AIIMS 1995 Chemistry Question Paper with Answer and Solution

(C) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_{atm} + h\rho g$. Given that atmospheric pressure $P_{atm} = H\rho g$,we have $P_1 = (H + h)\rho g$.
At the surface,the pressure is $P_2 = P_{atm} = H\rho g$.
Let the radius of the bubble at the bottom be $r$. Its volume is $V_1 = \frac{4}{3}\pi r^3$.
At the surface,the radius becomes $2r$,so the volume is $V_2 = \frac{4}{3}\pi (2r)^3 = 8 \times \frac{4}{3}\pi r^3 = 8V_1$.
Assuming the temperature remains constant,we apply Boyle's Law: $P_1V_1 = P_2V_2$.
Substituting the values: $(H + h)\rho g \times V_1 = H\rho g \times 8V_1$.
Dividing both sides by $\rho g V_1$,we get $H + h = 8H$.
Therefore,$h = 7H$.

(C) For the mass to remain in contact with the surface,the downward acceleration of the surface must not exceed the acceleration due to gravity $g$.
At the highest point of the $SHM$,the acceleration is directed downwards and its magnitude is $a_{max} = \omega^2 A$.
To maintain contact,we require $a_{max} \leq g$.
Thus,$\omega^2 A = g$.
Substituting $\omega = 2\pi n$,where $n$ is the frequency:
$(2\pi n)^2 A = g$
$4\pi^2 n^2 A = g$
$n^2 = \frac{g}{4\pi^2 A}$.
Given $g = 10 \, m/s^2$,$A = 1 \, cm = 0.01 \, m$,and $\pi^2 \approx 10$:
$n^2 = \frac{10}{4 \times 10 \times 0.01} = \frac{10}{0.4} = 25$.
$n = \sqrt{25} = 5 \, Hz$.

(B) For the light ray to retrace its path after reflection from the silvered surface,it must strike the silvered surface normally (at an angle of incidence $0^{\circ}$).
From the geometry of the prism,the angle of refraction $r$ at the first surface is related to the prism angle $A$ by $r = A$.
Applying Snell's law at the first surface:
$1 \cdot \sin i = \mu \cdot \sin r$
Given that the angle of incidence $i = 2A$ and the angle of refraction $r = A$,we substitute these values:
$\sin(2A) = \mu \sin A$
Using the trigonometric identity $\sin(2A) = 2 \sin A \cos A$:
$2 \sin A \cos A = \mu \sin A$
Dividing both sides by $\sin A$ (assuming $A \neq 0$):
$\mu = 2 \cos A$

(B) The amplitudes of the waves are $a_1 = 10 \mu m,$ $a_2 = 4 \mu m,$ and $a_3 = 7 \mu m.$ The phase difference between the $1^{st}$ and $2^{nd}$ wave is $\frac{\pi}{2},$ and between the $2^{nd}$ and $3^{rd}$ wave is $\frac{\pi}{2}.$ Therefore,the phase difference between the $1^{st}$ and $3^{rd}$ wave is $\pi.$
Combining the $1^{st}$ and $3^{rd}$ waves,their resultant amplitude $A_1$ is given by:
$A_1 = \sqrt{a_1^2 + a_3^2 + 2a_1a_3 \cos(\pi)} = \sqrt{10^2 + 7^2 + 2(10)(7)(-1)} = \sqrt{100 + 49 - 140} = \sqrt{9} = 3 \mu m.$
This resultant $A_1$ is in the direction of the $1^{st}$ wave.
Now,combining this resultant $A_1$ with the $2^{nd}$ wave (which has a phase difference of $\frac{\pi}{2}$ relative to the $1^{st}$ wave),the final resultant amplitude $A$ is:
$A = \sqrt{A_1^2 + a_2^2 + 2A_1a_2 \cos(\frac{\pi}{2})} = \sqrt{3^2 + 4^2 + 2(3)(4)(0)} = \sqrt{9 + 16} = \sqrt{25} = 5 \mu m.$

(B) Let the angle of incidence be $i = 2A$ and the angle of refraction be $r_1$.
According to Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r_1} = \frac{\sin 2A}{\sin r_1}$.
Since the light ray returns back through the same path after reflection at the silvered surface,it must strike the silvered surface normally.
This implies the angle of refraction at the second surface is $r_2 = 0$.
For a prism,$A = r_1 + r_2$. Since $r_2 = 0$,we have $r_1 = A$.
Substituting $r_1 = A$ into the Snell's Law equation:
$\mu = \frac{\sin 2A}{\sin A}$
Using the trigonometric identity $\sin 2A = 2 \sin A \cos A$:
$\mu = \frac{2 \sin A \cos A}{\sin A} = 2 \cos A$.

