JEE Advanced 2015 Paper 1 Question Paper with Solutions

Case I

Reflection from mirror

${1 \over f} = {1 \over v} + {1 \over u} \Rightarrow {1 \over { - 10}} = {1 \over v} + {1 \over { - 15}}$

$\Rightarrow v = - 30$

For lens ${1 \over f} = {1 \over v} - {1 \over u}$

${1 \over {10}} = {1 \over v} - {1 \over { - 20}}$

$v = 20$

$\left| {{M_1}} \right| = \left| {{{{v_1}} \over {{u_1}}}} \right|\left| {{{{v_2}} \over {{u_2}}}} \right|$

$= \left( {{{30} \over {15}}} \right)\left( {{{20} \over {20}}} \right)$

$= 2 \times 1 = 2$ (in air)

Case II :

Now, consider the setup placed in a medium of refractive index ${\mu {'_1}}$ = 7/6. The focal length of the mirror does not change. Thus, the distance of the image formed by the mirror and its magnification does not change. The focal length of the lens changes. The refractive index of the lens material is $\mu$₂ = 1.5. Apply lens maker's formula to get the new focal length of the lens

${1 \over {f'}} = \left( {{{{\mu _2} - \mu {'_1}} \over {\mu {'_1}}}} \right)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]$

$= {{{\mu _2} - \mu {'_1}} \over {\mu {'_1}}}{{{\mu _1}} \over {{\mu _2} - {\mu _1}}}{{{\mu _2} - {\mu _1}} \over {{\mu _1}}}\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]$

$= \left( {{{{\mu _2} - \mu {'_1}} \over {{\mu _2} - {\mu _1}}}} \right)\left( {{{{\mu _1}} \over {\mu {'_1}}}} \right){1 \over f}$

$= \left( {{{1.5 - 7/6} \over {1.5 - 1}}} \right)\left( {{1 \over {7/6}}} \right){1 \over {10}} = {2 \over {35}}$. ..... (1)

Again using lens formula, ${1 \over v} - {1 \over u} = {1 \over {{f_l}'}}$

${1 \over v} - {1 \over { - 20}} = {2 \over {35}} \Rightarrow {1 \over v} = {2 \over {35}} - {1 \over {20}} = {1 \over {140}}$

$\therefore$ v = 140 cm

Magnification, ${m_2}' = {v \over u} = {{140} \over { - 20}} = - 7$

Magnification produced by the combination,

${M_2} = {m_1} \times {m_2}' = ( - 2) \times ( - 7) = 14$

$\therefore$ $\left| {{{{M_2}} \over {{M_1}}}} \right| = {{14} \over 2} = 7$

Given situation is shown in the figure. Let acceleration due to gravity at the surface of the planet be g. At height h above planet's surface v = 0.

According to question, acceleration due to gravity of the planet at height h above its surface becomes g/4.

${g_h} = {g \over 4} = {g \over {{{\left( {1 + {h \over R}} \right)}^2}}}$

$4 = {\left( {1 + {h \over R}} \right)^2} \Rightarrow 1 + {h \over R} = 2$

${h \over R} = 1 \Rightarrow h = R$.

So, velocity of the bullet becomes zero at h = R.

Also, ${v_{esc}} = v\sqrt N \Rightarrow \sqrt {{{2GM} \over R}} = v\sqrt N$ ...... (i)

Applying energy conservation principle,

Energy of bullet at surface of earth = Energy of bullet at highest point

${{ - GMm} \over R} + {1 \over 2}m{v^2} = {{ - GMm} \over {2R}}$

${1 \over 2}m{v^2} = {{GMm} \over {2R}}$ $\therefore$ $v = \sqrt {{{GM} \over R}}$

Putting this value in eqn. (i), we get

$\sqrt {{{2GM} \over R}} = \sqrt {{{NGM} \over R}}$

$\therefore$ N = 2

$E = {\lambda \over {2\pi {\varepsilon _0}r}}$

Case 1 : If q is shifted towards right by x, we get

$F = {F_2} - {F_1} = {{\lambda q} \over {2\pi {\varepsilon _0}}}\left( {{1 \over {{d \over 2} - x}} - {1 \over {{d \over 2} + x}}} \right)$ (towards left)

Case 2 : If $-$q is shifted towards right by x

$F = {F_2} - {F_1} = {{\lambda q} \over {2\pi {\varepsilon _0}}}\left( {{1 \over {{d \over 2} - x}} - {1 \over {{d \over 2} + x}}} \right)$ (towards right)

Thus, +q exhibits SHM while $-$q continues to move towards rightwards.

