IIT JEE 2002 Chemistry Question Paper with Answer and Solution

(C) Initial energy of the system is $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2$.
When the capacitors are connected in parallel,the common potential is $V = \frac{CV_1 + CV_2}{2C} = \frac{V_1 + V_2}{2}$.
Final energy of the system is $U_f = \frac{1}{2}(2C)V^2 = C \left( \frac{V_1 + V_2}{2} \right)^2 = \frac{1}{4}C(V_1 + V_2)^2$.
The decrease in energy is $\Delta U = U_i - U_f = \frac{1}{2}C(V_1^2 + V_2^2) - \frac{1}{4}C(V_1 + V_2)^2$.
Simplifying this,$\Delta U = \frac{1}{4}C [2V_1^2 + 2V_2^2 - (V_1^2 + V_2^2 + 2V_1V_2)] = \frac{1}{4}C(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4}C(V_1 - V_2)^2$.

(C) Polyphosphates (such as sodium hexametaphosphate or $STPP$) are used as water softening agents because they form soluble complexes with metal cations like $Ca^{2+}$ and $Mg^{2+}$ present in hard water,thereby preventing them from forming precipitates with soap or detergents.

(C) The slope of $QP$ is $0$ (since $P$ and $Q$ lie on the $x$-axis,specifically the negative $x$-axis).
The slope of $QR$ is $\frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$.
This means the angle made by $QR$ with the positive $x$-axis is $60^\circ$.
Since $QP$ lies along the negative $x$-axis,the angle $\angle PQR$ is $180^\circ - 60^\circ = 120^\circ$.
The angle bisector of $\angle PQR$ will make an angle of $120^\circ - 60^\circ = 60^\circ$ with $QP$ (the negative $x$-axis).
Thus,the angle made by the bisector with the positive $x$-axis is $180^\circ - 60^\circ = 120^\circ$.
The equation of a line passing through the origin $(0, 0)$ with slope $m = \tan(120^\circ) = -\sqrt{3}$ is $y = -\sqrt{3}x$,which simplifies to $\sqrt{3}x + y = 0$.

(C) The scalar triple product is defined as $[\vec{U} \, \vec{V} \, \vec{W}] = \vec{U} \cdot (\vec{V} \times \vec{W})$.
First,calculate the cross product $\vec{V} \times \vec{W}$:
$\vec{V} \times \vec{W} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix} = \hat{i}(3 - 0) - \hat{j}(6 - (-1)) + \hat{k}(0 - 1) = 3\hat{i} - 7\hat{j} - \hat{k}$.
Let $\vec{A} = \vec{V} \times \vec{W} = 3\hat{i} - 7\hat{j} - \hat{k}$.
Then $[\vec{U} \, \vec{V} \, \vec{W}] = \vec{U} \cdot \vec{A} = |\vec{U}| |\vec{A}| \cos \theta$,where $\theta$ is the angle between $\vec{U}$ and $\vec{A}$.
Since $\vec{U}$ is a unit vector,$|\vec{U}| = 1$.
Thus,the scalar triple product is $|\vec{A}| \cos \theta$.
The maximum value occurs when $\cos \theta = 1$,which is $|\vec{A}|$.
$|\vec{A}| = \sqrt{3^2 + (-7)^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}$.
Therefore,the maximum value is $\sqrt{59}$.

(C) According to Charles's Law,for a fixed amount of gas at constant pressure,the volume is directly proportional to the absolute temperature: $V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 22.4 \ L$,$T_1 = 273 \ K$,$T_2 = 373 \ K$.
We need to find $V_2$:
$V_2 = V_1 \times \frac{T_2}{T_1} = 22.4 \ L \times \frac{373 \ K}{273 \ K} \approx 30.6 \ L$.
Thus,the plot showing the point $(30.6 \ L, 373 \ K)$ is correct.

