WBJEE 2009 Physics Question Paper with Answer and Solution

(A) The relationship between kinetic energy $(K)$ and linear momentum $(p)$ is given by $K = \frac{p^2}{2m}$,which implies $p = \sqrt{2mK}$.
Given masses are $m_1 = m$ and $m_2 = 4m$.
Given kinetic energies are $K_1$ and $K_2$ such that $\frac{K_1}{K_2} = \frac{2}{1}$.
The ratio of their linear momenta is $\frac{p_1}{p_2} = \sqrt{\frac{2m_1 K_1}{2m_2 K_2}} = \sqrt{\frac{m_1}{m_2} \times \frac{K_1}{K_2}}$.
Substituting the values: $\frac{p_1}{p_2} = \sqrt{\frac{m}{4m} \times \frac{2}{1}} = \sqrt{\frac{1}{4} \times 2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio of their linear momenta is $\frac{1}{\sqrt{2}}$.

(D) The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$.
Let the initial kinetic energy be $K_i$ and the final kinetic energy be $K_f$. Given that the kinetic energy changes by $20 \%$,we have $K_f = 1.20 K_i$.
Since $K \propto p^2$,we can write $\frac{K_f}{K_i} = \left( \frac{p_f}{p_i} \right)^2$.
Substituting the values,$1.20 = \left( \frac{p_f}{p_i} \right)^2$.
Taking the square root on both sides,$\frac{p_f}{p_i} = \sqrt{1.20} \approx 1.0954$.
The percentage change in momentum is given by $\frac{p_f - p_i}{p_i} \times 100 = (1.0954 - 1) \times 100 = 9.54 \%$.
Since $9.54 \%$ is not among the given options,the correct answer is $D$.

(B) From the ideal gas equation,$PV = nRT$,we can express the gas constant $R$ as $R = \frac{PV}{nT}$.
Here,$P$ is pressure $(N m^{-2})$,$V$ is volume $(m^3)$,$n$ is the amount of substance $(mol)$,and $T$ is temperature $(K)$.
The unit of $PV$ is $(N m^{-2}) \times (m^3) = N m = J$ (Joule).
Therefore,the unit of $R$ is $\frac{J}{mol \times K} = J K^{-1} mol^{-1}$.

(C) When two springs with force constants $K_1$ and $K_2$ are connected in series (end to end),the total extension $x$ is the sum of individual extensions $x_1$ and $x_2$.
For a force $F$ applied to the system,$x_1 = F/K_1$ and $x_2 = F/K_2$.
The total extension is $x = x_1 + x_2 = F/K_1 + F/K_2$.
If $K$ is the equivalent force constant,then $x = F/K$.
Therefore,$F/K = F/K_1 + F/K_2$,which simplifies to $1/K = 1/K_1 + 1/K_2$.
Solving for $K$,we get $K = \frac{K_1 K_2}{K_1 + K_2}$.

(D) The force constant $k$ of a spring is inversely proportional to its length $\ell$,i.e.,$k \propto \frac{1}{\ell}$ or $k \ell = \text{constant}$.
When a spring of length $\ell$ and force constant $k$ is cut into two equal halves,the length of each new part becomes $\ell' = \frac{\ell}{2}$.
Let the force constant of each new part be $k'$.
Using the relation $k \ell = k' \ell'$,we get:
$k \ell = k' \left( \frac{\ell}{2} \right)$
$k = \frac{k'}{2}$
$k' = 2k$.
Therefore,the force constant of each half is $2k$.