1st Particle

P = 0 at x = a $\Rightarrow$ 'a' is the amplitude of oscillation 'A₁'.

At x = 0, P = b (at mean position)

$\Rightarrow m{v_{\max }} = b \Rightarrow {v_{\max }} = {b \over m}$

${E_1} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{b \over m}} \right]^2} = {{{b^2}} \over {2m}}$

${A_1}{\omega _1} = {v_{\max }} = {b \over m}$

$\Rightarrow {\omega _1} = {b \over {ma}} = {1 \over {m{n^2}}}(A = a,\,{a \over b} = {n^2})$

2nd Particle

P = 0 at x = R $\Rightarrow$ A₂ = R

At x = 0, P = R $\Rightarrow$ ${v_{\max }} = {R \over m}$

${E_2} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{R \over m}} \right]^2} = {{{R^2}} \over {2m}}$

${A_2}{\omega _2} = {R \over m} \Rightarrow {\omega _2} = {R \over {mR}} = {1 \over m}$

(b) ${{{\omega _2}} \over {{\omega _1}}} = {{1/m} \over {1/m{n^2}}} = {n^2}$

(c) ${\omega _1}{\omega _2} = {1 \over {m{n^2}}} \times {1 \over m} = {1 \over {{m^2}{n^2}}}$

(d) ${{{E_1}} \over {{\omega _1}}} = {{{b^2}/2m} \over {1/m{n^2}}} = {{{b^2}{n^2}} \over 2} = {{{a^2}} \over {2{n^2}}} = {{{R^2}} \over 2}$

${{{E_2}} \over {{\omega _2}}} = {{{R^2}/2m} \over {1/m}} = {{{R^2}} \over 2}$

$\Rightarrow {{{E_1}} \over {{\omega _1}}} = {{{E_2}} \over {{\omega _2}}}$

Stopping potential (V₀) is given by

$e{V_0} = {{hc} \over \lambda } - \phi$

Graph between V₀ and $\lambda$ :

$e{V_0} + \phi = {{hc} \over \lambda }$

$(e{V_0} + \phi )\lambda = hc$

$(e{V_0} + \phi )\lambda$ = constant

Here, both e and $\phi$ are also constant. It represents a hyperbola.

For, V₀ = 0, $\lambda = {{constant} \over \phi } =$ constant

So correct option is (a).

Graph between V₀ and ${1 \over \lambda }$ :

${V_0} = \left( {{{hc} \over e}} \right)\left( {{1 \over \lambda }} \right) - \left( {{\phi \over e}} \right)$

It represents a straight line with slope $\left( {{{hc} \over e}} \right)$ and intercept $\left( { - {\phi \over e}} \right)$ on V₀ axis.

Power, $P = (\sigma {T^4}A) = \sigma {T^4}(4\pi {R^2})$

or, $P \propto {T^4}{R^2}$ ..... (i)

According to Wien's law,

$\lambda \propto {1 \over T}$

($\lambda$ is the wavelength at which peak occurs)

$\therefore$ Eq. (i) will become,

$P \propto {{{R^2}} \over {{\lambda ^4}}}$

or, $\lambda \propto {\left[ {{{{R^2}} \over P}} \right]^{1/4}}$

$\Rightarrow {{{\lambda _A}} \over {{\lambda _B}}} = {\left[ {{{{R_A}} \over {{R_B}}}} \right]^{1/2}}{\left[ {{{{P_B}} \over {{P_A}}}} \right]^{1/4}}$

$= {[400]^{1/2}}{\left[ {{1 \over {{{10}^4}}}} \right]^{1/4}} = 2$

The dimensions of Planck's constant is $h = [{M^1}{L^3}{T^{ - 2}}]$

Speed of light is $c = [{L^1}{T^{ - 1}}]$

Gravitational constant is $G = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

Let $L \propto {h^x}{c^y}{G^z}$.

