AP EAMCET 2013 Chemistry Question Paper with Answer and Solution

(D) The common chord of the two circles $S_1 = x^2+y^2+4x-6y+c=0$ and $S_2 = x^2+y^2-6x+4y-12=0$ is given by $S_1 - S_2 = 0$.
$(x^2+y^2+4x-6y+c) - (x^2+y^2-6x+4y-12) = 0$
$10x - 10y + c + 12 = 0$
Since the first circle bisects the circumference of the second circle,the common chord must pass through the center of the second circle.
The center of the second circle $x^2+y^2-6x+4y-12=0$ is $(-g, -f) = (3, -2)$.
Substituting $(3, -2)$ into the equation of the common chord:
$10(3) - 10(-2) + c + 12 = 0$
$30 + 20 + c + 12 = 0$
$62 + c = 0$
$c = -62$

(A) The given equation of the circle is $x^2+y^2-16x-12y+64=0$. The center is $(8, 6)$ and the radius is $r = \sqrt{8^2+6^2-64} = \sqrt{64+36-64} = 6$.
$(i)$ The equation of the polar of $(x_1, y_1)$ w.r.t. $x^2+y^2+2gx+2fy+c=0$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.
For $(-5, 1)$,it is $x(-5)+y(1)-8(x-5)-6(y+1)+64=0$ $\Rightarrow -5x+y-8x+40-6y-6+64=0$ $\Rightarrow -13x-5y+98=0$ $\Rightarrow 13x+5y=98$. This matches $(D)$.
$(ii)$ The equation of the tangent at $(x_1, y_1)$ is $xx_1+yy_1-8(x+x_1)-6(y+y_1)+64=0$.
For $(8, 0)$,it is $x(8)+y(0)-8(x+8)-6(y+0)+64=0$ $\Rightarrow 8x-8x-64-6y+64=0$ $\Rightarrow -6y=0$ $\Rightarrow y=0$. This matches $(A)$.
$(iii)$ The normal at any point on a circle passes through the center $(8, 6)$. The normal at $(2, 6)$ passes through $(2, 6)$ and $(8, 6)$. Since the $y$-coordinates are the same,the equation is $y=6$. This matches $(B)$.
$(iv)$ The diameter passes through the center $(8, 6)$ and the point $(8, 12)$. Since the $x$-coordinates are the same,the equation is $x=8$. This matches $(E)$.
Thus,the correct match is $(i)-(D), (ii)-(A), (iii)-(B), (iv)-(E)$.

(C) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.
Let the chord be $y = mx + c$.
The points of intersection of the line $y = mx + c$ and the parabola $y^2 = 8x$ are obtained by substituting $x = (y - c)/m$ into the parabola equation: $y^2 = 8(y - c)/m \Rightarrow my^2 - 8y + 8c = 0$.
Let the points be $P(x_1, y_1)$ and $Q(x_2, y_2)$. Then $y_1 + y_2 = 8/m$ and $y_1y_2 = 8c/m$.
The midpoint of the chord is $M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
The circle with the chord as diameter has radius $R = 4$.
The distance from the center of the circle to the axis of the parabola $(y = 0)$ is equal to the radius $4$,so the $y$-coordinate of the center is $4$ (or $-4$).
Thus,$\frac{y_1+y_2}{2} = 4 \Rightarrow y_1 + y_2 = 8$.
Since $y_1 + y_2 = 8/m$,we have $8/m = 8$,which implies $m = 1$.

(C) Given expression is $E = \frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$.
Using the binomial expansion $(1+ax)^n \approx 1+nax$ for small $x$:
$(1-2 x)^{-1} \approx 1 + (-1)(-2x) = 1 + 2x$.
$(1-3 x)^{-2} \approx 1 + (-2)(-3x) = 1 + 6x$.
$(1-4 x)^{-3} \approx 1 + (-3)(-4x) = 1 + 12x$.
Substituting these into the expression:
$E \approx \frac{(1+2x)(1+6x)}{1+12x} = \frac{1 + 6x + 2x + 12x^2}{1+12x}$.
Neglecting $x^2$ and higher powers:
$E \approx \frac{1+8x}{1+12x} = (1+8x)(1+12x)^{-1}$.
Using the binomial expansion again:
$E \approx (1+8x)(1-12x) = 1 - 12x + 8x - 96x^2$.
Neglecting $x^2$:
$E \approx 1 - 4x$.

(A) The equation of a chord of a conic $S=0$ with mid-point $(x_1, y_1)$ is given by $T=S_1$.
Given ellipse: $S: x^2+4y^2-2x+20y=0$.
Mid-point $(x_1, y_1) = (2, -4)$.
$T = x(x_1) + 4y(y_1) - (x+x_1) + 10(y+y_1) = x(2) + 4y(-4) - (x+2) + 10(y-4) = 2x - 16y - x - 2 + 10y - 40 = x - 6y - 42$.
$S_1 = (x_1)^2 + 4(y_1)^2 - 2(x_1) + 20(y_1) = (2)^2 + 4(-4)^2 - 2(2) + 20(-4) = 4 + 64 - 4 - 80 = -16$.
Equating $T=S_1$:
$x - 6y - 42 = -16$.
$x - 6y = 42 - 16$.
$x - 6y = 26$.