(C) The terminal velocity $v_T$ of a sphere of radius $r$ and density $\sigma$ falling in a liquid of density $\rho$ and viscosity $\eta$ is given by $v_T = \frac{2r^2(\sigma - \rho)g}{9\eta}$.
Since the masses are equal,$m = \frac{4}{3}\pi r^3 \sigma$,which implies $\sigma \propto \frac{1}{r^3}$.
Substituting $\sigma = \frac{m}{\frac{4}{3}\pi r^3}$ into the terminal velocity formula:
$v_T = \frac{2r^2}{9\eta} \left( \frac{m}{\frac{4}{3}\pi r^3} - \rho \right)g = \frac{2g}{9\eta} \left( \frac{3m}{4\pi r} - r^2\rho \right)$.
For a liquid of infinite column where the viscous force dominates or assuming the density of the sphere is much larger than the liquid $(\sigma \gg \rho)$,we have $v_T \propto r^2 \sigma$.
Since $r^3 \sigma = \text{constant}$,then $\sigma \propto r^{-3}$.
Therefore,$v_T \propto r^2 \cdot r^{-3} = r^{-1} = \frac{1}{r}$.
Thus,$\frac{v_1}{v_2} = \frac{r_2}{r_1}$.

(D) The Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{dD/D}{dL/L}$.
Given $\sigma = 0.5$,we have $0.5 = -\frac{dD/D}{dL/L}$,which implies $\frac{dD}{D} = -0.5 \frac{dL}{L}$.
The cross-sectional area $A$ is given by $A = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
Taking the logarithmic derivative,$\frac{dA}{A} = 2 \frac{dD}{D}$.
Substituting the expression for $\frac{dD}{D}$,we get $\frac{dA}{A} = 2 (-0.5 \frac{dL}{L}) = -\frac{dL}{L}$.
Given that the area decreases by $4 \%$,we have $\frac{dA}{A} = -0.04$.
Therefore,$-0.04 = -\frac{dL}{L}$,which means $\frac{dL}{L} = 0.04$ or $4 \%$.
Thus,the percentage increase in length is $4 \%$.

(D) In projectile motion,the horizontal component of velocity remains constant throughout the flight because there is no acceleration acting in the horizontal direction.
Given the initial velocity is $u$ and the angle of projection is $\theta = 60^{\circ}$.
The horizontal component of the initial velocity is given by $u_x = u \cos(\theta)$.
Substituting the values,we get $u_x = u \cos(60^{\circ})$.
Since $\cos(60^{\circ}) = 1/2$,the horizontal component is $u_x = u \times (1/2) = u/2$.
At the maximum height,the vertical component of velocity becomes zero,but the horizontal component remains unchanged.
Therefore,the horizontal component of the velocity at the maximum height is $u/2$.

(C) The initial kinetic energy of the particle is given by $K = \frac{1}{2} m v^2$,where $v$ is the initial velocity.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,and the velocity of the particle is equal to the horizontal component,which is $v_x = v \cos \theta$.
Given $\theta = 60^{\circ}$,the velocity at the highest point is $v_x = v \cos 60^{\circ} = \frac{v}{2}$.
The kinetic energy at the highest point $K'$ is given by $K' = \frac{1}{2} m (v_x)^2$.
Substituting the value of $v_x$,we get $K' = \frac{1}{2} m (\frac{v}{2})^2 = \frac{1}{2} m (\frac{v^2}{4}) = \frac{1}{4} (\frac{1}{2} m v^2)$.
Since $K = \frac{1}{2} m v^2$,we have $K' = \frac{K}{4}$.

(C) Step $1$: Calculate heat required to bring $5 \ g$ of ice from $-20^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_{ice} \cdot c_{ice} \cdot \Delta T = 5 \ g \times 0.5 \ cal \ g^{-1} (^{\circ} C)^{-1} \times 20^{\circ} C = 50 \ cal$.
Step $2$: Calculate heat required to melt $5 \ g$ of ice at $0^{\circ} C$: $Q_2 = m_{ice} \cdot L_f = 5 \ g \times 80 \ cal \ g^{-1} = 400 \ cal$.
Total heat required to convert ice to water at $0^{\circ} C$ is $Q_{total} = 50 + 400 = 450 \ cal$.
Step $3$: Calculate heat released by $19 \ g$ of water cooling from $30^{\circ} C$ to $0^{\circ} C$: $Q_{released} = m_{water} \cdot c_{water} \cdot \Delta T = 19 \ g \times 1 \ cal \ g^{-1} (^{\circ} C)^{-1} \times 30^{\circ} C = 570 \ cal$.
Since $Q_{released} > Q_{total}$,the ice melts completely and the final temperature $T_f$ will be above $0^{\circ} C$.
Step $4$: Apply the principle of calorimetry: Heat lost by water = Heat gained by ice.
$19 \times 1 \times (30 - T_f) = 450 + 5 \times 1 \times (T_f - 0)$.
$570 - 19 T_f = 450 + 5 T_f$.
$120 = 24 T_f$.
$T_f = 5^{\circ} C$.

