IIT-JEE 2010 Paper 1 Question Paper with Solutions

Consider the refraction at the face AB. Snell's law,

$\sin i/\sin r = \sin 60^\circ /\sin r = \sqrt 3$,

gives $r = 30^\circ$.

The geometry in BCQP gives

$\angle BPQ = 30^\circ + 90^\circ = 120^\circ$

$\angle CQP = 360^\circ - (135^\circ + 60^\circ + 120^\circ ) = 45^\circ$.

Thus, the angle of incidence at Q is ${i_1} = 45^\circ$. The critical angle for prism to air refraction is given by $\sin {i_c} = 1/\sqrt 3$. Since $\sin {i_1} = 1/\sqrt 2 > 1\sqrt 3$, we get ${i_1} > {i_c}$ i.e., the angle of incidence is greater than the critical angle. Thus, the ray undergoes total internal reflection at Q. The laws of reflection gives ${r_1} = {i_1} = 45^\circ$. In triangle QRD, $\angle QRD = 60^\circ$ and hence the angle of incidence at R is ${i_2} = 30^\circ$.

Applying Snell's law at face AD, we get

$\sqrt 3 \times \sin 30^\circ = 1 \times \sin e$

or, $\sqrt 3 \times {1 \over 2} = \sin e$

$\sin e = {{\sqrt 3 } \over 2}$

or, $e = {\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) = 60^\circ$

From figure,

The angle between the incident ray and the emergent ray is 90$^\circ$.

Hence, option (c) is correct and option (d) is incorrect.

Note : Angle between incident and emergent rays is the same as the angle between the two faces = 90$^\circ$.

For $|x| > {X_0}$, $V = {V_0}$ = constant

Force $= - {{dV} \over {dx}} = 0$

Hence, acceleration of the particle is zero for $|x| > {X_0}$.

The power of the bulb is

$P = {{{V^2}} \over R}$

Therefore,

$100 = {{{V^2}} \over {{R_{100}}}} \Rightarrow {1 \over {{R_{100}}}} = {{100} \over {{V^2}}}$

where R₁₀₀ is the resistance (at any temperature) corresponds to 100 W. Similarly,

$60 = {{{V^2}} \over {{R_{60}}}} \Rightarrow {1 \over {{R_{60}}}} = {{60} \over {{V^2}}}$ and $40 = {{{V^2}} \over {{R_{40}}}} \Rightarrow {1 \over {{R_{40}}}} = {{40} \over {{V^2}}}$

From these equations, we get

${P_{100}} > {P_{60}} > {P_{40}} \Rightarrow {1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$

Impedance of the circuit,

$Z = \sqrt {{{({X_C})}^2} + {{(R)}^2}} = \sqrt {{{\left( {{1 \over {\omega C}}} \right)}^2} + {R^2}}$

As $\omega$ increases, Z decreases.

Current in the circuit, $I = {{{V_0}} \over Z}$

When $\omega$ is increased, the impedance of the circuit decreases and the current through the bulb increases. Therefore the bulb glows brighter.

The kinetic energy of the particle cannot be negative. The total energy E is the sum of the kinetic energy K(x) and potential energy V(x) i.e.,

E = K(x) + V(x). ....... (1)

From the given figure, V(x) $\ge$ 0 for all x. If E $\le$ 0 for some x then kinetic energy K(x) = E $-$ V(x) $\le$ 0 for those x and hence the motion of the particle is not allowed for those x. On the other hand, if E $\ge$ V₀ then K(x) = E $-$ V(x) $\ge$ 0 for all x and particle is allowed to move for all x, including infinity (in this case the particle will escape to infinity). Thus, for the particle to have a periodic motion, V₀ > E > 0. In this case, particle is allowed at points where

K(x) = E $-$ V(x) = E $-$ $\alpha$x⁴ $\ge$ 0 i.e.,

| x | $\le$ (E/$\alpha$)^1/4.

As $V = a{x^4}$

$[\alpha ] = {{[V]} \over {[{x^4}]}} = {{[M{L^2}{T^{ - 2}}]} \over {[{L^4}]}} = [M{L^{ - 2}}{T^{ - 2}}]$

By method of dimensions,

$\left[ {{1 \over A}\sqrt {{m \over \alpha }} } \right] = {{{{[M]}^{1/2}}} \over {[L]{{[M{L^{ - 2}}{T^{ - 2}}]}^{1/2}}}} = [T]$

Only option (b) has the dimensions of time.