$[L] = {[{M^1}{L^2}{T^{ - 1}}]^x}{[{L^1}{T^{ - 1}}]^y}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^z}$

Comparing, we get

$\left. {\matrix{ \hfill {x - z = 0} \cr \hfill {2x + y + 3z = 1} \cr \hfill { - x - y - 2z = 0} \cr } } \right\}$

Solving, we get

$x = z$

$y + 5x = 1$

$- y - 3x = 0$

or, $2x = 1$

$x = {1 \over 2} = z$

$y = - {3 \over 2}$

Therefore,

$L \propto {h^{1/2}}{C^{ - 3/2}}{G^{1/2}}$

$L \propto \sqrt h$

$L \propto \sqrt G$

Let $M \propto {h^a}{C^b}{G^c}$

$[M] = {[{M^1}{L^2}{T^{ - 1}}]^a}{[{L^1}{T^{ - 1}}]^b}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^c}$

Comparing, we get $a - c = 1$.

$\left. \matrix{ 2a + b + 3c = 0 \hfill \cr - a - b - 2c = 0 \hfill \cr} \right\}a + c = 0$

Therefore,

$2a = 1$

$a = {1 \over 2}$

$c = - {1 \over 2}$

$b = {1 \over 2}$

Therefore,

$M \propto {h^{1/2}}{c^{1/2}}{G^{ - 1/2}}$

$M \propto \sqrt c$

Ring has mass M and radius R. Initial angular speed of ring = $\omega$. Two point masses, each of mass are at rest at O. Initial angular momentum of ring and point masses system,

${L_i} = {I_R}\omega + {I_m}\omega + {I_m}\omega$

$= M{R^2}\omega + 0 + 0 = M{R^2}\omega$

After some time, situation is changed as shown in the figure.

Angular speed of the system, $\omega ' = {8 \over 9}\omega$; $OA = {{3R} \over 5}$; OB = r = ?

Moment of inertia about O of point mass at A,

${I_A} = {M \over 8} \times {{9{R^2}} \over {25}}$

Moment of inertia about O of point mass at B, ${I_B} = {M \over 8}{r^2}$

Fina angular momentum of the system

${L_f} = M{R^2}\omega ' + {I_A}\omega ' + {I_B}\omega '$

$= M{R^2} \times {{8\omega } \over 9} + {M \over 8} \times {{9{R^2}} \over {25}} \times {{8\omega } \over 9} + {M \over 8}{r^2} \times {{8\omega } \over 9}$

As there is no external torque acting on the system so its angular momentum will be conserved, L_i = L_f

$M{R^2}\omega = M{R^2} \times {{8\omega } \over 9} + {M \over 8} \times {{9{R^2}} \over {25}} \times {{8\omega } \over 9} + {M \over 8}{r^2} \times {{8\omega } \over 9}$

${R^2} = {{8{R^2}} \over 9} + {{{R^2}} \over {25}} + {{{r^2}} \over 9} \Rightarrow {{{r^2}} \over 9} = {{16} \over {225}}{R^2}$ $\therefore$ $r = {4 \over 5}R$

Resistance of a wire, $R = {{\rho l} \over A}$

For iron (Fe) bar,

$\rho$ = 10^$-$7 $\Omega$m, l = 50 mm = 50 $\times$ 10^$-$3 m

A = (2 mm) $\times$ (2 mm) = 4 mm² = 4 $\times$ 10^$-$6 m²

${R_1} = {{{{10}^{ - 7}} \times 50 \times {{10}^{ - 3}}} \over {4 \times {{10}^{ - 6}}}} = 1250 \times {10^{ - 6}}$ $\Omega$ = 1250 $\mu$W

For aluminium (Al) bar,

$\rho$ = 2.7 $\times$ 10^$-$8 $\Omega$ m, l = 50 mm = 50 $\times$ 10^$-$3 m

A = (7² $-$ 2²) mm² = 45 mm² = 45 $\times$ 10^$-$6 m²

$\therefore$ ${R_2} = {{2.7 \times {{10}^{ - 8}} \times 50 \times {{10}^{ - 3}}} \over {45 \times {{10}^{ - 6}}}}$

$= {{27 \times 50} \over {45}} \times {10^{ - 6}} = 30 \times {10^{ - 6}}$ $\Omega$ = 30 $\mu$$\Omega$

Potential difference across both bars (resistors) is same so they are in parallel combination.