(C) The potential energy of the water at the top is converted into heat energy at the bottom.
Using the principle of conservation of energy: $mgh = J \cdot ms \Delta t$,where $J$ is the mechanical equivalent of heat ($J = 4.2 \ J/cal$ or $4186 \ J/kg^{\circ}C$).
Here,$s$ is the specific heat capacity of water,$s = 4186 \ J/kg^{\circ}C$.
Rearranging for $\Delta t$: $\Delta t = \frac{gh}{s}$.
Substituting the values: $\Delta t = \frac{9.8 \times 50}{4186} \approx 0.117^{\circ} C$.

(A) For two rods of equal length $\ell$ and cross-sectional area $A$ joined in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual thermal resistances $R_1$ and $R_2$.
$R_{eq} = R_1 + R_2$
Since $R = \frac{\ell}{KA}$,we have:
$\frac{2\ell}{K_{eff} A} = \frac{\ell}{K_1 A} + \frac{\ell}{K_2 A}$
Canceling $\frac{\ell}{A}$ from both sides:
$\frac{2}{K_{eff}} = \frac{1}{K_1} + \frac{1}{K_2}$
Given $K_1 = 3$ and $K_2 = 4$:
$\frac{2}{K_{eff}} = \frac{1}{3} + \frac{1}{4} = \frac{4+3}{12} = \frac{7}{12}$
$K_{eff} = \frac{24}{7} \approx 3.43$ units.

(C) At high altitudes,the atmospheric pressure is significantly lower than at sea level.
Since the boiling point of a liquid depends on the external pressure,a decrease in atmospheric pressure leads to a decrease in the boiling point of water.
Consequently,water boils at a temperature lower than $100 \, ^\circ\text{C}$ at high altitudes.
Because the water boils at a lower temperature,it does not provide enough heat to cook the rice properly,making the cooking process difficult.

(D) The work done by a variable force is equal to the area under the force-displacement $(F-x)$ graph.
To find the work done in the $1^{\text{st}}$ meter, we calculate the area of the triangle formed between $x = 0$ and $x = 1$.
The base of the triangle is $1 \,m$ and the height is $5 \,N$.
$\text{Work done} = \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Work done} = \frac{1}{2} \times 1 \,m \times 5 \,N = 2.5 \,J$.

(C) The angle of dip $\theta$ is related to the vertical component $B_V$ and the horizontal component $B_H$ of the Earth's magnetic field by the formula: $\tan \theta = \frac{B_V}{B_H}$.
Given that the horizontal component $B_H$ is $\frac{1}{\sqrt{3}}$ times the vertical component $B_V$,we have $B_H = \frac{1}{\sqrt{3}} B_V$.
Substituting this into the formula: $\tan \theta = \frac{B_V}{\frac{1}{\sqrt{3}} B_V} = \sqrt{3}$.
Since $\tan 60^{\circ} = \sqrt{3}$,the angle of dip is $\theta = 60^{\circ}$.

(C) Given: $I_{rms} = 10 \, A$ and $R = 12 \, \Omega$.
The peak current $I_0$ is given by the relation $I_0 = I_{rms} \times \sqrt{2}$.
$I_0 = 10 \times 1.414 = 14.14 \, A$.
The maximum potential difference $V_0$ across the resistor is given by Ohm's law: $V_0 = I_0 \times R$.
$V_0 = 14.14 \times 12 = 169.68 \, V$.

(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Here,the electron jumps from $n_i = 2$ to $n_f = 1$.
Substituting the values: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$.
$\frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$.
Therefore,$\lambda = \frac{4}{3R}$.

