IIT-JEE 2008 Paper 1 Question Paper with Solutions

The cut-off wavelength is given by ${\lambda _{\min }} = {{hc} \over {eV}}$

The cut-off wavelength depends on the energy eV of the accelerated electrons and is independent of the atomic number of target. Thus, greater the accelerating voltage for electrons, higher will be kinetic energy it attains before striking the target, higher will be the frequency of X-rays and smaller will be wavelength.

${{hc} \over \lambda } = {E_4} - {E_3}$

${{hc} \over \lambda } = - 13.68 \times {2^2}\left( {{1 \over {{4^2}}} - {1 \over {{3^2}}}} \right)$

${{hc} \over \lambda } = 13.6 \times {2^2}\left( {{1 \over 9} - {1 \over {16}}} \right)$

${{4.1356 \times {{15}^{ - 15}}\,eV.s \times 3 \times {{10}^8}\,m{s^{ - 1}}} \over 1} = 2.644$

$\Rightarrow \lambda \cong 4.8\, * \,{10^{ - 7}}\,m$

If $d=\lambda$, the maximum path difference (path difference is given by $d \sin\theta$) will be less than $\lambda$. So there will be only central maximum on the screen, because in the equation $d\sin\theta=n\lambda,n$ can take only one value.

If $\lambda < d < 2\lambda$, then the maximum path difference will be less than 2$\lambda$. So there will be two more maximum on the screen in addition to the central maximum.

Intensity of dark fringes becomes zero when intensities at the two slits are equal. Initial intensity at both the slits is unequal so there will some brightness at dark fringes. Hence, when intensity of both slits is made equal, the intensity at dark fringes on screen will reduces to zero.

For H atom,

${E_1} = - {{13.6} \over {{1^2}}} = - 13.6\,eV$

${E_2} = - {{13.6} \over {{2^2}}} = - 3.4\,eV$

Energy released by H atom = E$_2$ $-$ E$_1$

$= - 3.4 - ( - 13.6) = 10.2\,eV$

This energy will be absorbed by He atom. Thus, for He atom

$10.2 = - 13.6 \times {2^2}\left( {{1 \over {{2^2}}} - {1 \over {{n^2}}}} \right)$

$0.1875 = {1 \over 4} - {1 \over {{n^2}}}$

${n^2} = 16$

$n = 4$

Velocity along the plane just before collision,

${v^2} - {u^2} = 2gs \Rightarrow v = \sqrt {2g \times 3} = \sqrt {60} \,m/s$

Velocity along the plane just after collision

${v_B} - v\cos 30 = \sqrt {45} \,m{s^{ - 1}}$

When the temperature of a metal increases, its resistance will increase.

Therefore statement-2 is correct.

For meter bridge, when null point N is obtained, we get

${X \over l} = {R \over {100 - l}}$

When the unknown resistance is put inside an enclosure, maintained at a high temperature, then X increases. To maintain the ratio of null point, $l$ should also increase. But if we want to keep the null point at the initial position (i.e., if we want to change the value of $l$) to maintain the ratio, R should be increased.

Therefore, statement - 1 is false.

We know that $P = {{{V^2}} \over R}$

If potential is constant we have

$P \propto {1 \over R}$

Case I

Hence, this is a clear case of wheat stone bridge R$_1$ = 1 $\Omega$

Case II

Equivalent resistance (R$_2$)

$\therefore$ ${1 \over {{R_2}}} = {1 \over 2} + {1 \over 1} + {1 \over 2} = {{1 + 2 + 1} \over 2} = {4 \over 2} = 2\,\Omega$

Hence, ${R_2} = {1 \over 2} = 0.5\,\Omega$

It is clear that the equivalent resistance (R$_2$) will be less than 1 $\Omega$.

Case III

Hence, R$_3$ = 2 $\Omega$

Since, R$_2$ < R$_1$ < R$_2$

$\therefore$ P$_2$ > P$_1$ > P$_3$ ($\because$ $P \propto {1 \over R}$)

Buoyancy force = Weight of fluid displaced

= (mass of fluid displaced)g

$= V{\rho _l}g$ .... (i)

Where V = Volume of fluid displaced

= Volume of the bubble

$PV = nRT$

$\Rightarrow V = {{nRT} \over P} = {{nRT} \over {{P_0}(H - y){\rho _l}g}}$ .... (ii)

Where P is the pressure of the bubble at an arbitrary location at a distance 'y' from the bottom.