On the planet,

${g_p} = {{G{M_p}} \over {R_p^2}} = {G \over {R_p^2}}\left( {{4 \over 3}\pi R_p^3{\rho _p}} \right) = {4 \over 3}G\pi R_p^{}{\rho _p}$

On the earth,

${g_e} = {{G{M_e}} \over {R_e^2}} = {G \over {R_e^2}}\left( {{4 \over 3}\pi R_e^3{\rho _e}} \right) = {4 \over 3}G\pi R_e^{}{\rho _e}$

$\therefore$ ${{{g_p}} \over {{g_e}}} = {{R_p^{}{\rho _p}} \over {R_e^{}{\rho _e}}}$ or ${{{R_p}} \over {{R_e}}} = {{g_p^{}{\rho _p}} \over {g_e^{}{\rho _e}}}$ ..... (i)

On the planet, ${v_p} = \sqrt {2{g_p}{R_p}}$

On the earth, ${v_e} = \sqrt {2{g_e}{R_e}}$

$\therefore$ ${{{v_p}} \over {{v_e}}} = \sqrt {{{{g_p}{R_p}} \over {{g_e}{R_e}}}} = {{{g_p}} \over {{g_e}}}\sqrt {{{{\rho _e}} \over {{\rho _p}}}}$ (Using (i))

Here, ${\rho _p} = {2 \over 3}{\rho _e}$, ${g_p} = {{\sqrt 6 } \over {11}}{g_e}$

$\therefore$ ${{{v_p}} \over {{v_e}}} = {{\sqrt 6 } \over {11}}\sqrt {{3 \over 2}}$

or, ${v_p} = 11 \times {{\sqrt 6 } \over {11}} \times \sqrt {{3 \over 2}}$ ($\because$ v_e = 11 km s^$-$1 (Given))

= 3 km s^$-$1

We need to find gravitational potential energy of a unit mass placed at the point P.

The surface mass density of the annular disc is

$\sigma = M/(16\pi {R^2} - 9\pi {R^2}) = M/(7\pi {R^2})$.

Consider a small ring of radius r and thickness dr. The mass of the ring is dm = 2$\pi$r$\sigma$dr. As distance of any point of the ring from P is same, the potential at P due to the ring is

${V_P} = - {{G(2\pi r\sigma dr)} \over {\sqrt {16{R^2} + {r^2}} }}$

Integrate from r = 3R to r = 4R to get the potential energy of the unit mass placed at P

${V_P} = \int\limits_{3R}^{4R} { - {{GdM} \over {\sqrt {{{(4R)}^2} + {{(r)}^2}} }} = - {{GM2\pi } \over {7\pi {R^2}}}\int\limits_{3R}^{4R} {{{rdr} \over {\sqrt {16{R^2} + {r^2}} }}} }$

Solving, we get

${V_P} = - {{GM2\pi } \over {7\pi {R^2}}}\left[ {\sqrt {16{R^2} + {r^2}} } \right]_{3R}^{4R} = - {{2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)$

Work done in moving a unit mass from P to $\infty$ = V_$\infty$ $-$V_P

$= 0 - \left( {{{ - 2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)} \right) = {{2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)$

In the final state, ice-water mixture is in equilibrium. Thus, the temperature of m grams of ice is raised from $-$5 $^\circ$C to 0 $^\circ$C. The heat absorbed in this process is

Q₁ = mS$\Delta$T. ...... (1)

The state of m₁ = 1 g of ice is changed from solid to liquid. The heat absorbed in the melting process is

Q₂ = m₁L. ........ (2)

The heat supplied is Q = 420 J. By energy conservation, Q = Q₁ + Q₂. Substitute Q₁ and Q₂ from equations (1) and (2) to get

$m = {{Q - {m_1}L} \over {S\Delta T}} = {{420 - ({{10}^{ - 3}})(3.36 \times {{10}^5})} \over {(2100)(5)}}$

$= 8 \times {10^{ - 3}}$ kg = 8 g.

