IIT-JEE 2009 Paper 2 Question Paper with Solutions

The segments PQ and PR, in the triangle shown here, cannot produce $\overrightarrow B$ at point P since point P lies on them. Here, only QR creates $\overrightarrow B$ at point P.

Applying ${B_P} = {{{\mu _0}i} \over {4\pi R}}(\cos {\phi _1} + \cos {\phi _2})$, where ${\phi _1} = 53^\circ$ and ${\phi _2} = 37^\circ$, we get the magnitude of the magnetic field as follows:

${B_P} = {{{\mu _0}I} \over {4\pi (4\pi \sin 37^\circ )}}(\cos 53^\circ + \cos 37^\circ )$

$= {{{\mu _0}I} \over {16\pi x(3/5)}}\left( {{{3x} \over {5x}} + {{4x} \over {5x}}} \right)$

$= {{5{\mu _0}I} \over {48\pi x}}\left( {{7 \over 5}} \right) = {{7{\mu _0}I} \over {48\pi x}} = k\left( {{{{\mu _0}I} \over {48\pi x}}} \right)$

where $k = 7$.

The distance between the successive nodes is $\lambda/2$. Therefore,

${\lambda \over 2} = {V \over {2v}} = {1 \over {2 \times 100}}\sqrt {{{Tl} \over m}} = {1 \over {200}}\sqrt {{{0.5 \times 0.2} \over {{{10}^{ - 3}}}}} = {1 \over {20}}$ m = 5 cm

(A)$\to$(P), (S)

Since the path difference is

$S_1P_0=S_2P_0=0$

the phase difference becomes

$\delta(P_0)=0$

The intensity at any point is

$I = {I_{\max }}{\cos ^2}\left( {{8 \over 2}} \right)$

Now, $I({P_0}) = {I_{\max }}$ [$\because$ $\delta ({P_0}) = 0$]

$I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi \over 4}} \right) = {{{I_{\max }}} \over 2}$

Therefore, $I({P_0}) > I({P_1})$

(B)$\to$(Q)

In this case, the path difference at P$_1$ is

${P_1} = {\lambda \over 4} - (\mu - 1)t = {\lambda \over 4} - {\lambda \over 4} = 0$

Therefore, $\delta ({P_1}) = 0$.

(C)$\to$(T)

In this case, the path difference at P$_1$ is

${P_1} = {\lambda \over 4} - (\mu - 1)t = {\lambda \over 4} - {\lambda \over 2} = {{ - \lambda } \over 4}$

Therefore, the phase difference at P$_1$ is

${P_1} = {{2\pi } \over \lambda } \times \left( {{{ - \lambda } \over 4}} \right) = - {\pi \over 2}$

Therefore, $I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi \over 4}} \right) = {{{I_{\max }}} \over 2}$

The path difference at P$_2$ is

${P_2} = {\lambda \over 3} - (\mu - 1)t = {\lambda \over 3} - {\lambda \over 2} = {{ - \lambda } \over 6}$

Therefore, the phase difference at P$_1$ is

${P_2} = {{2\pi } \over \lambda } \times \left( {{{ - \lambda } \over 6}} \right) = - {\pi \over 3}$

Therefore, $I({P_2}) = {I_{\max }}{\cos ^2}\left( {{\pi \over 6}} \right) = {3 \over 4}{I_{\max }}$

Therefore, $I({P_2}) > I({P_1})$.

(D)$\to$(R), (S), (T)

In this case, the path difference at P$_1$ is

${P_1} = {\lambda \over 4} - (\mu - 1)t = {\lambda \over 4} - {{3\lambda } \over 4} = {{ - \lambda } \over 2}$

Therefore, the phase difference at P$_1$ is

${P_1} = {{2\pi } \over \lambda } \times \left( { - {\lambda \over 2}} \right) = - \pi$

Therefore, $I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi \over 2}} \right) = 0$

The path difference at P$_0$ is

${P_0} = 0 - (\mu - 1)t = {{ - 3\lambda } \over 4}$

Therefore, the phase difference at P$_0$ is

${P_0} = {{2\pi } \over \lambda } \times \left( {{{ - 3\lambda } \over 4}} \right) = {{ - 3\lambda } \over 2}$