Put the value of temperature from eq. (i)

$V = {{nR} \over {[{P_0} + (H - y){\rho _l}g]}} \times {{{T_0}{{[{P_0} + (H - y){\rho _l}g]}^{2/5}}} \over {{{[{P_0} + H{\rho _l}g]}^{2/5}}}}$

$= {{nR{T_0}} \over {{{[{P_0} + (H - y){\rho _l}g]}^{3/5}}{{[{P_0} + H{\rho _l}g]}^{2/5}}}}$ .... (iii)

From eq. (i) and (iii) Buoyance force

$= {{nR{T_0}{\rho _1}g} \over {{{[{P_0} + (H - y){\rho _l}g]}^{3/5}}{{[{P_0} + H{\rho _l}g]}^{2/5}}}}$

When binding energy per nucleon increases for a nuclear process, energy is released.

(a) For 1 < A < 50, on fusion, mass number of the resulting nucleus will be less than 100. No energy will be released.

(b) For 51 < A < 100, on fusion, mass number of the resulting nucleus will be between 100 and 200. As BE increases, energy will be released.

(c) On fission for 100 < A < 200, the mass number for fission nuclei will be between 50 and 100. As BE decreases, no energy will be released.

(d) On fission for 200 < A < 260, the mass number for the fission nuclei will be between 100 and 130. Since BE will increase, energy will be released.

For minimum deviation, the ray in the prism is parallel to base of the prism. This condition does not depend on the colour (i.e., wavelength of incident radiation. So, both the cases, by geometry, $r=30^\circ$.

To determine the ratio of the kinetic energy of the $n=2$ electron for the H atom to that of the He$^+$ ion, we need to consider the Bohr model of the atom. According to the Bohr model, the kinetic energy (KE) of an electron in a hydrogen-like ion in the nth orbit can be given by:

$KE_n = \frac{1}{2} m v_n^2 = \frac{Z^2 e^4 m}{8 \epsilon_0^2 h^2 n^2}$

where:

• $m$ is the mass of the electron
• $Z$ is the atomic number of the nucleus
• $e$ is the charge of the electron
• $\epsilon_0$ is the permittivity of free space
• $h$ is Planck's constant
• $n$ is the principle quantum number, which is 2 in this case

For a hydrogen atom (H) ($Z = 1$) in the $n = 2$ state:

$KE_{H (n=2)} = \frac{1^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{e^4 m}{32 \epsilon_0^2 h^2}$

For a singly ionized helium ion (He$^+$) ($Z = 2$) in the $n = 2$ state:

$KE_{He^+ (n=2)} = \frac{2^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{4 e^4 m}{32 \epsilon_0^2 h^2} = \frac{e^4 m}{8 \epsilon_0^2 h^2}$

To find the ratio of the kinetic energy of the $n=2$ electron for the H atom to that of the He$^+$ ion, we divide the kinetic energies:

$\text{Ratio} = \frac{KE_{H (n=2)}}{KE_{He^+ (n=2)}} = \frac{\frac{e^4 m}{32 \epsilon_0^2 h^2}}{\frac{e^4 m}{8 \epsilon_0^2 h^2}} = \frac{1}{4}$

Therefore, the ratio of the kinetic energy of the $n=2$ electron for the H atom to that of the He$^+$ ion is:

Option A: $\frac{1}{4}$

$v_C^2 - v_B^2 = 2as$

$v_C^2 - 45 = 2 \times 10 \times 3$

${v_c} = \sqrt {60 + 45} = \sqrt {105} \,m{s^{ - 1}}$ is the velocity of block just before leaving incline.

For $r\ge R$,

Force on test mass m is $F=m\times |E_g|$

where, $E_g$ = Gravitational field intensity at the point of observation

$\therefore$ ${{m{v^2}} \over r} = m \times \left[ {{{GM} \over {{r^2}}}} \right]$,

Where, M = Total mass of spherical system

$\therefore$ $v \propto {1 \over {\sqrt r }}$

For $r < R$,

$F' = m|E{'_g}|$

$\therefore$ ${{m{V^2}} \over r} = m\left[ {{{GM} \over {{R^3}}} \times r} \right]$

Hence, $V \propto r$

Time period (T) $= 2\pi \sqrt {{l \over g}}$

or ${t \over n} = 2\pi \sqrt {{l \over g}}$

$\therefore$ $g = {{(4{\pi ^2})({n^2})l} \over {{t^2}}}$

% error in $g = {{\Delta g} \over g} \times 100$

$= \left( {{{\Delta l} \over l} \times {{2\Delta l} \over l}} \right) \times 100$

${E_I} = \left( {{{0.1} \over {64}} + {{2 \times 0.1} \over {128}}} \right) \times 100 = 0.3125\%$

${E_{II}} = \left( {{{0.1} \over {64}} + {{2 \times 0.1} \over {64}}} \right) \times 100 = 0.46875\%$

${E_{III}} = \left( {{{0.1} \over {20}} + {{2 \times 0.1} \over {36}}} \right) \times 100 = 1.055\%$

Hence, $\mathrm{E_I}$ is minimum.