From the given figure, T_c(B) is a monotonically decreasing function of B. Thus, B₂ > B₁ implies T_c(B₂) < T_c(B₁). Hence resistance with B₂ will become zero at lower temperature in comparison to B₁.

The sheet is of square shape with thickness t, width $\omega$ = L, and length l = L. The resistance between the two opposite faces is given by

$R = {{\rho l} \over A} = {{\rho l} \over {wt}} = {{\rho L} \over {Lt}} = {\rho \over t}$.

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

or, ${u \over v} - 1 = {u \over f}$

or, ${u \over v} = \left( {{{u + f} \over f}} \right)$

$\therefore$ $m = {v \over u} = \left( {{f \over {u + f}}} \right)$

${{{m_{25}}} \over {{m_{50}}}} = {{\left( {{{20} \over { - 25 + 20}}} \right)} \over {\left( {{{20} \over { - 50 + 20}}} \right)}} = 6$

Consider an small element AB of length dl of the circle of radius R subtending an angle $\theta$ at the centre O.

If T is the tension in the wire, then force towards the centre will be equal to $2T\sin \left( {{\theta \over 2}} \right)$ which is balanced by outward magnetic force on the current carrying element $( = IdlB)$

$2T\sin \left( {{\theta \over 2}} \right) = IdlB$

For small angle $\theta$, $\sin {\theta \over 2} \approx {\theta \over 2}$

or, $T = {{IBdl} \over \theta } = IBR$ ($\because$ $\theta = {{dl} \over R}$)

$= {{IBL} \over {2\pi }}$ ($\because$ $R = {L \over {2\pi }}$)

Here, m₁ = 1 kg, m₂ = 5 kg

u₁ = u, u₂ = 0

v₁ = $-$2 m s^$-$1, v₂ = v

By the law of conservation of linear momentum, we get

${m_1}{u_1} + {m_2}{u_2} = {m_1}{u_1} + {m_2}{u_2}$

$1 \times u + 5 \times 0 = 1 \times ( - 2) + 5 \times v$

$u = 5v - 2$ ...... (i)

By the definition of coefficient of restitution,

$e = {{{v_2} - {v_1}} \over {{u_1} - {u_2}}}$

For a perfectly elastic collision, e = 1

$\therefore$ $1 = {{v + 2} \over u}$ or $u = v + 2$ ..... (ii)

Solving equations (i) and (ii), we get

u = 3 m s^$-$1, v = 1 m s^$-$1

Before collision,

Total momentum of the system = 1 $\times$ 3 + 5 $\times$ 0 = 3 kg m s^$-$1

After collision,

Total momentum of the system = 1 $\times$ ($-$2) + 5 $\times$ (1) = 3 kg m s^$-$1

Hence, option (a) is correct.

Momentum of 5 kg mass after collision = 5 $\times$ 1 = 5 kg m s^$-$1.

Hence, option (b) is incorrect.

Velocity of centre mass is ${v_{cm}} = {{1 \times 3 + 5 \times 0} \over {1 + 5}} = {1 \over 2}$ m s ^$-$1

Kinetic energy of the centre of mass

$= {1 \over 2}{m_{system}}v_{CM}^2 = {1 \over 2} \times 6 \times {\left( {{1 \over 2}} \right)^2} = 0.75$ J

Hence, option (c) is correct.

Before collision,

Total kinetic energy of the system

$= {1 \over 2} \times 1 \times {3^2} + {1 \over 2} \times 5 \times {0^2} = 4.5$ J

After collision,

Total kinetic energy of the system

$= {1 \over 2} \times 1 \times {( - 2)^2} + {1 \over 2} \times 5 \times {(1)^2} = 4.5$ J

Hence, option (d) is incorrect.

When mass m is pulled by a force F, the wire elongation x, length l, cross-sectional area A, and Young's modulus of wire material Y are related by $Y = {{F/A} \over {x/l}}$ i.e.,

$F = (Y\,A/l)x$.

The restoring force by the wire is equal but opposite to F i.e., F_r = $-$F. Apply Newton's second law to get

$m{d^2}x/d{t^2} = - (YA/l)x = - {\omega ^2}x$.

This equation represents SHM with an angular frequency $\omega = \sqrt {YA/(lm)}$. Substitute the values to get $Y = {\omega ^2}lm/A = 4 \times {10^9}$ N/m².