(A) The reaction of an internal alkyne with $Hg^{2+}/H^{+}$ (Kuccherov reaction) involves the hydration of the alkyne to form a ketone.
For the unsymmetrical alkyne $Ph-C \equiv C-CH_3$,the hydration can occur at either carbon of the triple bond.
However,the formation of the ketone is governed by the stability of the intermediate carbocation or the electronic effects of the substituents.
In this case,the reaction proceeds to form $Ph-CO-CH_2-CH_3$ (propiophenone derivative) as the major product,which corresponds to the structure in image $287-$a474.

(B) The range of the expression $a\cos x + b\sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
For the equation $7\cos x + 5\sin x = 2k + 1$,the range of $7\cos x + 5\sin x$ is $[-\sqrt{7^2 + 5^2}, \sqrt{7^2 + 5^2}] = [-\sqrt{74}, \sqrt{74}]$.
Since $\sqrt{74} \approx 8.6$,the condition for a solution is $-\sqrt{74} \le 2k + 1 \le \sqrt{74}$.
Substituting the approximate value: $-8.6 \le 2k + 1 \le 8.6$.
Subtracting $1$ from all parts: $-9.6 \le 2k \le 7.6$.
Dividing by $2$: $-4.8 \le k \le 3.8$.
The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.
Counting these,we find there are $8$ such values.

(C) Let the point $P$ be $(x_1, y_1)$. The equation of the tangent at $P$ to the circle $x^2 + y^2 + 6x + 6y - 2 = 0$ is given by $x x_1 + y y_1 + 3(x + x_1) + 3(y + y_1) - 2 = 0$.
Since the point $Q$ lies on the $y$-axis,its $x$-coordinate is $0$.
Given that $Q$ lies on the line $5x - 2y + 6 = 0$,substituting $x = 0$ gives $5(0) - 2y + 6 = 0$,which implies $y = 3$. Thus,$Q = (0, 3)$.
Since $Q$ lies on the tangent,it must satisfy the tangent equation: $0(x_1) + 3(y_1) + 3(0 + x_1) + 3(3 + y_1) - 2 = 0$.
Simplifying this,we get $3y_1 + 3x_1 + 9 + 3y_1 - 2 = 0$,or $3x_1 + 6y_1 + 7 = 0$.
The length $PQ$ is given by $\sqrt{(x_1 - 0)^2 + (y_1 - 3)^2} = \sqrt{x_1^2 + y_1^2 - 6y_1 + 9}$.
Since $P$ lies on the circle,$x_1^2 + y_1^2 = 2 - 6x_1 - 6y_1$.
Substituting this into the expression for $PQ^2$: $PQ^2 = (2 - 6x_1 - 6y_1) - 6y_1 + 9 = 11 - 6x_1 - 12y_1 = 11 - 2(3x_1 + 6y_1)$.
Substituting $3x_1 + 6y_1 = -7$,we get $PQ^2 = 11 - 2(-7) = 11 + 14 = 25$.
Therefore,$PQ = 5$.

(D) Let the equation of the tangent to the parabola $y^2 = 8x$ be $y = mx + \frac{2}{m}$,where $m \neq 0$.
This line is also a tangent to the hyperbola $xy = -1$,which can be written as $y = -\frac{1}{x}$.
Substituting $y = mx + \frac{2}{m}$ into $xy = -1$,we get $x(mx + \frac{2}{m}) = -1$,which simplifies to $mx^2 + \frac{2}{m}x + 1 = 0$.
For the line to be a tangent,the discriminant of this quadratic equation must be zero:
$D = (\frac{2}{m})^2 - 4(m)(1) = 0$
$\frac{4}{m^2} - 4m = 0$
$4 = 4m^3$ $\Rightarrow m^3 = 1$ $\Rightarrow m = 1$.
Substituting $m = 1$ into the tangent equation $y = mx + \frac{2}{m}$,we get $y = 1x + \frac{2}{1}$,which is $y = x + 2$.