Equivalent resistance between P and Q is given by

$R = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {{1250 \times 30} \over {1250 + 30}} = {{125 \times 30} \over {128}} = {{1875} \over {64}}\mu \Omega$.

Suppose mass and radius of each disc are m and R respectively. Also potential energy at points B and D is zero i.e., they are on reference line.

Given final kinetic energy for each disc is same, say it is K.

Applying energy conservation principle,

For surface AB,

${1 \over 2}{I_2}\omega _1^2 + mg \times 30 = K$ ..... (i)

For surface CD,

${1 \over 2}{I_2}\omega _2^2 + mg \times 27 = K$ ..... (ii)

From eqns. (i) and (ii), we get

${1 \over 2}{I_2}\omega _1^2 + mg \times 30 = {1 \over 2}{I_2}\omega _2^2 + mg \times 27$ ...... (iii)

Here, ${\omega _1} = {{{v_1}} \over R}$, ${\omega _2} = {{{v_2}} \over R}$, v₁ = 3 m s^$-$1, v₂ = ?

I₁ = I₂ = Moment of inertia of disc about the point of contact $= {1 \over 2}m{R^2} + m{R^2} = {3 \over 2}m{R^2}$

From eqn. (iii), ${1 \over 2}\left( {{3 \over 2}m{R^2}} \right) \times {\left( {{3 \over R}} \right)^2} + m \times 10 \times 30$

$= {1 \over 2}\left( {{3 \over 2}m{R^2}} \right) \times {\left( {{{{v_2}} \over R}} \right)^2} + m \times 10 \times 27$

${{27} \over 4} + 300 = {3 \over 4}v_2^2 + 270$

${3 \over 4}v_2^2 = {{27} \over 4} + 30 \Rightarrow 3v_2^2 = 147$

$v_2^2 = 49$ $\therefore$ v₂ = 7 m s^$-$1

$\mu ({S_2}P) - {S_1}P = m\lambda$

$\Rightarrow \mu \sqrt {{d^2} + {x^2}} - \sqrt {{d^2} + {x^2}} = m\lambda$

$\Rightarrow (\mu - 1)\sqrt {{d^2} + {x^2}} = m\lambda$

$\Rightarrow \left( {{4 \over 3} - 1} \right)\sqrt {{d^2} + {x^2}} = m\lambda$

or, $\sqrt {{d^2} + {x^2}} = 3m\lambda$

Squaring this equation we get,

${x^2} = 9{m^2}{\lambda ^2} - {d^2}$

$\Rightarrow {p^2} = 9$ or $p = 3$

In given Vernier callipers, each 1 cm is equally divided into 8 main scale divisions (MSD). Thus, 1 MSD = ${1 \over 8}$ = 0.125 cm. Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD) i.e., 4 MSD = 5 VSD. Thus, 1 VSD = 4/5 MSD = 0.8 $\times$ 0.125 = 0.1 cm. The least count of the Vernier callipers is given by

LC = 1 MSD $-$ 1 VSD = 0.125 $-$ 0.1 = 0.025 cm.

In screw gauge, let l be the distance between two adjacent divisions on the linear scale. The pitch p of the screw gauge is the distance travelled on the linear scale when it makes one complete rotation. Since circular scale moves by two divisions on the linear scale when it makes one complete rotation, we get p = 2l. The least count of the screw gauge is defined as ratio of the pitch to the number of divisions on the circular scale (n) i.e.,

$LC' = {p \over n} = {{2l} \over {100}} = {l \over {50}}$ ....... (1)

If p = 2 LC = 2(0.025) = 0.05 cm, then $l = {p \over 2} = 0.025$ cm. Substitute l in equation (1) to get the least count of the screw gauge

$LC' = {{0.025} \over {50}} = 5 \times {10^{ - 4}}$ cm = 0.005 mm.

If l = 2 LC = 2(0.025) = 0.05 cm then equation (1) gives

$LC' = {{0.05} \over {50}} = 1 \times {10^{ - 3}}$ cm = 0.01 mm.