Therefore, $I({P_0}) = {I_{\max }}{\cos ^2}{{3\pi } \over 4} = {{{I_{\max }}} \over 2}$

Now, the path difference at P$_1$ is

${P_1} = {\lambda \over 4} - (\mu - 1)t = {\lambda \over 4} - {{3\lambda } \over 4} = - {\lambda \over 2}$

The phase difference at P$_1$ is

${P_1} = {{2\pi } \over \lambda } \times \left( {{{ - \lambda } \over 2}} \right) = - \pi$

Therefore, $I({P_1}) = {I_{\max }}\cos \left( {{\pi \over 2}} \right) = 0$

We have

${P_0}{V_0} = {P_1}{V_1}$ ..... (1)

and the pressure at equilibrium is

$P = {P_0} - \rho gh$

$= (1.0 \times {10^5}) - [({10^3})(10)(200 \times {10^{ - 3}})]$

$= (98 \times {10^3})$ N/m$^2$

Substituting in Eq. (1), we get

${10^5}[A(500 - H)] = 98 \times {10^3}[A(500 - 200)]$

where A is the cross-section of water column.

Now, if H = 206 mm, we conclude that the level of water falls down by 6 mm. Hence, the correct answer is 6.

The sharpness of resonance decreases with increasing column length.

The first resonance happens when length of air column $\approx$ $\lambda/2$. The prongs of the tuning fork are always kept in a vertical plane and amplitude of vibration is typically in the mm range.

From the figure, we have

${\lambda \over 4} = {l_1} + e$

where e is the end-correction. Therefore,

${l_1} + \left( {{\lambda \over 4} - e} \right) < {\lambda \over 4}$

We have

$E = \left( {{{hc} \over \lambda }} \right)$ J $\Rightarrow E = \left( {{{1240} \over {\lambda \,nm}}} \right)$ eV

Therefore,

$\lambda_1$ = 550 nm, $E_1$ = 2.25 eV

$\lambda_2$ = 450 nm, $E_2$ = 2.75 eV

$\lambda_3$ = 350 nm, $E_3$ = 3.5 eV

Also,

$\phi_p=2$ eV, all $\lambda$'s cause emissions.

$\phi_q=2.5$ eV, last two $\lambda$'s cause emissions.

$\phi_r=3$ eV, only the last $\lambda$ causes emissions.

That is, $I_p > I_q > I_r$.

Let $V_1$ and $V_2$ be the velocities of blocks A and B immediately after the elastic collision:

${v_1} = \left( {{{{m_1} - {m_2}} \over {{m_1} + {m_2}}}} \right)m = \left( {{{m - 2m} \over {m + 2m}}} \right) \times 9 = - 3$ m/s

${v_2} = \left( {{{2{m_1}} \over {{m_1} + {m_2}}}} \right)m = \left( {{{2m} \over {m + 2m}}} \right) \times 9 = 6$ m/s

After the perfectly inelastic collision between blocks B and C, let $v$ be the common velocity. Applying centre of mass concept, we get

$2m{v_2} = (2m + m)v$

$\Rightarrow v = {2 \over 3} \times 6 = 4$ m/s

(A)$\to$(P), (Q), (T)

$\bullet$ Case (P) : $U=\frac{1}{2}CV^2$

$\bullet$ Case (Q) : Since the work is done on the system, the internal energy increases.

$\bullet$ Case (T) : time varying $\overrightarrow B$ creates an induced $\overrightarrow E$ causing current to flow. This dissipates heat in the loop: $H=L^2(Rt)$.

(B)$\to$(Q) : As explained above.

(C)$\to$(S)

$\bullet$ Case (S) : Mass defect is converted into energy which is released.

(D)$\to$(S) : As explained above.

We have,

$\overrightarrow J = \overrightarrow r \times \overrightarrow F$

$|\overrightarrow J | = rF\sin 180^\circ = 0$

Therefore, the angular momentum is conserved, that is,

$mvr\sin \theta =$ Constant

where $\theta$ is the angle between the velocity $\overrightarrow v$ and position vector $\overrightarrow r$ of $-$q with respect to Q.