Given, PT$^2$ = constant ..... (i)

For an ideal gas,

$\frac{\mathrm{PV}}{\mathrm{T}}$ = constant ...... (ii)

From above two equation, after eliminating P.

$\frac{\mathrm{V}}{\mathrm{T^3}}$ = Constant

$\Rightarrow$ V = KT$^3$, where $k$ = constant.

$\frac{d\mathrm{V}}{\mathrm{V}}=3\frac{d\mathrm{T}}{\mathrm{T}}$

$\Rightarrow dV = \left( {{3 \over T}} \right)VdT$ ..... (iii)

Change in volume due to thermal expansion is given by $dV=Vydt$ ..... (iv)

Where, $y$ = coefficient of volume expansion

From equation (iii) and (iv), we have,

$VydT = \left( {{3 \over T}} \right)VdT$

$\Rightarrow y = {3 \over T}$

Since the process is adiabatic,

$P{V^\gamma }$ = constant for gas inside bubble.

Thus, $P{V^{(1 - \gamma )}}.\,{T^\gamma } =$ constant

$\Rightarrow {P^{1 - \gamma }}_{bottom}{T^\gamma }_{bottom} = P_y^{1 - \gamma }T_y^\gamma =$ constant

$\Rightarrow {({P_0} + {\rho _1}gH)^{ - 2/3}}(T_0^{5/3}) = ({P_0} + {\rho _1}g{(H - y)^{ - 2/3}}){T^{5/3}}$

$\Rightarrow {T_0} = {\left[ {{{{P_0} + {\rho _1}g(H - y)} \over {{P_0} + {\rho _1}gH}}} \right]^{2/5}}$

Volume flow rate (V) of an incompressible fluid in the steady flow remains constant.

$V=a\times v$

Where $a$ = area of cross-section and $v$ = velocity

If '$v$' decreases, '$a$' increases and vice-versa.

When steam of water moves up, its speed (v) decreases and therefore 'a' increases, i.e., the water spreads out like a fountain. When steam of water from hose pipe moves down, its speed increases and as a result, area of cross-section decreases.

Therefore, statement - 1 is true.

Statement - 2 is the correct explanation of statement - 1.

${v_V} = v\sin 30^\circ \cos 30^\circ - v\cos 30^\circ \cos 60^\circ = 0$

The acceleration of a body rolling down an inclined plane in given by

$a = {{g\sin \theta } \over {1 + {I \over {M{R^2}}}}}$

For hollow cylinder,

${I \over {M{R^2}}} = {{M{R^2}} \over {M{R^2}}} = 1$

For solid cylinder,

${1 \over {M{R^2}}} = {{{1 \over 2}M{R^2}} \over {M{R^2}}} = {1 \over 2}$

$\therefore$ Acceleration of the solid cylinder is more than that of hollow cylinder and therefore solid cylinder will reach the bottom of the inclined plane first.

$\therefore$ Statement - 1 is False.

Statement - 2 : In case of rolling, there will be no heat losses. Therefore, total mechanical energy remains conserved. The potential energy, therefore gets converted into kinetic energy. In both the cases, since the initial potential energy is same, the final kinetic energy will also be same. Therefore, statement - 2 is correct.

As the bubble moves upwards, besides the buoyancy force (the cause of which is pressure difference), only the force of gravity and the force of viscosity will act.

1^st Method : The normal force exerted by the astronaut on the orbiting space station is zero. Therefore, the apparent weight of astronaut in the orbiting space station is zero. Astronaut is called in a state of weightlessness. Why because astronaut as well as space ship are free falling bodies. Statement - I is true, Statement - 2 is true and Statement - 2 is the correct explanation of Statement - 1.

2^nd method : For the body to follow circular path, there must be a centripetal force. Here the astronaut is inside the satellites which is revolving around the earth under the influence of the earth's gravitation. Thus, the earth's gravitation acts as a centripetal force and the net force on the astronaut is zero.

Statement - 2 is right explanation of 1.