(A) By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality,for positive real numbers $a_1, a_2, \dots, a_{n-1}, 2a_n$:
$\frac{a_1 + a_2 + \dots + a_{n-1} + 2a_n}{n} \geq (a_1 \cdot a_2 \cdot \dots \cdot a_{n-1} \cdot 2a_n)^{1/n}$
Given that the product $a_1 \cdot a_2 \cdot \dots \cdot a_n = c$,we can substitute this into the inequality:
$\frac{a_1 + a_2 + \dots + a_{n-1} + 2a_n}{n} \geq (2 \cdot a_1 \cdot a_2 \cdot \dots \cdot a_n)^{1/n}$
$\frac{a_1 + a_2 + \dots + a_{n-1} + 2a_n}{n} \geq (2c)^{1/n}$
$a_1 + a_2 + \dots + a_{n-1} + 2a_n \geq n(2c)^{1/n}$
Thus,the minimum value is $n(2c)^{1/n}$.

(B) Let the two equal charges be $q$. The potential energy $U$ of charge $Q$ at a position $x$ is given by the sum of potentials due to the two charges: $U(x) = k q Q [\frac{1}{a-x} + \frac{1}{a+x}]$,where $k = \frac{1}{4 \pi \epsilon_0}$.
Simplifying the expression: $U(x) = k q Q [\frac{a+x+a-x}{a^2-x^2}] = k q Q [\frac{2a}{a^2-x^2}]$.
The potential energy at the origin $(x=0)$ is $U(0) = k q Q [\frac{2a}{a^2}] = \frac{2 k q Q}{a}$.
The change in potential energy is $\Delta U = U(x) - U(0) = k q Q [\frac{2a}{a^2-x^2} - \frac{2}{a}] = 2 k q Q a [\frac{1}{a^2-x^2} - \frac{1}{a^2}]$.
$\Delta U = 2 k q Q a [\frac{a^2 - (a^2-x^2)}{a^2(a^2-x^2)}] = 2 k q Q a [\frac{x^2}{a^2(a^2-x^2)}]$.
For small displacement $x$,$a^2-x^2 \approx a^2$. Thus,$\Delta U \approx 2 k q Q a [\frac{x^2}{a^4}] = \frac{2 k q Q}{a^3} x^2$.
Therefore,$\Delta U \propto x^2$.

(B) When a charged particle enters a perpendicular magnetic field,it follows a circular path of radius $r$ given by the balance between the magnetic force and the centripetal force:
$qvB = \frac{mv^2}{r}$
Rearranging for velocity,we get:
$v = \frac{qBr}{m}$
For the particle to just enter the region $x > b$,the radius of its circular path must be at least equal to the width of the magnetic field region,which is $(b - a)$.
Thus,the minimum radius required is $r_{\min} = b - a$.
Substituting this into the velocity equation:
$v_{\min} = \frac{qB(b - a)}{m}$
Therefore,the correct option is $B$.

(A) The current $I$ flows in the negative $z-$ direction,i.e.,along $-\hat k$. The position vector of the point $P(x, y)$ is $\vec r = x\hat i + y\hat j$.
Using the Biot-Savart law or the right-hand rule for a long straight wire,the magnetic field $\vec B$ is given by $\vec B = \frac{\mu_0 I}{2\pi r^2} (\hat k \times \vec r)$.
Substituting $\vec r = x\hat i + y\hat j$ and the current direction $-\hat k$,the magnetic field is $\vec B = \frac{\mu_0 I}{2\pi r^2} ((-\hat k) \times (x\hat i + y\hat j))$.
Since $\hat k \times \hat i = \hat j$ and $\hat k \times \hat j = -\hat i$,we get $\vec B = \frac{\mu_0 I}{2\pi r^2} (-(x\hat j - y\hat i)) = \frac{\mu_0 I}{2\pi r^2} (y\hat i - x\hat j)$.
Substituting $r^2 = x^2 + y^2$,we get $\vec B = \frac{\mu_0 I (y\hat i - x\hat j)}{2\pi (x^2 + y^2)}$.