While the parabola is moving right side, the bead experiences a force with a magnitude $ma$ to the left. That is, at equilibrium,

$N\cos \theta = mg$

$N\sin \theta = ma$

Now, $\tan \theta = {a \over g}$

Also, $\tan \theta = {{dy} \over {dx}} = 2kx \Rightarrow x = {a \over {2gk}}$

The steady rate of flow of heat is

${{\Delta Q} \over {\Delta t}} = kA\left( {{{\Delta T} \over {\Delta x}}} \right)$

Therefore, ${{kA(400 - 0^\circ )} \over {\lambda x}} = {m_{ice}}\,{l_{ice}}$

${{kA(400 - 100)} \over {(10 - \lambda )x}} = {m_{water}}\,{l_{water}}$

where ${m_{ice}}$ is the mass of ice melted per unit times and ${m_{water}}$ is the mass of water evaporated per unit times. It is given that

${m_{ice}} = {m_{water}}$

Therefore, ${{kA \times 400} \over {\lambda x({l_{ice}})}} = {{kA \times 300} \over {(10 - \lambda )x\,{l_{water}}}}$

$\Rightarrow 4(10 - \lambda ) \times 540 = 3\lambda \times 80$

$\Rightarrow 5400 - 540\lambda = 60\lambda$

$\Rightarrow \lambda = 9$

We have

$\overrightarrow {{V_C}} = \overrightarrow {{V_{CM}}} + \overrightarrow {r\omega } = 2\overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_B}} = \overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_A}} = 0$

Therefore,

$\overrightarrow {{V_C}} - \overrightarrow {{V_B}} = \overrightarrow {r\omega } = \overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_B}} - \overrightarrow {{V_A}} = \overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_C}} - \overrightarrow {{V_A}} = \overrightarrow {{V_{CM}}} + \overrightarrow {r\omega } = 2\overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_B}} - \overrightarrow {{V_C}} = - \overrightarrow {r\omega }$

For pure rolling, we have

$|\overrightarrow {{V_{CM}}} | = |\overrightarrow {r\omega } |$

Therefore,

$|\overrightarrow {{V_C}} - \overrightarrow {{V_A}} | = 2{V_{CM}}$

$\Rightarrow 2|\overrightarrow {{V_B}} - \overrightarrow {{V_C}} | = 2r\omega = 2{V_{CM}}$

We have,

${P_A} = {P_0} + {{4S} \over {{r_A}}} = 8 + \left[ {{{4(0.04)} \over {2 \times {{10}^{ - 2}}}}} \right] = 16$ N/m$^2$

${P_B} = {P_0} + {{4S} \over {{r_B}}} = 9 + \left[ {{{4(0.04)} \over {4 \times {{10}^{ - 2}}}}} \right] = 12$ N/m$^2$

Now, $PV = nRT$

Therefore, ${{{n_B}} \over {{n_A}}} = {{{P_B}{V_B}} \over {{P_A}{V_A}}} - {{{P_B}r_B^3} \over {{P_A}r_A^3}} = {{12} \over {16}} \times {{64} \over 8} = 6$

Applying Gauss's theorem, we get

$E(4\pi {r^2}) = {{{q_{encl}}} \over {{t_0}}} = {1 \over t}\int\limits_0^r {k{x^2}(4\pi {x^2})dx}$

Now, $E{r^2} = {k \over {{t_0}}}\int\limits_0^r {{x^{2 + a}}dx = {k \over {{t_0}}}\left( {{{{r^{3 + a}}} \over {3 + a}}} \right)}$

Therefore, $E = {k \over {{t_0}}}{{{r^{1 + a}}} \over {3 + a}}$

That is, $E \propto {r^{1 + a}}$

Now, $E\left( {{R \over 2}} \right) = {1 \over 8}E(R)$

Therefore, ${\left( {{R \over 2}} \right)^{1 + a}} = {1 \over 8}{(R)^{1 + a}} \Rightarrow 8 = {2^{1 + a}}$

where $1 + a = 3$ and hence $a = 2$.

Since the restoring force is same in both springs (which are being in series), we have

${k_1}{x_1} = {k_2}{x_2}$

It is given that

${x_1} + {x_2} = A$

$\Rightarrow {x_1} = {{A{k_2}} \over {{k_1} + {k_2}}}$

We observe the following:

$\bullet$ For the path B $\to$ C $\to$ D, the volume (V) decreases; $\Delta$W < 0.