(B) The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here $a^2=25$ and $b^2=16$.
For an ellipse,$b^2=a^2(1-e^2)$,so $16=25(1-e^2)$,which gives $e^2=1-\frac{16}{25}=\frac{9}{25}$,so $e=\frac{3}{5}$.
The foci of the ellipse are $(\pm ae, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$.
The equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{b^2}=1$. Here $a^2=4$,so $a=2$.
Since the foci of the hyperbola coincide with the foci of the ellipse,the foci of the hyperbola are $(\pm 3, 0)$.
For a hyperbola,the foci are $(\pm ae_1, 0)$,so $ae_1=3$. Since $a=2$,we have $2e_1=3$,which means $e_1=\frac{3}{2}$.
For a hyperbola,$b^2=a^2(e_1^2-1)$.
Substituting the values,$b^2=4((\frac{3}{2})^2-1) = 4(\frac{9}{4}-1) = 4(\frac{5}{4}) = 5$.
Thus,$b^2=5$.

(B) Given that $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$.
Substituting $x=9$ into the hyperbola equation:
$81-y^2=9$
$y^2=72$
$y = \pm 6\sqrt{2}$.
Thus,the points of contact are $(9, 6\sqrt{2})$ and $(9, -6\sqrt{2})$.
Differentiating $x^2-y^2=9$ with respect to $x$:
$2x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{y}$.
At point $(9, 6\sqrt{2})$,the slope $m = \frac{9}{6\sqrt{2}} = \frac{3}{2\sqrt{2}}$.
The equation of the tangent at $(9, 6\sqrt{2})$ is:
$y - 6\sqrt{2} = \frac{3}{2\sqrt{2}}(x-9)$
$2\sqrt{2}y - 24 = 3x - 27$
$3x - 2\sqrt{2}y - 3 = 0$.
At point $(9, -6\sqrt{2})$,the slope $m = \frac{9}{-6\sqrt{2}} = -\frac{3}{2\sqrt{2}}$.
The equation of the tangent at $(9, -6\sqrt{2})$ is:
$y + 6\sqrt{2} = -\frac{3}{2\sqrt{2}}(x-9)$
$2\sqrt{2}y + 24 = -3x + 27$
$3x + 2\sqrt{2}y - 3 = 0$.
Comparing with the options,$3x + 2\sqrt{2}y - 3 = 0$ is option $B$.

(B) We have the limit $L = \lim _{x \rightarrow 0} \frac{\tan ^3 x-\sin ^3 x}{x^5}$.
Using the Taylor series expansions: $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$ and $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$.
Factorizing the numerator: $\tan^3 x - \sin^3 x = (\tan x - \sin x)(\tan^2 x + \tan x \sin x + \sin^2 x)$.
Now,$\tan x - \sin x = \sin x (\sec x - 1) = \sin x \left(\frac{1-\cos x}{\cos x}\right) = \sin x \cdot \frac{2 \sin^2(x/2)}{\cos x}$.
As $x \rightarrow 0$,$\sin x \approx x$,$\sin(x/2) \approx x/2$,and $\cos x \approx 1$.
So,$\tan x - \sin x \approx x \cdot \frac{2(x/2)^2}{1} = \frac{x^3}{2}$.
Substituting this into the limit expression:
$L = \lim _{x \rightarrow 0} \frac{(\frac{x^3}{2})(\tan^2 x + \tan x \sin x + \sin^2 x)}{x^5}$.
$L = \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{\tan^2 x + \tan x \sin x + \sin^2 x}{x^2}$.
Since $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$L = \frac{1}{2} (1^2 + 1 \cdot 1 + 1^2) = \frac{1}{2} (3) = \frac{3}{2}$.

(B) $PHBV$ and $Dextron$ are examples of biodegradable polymers.
Biodegradable polymers are those that disintegrate over time through enzymatic hydrolysis and oxidation.
$PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is used in orthopaedic devices and controlled drug release.
$Dextron$ (a copolymer of glycolic acid and lactic acid) is used for stitching wounds after surgery.
Thus,the correct pair is $PHBV$ and $Dextron$.

(B) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, m\}$.
The mean $\bar{X} = E[X] = \sum_{n=1}^{m} n \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n = \frac{1}{m} \cdot \frac{m(m+1)}{2} = \frac{m+1}{2}$.
The variance is given by $\operatorname{Var}(X) = E[X^2] - (E[X])^2$.
$E[X^2] = \sum_{n=1}^{m} n^2 \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m+1)(2m+1)}{6} = \frac{(m+1)(2m+1)}{6}$.
Therefore,$\operatorname{Var}(X) = \frac{(m+1)(2m+1)}{6} - \left(\frac{m+1}{2}\right)^2$.
$\operatorname{Var}(X) = \frac{m+1}{2} \left[ \frac{2(2m+1)}{3} - \frac{m+1}{2} \right] = \frac{m+1}{2} \left[ \frac{8m+4-3m-3}{6} \right] = \frac{m+1}{2} \cdot \frac{5m+1}{6}$ (Wait,let's re-calculate).
$\operatorname{Var}(X) = \frac{(m+1)(2m+1)}{6} - \frac{(m+1)^2}{4} = \frac{m+1}{2} \left[ \frac{2(2m+1) - 3(m+1)}{6} \right] = \frac{m+1}{2} \left[ \frac{4m+2-3m-3}{6} \right] = \frac{m+1}{2} \cdot \frac{m-1}{6} = \frac{m^2-1}{12}$.

