AIPMT 2014 Chemistry Question Paper with Answer and Solution

(D) In a $P-V$ diagram,the work done in a cyclic process is equal to the net area enclosed by the cycle. The work is positive if the cycle is clockwise and negative if the cycle is anticlockwise.
From the figure,the cycle consists of two triangles: $\triangle A E D$ and $\triangle B E C$,where $E$ is the intersection point $(2 P_0, 1.5 V_0)$.
$1$. For the loop $\triangle A E D$ (clockwise):
Work $W_1 = + \text{Area of } \triangle A E D = + \frac{1}{2} \times \text{base} \times \text{height} = + \frac{1}{2} \times (2 V_0 - V_0) \times (2 P_0 - P_0) = + \frac{1}{2} P_0 V_0$.
$2$. For the loop $\triangle B E C$ (anticlockwise):
Work $W_2 = - \text{Area of } \triangle B E C = - \frac{1}{2} \times \text{base} \times \text{height} = - \frac{1}{2} \times (2 V_0 - V_0) \times (3 P_0 - 2 P_0) = - \frac{1}{2} P_0 V_0$.
The net work done by the system is $W_{\text{net}} = W_1 + W_2 = + \frac{1}{2} P_0 V_0 - \frac{1}{2} P_0 V_0 = 0$.

(B) Given:
Efficiency of the transformer,$\eta = 90\% = 0.9$
Input power,$P_{in} = 3\, kW = 3000\, W$
Primary voltage,$V_{p} = 200\, V$
Secondary current,$I_{s} = 6\, A$
$1$. To find the current in the primary coil $(I_{p})$:
Since $P_{in} = V_{p} \times I_{p}$,we have:
$I_{p} = \frac{P_{in}}{V_{p}} = \frac{3000\, W}{200\, V} = 15\, A$
$2$. To find the voltage across the secondary coil $(V_{s})$:
Efficiency $\eta = \frac{P_{out}}{P_{in}} = \frac{V_{s} \times I_{s}}{P_{in}}$
$0.9 = \frac{V_{s} \times 6\, A}{3000\, W}$
$V_{s} = \frac{0.9 \times 3000}{6} = \frac{2700}{6} = 450\, V$
Thus,the secondary voltage is $450\, V$ and the primary current is $15\, A$.

(A) Let the upthrust of air be $F_a$.
For the downward motion of the balloon with mass $m$ and acceleration $a$:
$mg - F_a = ma$ --- $(1)$
From this,we get $F_a = m(g - a)$.
Now,let the mass removed be $\Delta m$. The new mass of the balloon is $(m - \Delta m)$.
For the upward motion with acceleration $a$:
$F_a - (m - \Delta m)g = (m - \Delta m)a$
$F_a = (m - \Delta m)(g + a)$ --- $(2)$
Equating the expressions for $F_a$ from $(1)$ and $(2)$:
$m(g - a) = (m - \Delta m)(g + a)$
$mg - ma = mg + ma - \Delta m(g + a)$
$\Delta m(g + a) = 2ma$
$\Delta m = \frac{2ma}{g + a}$

(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_0 \right)$,where $T_0$ is the surrounding temperature.
For the first interval: $\frac{70 - 60}{5} = K \left( \frac{70 + 60}{2} - T_0 \right) \Rightarrow 2 = K(65 - T_0)$ --- $(1)$
For the second interval: $\frac{60 - 54}{5} = K \left( \frac{60 + 54}{2} - T_0 \right) \Rightarrow 1.2 = K(57 - T_0)$ --- $(2)$
Dividing $(1)$ by $(2)$: $\frac{2}{1.2} = \frac{65 - T_0}{57 - T_0} \Rightarrow \frac{5}{3} = \frac{65 - T_0}{57 - T_0}$.
$5(57 - T_0) = 3(65 - T_0) \Rightarrow 285 - 5T_0 = 195 - 3T_0$.
$285 - 195 = 5T_0 - 3T_0 \Rightarrow 90 = 2T_0 \Rightarrow T_0 = 45\,^{\circ}C$.

(A) The given graph shows the $V-I$ characteristic curve for a solar cell in the fourth quadrant.
In this quadrant,the device acts as a source of power.
Point $A$ lies on the voltage axis where the current is zero,which corresponds to the open circuit voltage $(V_{oc})$.
Point $B$ lies on the current axis where the voltage is zero,which corresponds to the short circuit current $(I_{sc})$.
Therefore,the correct statement is that it is the $V-I$ characteristic for a solar cell where point $A$ represents open circuit voltage and point $B$ represents short circuit current.

(C) When small drops coalesce into a larger drop,the total surface area decreases,which results in the release of energy.
Let $n$ be the number of small drops of radius $r$. The volume is conserved,so $V = n \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi R^3$,which implies $n = \frac{R^3}{r^3}$.
The initial surface area $A_i = n (4 \pi r^2) = \left( \frac{R^3}{r^3} \right) (4 \pi r^2) = 4 \pi R^3 / r$.
The final surface area $A_f = 4 \pi R^2$.
The change in surface area $\Delta A = A_i - A_f = 4 \pi R^3 / r - 4 \pi R^2 = 4 \pi R^2 \left( \frac{R}{r} - 1 \right)$.
Energy released $\Delta E = T \Delta A = T (4 \pi R^2) \left( \frac{R}{r} - 1 \right) = T (4 \pi R^3) \left( \frac{1}{r} - \frac{1}{R} \right)$.
Since $V = \frac{4}{3} \pi R^3$,we have $4 \pi R^3 = 3V$.
Substituting this,$\Delta E = 3VT \left( \frac{1}{r} - \frac{1}{R} \right)$.

(A) The equation of the trajectory of a projectile is given by $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
For two trajectories to be identical,the coefficients of $x$ and $x^2$ must be the same.
Since the angle $\theta$ is the same,$\tan \theta$ is constant.
For the $x^2$ term to be identical,$\frac{g_e}{u_e^2} = \frac{g_p}{u_p^2}$,where $g_e$ and $u_e$ are gravity and velocity on Earth,and $g_p$ and $u_p$ are gravity and velocity on the planet.
Given $u_e = 5\, ms^{-1}$,$u_p = 3\, ms^{-1}$,and $g_e = 9.8\, ms^{-2}$.
Substituting the values: $\frac{9.8}{5^2} = \frac{g_p}{3^2}$.
$g_p = 9.8 \times \frac{9}{25} = 9.8 \times 0.36 = 3.528\, ms^{-2}$.
Rounding to the nearest provided option,we get $3.5\, ms^{-2}$.