(A)→(P), (Q), (R), (T); (B)→(Q), (S); (C)→(P), (Q); (D)→(P), (R), (T)

The nuclear fusion is responsible for energy production in stars via fusion of hydrogen nuclei into helium nuclei. In sun, the fusion takes place dominantly by proton-proton cycle, $4_1^1H \to _2^4He + 2{e^ + } + 2\nu + 2\gamma$. The neutrino ($\nu$) and $\gamma$-rays emissions are parts of this fusion reaction.

The uranium based fission reactions involve absorption of thermal neutrons by ${}_{92}^{235}$U nuclei to produce the highly fissionable ${}_{92}^{236}$U nuclei. This nuclei then fissions into two parts e.g., ${}_{92}^{236}$U $\to$ ${}_{63}^{137}$I + ${}_{39}^{97}$Y + 2n. The fission fragments are unstable and undergo $\beta$-decay to reduce their neutron to proton ratio. The fragments are generally formed in excited states and consequently emit $\gamma$-rays. The heavy water (D₂O) is used as a moderator to slow down the fast moving neutrons.

In $\beta$-decay, a neutron is converted into a proton. In this process, an electron and an antineutrino are created and emitted from the nucleus, n $\to$ p + e^$-$ + $\overline v$. The $\beta$-decay in ${}_{27}^{60}$Co $\to$ ${}_{28}^{60}$Ni + e^$-$ + $\overline v$. The daughter nuclei ${}_{28}^{60}$Ni is formed in excited state and comes to ground state by $\gamma$-ray emission.

The $\gamma$-rays are high energy electromagnetic rays. These rays are generally emitted when a nuclei in excited state (high energy) makes a transition to a lower state (low energy).

The force on a conducting element of length $d\overrightarrow l$, carrying a current I in a magnetic field $\overrightarrow B$, is given by d$\overrightarrow F$ = I d$\overrightarrow l$ $\times$ $\overrightarrow B$. If the field is uniform, the total force on the conductor is given by

$\overrightarrow F = \int_a^g {Id\overrightarrow l \times \overrightarrow B = I\left( {\int_a^g {d\overrightarrow l } } \right) \times \overrightarrow B = I\,\overrightarrow {ag} \times \overrightarrow B }$

Note that $\overrightarrow B$ is taken out of the integral sign because it is constant. The vector $\overrightarrow {ag} = 2(L + R)\widehat x$. The magnetic forces on the conductor in the given cases are

Case (A) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat z)$

$= - 2IB(L + R)\widehat y$

Case (B) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat x) = \overrightarrow 0$

Case (C) : $\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat y)$

$= 2IB(L + R)\widehat z$

The internal energy of one mole of an ideal gas at temperature T is given by $U = {f \over 2}RT$, where f is the degrees of freedom of the gas molecule. The degress of freedom of the gas molecule. The degrees of freedom for hydrogen (diatomic) and helium (monatomic) gases are f_H2 = 5 and f_He = 3, respectively. Thus, ${U_{{H_2}}} = {5 \over 2}RT$ and ${U_{He}} = {3 \over 2}RT$. The total internal energy of the gas mixture is

${U_{total}} = {U_{{H_2}}} + {U_{He}} = {5 \over 2}RT + {3 \over 2}RT = 4RT$.

The mixture contains two moles of the gases. The internal energy per mole of the mixture is ${U_{mix}} = {U_{total}}/2 = 2RT$.

The specific heat at constant volume is given by ${C_v} = dU/dT$. Thus, the specific heats at constant volume for helium and the mixture are

${C_{v,He}} = d{U_{He}}/dT = {3 \over 2}R$, and

${C_{v,mix}} = d{U_{mix}}/dT = 2R$

The specific heats at constant pressure, ${C_p} = {C_v} + R$, for these gases are

${C_{p,He}} = {C_{v,He}} + R = {5 \over 2}R$, and

${C_{p,mix}} = {C_{v,mix}} + R = 3R$.