(C) Given the equation: $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$.
Taking $LCM$ on the left side: $\frac{(b+c) + (a+c)}{(a+c)(b+c)} = \frac{3}{a+b+c}$.
$\frac{a+b+2c}{ab+ac+bc+c^2} = \frac{3}{a+b+c}$.
Cross-multiplying: $(a+b+2c)(a+b+c) = 3(ab+ac+bc+c^2)$.
$(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3ab + 3ac + 3bc + 3c^2$.
$a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3ab + 3ac + 3bc + 3c^2$.
$a^2 + b^2 - ab = c^2$.
Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$.
Comparing the two equations: $a^2 + b^2 - ab = a^2 + b^2 - 2ab \cos C$.
$-ab = -2ab \cos C$.
$\cos C = \frac{1}{2}$.
Therefore,$\angle C = 60^{\circ}$.

(B) Given that,$\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$.
Taking common denominators on the left side:
$\frac{(c+a) + (b+c)}{(b+c)(c+a)} = \frac{3}{a+b+c}$
$\frac{a+b+2c}{bc + ab + c^2 + ac} = \frac{3}{a+b+c}$
Cross-multiplying:
$(a+b+2c)(a+b+c) = 3(bc + ab + c^2 + ac)$
$(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$
$a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$
Subtracting $3ac + 3bc$ from both sides:
$a^2 + b^2 + 2ab + 2c^2 = 3ab + 3c^2$
$a^2 + b^2 - c^2 = ab$
Using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$:
$\cos C = \frac{ab}{2ab} = \frac{1}{2}$
Since $\cos C = \frac{1}{2}$,we have $C = 60^{\circ}$.

(A) We know that the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these into the expression $r_1 r_2 + r_2 r_3 + r_3 r_1$:
$= \frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}$
$= \frac{\Delta^2(s-c + s-a + s-b)}{(s-a)(s-b)(s-c)}$
Since $2s = a+b+c$,we have $s-a+s-b+s-c = 3s - (a+b+c) = 3s - 2s = s$.
Also,we know that $\Delta^2 = s(s-a)(s-b)(s-c)$,so $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.
Substituting these values:
$= \frac{\Delta^2 \cdot s}{\frac{\Delta^2}{s}} = s^2$
Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r}$,so $s^2 = \frac{\Delta^2}{r^2}$.

(B) Given the matrix $A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$.
Since $A$ satisfies the equation $x^2 + 4x - p = 0$,we have $A^2 + 4A - pI = 0$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} (-8)(-8) + (5)(2) & (-8)(5) + (5)(4) \\ (2)(-8) + (4)(2) & (2)(5) + (4)(4) \end{bmatrix} = \begin{bmatrix} 64 + 10 & -40 + 20 \\ -16 + 8 & 10 + 16 \end{bmatrix} = \begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix}$.
Next,calculate $4A$:
$4A = 4 \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix}$.
Now,substitute these into the equation $A^2 + 4A - pI = 0$:
$\begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix} + \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} p & 0 \\ 0 & p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Performing the matrix addition and subtraction:
$\begin{bmatrix} 74 - 32 - p & -20 + 20 - 0 \\ -8 + 8 - 0 & 26 + 16 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
$\begin{bmatrix} 42 - p & 0 \\ 0 & 42 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Comparing the elements,we get $42 - p = 0$,which implies $p = 42$.

(A) Given $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
We calculate the powers of $A$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.
By observation,$A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.
Now,check option $A$: $nA - (n-1)I = n \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} - (n-1) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$= \begin{bmatrix} n & n \\ 0 & n \end{bmatrix} - \begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix} = \begin{bmatrix} n-(n-1) & n-0 \\ 0-0 & n-(n-1) \end{bmatrix} = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} = A^n$.
Thus,the correct relation is $A^n = nA - (n-1)I$.

(D) Let $\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$.
Apply operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ 2 & 3 & 4 \\ 6 & 8 & 10\end{array}\right|$.
Now,apply $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$:
$\Delta = \left|\begin{array}{ccc}x+2 & 1 & 3 \\ 2 & 1 & 2 \\ 6 & 2 & 4\end{array}\right|$.
Expanding along $R_1$:
$\Delta = (x+2)(4-4) - 1(8-12) + 3(4-6)$
$\Delta = (x+2)(0) - 1(-4) + 3(-2)$
$\Delta = 0 + 4 - 6 = -2$.

(D) The given system of equations is:
$3x + 2y + z = 6$
$3x + 4y + 3z = 14$
$6x + 10y + 8z = a$
Let $A = \begin{bmatrix} 3 & 2 & 1 \\ 3 & 4 & 3 \\ 6 & 10 & 8 \end{bmatrix}$ and $B = \begin{bmatrix} 6 \\ 14 \\ a \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 3(32 - 30) - 2(24 - 18) + 1(30 - 24) = 3(2) - 2(6) + 6 = 6 - 12 + 6 = 0$.
Since $|A| = 0$,the system has either no solution or infinite solutions.
For infinite solutions,we must have $(\text{adj } A) \cdot B = 0$.
The cofactor matrix of $A$ is:
$C_{11} = 2, C_{12} = -6, C_{13} = 6$
$C_{21} = -6, C_{22} = 18, C_{23} = -18$
$C_{31} = 2, C_{32} = -6, C_{33} = 6$
Thus,$\text{adj } A = \begin{bmatrix} 2 & -6 & 2 \\ -6 & 18 & -6 \\ 6 & -18 & 6 \end{bmatrix}$.
Now,$(\text{adj } A) \cdot B = \begin{bmatrix} 2 & -6 & 2 \\ -6 & 18 & -6 \\ 6 & -18 & 6 \end{bmatrix} \begin{bmatrix} 6 \\ 14 \\ a \end{bmatrix} = \begin{bmatrix} 12 - 84 + 2a \\ -36 + 252 - 6a \\ 36 - 252 + 6a \end{bmatrix} = \begin{bmatrix} 2a - 72 \\ 216 - 6a \\ 6a - 216 \end{bmatrix}$.
Setting this to zero vector,we get $2a - 72 = 0$,which implies $a = 36$.