(B) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance.
Let the initial charge be $q_i = q$ and the final charge be $q_f = q + 2$.
The initial energy is $U_i = \frac{q^2}{2C}$ and the final energy is $U_f = \frac{(q+2)^2}{2C}$.
Given that the energy increases by $21 \%$,we have $U_f = U_i + 0.21 U_i = 1.21 U_i$.
Substituting the expressions for energy: $\frac{(q+2)^2}{2C} = 1.21 \times \frac{q^2}{2C}$.
Canceling $\frac{1}{2C}$ from both sides,we get $(q+2)^2 = 1.21 q^2$.
Taking the square root of both sides: $q + 2 = 1.1 q$.
Rearranging the terms: $1.1 q - q = 2$,which gives $0.1 q = 2$.
Therefore,$q = \frac{2}{0.1} = 20 \ C$.

(B) Let the capacitance of each capacitor be $C$.
When $4$ capacitors are connected in series,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{4}{C}$,which implies $C_1 = \frac{C}{4}$.
When $4$ capacitors are connected in parallel,the equivalent capacitance $C_2$ is given by $C_2 = C + C + C + C = 4C$.
Now,the ratio $\frac{C_1}{C_2}$ is calculated as $\frac{C/4}{4C} = \frac{1}{16}$.

(A) For two identical cells of emf $E$ and internal resistance $r$ connected in parallel,the equivalent emf $E_{eq} = E$ and the equivalent internal resistance $r_{eq} = r / 2$.
The current $I$ flowing through the external resistance $R$ is given by $I = \frac{E}{R + r_{eq}} = \frac{E}{R + r/2} = \frac{2E}{2R + r}$.
The power $P$ developed across the external resistance $R$ is $P = I^2 R = \left( \frac{2E}{2R + r} \right)^2 R$.
To maximize power,we set $\frac{dP}{dR} = 0$,which leads to the condition $R = r_{eq}$.
Substituting $r_{eq} = r / 2$,we get $R = r / 2$.

(D) The power delivered to the factory is $P_{out} = 10 \text{ kW} = 10000 \text{ W}$ at a voltage $V = 200 \text{ V}$.
The current $I$ flowing through the cable is given by $I = \frac{P_{out}}{V} = \frac{10000}{200} = 50 \text{ A}$.
The power loss in the cable due to its resistance $R = 0.2 \Omega$ is $P_{loss} = I^2 R = (50)^2 \times 0.2 = 2500 \times 0.2 = 500 \text{ W}$.
The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 10000 + 500 = 10500 \text{ W}$.
The efficiency of transmission is $\eta = \frac{P_{out}}{P_{in}} \times 100 = \frac{10000}{10500} \times 100 \approx 95.24 \%$.
Rounding to the nearest given option, the efficiency is $95 \%$.

(C) The mass of the copper wire remains constant. Since $Mass = \text{Volume} \times \text{Density} = (A \times \ell) \times \sigma$,where $A = \pi r^2$ is the cross-sectional area,$\ell$ is the length,and $\sigma$ is the density.
For two wires of the same mass: $\pi r_1^2 \ell_1 \sigma = \pi r_2^2 \ell_2 \sigma \implies \frac{\ell_1}{\ell_2} = \left(\frac{r_2}{r_1}\right)^2$.
Given diameters $d_1 = 1 \ mm$ and $d_2 = 2 \ mm$,the radii ratio is $\frac{r_2}{r_1} = \frac{2}{1} = 2$. Thus,$\frac{\ell_1}{\ell_2} = (2)^2 = 4$.
The resistance $R$ is given by $R = \rho \frac{\ell}{A} = \rho \frac{\ell}{\pi r^2}$.
Therefore,the ratio of resistances is $\frac{R_1}{R_2} = \frac{\ell_1}{\ell_2} \times \left(\frac{r_2}{r_1}\right)^2 = 4 \times (2)^2 = 4 \times 4 = 16$.
Thus,the ratio is $16: 1$.