The ratio of specific heats, $\gamma = {C_p}/{C_v}$, are ${\gamma _{He}} = 5/3$ and ${\gamma _{mix}} = 3/2$. The speed of sound, in a gas of molecular mass M, is given by ${v_s} = \sqrt {\gamma RT/M}$. The molecular mass of the gas mixture is

${M_{mix}} = {{{n_{{H_2}}}{M_{{H_2}}} + {n_{He}}{M_{He}}} \over {{n_{{H_2}}} + {n_{He}}}}$

$= {{(1)(2) + (1)(4)} \over {1 + 1}} = 3$ g/mol,

where n_H2 = 1 and n_He = 1 are the number of moles of hydrogen and helium in the gas mixture. The ratio of the speeds of sound in the gas mixture and helium is

${{{v_{s,mix}}} \over {{v_{s,He}}}} = {{\sqrt {{\gamma _{mix}}RT/{M_{mix}}} } \over {\sqrt {{\gamma _{He}}RT/{M_{He}}} }} = \sqrt {{{{\gamma _{mix}}} \over {{\gamma _{He}}}}{{{M_{He}}} \over {{M_{mix}}}}}$

$= \sqrt {{{(3/2)(4)} \over {(5/3)(3)}}} = \sqrt {{6 \over 5}}$

The rms speed of the atoms/molecules is given by ${v_{rms}} = \sqrt {3RT/M}$. The ratio of the rms speed of helium atoms to that of hydrogen molecules is

${{{v_{rms,He}}} \over {{v_{rms,{H_2}}}}} = {{\sqrt {3RT/{M_{He}}} } \over {\sqrt {3RT/{M_{{H_2}}}} }} = \sqrt {{{{M_{{H_2}}}} \over {{M_{He}}}}} = \sqrt {{2 \over 4}} = \sqrt {{1 \over 2}}$.

Energy of the incident photon $= hf = {{hc} \over \lambda } = {{1242} \over {90}} = 13.8$ eV. Since after ionisation, electron is ejected with some kinetic energy. By energy conservation, we get

Energy (photon) = Kinetic energy (electron) + $\Delta$E

Transition energy from nth orbit to n $\to$ $\infty$. Therefore,

13.8 = 10.4 + $\Delta$E

$\Rightarrow$ $\Delta$E = 3.4 eV

From Bohr's theory,

${E_n} = {{ - 13.6} \over {{n^2}}} = - 3.4 \Rightarrow n = 2$

Apply the formula for the refraction at the spherical surface of S₁, ${{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{\mu _2} - {\mu _1}} \over R}$, to get

${{1.0} \over v} - {{1.5} \over {( - 50)}} = {{1 - 1.5} \over {( - 10)}}$.

Solve to get the image distance v = 50 cm (point Q in the figure).

The image Q acts as an object for refraction at the spherical surface of S₂. The object distance is $u = - (d - 50)$. The image distance is v = $\infty$ (because rays are parallel in S₂). Apply the formula for refraction at the spherical surface of S₂, to get

${{1.5} \over \infty } - {1 \over { - (d - 50)}} = {{1.5 - 1} \over {(10)}}$.

Solve to get d = 70 cm.

ANBP is cross-section of a cylinder of length L. The line charge passes through the centre O and perpendicular to paper.

$AM = {a \over 2}$, $MO = {{\sqrt 3 a} \over 2}$

$\therefore$ $\angle AOM = {\tan ^{ - 1}}\left( {{{AM} \over {OM}}} \right)$

$= {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) = 30^\circ$

Electric flux passing from the whole cylinder

${\phi _1} = {{{q_{in}}} \over {{\varepsilon _0}}} = {{\lambda L} \over {{\varepsilon _0}}}$

$\therefore$ Electric flux passing through ABCD plane surface (shown only AB) = Electric flux passing through cylindrical surface ANB

$= \left( {{{60^\circ } \over {360^\circ }}} \right)({\phi _1}) = {{\lambda L} \over {6{\varepsilon _0}}}$

$\therefore$ n = 6

Half life of radioactive material = T years

Let amount of radioactive material as fuel at the beginning be N₀ and corresponding power produced by it be P₀.

According to question,

Power requirement of the village

= 12.5% of ${P_0} = {{{P_0}} \over 8}$

Since, after each T year, power will be half, i.e.,

{P_0}\buildrel T \over \longrightarrow {{{P_0}} \over 2}\buildrel T \over \longrightarrow {{{P_0}} \over 4}\buildrel T \over \longrightarrow {{{P_0}} \over 8}

Total time upto which the plant can meet the village's need = 3T years = nT years

$\therefore$ n = 3

Carbonate ores are
(P) Siderite : FeCO₃
(Q) Malachite : CuCO₃.Cu(OH)₂
(S) Calamine : ZnCO₃

Sulphide ore is (T) Argentite : Ag₂S.