(C) We use the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} (AB - \sqrt{1-A^2} \sqrt{1-B^2})$.
Given $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$.
Here $A = \frac{5}{13}$ and $B = \frac{3}{5}$.
Then $\sqrt{1-A^2} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
And $\sqrt{1-B^2} = \sqrt{1-\left(\frac{3}{5}\right)^2} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Substituting these values into the formula:
$\cos ^{-1} x = \cos ^{-1} \left( \frac{5}{13} \cdot \frac{3}{5} - \frac{12}{13} \cdot \frac{4}{5} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{15}{65} - \frac{48}{65} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{-33}{65} \right)$.
Therefore,$x = \frac{-33}{65}$.

(D) We know that $\operatorname{coth}^{-1}(x) = \tanh^{-1}\left(\frac{1}{x}\right)$ for $|x| > 1$.
Therefore,$\operatorname{coth}^{-1}(2) = \tanh^{-1}\left(\frac{1}{2}\right)$.
Substituting this into the expression,we get:
$\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2) = \tanh ^{-1}\left(\frac{1}{2}\right)+\tanh ^{-1}\left(\frac{1}{2}\right) = 2 \tanh ^{-1}\left(\frac{1}{2}\right)$.
Using the formula $\tanh ^{-1}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$,we have:
$2 \tanh ^{-1}\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} \log \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)$.
$= \log \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right) = \log 3$.

(D) For the expression $\log_{10} ((1.6)^{1-x^2} - (0.625)^{6(1+x)})$ to be defined in $R$,the argument of the logarithm must be strictly greater than $0$.
$(1.6)^{1-x^2} - (0.625)^{6(1+x)} > 0$
$(1.6)^{1-x^2} > (0.625)^{6(1+x)}$
Since $1.6 = \frac{8}{5}$ and $0.625 = \frac{5}{8} = (\frac{8}{5})^{-1}$,we have:
$(\frac{8}{5})^{1-x^2} > ((\frac{8}{5})^{-1})^{6(1+x)}$
$(\frac{8}{5})^{1-x^2} > (\frac{8}{5})^{-6(1+x)}$
Since the base $\frac{8}{5} > 1$,the inequality direction remains the same:
$1 - x^2 > -6(1 + x)$
$1 - x^2 > -6 - 6x$
$x^2 - 6x - 7 < 0$
$(x - 7)(x + 1) < 0$
Thus,$x \in (-1, 7)$.

(D) Given,$f(x)=\cos \left(\frac{x}{3}\right)+\sin \left(\frac{x}{2}\right)$.
We know that the period of $\cos(ax)$ and $\sin(ax)$ is $\frac{2\pi}{|a|}$.
For the term $\cos \left(\frac{x}{3}\right)$,the period $T_1 = \frac{2\pi}{1/3} = 6\pi$.
For the term $\sin \left(\frac{x}{2}\right)$,the period $T_2 = \frac{2\pi}{1/2} = 4\pi$.
The period of the sum of two periodic functions is the Least Common Multiple $(LCM)$ of their individual periods.
Therefore,the period of $f(x) = \text{LCM}(6\pi, 4\pi)$.
Since $6\pi = 2 \times 3\pi$ and $4\pi = 2 \times 2\pi$,the $\text{LCM}(6\pi, 4\pi) = 12\pi$.
Thus,the period of $f(x)$ is $12\pi$.

(A) Given,$f(x) = (p - x^n)^{1/n}$,where $p > 0$.
To find $f[f(x)]$,we substitute $f(x)$ into the function $f$:
$f[f(x)] = f((p - x^n)^{1/n})$
$= (p - ((p - x^n)^{1/n})^n)^{1/n}$
$= (p - (p - x^n))^{1/n}$
$= (p - p + x^n)^{1/n}$
$= (x^n)^{1/n}$
$= x$

(B) Given the functional equation $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$.
We know that the continuous function satisfying this property is of the form $f(x) = a^x$.
Given $f(2) = 9$,we have $a^2 = 9$.
Since $f$ is a non-zero function,$a^2 = 3^2$,which implies $a = 3$ (as $a$ must be positive for $f(x) = a^x$ to be defined for all real $x$).
Thus,$f(x) = 3^x$.
Now,we need to find $f(6)$.
$f(6) = 3^6$.