As the particle enters the magnetic field, a force acts on it due to the magnetic field which moves the particle in a circular path of radius.

$r={{mv} \over {qB}}$

For the particle entering in region III, r > $l$

$\Rightarrow {{mv} \over {qB}} > l$

$\Rightarrow v > qlB/m$

For maximum path length in region II, r = $l$

$\therefore$ $l = {{mv} \over {qB}}$

$\Rightarrow v = {{qlB} \over m}$

[$\because$ ${v_{\max }} = {{qBl} \over m},T = {{2\pi r} \over v} = {{2\pi m} \over {q.B}}$]

The time taken by the particle to move in region II before coming back in region I is given by $t=\pi m$, which is independent of v.

Both NH$_3$ and PH$_3$ have one lone pair of electron on it. The lone pair in case of ammonia occupy sp$^3$ orbital is directional.

Hence, the lone pair in case of ammonia is easily available and can be easily donated. But for PH$_3$ the lone pair of electrons occupy spherical s-orbital which is less directional and thus its lone pair is not easily available neither it can be donated easily.

Also the lone pair of electrons in sp$^3$ orbital has lesser s-character and less attracted by nucleus and available for easy donation than in PH$_3$ with pure s orbital, the lone pair in which lone pair are more strongly attracted by nucleus. As we go down a group, the p-character in bond pair increases and the s-character in lone pair increases.

Thus ammonia is a better base than PH$_3$.

Due to greater solubility and nature to be prone to microbial action, nitrates are less abundant in earth's crust. Nitrates are soluble. In nitrates, nitrogen has maximum oxidation state and cannot be further oxidised.

Bromine orientation is controlled by its mesomeric effect. Its stabilisation of arenium ion by +M activity.

In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile at para position. As a result, only mono substitution of bromobenzene takes place.

In p-block elements, inert pair effect is observed, due to which lower oxidation state becomes more stable on going down the group. In 14^th group (C, Si, Ge, Sn, Pb) Pb is placed at lower position than Sn, that's why lower oxidation state is more stable for lead (Pb$^{2+}$) and Sn$^{+4}$ is more stable than Pb$^{+4}$. So, Pb$^{+4}$ compounds are stronger oxidising agents than Sn$^{+4}$ compounds. Lower oxidation for the group 14^th elements are more stable for the heavier members.

Reaction involving leaching of silver ores is reversible add, so to get the maximum yield of the product the reaction is carried out in presence of oxygen. Also, oxygen oxidises silver which then produces soluble complex with CN$^-$ ions.

4Ag + 8NaCN + 2H$_2$O + O$_2$ $\to$ 4Na[Ag(CN)$_2$] + 4NaOH

The correct statements concerning the structures E, F and G are:

(B) E, F and G are tautomers.

E is in keto form (C=O) and F and G are in enol form (C=C$-$OH)

(C) F and G are geometrical isomers. F is Z isomer and E is E isomer.

(D) F and G are diastereomers. They are not mirror images of each other and non-superimposable.

Hyperconjugation effect is one of the type of permanent effect in which delocalisation of $\sigma$ electrons of C-H bond of an alkyl group directly attached to an atom of the unsaturated system or to an atom with an unshared p orbital. The main role of hyperconjugation is to stabilise the carbocation because it helps in the dispersal of positive charge. Higher the number of alkyl groups attached to a positively charged carbon atom, greater is the hyperconjugative effect and stability of carbonation. It generally involves the interaction of electrons in a C-H sigma orbital with nearby non-bonding p or anti-bonding $\sigma^*$ or $\pi^*$ orbitals to provide extended conjugation which further increases the stability of the carbocation. Sometimes, low lying anti-bonding $\sigma^*$ also interact with filled orbitals of lone pair character. This type of conjugation is termed as negative hyperconjugation.

With increase in the atomic number, the proton-proton electrostatic repulsion begins to overcome attractive forces involving proton and neutrons in heavier nuclides. The stability relationship can be represented by a line with a slope of 45$^\circ$, i.e., the maximum stability is attained when N = Z. Right of the curve a radioactive nuclide would be neutron rich and would decay by $\beta$-emission to produce a daughter nucleus with a lower n/p ratio. For heavier nuclides, p-p repulsions start to offset the attractive forces and an excess of neutrons over protons, is required for stability.

White phosphorus on reaction with NaOH gives PH$_3$ as one of the products.

P$_4$ + 3NaOH + 3H$_2$O $\to$ 3NaH$_2$PO$_2$ + PH$_3$

The oxidation state of P in P$_4$, NaH$_2$PO$_2$ and PH$_3$ is 0, +1 and $-$3 respectively. It is an example of disproportionation reaction.