Hydroxide ion is present in
(Q) Malachite : CuCO₃.Cu(OH)₂
(R) Bauxite : Al₂O₃.2H₂O or AlO_x (OH)_3–2x where 0 < x < 1

Oxide ore is bauxite (R) only

Note : ( Remember all those ores names. Any one of those can be asked in the exam.)

Oxides Ores :

(1) ZnO - Zincite

(2) Fe₂O₃ - Haematite

(3) Fe₃O₄ - Magnetite (FeO + Fe₂O₃ mixture)

(4) Fe₂O₃ . 3H₂O - Limonite

(5) MnO₂ - Pyrolusite

(6) Cu₂O - Cuprite or Ruby Copper

(7) TiO₂ - Rutile

(8) FeCr₂O₄ - Chromite (FeO + Cr₂O₃)

(9) FeTiO₃ - Illmenite (FeO + TiO₂)

(10) Na₂B₄O₇ . 10H₂O - Borax or Tincal

(11) U₃O₈ - Pitch Blende

(12) SnO₂ - Tin Stone or Cassiterite

(13) Ca₂B₆O₁₁ . 5H₂O - Colemanite (2 Cao + 3 B₂O₃)

(14) Al₂O₃ . 2H₂O - Bauxite

(15) Al₂O₃ . H₂O - Diaspore

(16) Al₂O₃ - Corundum

Sulphides Ores :

(1) ZnS - Zinc Blende or Sphalerite

(2) PbS - Galena

(3) Ag₂S - Argentite or Silver Glance

(4) HgS - Cinnabar

(5) Cu₂S - Chalcocite or Copper glance

(6) CuFeS₂ - Copper pyrites or Chalco pyrites (Cu₂S + Fe₂S₃ mixture)

(7) FeS₂ - Iron pyrites or Fool's Gold

(8) 3Ag₂S . Sb₂S₂ - Pyrargyrite or ruby silver

Halides Ores :

(1) NaCl - Rock Salt

(2) KCl - Sylvine

(3) Na₃AlF₆ - Cryolite [3NaF + AlF₆]

(4) CaF₂ - Fluorspar

(5) KCl . MgCl₂ . 6H₂O - Carnalite

(6) AgCl - Horn Silver

Carbonates Ores :

(1) CaCO₃ - Limestone

(2) MgCO₃ - Magnesite

(3) CaCO₃ . MgCO₃ - Dolomite

(4) ZnCO₃ - Calamine

(5) PbCO₃ - Cerrusite

(6) FeCO₃ - Siderite

(7) CuCO₃ . Cu(OH)₂ or Cu₂CO₃(OH)₂ - Malachite or Basic Copper Carbonates

(8) 2 CuCO₃ . Cu(OH)₂ - Azurite

Sulphates Ores :

(1) CuSO₄ . 2H₂O - Gypsum

(2) MgSO₄ . 7H₂O - Epson Salt

(3) Na₂SO₄ . 10 H₂O - Glauber's Salt

(4) PbSO₄ - Anglesite

(5) ZnSO₄ . 7H₂O - White Vitriol

(6) FeSO₄ . 7H₂O - Green Vitriol

(7) CuSO₄ . 5H₂O - Blue Vitriol or Chalcanthite

Nitrate Ores :

(1) KNO₃ - Indian Saltpetre

(2) NaNO₃ - Chile Saltpetre

Arsenides Ores :

(1) NiAs - Kupfernickel

(2) NiAsS - Nickel glance

Fe³⁺ is reduced to Fe²⁺ by H₂O₂/NaOH and Na₂O₂/H₂O.

$2F{e^{3 + }} + {H_2}{O_2} + O{H^ - } \to 2F{e^{2 + }} + 2{H_2}O + {O_2}$

$N{a_2}{O_2} + {H_2}O \to {H_2}{O_2} + NaOH$

In D-(+)-glucose the OH group attached to the last stereocentre is on the right hand side while in L-($-$)-glucose it is on left hand side. The structure of L-($-$)-glucose is

For ccp, Z = 4 = no. of O-atoms

No. of octahedral voids = 4

No. of tetrahedral voids = 2 $\times$ 4 = 8

No. of Al³⁺ ions = m $\times$ 4

No. of Mg²⁺ ions = n $\times$ 8

Thus, the formula of the mineral is Al_4m Mg_8nO₄

4m(+3) + 8n(+2) + 4($-$2) = 0

12m + 16n $-$ 8 = 0

4(3m + 4n $-$ 2) = 0

3m + 4n = 2

Possible values of m and n are ${1 \over 2}$ and ${1 \over 8}$ respectively.