$\Delta U < 0 \Rightarrow \Delta Q < 0$

$\bullet$ For the process A $\to$ B $\to$ C, $\Delta$W is the area of the semicircle and $\Delta W \ne 0$.

$\bullet$ For the process A $\to$ B (semicircle), it cannot be an isothermal process (whose PV-diagram is a rectangular hyperbola).

$\bullet$ For (clockwise) process A $\to$ B $\to$ C $\to$ D $\to$ A, $\Delta$W is the area enclosed, which is positive.

We have,

$a = \left( {{{{m_1} - {m_2}} \over {{m_1} + {m_2}}}} \right)g = \left( {{{0.72 - 0.36} \over {0.72 - 0.36}}} \right) \times 10 = {g \over 3} = {{10} \over 3}$

$T = {{2{m_1}{m_2}g} \over {{m_1} + {m_2}}} = {{2 \times 0.72 \times 0.36 \times 10} \over {0.72 + 0.36}} = 4.8$ N

$s = {1 \over 2}a{t^2} = {1 \over 2} \times {{10} \over 3} \times {1^2} = {5 \over 3}$ m

The work done by the rope on 0.36 kg is

$W = Ts\cos 0^\circ = 4.8 \times {5 \over 3} = + 8$ J

Induced emf is same in both the rings:

$I = {e \over R} = {{\rho A} \over {\rho l}}$

$I \propto {1 \over \rho } \Rightarrow q \propto {1 \over \rho }$ .... (1)

Impulse is

$J = \int {Bil\,dt = mv = Bl\int {I\,dl = mv} }$

That is, $J = Blq = mv \Rightarrow v\left( {{q \over m}} \right)$ ..... (2)

and ${v^2} \propto h$ ...... (3)

From Eqs. (1), (2) and (3), we get

$mv \propto {1 \over \rho }$

$m\sqrt h \propto {1 \over \rho }$

$m\rho \propto {1 \over {\sqrt h }}$

Since ${h_A} > {h_B}$ and for ${m_A} = {m_B},\rho \sqrt h =$ Constant. Therefore, ${\rho _A} < {\rho _B}$. Also if ${m_A} < {m_B}$ and ${\rho _A} < {\rho _B}$,

${m_A}{\rho _A} < {m_B}{\rho _B}$

$\Rightarrow {h_A} > {h_B}$ (already given)

The restoring torque is

$J = - 2 \times kx\left( {{L \over 2}} \right)\cos \theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

Now, $x = {L \over 2}\sin \theta$

Therefore, $J = - k\left( {{{{L^2}} \over 2}} \right)\sin \theta \cos \theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

$\Rightarrow \left( {{{ - k{L^2}} \over 4}} \right)\sin 2\theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

For small $\theta$, $\sin 2\theta = 2\theta$. Therefore,

${{ - k{L^2}\theta } \over 2} = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

where $I = {{M{L^2}} \over {12}}$. Therefore,

${{{d^2}\theta } \over {d{t^2}}} = \left( {{{ - 6k} \over M}} \right)\theta = - {\omega ^2}\theta$ (SHM)

$\Rightarrow \omega = \sqrt {{{6k} \over M}}$

Hence, the frequency of oscillation is

${\omega \over {2\pi }} = {1 \over {2\pi }}\sqrt {{{6k} \over M}}$

Reducing sugars contain cyclic hemiacetal or hemiketal groups in equilibrium with the open chain from having a free $-$CHO or $-$C = O group. Non-reducing sugars contain stable acetal or ketal structures their cyclic structures cannot be opened into an open chain form having a free carbonyl group. In X the glyco-sidic linkage is in between two anomeric carbon atoms while in Y it is only with one anomeric carbon, the other one is free. So, X is non-reducing while Y is reducing.

The hydride shift from C-2 will lead to a resonance stabilized secondary carbocation (conjugated with oxygen).

The structure is CuSO$_4$ . 5H$_2$O $\to$ [Cu(H$_2$O)$_4$]SO$_4$ . H$_2$O

so, H$_2$O molecules directly attached to Cu are 4.

