IIT JEE 2006 Physics Question Paper with Answer and Solution

(A) For a monatomic ideal gas, the molar heat capacity $C$ for a polytropic process $PV^x = \text{constant}$ is given by $C = C_V + \frac{R}{1-x}$.
Given the process is $P/V = 1$, which implies $P = V^1$, or $PV^{-1} = \text{constant}$.
Comparing this with $PV^x = \text{constant}$, we get $x = -1$.
For a monatomic gas, the molar heat capacity at constant volume is $C_V = \frac{3R}{2}$.
Substituting the values into the formula: $C = \frac{3R}{2} + \frac{R}{1 - (-1)}$.
$C = \frac{3R}{2} + \frac{R}{2} = \frac{4R}{2} = 2R$.

(B) The formula for Young's modulus is $Y = \frac{FL}{Ae} = \frac{4FL}{\pi D^2 e}$.
Given: $L = 2 \,m$, $F = 1.0 \times 9.8 \,N$, $e = 0.8 \times 10^{-3} \,m$, $\Delta e = 0.05 \times 10^{-3} \,m$, $D = 0.4 \times 10^{-3} \,m$, $\Delta D = 0.01 \times 10^{-3} \,m$.
First, calculate the value of $Y$:
$Y = \frac{4 \times 9.8 \times 2}{\pi \times (0.4 \times 10^{-3})^2 \times (0.8 \times 10^{-3})} = \frac{78.4}{\pi \times 0.16 \times 10^{-6} \times 0.8 \times 10^{-3}} \approx 1.95 \times 10^{11} \,N/m^2 \approx 2.0 \times 10^{11} \,N/m^2$.
Now, calculate the relative uncertainty $\frac{\Delta Y}{Y}$:
$\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta D}{D} + \frac{\Delta e}{e}$.
Since $F$ and $L$ are exact, $\frac{\Delta F}{F} = 0$ and $\frac{\Delta L}{L} = 0$.
$\frac{\Delta Y}{Y} = 2 \left( \frac{0.01}{0.4} \right) + \left( \frac{0.05}{0.8} \right) = 2(0.025) + 0.0625 = 0.05 + 0.0625 = 0.1125$.
Absolute uncertainty $\Delta Y = 0.1125 \times Y = 0.1125 \times 1.95 \times 10^{11} \approx 0.22 \times 10^{11} \,N/m^2$.
Rounding to one decimal place, $\Delta Y \approx 0.2 \times 10^{11} \,N/m^2$.
Thus, $Y = (2.0 \pm 0.2) \times 10^{11} \,N/m^2$.

(A) In a resonance column experiment, the tuning fork is held above the open end of the tube. To ensure that the sound waves propagate effectively down the tube, the prongs of the tuning fork are kept in a vertical plane.
For a tube closed at one end, the resonance occurs when the length of the air column $L$ satisfies the condition $L + e = (2n - 1) \frac{\lambda}{4}$, where $e$ is the end correction and $n = 1, 2, 3, ...$.
The first resonance occurs at $L_1 + e = \frac{\lambda}{4}$ and the second resonance occurs at $L_2 + e = \frac{3\lambda}{4}$.
Subtracting these, we get $L_2 - L_1 = \frac{\lambda}{2}$.
Thus, the difference between the lengths of the two resonating air columns is equal to half the wavelength of sound in air.

(B) Consider a vertical rectangular strip of height $dx$ at a depth $x$ from the free surface of the water. The width of this strip is the diameter of the beaker,which is $2R$.
The pressure at depth $x$ is $P(x) = P_0 + \rho g x$.
The force exerted by the pressure on this strip is $dF_p = P(x) \cdot (2R) dx = (P_0 + \rho g x) 2R dx$.
Integrating this from $x = 0$ to $x = h$,the total force due to pressure is $F_p = \int_0^h (P_0 + \rho g x) 2R dx = 2R [P_0 x + \frac{1}{2} \rho g x^2]_0^h = 2 P_0 R h + R \rho g h^2$.
Additionally,there is a force due to surface tension acting along the top edge of the section. The length of the section at the surface is $2R$,so the force due to surface tension is $F_T = T \cdot (2R) = 2RT$.
Since the surface tension force acts in the opposite direction to the pressure force,the net magnitude of the force is $F = |F_p - F_T| = |2 P_0 R h + R \rho g h^2 - 2 RT|$.

(C) The given arrangement of charges is an electric dipole placed along the $z$-axis,with the positive charge at $(0, 0, a/2)$ and the negative charge at $(0, 0, -a/2)$.
The electric potential $V$ at any point $(x, y, z)$ due to an electric dipole is given by $V = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3}$.
For a dipole centered at the origin,the equatorial plane is the $xy$-plane $(z = 0)$.
At any point on the $xy$-plane,the distance from the positive charge is equal to the distance from the negative charge,so the potential $V$ is zero everywhere on the $xy$-plane.
The initial position is $(-a, 0, 0)$,which lies on the $xy$-plane,so $V_i = 0$.
The final position is $(0, a, 0)$,which also lies on the $xy$-plane,so $V_f = 0$.
The work done by the electric field $W_e$ is given by $W_e = -\Delta U = -q_0(V_f - V_i)$,where $q_0$ is the test charge.
Since $V_f = V_i = 0$,the work done $W_e = -q_0(0 - 0) = 0$.

(B) For a single electron system, the radius of the $n^{th}$ orbit is given by $r_n = a_0 \times \frac{n^2}{Z}$.
Given $Z=2$ for $He^{+}$.
For the initial orbit, $r_2 = 105.8 \ pm$:
$105.8 = 52.9 \times \frac{n_2^2}{2} \implies n_2^2 = \frac{105.8 \times 2}{52.9} = 4 \implies n_2 = 2$.
For the final orbit, $r_1 = 26.45 \ pm$:
$26.45 = 52.9 \times \frac{n_1^2}{2} \implies n_1^2 = \frac{26.45 \times 2}{52.9} = 1 \implies n_1 = 1$.
The transition is from $n=2$ to $n=1$.
The energy of the emitted photon is $\Delta E = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
$\Delta E = 2.2 \times 10^{-18} \times (2)^2 \times \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 2.2 \times 10^{-18} \times 4 \times \left( 1 - 0.25 \right) = 8.8 \times 10^{-18} \times 0.75 = 6.6 \times 10^{-18} \ J$.
Using $\Delta E = \frac{hc}{\lambda}$, we get $\lambda = \frac{hc}{\Delta E} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 10^{-18}} = 3 \times 10^{-8} \ m = 30 \ nm$.