It is given that T_c(0) = 100 K and T_c(7.5) = 75 K. Since T_c(B) is a monotonically decreasing function of B, T_c(5) < T_c(0) and T_c(5) > T_c(7.5). Thus, 75 K < T_c(5) < 100 K.

Let car B be the observer (moving towards S).

The frequency observed is

${f_1} = {f_0}\left( {{{c + v} \over c}} \right)$

When sound gets reflected, the frequency observed by source S is

${f_2} = {f_1}\left( {{c \over {c - v}}} \right)$

where v is the speed of car and c is the speed of sound. Therefore,

${f_2} = {f_0}\left( {{{c + v} \over {c - v}}} \right)$

Now, $d{f_x} = {f_0}\left[ {{{(c - v)dv - (c + v)( - dv)} \over {{{(c - v)}^2}}}} \right]$

$= {{2{f_0}c\,dv} \over {{{(c - v)}^2}}}$

That is,

${{2{f_0}c\,dv} \over {{{(c - v)}^2}}} = \left( {{{1.2} \over {100}}} \right){f_0}$

$\Rightarrow dv = {{1.2} \over {100}} \times {{{{(c - v)}^2}} \over {2c}}$

Since, v << c, we get c $-$ v $\simeq$ c.

Therefore,

$dv = {{1.2} \over {100}} \times {c \over 2} = 1.98$ m/s

$= 1.98 \times {{18} \over 5}$ km/h = 7 km/h.

To verify Ohm's law, we need to measure the voltage across the test resistance R_T and current passing through it. The voltage can be measured by connecting a high resistance R₁ in series with galvanometer. This combination becomes a voltmeter and should be connected in parallel to R_T. The current can be measured by connecting a low resistance R₂ (shunt) in parallel with galvanometer. This combination becomes an ammeter and should be connected in series to measure the current through R_T.

Number of electric field lines originating from Q₁ is more than terminating at Q₂.

$\therefore$ $|{Q_1}| > |{Q_2}|$

Here, Q₁ is positive while Q₂ is negative.

The electric field at a distance x towards the right of Q₂ is given by

$|\overrightarrow E | = {1 \over {4\pi { \in _0}}}{{{Q_1}} \over {{{(d + x)}^2}}} - {1 \over {4\pi { \in _0}}}{{{Q_2}} \over {{x^2}}}$,

where d is the separation between Q₁ and Q₂. Since ${Q_1} > {Q_2}$, $|\overrightarrow E |$ becomes zero for some finite x.

Let stars A and B are rotating about their centre of mass with angular velocity $\omega$.

Let distance of stars A and B from the centre of mass be r_A and r_B respectively as shown in the figure.

Total angular momentum of the binary stars about the centre of mass is

$L = {M_A}r_A^2\omega + {M_B}r_B^2\omega$

Angular momentum of the star B about centre of mass is

${L_B} = {M_B}r_B^2\omega$

$\therefore$ ${L \over {{L_B}}} = {{({M_A}r_A^2 + {M_B}r_B^2)\omega } \over {{M_B}r_B^2\omega }} = \left( {{{{M_A}} \over {{M_B}}}} \right){\left( {{{{r_A}} \over {{r_B}}}} \right)^2} + 1$

Since ${M_A}{r_A} = {M_B}{r_B}$

or, ${{{r_A}} \over {{r_B}}} = {{{M_B}} \over {{M_A}}}$

$\therefore$ ${L \over {{L_B}}} = {{{M_B}} \over {{M_A}}} + 1 = {{11{M_S}} \over {2.2{M_S}}} + 1 = {{11 + 2.2} \over {2.2}} = 6$

Relative error in measurement of time,

${{\Delta t} \over t} = {{1\,s} \over {40\,s}} = {1 \over {40}}$

Time period, $T = {{40\,s} \over {20}} = 2\,s$

Error in measurement of time period,

$\Delta T = T \times {{\Delta t} \over t} = 2\,s \times {1 \over {40}} = 0.05\,s$

The time period of simple pendulum is

$T = 2\pi \sqrt {{l \over g}}$

or, ${T^2} = {{4{\pi ^2}l} \over g}$

or, $g = {{4{\pi ^2}l} \over {{T^2}}}$

$\therefore$ ${{\Delta g} \over g} = {{2\Delta T} \over T} = 2 \times {1 \over {40}} = {1 \over {20}}$ ($\because$ ${{\Delta T} \over T} = {{\Delta t} \over t}$)

