AIEEE 2002 Chemistry Question Paper with Answer and Solution

(B) Let the two charges $Q$ be at points $A$ and $B$,separated by a distance $x$. The charge $q$ is placed at point $C$,the midpoint of $AB$,so $AC = CB = x/2$.
For the system to be in equilibrium,the net force on each charge must be zero.
Consider the equilibrium of charge $Q$ at point $A$. The force exerted by charge $Q$ at $B$ on $A$ is $F_{BA} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{x^2}$ (repulsive).
The force exerted by charge $q$ at $C$ on $A$ is $F_{CA} = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{(x/2)^2}$ (attractive,so $q$ must be negative).
For equilibrium,$F_{BA} + F_{CA} = 0$,so $\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{x^2} + \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{x^2/4} = 0$.
$\frac{Q^2}{x^2} + \frac{4qQ}{x^2} = 0 \Rightarrow Q + 4q = 0 \Rightarrow q = -\frac{Q}{4}$.

(B) For an ideal transformer,the relationship between the number of turns $(N)$ and the current $(I)$ is given by the inverse ratio: $\frac{N_{1}}{N_{2}} = \frac{I_{2}}{I_{1}}$.
Given: $N_{1} = 140$,$N_{2} = 280$,and $I_{1} = 4 \, A$.
Substituting the values into the formula: $\frac{140}{280} = \frac{I_{2}}{4}$.
Simplifying the fraction: $\frac{1}{2} = \frac{I_{2}}{4}$.
Solving for $I_{2}$: $I_{2} = \frac{4}{2} = 2 \, A$.

(B) The angular momentum $L_O$ of a particle about a point $O$ is defined as the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p} = m\vec{v}$.
Mathematically,$L_O = |\vec{r} \times \vec{p}| = m v d_{\perp}$,where $d_{\perp}$ is the perpendicular distance (lever arm) from the point $O$ to the line of motion of the particle.
From the given figure,the perpendicular distance from point $O$ to the line $PC$ is $l$.
Therefore,the angular momentum of the particle about $O$ is $L_O = mvl$.

(D) Since the inclined plane is frictionless,there is no rolling motion,and the objects will only slide down the plane.
The force acting along the inclined plane is $F = mg \sin \theta$.
According to Newton's second law,$F = ma$,so $ma = mg \sin \theta$.
Thus,the acceleration is $a = g \sin \theta$.
Since this acceleration depends only on the gravitational acceleration $g$ and the angle of inclination $\theta$,it is independent of the shape or mass of the object.
Therefore,the acceleration is the same for the solid sphere,the hollow sphere,and the ring.

(C) In $S.H.M.$,the kinetic energy of a particle at any displacement $x$ is given by:
$K.E. = \frac{1}{2} m \omega^{2} (a^{2} - x^{2})$
The potential energy of a particle at any displacement $x$ is given by:
$P.E. = \frac{1}{2} m \omega^{2} x^{2}$
where $a$ is the amplitude of the particle and $x$ is the displacement from the mean position.
At the mean position,the displacement $x = 0$.
Substituting $x = 0$ into the equations:
$K.E. = \frac{1}{2} m \omega^{2} a^{2}$ (which is the maximum value).
$P.E. = \frac{1}{2} m \omega^{2} (0)^{2} = 0$ (which is the minimum value).
Therefore,at the mean position,kinetic energy is maximum and potential energy is minimum.

(C) The total ratio sum is $9 + 1 + 3.5 = 13.5$.
Calculate the moles of each element in $108 \ g$ of the compound:
Moles of $C = \frac{(9 / 13.5) \times 108}{12} = \frac{0.6667 \times 108}{12} = 6 \ mol$.
Moles of $H = \frac{(1 / 13.5) \times 108}{1} = \frac{0.0741 \times 108}{1} = 8 \ mol$.
Moles of $N = \frac{(3.5 / 13.5) \times 108}{14} = \frac{0.2593 \times 108}{14} = 2 \ mol$.
Thus,the molecular formula is $C_6H_8N_2$.

(B) For an ideal transformer,the power input equals the power output,which implies $V_{1}I_{1} = V_{2}I_{2}$.
Since the voltage ratio is proportional to the turns ratio,$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$.
Combining these,we get the relationship $\frac{N_{1}}{N_{2}} = \frac{I_{2}}{I_{1}}$.
Given $N_{1} = 140$,$N_{2} = 280$,and $I_{1} = 4\,A$.
Substituting the values: $\frac{140}{280} = \frac{I_{2}}{4}$.
$\frac{1}{2} = \frac{I_{2}}{4}$.
$I_{2} = \frac{4}{2} = 2\,A$.

(B) The magnetic field $B$ produced by the long straight wire $1$ carrying current $i_1$ at a distance $r$ is given by the formula $B = \frac{\mu_0 i_1}{2\pi r}$.
This magnetic field is directed perpendicular to the plane containing the wires.
The force $dF$ on a small current element $i_2 \vec{dl}$ placed in a magnetic field $\vec{B}$ is given by $d\vec{F} = i_2 (\vec{dl} \times \vec{B})$.
The magnitude of this force is $dF = i_2 dl B \sin \alpha$,where $\alpha$ is the angle between the current element $dl$ and the magnetic field $B$.
Since the magnetic field $B$ is perpendicular to the plane of the wires,it is perpendicular to the current element $dl$ as well. Therefore,$\alpha = 90^\circ$ and $\sin 90^\circ = 1$.
Thus,the magnitude of the force is $dF = i_2 dl B = i_2 dl \left( \frac{\mu_0 i_1}{2\pi r} \right) = \frac{\mu_0 i_1 i_2}{2\pi r} dl$.

(C) The sum of the weight ratios is $9 + 1 + 3.5 = 13.5$.
To find the number of moles of each element in $108 \ g$ of the compound:
For $C$: $\text{Moles} = \frac{(9 / 13.5) \times 108}{12} = \frac{0.6667 \times 108}{12} = 6$.
For $H$: $\text{Moles} = \frac{(1 / 13.5) \times 108}{1} = \frac{0.0741 \times 108}{1} = 8$.
For $N$: $\text{Moles} = \frac{(3.5 / 13.5) \times 108}{14} = \frac{0.2593 \times 108}{14} = 2$.
Thus,the molecular formula is $C_6H_8N_2$.