(A) The magnetic field $\vec B$ due to a long straight wire carrying current $I$ in the negative $z$-direction (along $-\hat k$) at a point $\vec r = x\hat i + y\hat j$ is given by the Biot-Savart law or Ampere's law. The direction of the magnetic field is given by the right-hand thumb rule. For a current in the $-\hat k$ direction,the magnetic field lines are clockwise in the $xy$-plane.
The magnitude of the magnetic field is $B = \frac{\mu_0 I}{2\pi r}$,where $r = \sqrt{x^2 + y^2}$.
The unit vector in the radial direction is $\hat r = \frac{x\hat i + y\hat j}{r}$.
The magnetic field vector is $\vec B = B (\hat k \times \hat r) = \frac{\mu_0 I}{2\pi r} (\hat k \times \frac{x\hat i + y\hat j}{r}) = \frac{\mu_0 I}{2\pi r^2} (x(\hat k \times \hat i) + y(\hat k \times \hat j))$.
Since $\hat k \times \hat i = \hat j$ and $\hat k \times \hat j = -\hat i$,we get $\vec B = \frac{\mu_0 I}{2\pi (x^2 + y^2)} (x\hat j - y\hat i) = \frac{\mu_0 I (y\hat i - x\hat j)}{2\pi (x^2 + y^2)}$ is incorrect; let's re-evaluate: $\vec B = \frac{\mu_0 I}{2\pi (x^2 + y^2)} (y\hat i - x\hat j)$.
Wait,for current in $-\hat k$,the field is $\vec B = \frac{\mu_0 I}{2\pi r^2} (y\hat i - x\hat j)$. Thus,option $A$ is correct.

(A) The frequency of a vibrating string in a sonometer is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the number of antinodes,$L$ is the length of the wire,$T$ is the tension $(T = Mg)$,and $\mu$ is the linear mass density.
In the first case,$n_1 = 5$ and $T_1 = 9g$. Thus,$f = \frac{5}{2L} \sqrt{\frac{9g}{\mu}}$.
In the second case,$n_2 = 3$ and $T_2 = Mg$. Thus,$f = \frac{3}{2L} \sqrt{\frac{Mg}{\mu}}$.
Since the tuning fork is the same,the frequency $f$ remains constant. Equating the two expressions:
$\frac{5}{2L} \sqrt{\frac{9g}{\mu}} = \frac{3}{2L} \sqrt{\frac{Mg}{\mu}}$
$5 \sqrt{9g} = 3 \sqrt{Mg}$
$5 \times 3 \sqrt{g} = 3 \sqrt{Mg}$
$5 \sqrt{g} = \sqrt{Mg}$
Squaring both sides: $25g = Mg$
$M = 25 \, kg$.

(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path with radius $r = \frac{mv}{qB}$.
The magnetic field exists in the region $a \leq x \leq b$. The width of this region is $d = b - a$.
For the particle to emerge from the region $x > b$,the radius of the circular path $r$ must be greater than the width of the magnetic field region $d$.
If $r \leq (b - a)$,the particle will be deflected back into the region $x < a$ and will not reach $x > b$.
Therefore,the condition to just enter the region $x > b$ is $r = b - a$.
Substituting the expression for $r$,we get $\frac{mv}{qB} = b - a$.
Solving for $v$,we find the minimum velocity $v = \frac{q(b - a)B}{m}$.

(D) Initially, the block and the coin float together. The total weight is $W = (M_{block} + m_{coin})g$. According to Archimedes' principle, the buoyant force equals the total weight, so the volume of water displaced is $V_{displaced} = (M_{block} + m_{coin}) / \rho_{water}$.
When the coin falls into the water, it sinks to the bottom. The block now only supports its own weight $M_{block}g$. The new volume of water displaced by the block is $V'_{block} = M_{block} / \rho_{water}$.
The coin, now at the bottom, displaces water equal to its own volume $V_{coin} = m_{coin} / \rho_{coin}$.
Since the density of the coin $\rho_{coin}$ is much greater than the density of water $\rho_{water}$, the volume of water displaced by the coin when submerged $(m_{coin} / \rho_{coin})$ is less than the volume of water it displaced when floating on the block $(m_{coin} / \rho_{water})$.
Therefore, the total volume of water displaced decreases, causing the water level $h$ to decrease.
Since the block now supports less weight, it rises, meaning the submerged depth $l$ of the block decreases.