(a) is wrong statement. Impure copper is set as anode where copper is oxidized to Cu²⁺ and goes into electrolytic solutions.

(b) CuSO₄ is used as an electrolyte in purification process.

(c) Pure copper is deposited at cathode as : Cu²⁺ + 2e^$-$ $\to$ Cu : (At cathode)

(d) Less active metals like Ag, Au etc. settle down as anode mud.

The depression in freezing point is given by

$\Delta$T_f = K_f $\times$ m $\times$ i

0.0558 = 1.86 $\times$ 0.01 $\times$ i

i = 3

Therefore, one mole of complex gives three moles of ions in solution.

Hence, the complex is [Co(NH₃)₅Cl]Cl₂ and the number of Cl^$-$ ions inside the coordination sphere is 1.

Given :

X $\to$ Y; $\Delta$_rG$^\circ$ = $-$ 193 kJ mol^$-$1

M⁺ $\to$ M³⁺ + 2e^$-$; E$^\circ$ = $-$0.25 V

F = 96500 C mol^$-$1

Let 193 kJ is used for oxidising x moles of M⁺.

For 1 mole of M⁺,

$\Delta$G$^\circ$ = $-$nFE$^\circ$

= $-$2 $\times$ 96500 $\times$ ($-$0.25)

= 48250 J mol^$-$1 = 48.25 kJ mol^$-$1

Thus, no. of moles of M⁺ oxidized when one mole of X is converted to Y = ${{193} \over {48.25}} = 4$.

The structure of N₂O₃ is

Therefore, the total number of lone pairs is 8.

(A) $\to$ (R and T)

Freezing of water,

H₂O_(l) $\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{273\,K}^{1\,atm}}$ H₂O_(s)

The system is cooled i.e.; heat is released during the process

so, q < 0

$\mathop {Water}\limits_{(Less\,volume)}$ $\rightleftharpoons$ $\mathop {Ice}\limits_{(More\,volume)}$ + heat

Volume is increased i.e.; $\Delta$V = +ve.

w = $-$P$\Delta$V = $-$ve

i.e.; w < 0 (expansion)

Entropy of system is decreased, $\Delta$S_sys < 0.

$\Delta$U = q + w

As q < 0, w < 0 so, $\Delta$U < 0.

At equilibrium, $\Delta$G = 0.

(B) $\to$ (P, Q and S)

Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions,

w = 0, q = 0 so, $\Delta$U = 0

For expansion, $\Delta$S_sys > 0 as entropy increases.

$\Delta G = - nRT\ln {{{V_2}} \over {{V_1}}}$

For expansion, V₂ > V₁

$\Delta$G = $-$ve i.e.; $\Delta$G < 0.

(C) $\to$ (P, Q and S)

Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container.

q = 0 (isolated)

w = $-$P$\Delta$V

w = 0 ($\because$ $\Delta$V = 0)

$\Delta$S_sys > 0 (mixing of gases)

$\Delta$U = q + w = 0

$\Delta$G = $\Delta$H $-$ T$\Delta$S

$\Delta$G = q_p $-$ T$\Delta$S (at constant P, T)

$\Delta$G = 0 $-$ T$\Delta$S = $-$T$\Delta$S

$\Delta$G < 0 ($\because$ $\Delta$S_sys > 0)

(D) $\to$ (P, Q, S and T)

$\mathop {{H_{2(g)}}}\limits_{(300\,K)} \mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{Cool}^{Heat,\,1\,atm}} \mathop {{H_{2(g)}}}\limits_{(600\,K)}$

Internal energy (U), entropy (S) and free energy (G) are state functions which depend only upon the state of the system and do not depend upon the path by which the state is attained.

Thus, $\Delta$U = 0, $\Delta$S = 0 and $\Delta$G = 0

Work and heat are path functions but the same path is retraced so, q = 0 and w = 0.