(B) Given $f(x) = \frac{1}{1+\frac{1}{x}} = \frac{x}{x+1}$.
Then $g(x) = \frac{1}{1+\frac{1}{f(x)}} = \frac{1}{1+\frac{x+1}{x}} = \frac{1}{\frac{x+x+1}{x}} = \frac{x}{2x+1}$.
To find $g^{\prime}(x)$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u^{\prime} - u v^{\prime}}{v^2}$.
$g^{\prime}(x) = \frac{(2x+1)(1) - x(2)}{(2x+1)^2} = \frac{2x+1-2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}$.
Substituting $x = 2$,we get $g^{\prime}(2) = \frac{1}{(2(2)+1)^2} = \frac{1}{5^2} = \frac{1}{25}$.

(C) Given the equation: $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$
Multiplying by $\sqrt{xy}$,we get: $y+x=2\sqrt{xy}$
Squaring both sides: $(x+y)^2 = (2\sqrt{xy})^2$
$x^2+y^2+2xy = 4xy$
$x^2+y^2-2xy = 0$
$(x-y)^2 = 0$
This implies $x-y=0$,or $y=x$
Differentiating both sides with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x) = 1$

(D) Let $f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)$.
Multiply and divide by $(x-1)$:
$f(x) = \frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^2-1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^4-1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^8-1)(x^8+1)}{(x-1)} = \frac{x^{16}-1}{x-1}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule:
$\frac{d}{dx} \left[ \frac{x^{16}-1}{x-1} \right] = \frac{(x-1)(16x^{15}) - (x^{16}-1)(1)}{(x-1)^2} = \frac{16x^{16} - 16x^{15} - x^{16} + 1}{(x-1)^2} = \frac{15x^{16} - 16x^{15} + 1}{(x-1)^2}$.
Comparing this with the given expression $\left(15 x^p-16 x^q+1\right)(x-1)^{-2}$,we get $p=16$ and $q=15$.
Therefore,$(p, q) = (16, 15)$.

(A) Given the relation $p V^{1/4} = C$,where $C$ is a constant.
Taking the natural logarithm on both sides: $\ln p + \frac{1}{4} \ln V = \ln C$.
Differentiating both sides with respect to $V$: $\frac{1}{p} \frac{dp}{dV} + \frac{1}{4V} = 0$.
This gives $\frac{dp}{p} = -\frac{1}{4} \frac{dV}{V}$.
We are given that the percentage decrease in volume is $\frac{dV}{V} \times 100 = -\frac{1}{2} \%$.
Substituting this into the differential relation: $\frac{dp}{p} \times 100 = -\frac{1}{4} \left( \frac{dV}{V} \times 100 \right)$.
$\frac{dp}{p} \times 100 = -\frac{1}{4} \left( -\frac{1}{2} \% \right) = \frac{1}{8} \%$.
Thus,the percentage increase in pressure is $\frac{1}{8} \%$.

(B) Given the equation: $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$ $(i)$
Taking the differential of both sides:
$-\frac{2}{f^2} df = -\frac{1}{v^2} dv + \frac{1}{u^2} du$
Given that the errors in $u$ and $v$ are $p$,we have $dv = p$ and $du = p$.
Substituting these into the differential equation:
$-\frac{2}{f^2} df = -\frac{1}{v^2} p + \frac{1}{u^2} p$
$-\frac{2}{f^2} df = p \left( \frac{1}{u^2} - \frac{1}{v^2} \right)$
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
$-\frac{2}{f^2} df = p \left( \frac{1}{u} - \frac{1}{v} \right) \left( \frac{1}{u} + \frac{1}{v} \right)$
From equation $(i)$,$\frac{1}{u} - \frac{1}{v} = -\frac{2}{f}$. Substituting this:
$-\frac{2}{f^2} df = p \left( -\frac{2}{f} \right) \left( \frac{1}{u} + \frac{1}{v} \right)$
Dividing both sides by $-\frac{2}{f}$:
$\frac{df}{f} = p \left( \frac{1}{u} + \frac{1}{v} \right)$
Thus,the relative error in $f$ is $p \left( \frac{1}{u} + \frac{1}{v} \right)$.

(D) Let the height of the tower be $h$ and the distance of point $A$ from the base of the tower $(B)$ be $x$.
In $\triangle ABD$,$\tan 60^{\circ} = \frac{h}{x}$ $\Rightarrow h = x \sqrt{3}$ $\Rightarrow x = \frac{h}{\sqrt{3}}$.
The person moves $60 \ m$ perpendicular to $AB$ to reach point $C$. Thus,$AC = 60 \ m$ and $\angle CAB = 90^{\circ}$.
In $\triangle ABC$,the distance from $C$ to the base $B$ is $CB = \sqrt{AC^2 + AB^2} = \sqrt{60^2 + x^2}$.
In $\triangle CBD$,the angle of elevation is $45^{\circ}$,so $\tan 45^{\circ} = \frac{h}{CB} = 1$.
Therefore,$h = CB = \sqrt{3600 + x^2} \Rightarrow h^2 = 3600 + x^2$.
Substituting $x^2 = \frac{h^2}{3}$ into the equation: $h^2 = 3600 + \frac{h^2}{3}$.
$\frac{2h^2}{3} = 3600 \Rightarrow h^2 = 1800 \times 3 = 5400$.
$h = \sqrt{5400} = \sqrt{900 \times 6} = 30 \sqrt{6} \ m$.