(B) The equation is of the form $a \cos x + b \sin x = c$,which has a solution if and only if $-\sqrt{a^2 + b^2} \leq c \leq \sqrt{a^2 + b^2}$.
Here,$a = 7$,$b = 5$,and $c = 2k + 1$.
Thus,$-\sqrt{7^2 + 5^2} \leq 2k + 1 \leq \sqrt{7^2 + 5^2}$.
$-\sqrt{49 + 25} \leq 2k + 1 \leq \sqrt{49 + 25}$.
$-\sqrt{74} \leq 2k + 1 \leq \sqrt{74}$.
Since $\sqrt{74} \approx 8.602$,we have $-8.602 \leq 2k + 1 \leq 8.602$.
Subtracting $1$ from all sides: $-9.602 \leq 2k \leq 7.602$.
Dividing by $2$: $-4.801 \leq k \leq 3.801$.
The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.
The total number of such integral values is $8$.

(C) The primary requirement for an electrolyte in a salt-bridge is that the transport numbers (or ionic mobilities) of the cation and anion should be approximately equal.
In a $KNO_3$ solution,the ionic velocities of $K^+$ and $NO_3^-$ are nearly the same.
This ensures that the junction potential is minimized and the electrical neutrality is maintained efficiently across the two half-cells.
Therefore,option $(C)$ is correct.

(A) $(C_3H_8O)$ is a primary alcohol.
Since it oxidizes to an aldehyde $B$ $(C_3H_6O)$ that gives a positive Tollens' test (shining silver mirror),$B$ must be propanal.
$CH_3CH_2CH_2OH \xrightarrow{K_2Cr_2O_7/H^{+}} CH_3CH_2CHO$
$A$ (Propan$-1-$ol) $\rightarrow$ $B$ (Propanal)
Propanal reacts with semicarbazide $(NH_2NHCONH_2)$ in the presence of sodium acetate to form propanal semicarbazone $(C)$:
$CH_3CH_2CHO + NH_2NHCONH_2 \rightarrow CH_3CH_2CH=NNHCONH_2 + H_2O$
Thus,the structure of $C$ is $CH_3CH_2CH=NNHCONH_2$.

(C) The dissociation constant $(K_a)$ of a carboxylic acid is directly proportional to its acidity. The acidity is increased by the electron-withdrawing inductive effect ($-I$ effect) of substituents.
$1$. Comparing the substituents: $F$ has a stronger $-I$ effect than $Br$.
$2$. Comparing the position: The closer the substituent is to the $-COOH$ group,the stronger the $-I$ effect.
$3$. In $CH_3CHFCOOH$ and $CH_3CHBrCOOH$,the halogen is on the $\alpha$-carbon,exerting a strong $-I$ effect.
$4$. In $FCH_2CH_2COOH$ and $BrCH_2CH_2COOH$,the halogen is on the $\beta$-carbon,exerting a weaker $-I$ effect.
$5$. Between $FCH_2CH_2COOH$ and $BrCH_2CH_2COOH$,$Br$ has a weaker $-I$ effect than $F$.
Therefore,$BrCH_2CH_2COOH$ has the weakest $-I$ effect,making it the least acidic and thus having the smallest dissociation constant $(K_a)$.

(B) The boiling point depends on the intermolecular forces present in the compounds.
$CH_3CH_2CH_2COOH$ $(3)$ contains a carboxylic acid group,which forms strong intermolecular hydrogen bonds,often existing as dimers,leading to the highest boiling point.
$CH_3CH_2CH_2CH_2OH$ $(1)$ contains a hydroxyl group,which also forms intermolecular hydrogen bonds,but they are generally weaker than those in carboxylic acids.
$CH_3CH_2CH_2CHO$ $(2)$ contains an aldehyde group,which is polar but cannot form intermolecular hydrogen bonds,resulting in the lowest boiling point among the three.
Therefore,the correct order of boiling points is $3 > 1 > 2$.