e = A + \delta$
Substituting the values:
$45^{\circ} + 45^{\circ} = 60^{\circ} + \delta$
$90^{\circ} = 60^{\circ} + \delta$
$\delta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Thus,the angle of deviation is $30^{\circ}$.

(C) The Solvay process is an industrial method used for the large-scale production of sodium carbonate $(Na_2CO_3)$,commonly known as soda ash.

(C) The balanced chemical equation for the combustion of graphite is:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
From the stoichiometry,$1 \ mol$ of $C$ produces $1 \ mol$ of $CO_2$.
Since $1 \ mol$ of any substance contains $6.022 \times 10^{23}$ molecules,$1 \ mol$ of $C$ produces $6.022 \times 10^{23}$ molecules of $CO_2$.
Therefore,$0.1 \ mol$ of graphite will produce:
$0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$ molecules of $CO_2$.

(B) From Graham's law of diffusion,$\frac{r_{CH_4}}{r_X} = \sqrt{\frac{M_X}{M_{CH_4}}}$.
Given that $r_{CH_4} = 2 \cdot r_X$,we have $2 = \sqrt{\frac{M_X}{16}}$.
Squaring both sides,$4 = \frac{M_X}{16}$,so $M_X = 64 \ g/mol$.
The number of moles of gas $X$ in $32 \ g$ is $n = \frac{32}{64} = 0.5 \ mol$.
The number of molecules is $n \times N = 0.5 \times N = \frac{N}{2}$.

(C) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{n_2} - E_{n_1} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
Since $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$ $(\lambda = \frac{hc}{\Delta E})$.
To obtain the lowest wavelength,we need the transition with the highest energy difference $\Delta E$.
Comparing the transitions:
$(A)$ $n_2=\infty$ to $n_1=2$: $\Delta E \propto (1/4 - 0) = 0.25$
$(B)$ $n_2=4$ to $n_1=3$: $\Delta E \propto (1/9 - 1/16) = 0.0486$
$(C)$ $n_2=2$ to $n_1=1$: $\Delta E \propto (1/1 - 1/4) = 0.75$
$(D)$ $n_2=5$ to $n_1=3$: $\Delta E \propto (1/9 - 1/25) = 0.0711$
The transition $n_2=2$ to $n_1=1$ has the largest energy difference,therefore it emits radiation of the lowest wavelength.

(C) For a well-behaved wave function,the $BORN$ conditions are that $\psi$ must be finite,single-valued,and continuous. Therefore,the condition that $\psi$ must be infinite is incorrect.

(A) From the first law of thermodynamics,$\Delta U = Q + W$.
Since heat is provided to the system,$Q = +50 \ J$.
Since work is done on the system,$W = +10 \ J$.
Therefore,the change in internal energy is $\Delta U = 50 \ J + 10 \ J = 60 \ J$.