(B) Let $I_1 = \int \frac{d x}{x(\log x-2)(\log x-3)}$.
Substitute $t = \log x$,then $d t = \frac{1}{x} d x$.
The integral becomes $I_1 = \int \frac{d t}{(t-2)(t-3)}$.
Using partial fractions,$\frac{1}{(t-2)(t-3)} = \frac{A}{t-2} + \frac{B}{t-3}$.
$1 = A(t-3) + B(t-2)$.
For $t=3$,$B=1$. For $t=2$,$A=-1$.
So,$I_1 = \int \left( \frac{1}{t-3} - \frac{1}{t-2} \right) d t$.
$I_1 = \log |t-3| - \log |t-2| + C = \log \left| \frac{t-3}{t-2} \right| + C$.
Substituting $t = \log x$ back,we get $I = \log \left| \frac{\log x - 3}{\log x - 2} \right|$.

(D) Let $I = \int e^x \left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx$.
Using the trigonometric identities $1+\cos 2x = 2\cos^2 x$ and $\sin 2x = 2\sin x \cos x$,we get:
$I = \int e^x \left( \frac{2 + 2\sin x \cos x}{2\cos^2 x} \right) dx$
$I = \int e^x \left( \frac{2}{2\cos^2 x} + \frac{2\sin x \cos x}{2\cos^2 x} \right) dx$
$I = \int e^x (\sec^2 x + \tan x) dx$.
We know the standard integral form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
Here,let $f(x) = \tan x$,then $f'(x) = \sec^2 x$.
Therefore,$I = e^x \tan x + C$.

(C) Given the integral $I = \int_1^3 \frac{dx}{2+3x}$. Here,$n = 2$ and the interval $[1, 3]$ is divided into $n=2$ equal parts.
The step size $h = \frac{b-a}{n} = \frac{3-1}{2} = 1$.
The values of $x$ are $x_0 = 1, x_1 = 2, x_2 = 3$.
The corresponding values of $y = f(x) = \frac{1}{2+3x}$ are:
$y_0 = f(1) = \frac{1}{2+3(1)} = \frac{1}{5} = 0.2$
$y_1 = f(2) = \frac{1}{2+3(2)} = \frac{1}{8} = 0.125$
$y_2 = f(3) = \frac{1}{2+3(3)} = \frac{1}{11} \approx 0.0909$
According to Simpson's rule: $\int_a^b f(x) dx \approx \frac{h}{3} [y_0 + 4y_1 + y_2]$.
Substituting the values: $I \approx \frac{1}{3} [0.2 + 4(0.125) + 0.0909] = \frac{1}{3} [0.2 + 0.5 + 0.0909] = \frac{0.7909}{3} \approx 0.2636$.
Comparing with the options,$\frac{29}{110} \approx 0.2636$. Thus,the correct option is $C$.

(A) From the lens maker's formula:
$\frac{1}{f} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given that the refractive index of the glass lens is $\mu_g = 1.5$ and the refractive index of the medium is $\mu_m = 1.75$.
For a concave lens,the radii of curvature are $R_1 = -R$ and $R_2 = +R$.
Substituting these values into the formula:
$\frac{1}{f} = \left( \frac{1.5}{1.75} - 1 \right) \left( \frac{1}{-R} - \frac{1}{R} \right)$
$\frac{1}{f} = \left( \frac{1.5 - 1.75}{1.75} \right) \left( -\frac{2}{R} \right)$
$\frac{1}{f} = \left( \frac{-0.25}{1.75} \right) \left( -\frac{2}{R} \right)$
$\frac{1}{f} = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$
$f = +3.5 R$
The positive sign of the focal length indicates that the lens behaves as a convergent lens.

(B) For the eyepiece,the image distance $v_e = -25 \ cm$ and focal length $f_e = 5 \ cm$. Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:
$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5} \Rightarrow \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \Rightarrow u_e = -\frac{25}{6} \ cm$.
The distance of the image formed by the objective from the objective is $v_0 = L - |u_e| = 10.5 - \frac{25}{6} = \frac{63-25}{6} = \frac{38}{6} \ cm$.
For the objective,$f_0 = 1.9 \ cm$ and $v_0 = \frac{38}{6} \ cm$. Using the lens formula $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$:
$\frac{1}{38/6} - \frac{1}{u_0} = \frac{1}{1.9} \Rightarrow \frac{6}{38} - \frac{1}{u_0} = \frac{10}{19} \Rightarrow \frac{3}{19} - \frac{10}{19} = \frac{1}{u_0}$.
$\frac{1}{u_0} = -\frac{7}{19} \Rightarrow u_0 = -\frac{19}{7} \approx -2.71 \ cm$.
Thus,the object should be placed at a distance of $2.7 \ cm$ from the objective.

(A) disproportionation reaction is a special type of redox reaction in which the same element in a given oxidation state is simultaneously oxidized and reduced.
In the reaction $3Cl_{2\text{(g)}} + 6OH^-{_{\text{(aq)}}} \rightarrow ClO_3^-{_{\text{(aq)}}} + 5Cl^-{_{\text{(aq)}}} + 3H_2O_{\text{(l)}}$,the oxidation state of chlorine changes from $0$ in $Cl_2$ to $-1$ in $Cl^-$ (reduction) and to $+5$ in $ClO_3^-$ (oxidation). Thus,it is a disproportionation reaction.