(A) The resistance $R$ of a wire is given by $R = \rho \frac{\ell}{A}$.
Since the volume $V = A \ell$ remains constant when the wire is stretched,$A = \frac{V}{\ell}$.
Substituting this into the resistance formula,we get $R = \rho \frac{\ell^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto \ell^2$.
Given the new length $\ell_2 = 3 \ell_1$,the new resistance $R_2$ is related to the original resistance $R_1$ by the ratio $\frac{R_2}{R_1} = \left(\frac{\ell_2}{\ell_1}\right)^2$.
Substituting the values: $\frac{R_2}{5 \Omega} = (3)^2 = 9$.
Therefore,$R_2 = 5 \Omega \times 9 = 45 \Omega$.

(B) According to the quantization of charge,the total charge $Q$ is given by $Q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \ C)$.
Given $Q = 2 \ C$.
Rearranging the formula for $n$: $n = \frac{Q}{e}$.
Substituting the values: $n = \frac{2}{1.6 \times 10^{-19}} \ C$.
$n = 1.25 \times 10^{19}$ electrons.

(A) The relationship between current $I$ and charge $Q$ is given by $I = \frac{dQ}{dt}$,which implies $Q = \int I \ dt$.
Given $I = 3t^2 + 2t + 5$,we integrate this expression with respect to time from $t = 0$ to $t = 2 \ s$:
$Q = \int_{0}^{2} (3t^2 + 2t + 5) \ dt$
$Q = [t^3 + t^2 + 5t]_{0}^{2}$
$Q = (2^3 + 2^2 + 5(2)) - (0^3 + 0^2 + 5(0))$
$Q = (8 + 4 + 10) - 0$
$Q = 22 \ C$.

(D) The electrostatic force is a conservative force.
By definition,the work done by a conservative force in moving a charge along any closed path is always zero.
Since the charge $Q$ is moved once around a circle,which is a closed path,the total work done is $0$.

(C) According to Biot-Savart's law,the magnetic field $d \vec{B}$ due to a current element $I \vec{dl}$ at a position vector $\vec{r}$ is given by:
$d \vec{B} = \frac{\mu_0}{4 \pi} \frac{I (\vec{dl} \times \vec{r})}{r^3}$
Since $\vec{r} = r \hat{r}$,we can also write this as $d \vec{B} = \frac{\mu_0}{4 \pi} \frac{I (\vec{dl} \times \hat{r})}{r^2}$.
Comparing this with the given options,option $C$ represents the correct vector form.

(B) The magnetic field $B$ at the centre of a circular loop of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
In the electromagnetic unit (e.m.u.) system,the permeability of free space is defined such that $\frac{\mu_0}{4\pi} = 1$.
The magnetic field intensity $H$ is related to the magnetic field $B$ by $B = \mu_0 H$,which implies $H = \frac{B}{\mu_0}$.
Substituting $B = \frac{\mu_0 I}{2r}$ into the expression for $H$,we get $H = \frac{\mu_0 I / 2r}{\mu_0} = \frac{I}{2r}$.
However,in the e.m.u. system,the current $I$ is measured in abamperes,and the formula for the magnetic field at the centre of a circular loop is $H = \frac{2\pi I}{r}$ oersted.

(D) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ must be conserved on both sides of the equation.
Given reaction: ${ }_4^9 Be + { }_2^4 He \rightarrow { }_6^{12} C + x$
Let the particle $x$ be represented by ${ }_Z^A X$.
Conservation of mass number $(A)$: $9 + 4 = 12 + A \Rightarrow 13 = 12 + A \Rightarrow A = 1$.
Conservation of atomic number $(Z)$: $4 + 2 = 6 + Z \Rightarrow 6 = 6 + Z \Rightarrow Z = 0$.
The particle with mass number $1$ and atomic number $0$ is a neutron,denoted as ${ }_0^1 n$.