The bond formation in the given oxides of nitrogen is as follows.

Thus N-N bond exists in N$_2$O, N$_2$O$_3$ and N$_2$O$_4$.

Cyclic C$_5$H$_{10}$

For third structure, 2 $cis-trans$ and 1 optical isomer are possible. So, a total of 7 structures are.

The reaction for alkaline oxidative fusion is

2MnO$_2$ + 4KOH + O$_2$ $\to$ 2K$_2$MnO$_4$ + 2H$_2$O

In potassium manganite formed as product :

K$_2$MnO$_4$

2(+1) + $x$ + 4($-$2) = 0

$2+x-8=0$

$x-6=0\Rightarrow x=+6$

To find the numerical value for the enthalpy of combustion of the gas in kJ mol$^{-1}$, we first need to determine the total heat released by the combustion of the gas within the calorimeter. We then convert this amount of heat into per mole of the gas. Step 1: Calculate the total heat released, $q$.

The heat released, $q$, due to combustion in the calorimeter can be calculated using the formula:

$q = C \cdot \Delta T$

where:

• $C$ is the heat capacity of the calorimeter, and


• $\Delta T$ is the change in temperature.

In this problem:

• $C = 2.5 \text{ kJ K}^{-1}$


• $\Delta T = 298.45 \text{ K} - 298.0 \text{ K} = 0.45 \text{ K}$

Substituting these values into the equation gives:

$q = 2.5 \text{ kJ K}^{-1} \times 0.45 \text{ K} = 1.125 \text{ kJ}$

The total heat released by the process is therefore 1.125 kJ, where this amount of heat is a measure of energy released and absorbed by the calorimeter, therefore it is positive.

To convert the heat released into per mole of the gas, we first need to calculate the number of moles of the gas that was burnt. The number of moles, $n$, can be calculated from the mass of the gas and its molecular weight:

$n = \frac{\text{mass}}{\text{molecular weight}}$

In this problem:

• The mass of the gas = 3.5 g


• Molecular weight of the gas = 28 g mol$^{-1}$

Substituting these values gives:

$n = \frac{3.5 \text{ g}}{28 \text{ g mol}^{-1}} = 0.125 \text{ mol}$

The enthalpy of combustion per mole, $\Delta H$, is given by:

$\Delta H = \frac{q}{n}$

Substituting the values we obtained:

$\Delta H = \frac{1.125 \text{ kJ}}{0.125 \text{ mol}} = 9 \text{ kJ mol}^{-1}$

Therefore, the enthalpy of combustion of the gas is $-9 \text{ kJ mol}^{-1}$.

Note: The negative sign indicates that the process is exothermic (releases heat).

The reactions are as follows:

(A) 3Cu + 8HNO$_3$ (dil.) $\to$ 3Cu(NO$_3$)$_2$ + 4H$_2$O + 2NO

(B) Cu + 4HNO$_3$ (conc.) $\to$ Cu(NO$_3$)$_2$ + 4H$_2$O + 2NO$_2$

(C) 4Zn + 10HNO$_3$ (dil.) $\to$ 4Zn(NO$_3$)$_2$ + 5H$_2$O + N$_2$O

(D) Zn + HNO$_3$ (conc.) $\to$ Zn(NO$_3$)$_2$ + 2H$_2$O + 2NO$_2$

Tertiary amine will not react due to the bulkiness of trimethyl groups. Tertiary amines instead react with diborane to give simple addition compounds through symmetrical cleavage of diborane.

Given that

${K_a}({C_6}{H_5}COOH) = 1 \times {10^{ - 4}}$.

pH of 0.01 M ${C_6}{H_5}COONa$.

${K_h} = {{{K_w}} \over {{K_a}}} = {{0.01\,{h^2}} \over {1 - h}}$

$\Rightarrow {{{{10}^{ - 14}}} \over {{{10}^{ - 4}}}} = {{{{10}^{ - 2}}{h^2}} \over {1 - h}}$

$1 - h$ is approximately equal to 1.

[OH$^-$] = $0.01h$ = 0.01 $\times$ 10$^{-4}$ = 10$^{-6}$

[H$^+$] = 10$^{-8}$

pH = 8