Total vapour pressure $P = P_A^0{x_A}$

[$\because$ Solute is to be taken as non-volatile] $= 40 \times 0.9 = 36$ mm Hg

Van der Waals equation is $\left( {P + {{a{n^2}} \over {{V^2}}}} \right)(V - nb) = nRT$, where a is a constant whose value depends on the attraction between the gas molecules and b is the volume that is occupied by one mole of gas.

In option (A), it behaves similar to an ideal gas in the limit of large molar volumes. When the volume is very large then the attraction between the gas molecules can be neglected, so a $\approx$ 0 and the volume occupied by one mole of gas compared to large volume of container will also be neglected. So, b $\approx$ 0. Now, we can neglect a and b and the gas behaves ideally. Thus, option A is correct.

In option (B), it behaves similar to an ideal gas in limit of low pressure and high temperatures. Now, the pressure is very low. So, the attraction between the particles is negligible.

Van der Waals coefficients that are dependent on the identity of the gas but are independent of the temperature. Van der Waals coefficient depends on the attraction between the gas molecules (each of which is characteristic of a gas) and b is the volume that is occupied by one mole of the molecules. So, this option is correct. Since, molecules of Van der Waals gas exert force of attraction on each other; hence, the force which they exert on the walls of the container is less than the force exerted by it if the gas would have been ideal.

Thus, option (D) is correct.

For solution M, water is solute and ethanol is solvent.

$\because$ $\Delta {T_f} = {({K_f})_{ethanol}} \times m$

$2 \times {{0.1} \over {0.9 \times {{46} \over {1000}}}} = {{2000} \over {9 \times 46}} = 4.8$

$\therefore$ ${T_f} = T_f^0 - \Delta {T_f} = 155.7 - 4.8 = 150.9\,K$

We know that following are the cases when a compound is optically inactive:

1. When it does not have a chiral carbon atom e.g., Methyl bromide

2. When it has a plane or axis of symmetry e.g., Tartaric acid (these compounds are called as Meso compounds)

3. When it forms a racemic mixture e.g.: Equimolar mixture of d-Lactic acid and l-Lactic acid.

The given compound possesses an axis of symmetry

Rate constant for first order kinetics:

${k_1} = {{0.693} \over {{t_{1/2}}}} = {{0.693} \over {40\,{s^{ - 1}}}}$ ..... (i)

Rate constant for zero order kinetics:

${k_0} = {{{A_0}} \over {2{t_{1/2}}}} = {{1.386} \over {2 \times 20}}$ mol dm$^{-3}$ s$^{-1}$ ..... (ii)

Divide (i) by (ii)

${{{k_1}} \over {{k_0}}} = {{0.693} \over {40}} \times {{40} \over {1.386}} = {{0.693} \over {1.386}}$ mol$^{-1}$ dm$^{-3}$ = 0.5 mol$^{-1}$ dm$^{3}$

Weak monoacidic base, e.g., BOH is neutralised as follows:

BOH + HCl $\to$ BCl + H$_2$O

At the equivalence point, all BOH get converted into the salt. The concentration of H$^+$ (or pH of solution) is due to hydrolysis of the resultant salt (BCl, cationic hydrolysis here.)

Volume of HCl used up,

${V_a} = {{{N_b}{V_b}} \over {{N_a}}}$

${V_a} = {{2.5 \times 2 \times 15} \over {2 \times 5}}$

${V_a} = 7.5$

Concentration of salt,

$[BCl] = {{conc.\,of\,base} \over {total\,volume}}$

$= {{2 \times 2.5} \over {5(7.5 + 2.5)}} = {1 \over {10}} = 0.1$

${K_h} = {{C{h^2}} \over {1 - h}} = {{{K_w}} \over {{K_b}}} = {{{{10}^{ - 14}}} \over {{{10}^{ - 12}}}}$

$= {10^{ - 2}} = {{0.1 \times {h^2}} \over {1 - h}}$ ..... (i)

(h should be estimated whether that can be neglected or not.)

On calculating, $h = 0.27$ (significant, not negligible)

$[{H^ + }] = Ch = 0.1 \times 0.27 = 2.7 \times {10^{ - 2}}$ M

The product formed the reaction is used for chlorination of water bodies. The compound Na$_2$S$_2$O$_3$ gets oxidised by chlorine to form NaHSO$_3$. The balanced chemical reaction is,