(C) Let the amplitudes be represented as vectors: $\vec{A}_1 = 10\hat{i}$,$\vec{A}_2 = 4\hat{j}$,and $\vec{A}_3 = -7\hat{i}$ (since the phase difference is $\pi/2$ between successive waves).
The resultant amplitude vector $\vec{A}_R = \vec{A}_1 + \vec{A}_2 + \vec{A}_3$.
$\vec{A}_R = (10 - 7)\hat{i} + 4\hat{j} = 3\hat{i} + 4\hat{j}$.
The magnitude of the resultant amplitude is $A_R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, \mu m$.

(B) The equation of motion for a particle in simple harmonic motion is given by $x = A \sin(\omega t)$.
Given amplitude $A = 4\, cm$ and time period $T = 1.2\, s$.
To find the time taken to move from $x = 2\, cm$ to $x = 4\, cm$:
At $x = 2\, cm$,$2 = 4 \sin(\omega t_1) \implies \sin(\omega t_1) = 1/2 \implies \omega t_1 = \pi/6$.
At $x = 4\, cm$,$4 = 4 \sin(\omega t_2) \implies \sin(\omega t_2) = 1 \implies \omega t_2 = \pi/2$.
The time taken to go from $x = 2\, cm$ to $x = 4\, cm$ is $\Delta t = t_2 - t_1 = (\pi/2 - \pi/6) / \omega = (\pi/3) / (2\pi/T) = T/6$.
Since the particle returns from $x = 4\, cm$ to $x = 2\, cm$,the time taken for the return journey is also $T/6$.
Total minimum time $= T/6 + T/6 = T/3$.
Given $T = 1.2\, s$,total time $= 1.2 / 3 = 0.4\, s$.

(C) Let the depth of the lake be $h$ and the radius of the bubble at the bottom be $R$.
At the bottom, the total pressure is $P_{1} = P_{atm} + \rho gh = \rho gH + \rho gh = \rho g(H + h)$.
The volume at the bottom is $V_{1} = \frac{4}{3} \pi R^{3}$.
At the surface, the pressure is $P_{2} = P_{atm} = \rho gH$.
The radius of the bubble becomes $2R$, so the volume at the surface is $V_{2} = \frac{4}{3} \pi (2R)^{3} = 8 \cdot \frac{4}{3} \pi R^{3} = 8V_{1}$.
Assuming the temperature remains constant, we apply Boyle's Law: $P_{1}V_{1} = P_{2}V_{2}$.
$\rho g(H + h) V_{1} = \rho gH (8V_{1})$.
$H + h = 8H$.
$h = 7H$.
Thus, the depth of the lake is $7H$.

(B) Let the three amplitudes be represented as vectors: $\vec{A_1} = 10\hat{i}$,$\vec{A_2} = 4\hat{j}$,and $\vec{A_3} = -7\hat{i}$ (since they have a successive phase difference of $\pi/2$).
The resultant vector $\vec{A} = \vec{A_1} + \vec{A_2} + \vec{A_3} = (10 - 7)\hat{i} + 4\hat{j} = 3\hat{i} + 4\hat{j}$.
The magnitude of the resultant amplitude is $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,\mu m$.

(B) The energy difference between levels $C$ and $A$ is the sum of the energy differences between $C$ and $B$, and $B$ and $A$.
$E_C - E_A = (E_C - E_B) + (E_B - E_A)$
Using the relation $E = \frac{hc}{\lambda}$, we can write:
$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$
Dividing both sides by $hc$, we get:
$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$
$\frac{1}{\lambda_3} = \frac{\lambda_2 + \lambda_1}{\lambda_1 \lambda_2}$
Therefore, $\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.