(C) When chloroform $(CHCl_3)$ is heated with aqueous sodium hydroxide $(NaOH)$,it undergoes hydrolysis to form an unstable intermediate,methanetriol $(HC(OH)_3)$.
This intermediate loses a water molecule to form formic acid $(HCOOH)$.
Formic acid then reacts with the remaining $NaOH$ to produce sodium formate $(HCOONa)$.
The overall reaction is: $CHCl_3 + 4NaOH \rightarrow HCOONa + 3NaCl + 2H_2O$.

(C) The current gain $\beta$ of a transistor is defined as the ratio of the change in collector current to the change in base current: $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Given:
Change in base current $\Delta I_B = 140 \mu A - 45 \mu A = 95 \mu A = 95 \times 10^{-6} \text{ A}$.
Change in collector current $\Delta I_C = 4.0 \text{ mA} - 0.2 \text{ mA} = 3.8 \text{ mA} = 3.8 \times 10^{-3} \text{ A}$.
Substituting these values into the formula:
$\beta = \frac{3.8 \times 10^{-3}}{95 \times 10^{-6}} = \frac{3800 \times 10^{-6}}{95 \times 10^{-6}} = 40$.
Therefore,the current gain is $40$.

(C) Let the position vectors of the vertices be $\vec{a} = 2i+3j+4k$,$\vec{b} = 3i+4j+2k$,and $\vec{c} = 4i+2j+3k$.
The side vectors are:
$\vec{AB} = \vec{b} - \vec{a} = (3-2)i + (4-3)j + (2-4)k = i + j - 2k$.
$\vec{BC} = \vec{c} - \vec{b} = (4-3)i + (2-4)j + (3-2)k = i - 2j + k$.
$\vec{CA} = \vec{a} - \vec{c} = (2-4)i + (3-2)j + (4-3)k = -2i + j + k$.
The lengths of the sides are:
$|\vec{AB}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$.
$|\vec{BC}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
$|\vec{CA}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$.
Since $|\vec{AB}| = |\vec{BC}| = |\vec{CA}| = \sqrt{6}$,all three sides are equal in length.
Therefore,the triangle is an equilateral triangle.

(A) Let the direction ratios of the two lines $AB$ and $AC$ be $\vec{v_1} = \langle 1, -1, -1 \rangle$ and $\vec{v_2} = \langle 2, -1, 1 \rangle$ respectively.
Since the lines $AB$ and $AC$ lie in the plane $ABC$,the normal vector $\vec{n}$ to the plane is given by the cross product of the vectors along these lines:
$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 2 & -1 & 1 \end{vmatrix}$
$\vec{n} = \hat{i}(-1 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - (-2))$
$\vec{n} = \hat{i}(-2) - \hat{j}(3) + \hat{k}(1)$
$\vec{n} = -2\hat{i} - 3\hat{j} + \hat{k}$
Thus,the direction ratios of the normal are $\langle -2, -3, 1 \rangle$. Multiplying by $-1$,we get $\langle 2, 3, -1 \rangle$.
Therefore,the correct option is $A$.

(C) The equation of a plane passing through the point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = \langle a, b, c \rangle$ is given by $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Given the point is $(-1, 2, 3)$,the equation is $a(x+1) + b(y-2) + c(z-3) = 0$.
The normal makes equal angles $\alpha$ with the coordinate axes,so the direction cosines are $\cos \alpha, \cos \alpha, \cos \alpha$.
Since $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$,we have $3 \cos^2 \alpha = 1$,which implies $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Thus,the direction ratios $\langle a, b, c \rangle$ can be taken as $\langle 1, 1, 1 \rangle$.
Substituting these into the plane equation: $1(x+1) + 1(y-2) + 1(z-3) = 0$.
Simplifying,we get $x + 1 + y - 2 + z - 3 = 0$,which results in $x + y + z - 4 = 0$.

(B) Let the equation of the variable plane be $a(x - 1) + b(y - 2) + c(z - 3) = 0$,where $(a, b, c)$ are the direction ratios of the normal to the plane.
Let $P(x_0, y_0, z_0)$ be the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane.
The vector $\vec{OP} = (x_0, y_0, z_0)$ is perpendicular to the plane,so it is parallel to the normal vector $(a, b, c)$.
Thus,we can write $(a, b, c) = k(x_0, y_0, z_0)$ for some constant $k$.
Since $P$ lies on the plane,it satisfies the equation: $x_0(x_0 - 1) + y_0(y_0 - 2) + z_0(z_0 - 3) = 0$.
This simplifies to $x_0^2 - x_0 + y_0^2 - 2y_0 + z_0^2 - 3z_0 = 0$.
Rearranging the terms,we get $(x_0^2 - x_0 + \frac{1}{4}) + (y_0^2 - 2y_0 + 1) + (z_0^2 - 3z_0 + \frac{9}{4}) = \frac{1}{4} + 1 + \frac{9}{4} = \frac{14}{4} = 3.5$.
This is the equation of a sphere with center $(\frac{1}{2}, 1, \frac{3}{2})$ and radius $\sqrt{3.5}$.