(D) For a divergent (concave) lens,the focal length $f$ is taken as negative,so $f_{lens} = -f$ (where $f > 0$).
Given the object distance $u = -mf$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{lens}}$.
Substituting the values: $\frac{1}{v} - \frac{1}{-mf} = \frac{1}{-f}$.
$\frac{1}{v} + \frac{1}{mf} = -\frac{1}{f}$.
$\frac{1}{v} = -\frac{1}{f} - \frac{1}{mf} = -\frac{1}{f} \left(1 + \frac{1}{m}\right) = -\frac{1}{f} \left(\frac{m+1}{m}\right)$.
Therefore,$v = -f \left(\frac{m}{m+1}\right)$.
The linear magnification $M$ is given by $M = \frac{v}{u}$.
$M = \frac{-f \left(\frac{m}{m+1}\right)}{-mf} = \frac{1}{m+1}$.

(A) Given: Object height $h_o = 2.0 \,cm$,Object distance $u = -15 \,cm$,Focal length $f = -10 \,cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} + \frac{1}{-15} = \frac{1}{-10} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} = -\frac{1}{30}$.
So,$v = -30 \,cm$.
The magnification $m = -\frac{v}{u} = -\frac{-30}{-15} = -2$.
Image height $h_i = m \times h_o = -2 \times 2.0 = -4.0 \,cm$.
The negative sign indicates that the image is real and inverted. The size of the image is $4.0 \,cm$.

(D) The wavelength of light in a medium is given by the formula $\lambda_m = \frac{\lambda_0}{n}$,where $\lambda_0$ is the wavelength in vacuum (or air) and $n$ is the refractive index of the medium.
Given:
$\lambda_0 = 4200 Å$
$n = 4/3$
Substituting the values:
$\lambda_m = \frac{4200}{4/3} = 4200 \times \frac{3}{4} = 1050 \times 3 = 3150 Å$.
Therefore,the wavelength of the light in water is $3150 Å$.

(B) In the given circuit,the $p$-side of the diode is connected to a higher potential $(5 \text{ V})$ and the $n$-side is connected to a lower potential $(0 \text{ V})$.
Therefore,the diode is forward biased.
The forward bias resistance of the diode is $R_d = 25 \Omega$.
The external resistor is $R = 10 \Omega$.
The total resistance of the circuit is $R_{eq} = R_d + R = 25 \Omega + 10 \Omega = 35 \Omega$.
The potential difference across the circuit is $V = 5 \text{ V} - 0 \text{ V} = 5 \text{ V}$.
Using Ohm's law,the current $I$ in the circuit is given by $I = \frac{V}{R_{eq}} = \frac{5 \text{ V}}{35 \Omega} = \frac{1}{7} \text{ A}$.

(B) To convert a decimal number to binary,we repeatedly divide the number by $2$ and record the remainders.
$37 \div 2 = 18$ with a remainder of $1$
$18 \div 2 = 9$ with a remainder of $0$
$9 \div 2 = 4$ with a remainder of $1$
$4 \div 2 = 2$ with a remainder of $0$
$2 \div 2 = 1$ with a remainder of $0$
$1 \div 2 = 0$ with a remainder of $1$
Reading the remainders from bottom to top,the binary representation of $37$ is $(100101)_2$.
Counting the digits in $(100101)_2$,we find there are $6$ digits.
Therefore,the correct option is $B$.

(D) Electrical conductivity depends on the number of free electrons available for charge transport. Among the given options,$Copper$ $(Cu)$ is a transition metal with a very high density of free electrons and low resistivity,making it an excellent conductor of electricity. While $Gold$ and $Platinum$ are also good conductors,$Copper$ is widely used in electrical wiring due to its superior conductivity-to-cost ratio. $Silicon$ is a semiconductor,not a conductor.

(C) Let the intensity of each individual wave be $I_0$.
When two waves with the same frequency and amplitude superpose,the resultant intensity $I$ is given by the formula:
$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \delta$
Since the waves are identical,$I_1 = I_2 = I_0$.
Substituting these values:
$I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \delta$
$I = 2I_0 + 2I_0 \cos \delta$
$I = 2I_0 (1 + \cos \delta)$
Using the trigonometric identity $1 + \cos \delta = 2 \cos^2(\delta / 2)$:
$I = 2I_0 (2 \cos^2(\delta / 2))$
$I = 4I_0 \cos^2(\delta / 2)$
Therefore,the resultant intensity is proportional to $\cos^2(\delta / 2)$.