(C) Total number of ways to choose $2$ numbers from $8$ is ${}^8C_2 = \frac{8 \times 7}{2} = 28$.
Let the two numbers be $x$ and $y$ such that $x < y$. We want the smaller number $x$ to be less than $4$,i.e.,$x \in \{1, 2, 3\}$.
Case $I$: If $x = 1$,then $y$ can be any of the remaining $7$ numbers $\{2, 3, 4, 5, 6, 7, 8\}$. Number of ways $= 7$.
Case $II$: If $x = 2$,then $y$ can be any of the remaining $6$ numbers $\{3, 4, 5, 6, 7, 8\}$. Number of ways $= 6$.
Case $III$: If $x = 3$,then $y$ can be any of the remaining $5$ numbers $\{4, 5, 6, 7, 8\}$. Number of ways $= 5$.
Total favorable cases $= 7 + 6 + 5 = 18$.
Required probability $= \frac{18}{28} = \frac{9}{14}$.

(D) Total sample points,$n(S) = 6 \times 6 = 36$.
Favourable events for sum $\ge 10$ are:
$E = [(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)]$.
Total number of favourable events,$n(E) = 6$.
Required probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(A) Let $H$ be the event that the toss results in a head. The bag contains $2n+1$ coins in total.
There are $n$ coins with heads on both sides (biased) and $n+1$ fair coins.
The probability of picking a biased coin is $P(B) = \frac{n}{2n+1}$ and the probability of picking a fair coin is $P(F) = \frac{n+1}{2n+1}$.
If a biased coin is picked,the probability of getting a head is $1$.
If a fair coin is picked,the probability of getting a head is $\frac{1}{2}$.
Using the law of total probability:
$P(H) = P(H|B)P(B) + P(H|F)P(F)$
$\frac{31}{42} = (1) \times \frac{n}{2n+1} + \left(\frac{1}{2}\right) \times \frac{n+1}{2n+1}$
$\frac{31}{42} = \frac{2n + n + 1}{2(2n+1)} = \frac{3n+1}{4n+2}$
$31(4n+2) = 42(3n+1)$
$124n + 62 = 126n + 42$
$20 = 2n$
$n = 10$

(A) In a close packed structure ($hcp$ or $ccp$):
$(i)$ The number of octahedral voids is equal to the number of particles $(N)$ present in the close packing.
(ii) The number of tetrahedral voids is equal to $2 \times N$,where $N$ is the number of particles.
Given that there are '$X$' atoms,the number of octahedral voids is '$X$' and the number of tetrahedral voids is '$2X$'.
Therefore,the correct option is $A$.

(D) According to Raoult's law,the relative lowering of vapour pressure is given by: $\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{n_2}{n_1 + n_2}$
where,$p^{\circ} = \text{vapour pressure of pure water at } 100^{\circ} C = 760 \ mmHg$.
$p_s = \text{vapour pressure of solution at } 100^{\circ} C$.
$n_2 = \text{moles of glucose} = \frac{18}{180} = 0.1 \ mol$.
$n_1 = \text{moles of water} = \frac{180}{18} = 10 \ mol$.
Substituting these values: $\frac{760 - p_s}{760} = \frac{0.1}{10 + 0.1} = \frac{0.1}{10.1}$.
$760 - p_s = 760 \times \frac{0.1}{10.1} = \frac{76}{10.1} \approx 7.524 \ mmHg$.
$p_s = 760 - 7.524 = 752.476 \ mmHg \approx 752.4 \ mmHg$.

(B) Given: Concentration $(C) = 0.10 \ M$,Degree of ionization $(\alpha) = 4.0 \ \% = 0.04$.
For the dissociation of lactic acid: $CH_3CH(OH)COOH \rightleftharpoons CH_3CH(OH)COO^{-} + H^{+}$.
The equilibrium constant $K_a$ is given by the expression: $K_a = \frac{C\alpha^2}{1-\alpha}$.
Substituting the values: $K_a = \frac{0.1 \times (0.04)^2}{1 - 0.04}$.
$K_a = \frac{0.1 \times 0.0016}{0.96} = \frac{0.00016}{0.96} = \frac{1.6 \times 10^{-4}}{0.96} \approx 1.66 \times 10^{-4}$.

(A) Two solutions are isotonic if they have the same molar concentration of particles (osmolarity).
For $0.15 \ M \ NaCl$: $NaCl$ dissociates into $2$ ions ($Na^+$ and $Cl^-$). The concentration of particles $= 0.15 \times 2 = 0.30 \ M$.
For $0.1 \ M \ Na_2SO_4$: $Na_2SO_4$ dissociates into $3$ ions ($2Na^+$ and $SO_4^{2-}$). The concentration of particles $= 0.1 \times 3 = 0.30 \ M$.
Since both solutions have the same concentration of particles,they are isotonic.

(D) The kinetic energy $(KE)$ of $n$ moles of an ideal gas at temperature $T$ is given by $KE = n \times \frac{3}{2} RT$.
For $4 \text{ g}$ of $H_2$: Moles $n_1 = \frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ mol}$.
$KE_{H_2} = 2 \times \frac{3}{2} RT = 3RT$.
For $8 \text{ g}$ of $O_2$: Moles $n_2 = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ mol} = \frac{1}{4} \text{ mol}$.
$KE_{O_2} = \frac{1}{4} \times \frac{3}{2} RT = \frac{3}{8} RT$.
Ratio $KE_{H_2} : KE_{O_2} = 3RT : \frac{3}{8} RT = 1 : \frac{1}{8} = 8 : 1$.

(C) The quantum or wave mechanical model of an atom is based upon the dual nature of the electron,i.e.,the electron exhibits both particle and wave-like properties.
This model incorporates the de Broglie hypothesis regarding the wave nature of matter and the Heisenberg uncertainty principle.