Percentage error in determination of g is

${{\Delta g} \over g} \times 100 = {1 \over {20}} \times 100 = 5\%$

According to Wien's displacement law, $\lambda$_mT = constant

$\therefore$ ${({\lambda _m})_A}{T_A} = {({\lambda _m})_B}{T_B}$

or, ${{{T_A}} \over {{T_B}}} = {{{{({\lambda _m})}_B}} \over {{{({\lambda _m})}_A}}} = {{1500\,nm} \over {500\,nm}}$ or ${{{T_A}} \over {{T_B}}} = 3$ ...... (i)

According to Stefan Boltzmann law, rate of energy radiated by a black body

$E = \sigma A{T^4} = \sigma 4\pi {R^2}{T^4}$ [Here, $A = 4\pi {R^2}$]

$\therefore$ ${{{E_A}} \over {{E_B}}} = {\left( {{{{R_A}} \over {{R_B}}}} \right)^2}{\left( {{{{T_A}} \over {{T_B}}}} \right)^4} = {\left( {{{6\,cm} \over {18\,cm}}} \right)^2}{(3)^4}$ (Using (i))

$= 9$

Here, ${y_1} = 4\sin (2x - 6t)$

${y_2} = 3\sin \left( {2x - 6t - {\pi \over 2}} \right)$

The phase difference between two waves is $\phi = {\pi \over 2}$

The amplitude of the resultant wave is

$A = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi }$

$= \sqrt {{4^2} + {3^2} + 2 \times 4 \times 3 \times \cos {\pi \over 2}} = 5$

The de Broglie wavelength of a particle with momentum p is given by

$\lambda$ = h/p.

The momentum and kinetic energy of a particle of mass m are related by

$p = \sqrt {2mK}$.

The kinetic energy of a charge q, accelerated through potential V, is given by K = qV. Thus,

$\lambda = h/\sqrt {2mK} = h/\sqrt {2mqV}$,

which gives

${{{\lambda _p}} \over {{\lambda _\alpha }}} = \sqrt {{{2{m_\alpha }{q_\alpha }V} \over {2{m_p}{q_p}V}}} = \sqrt {{{2\,.\,4u\,.\,2e\,.\,100} \over {2\,.\,1u\,.\,1e\,.\,100}}}$

$= \sqrt 8 = 2.8 \approx 3$

The internal energy of one mole of an ideal gas at temperature T is given by U = 3RT/2. The process AB is isothermal i.e., T_A = T_B = T₀, which makes U_A = U_B.

The work done in the isothermal process AB is

W_AB = nRT ln(V_B/V_A)

= nRT₀ ln(4V₀/V₀) = p₀V₀ ln4.

Information regarding p and T at C can not be obtained from the given graph. Unless it is mentioned that line BC passes through origin or not.

Hence, the correct options are (a) and (b).

Note:

If we assume that the line BC pass through the origin. In the process BC, the slope V/T is constant. Thus, BC is an isobaric process (as V/T = nR/p, by the ideal gas equation). Thus,

p_C = p_B = RT_B / V_B

= RT₀/(4V₀) = (RT₀/V₀)4 = p₀/4.

Apply the ideal gas equation for the states A and C to get

${T_C} = {{{p_C}{V_C}} \over {{p_0}{V_0}}}{T_0} = {{({p_0}/4){V_0}} \over {{p_0}{V_0}}}{T_0} = {T_0}/4$.

In series : When the batteries are connected in series, we have

${J_1} = {\left( {{{2E} \over {R + 2}}} \right)^2}R$

In parallel : When the batteries are connected in parallel, we have

${J_2} = {\left( {{E \over {R + (1/2)}}} \right)^2}R$

It is given that,

${{{J_1}} \over {{J_2}}} = 2.25$

$\Rightarrow {4 \over {{{(R + 2)}^2}}} \times {{{{(2R + 1)}^2}} \over 4} = 2.25 \Rightarrow {{2R + 1} \over {R + 2}} = 1.5$

$\Rightarrow 2R + 1 = 1.5R + 3 \Rightarrow 0.5R = 2$

Therefore, $R = 4\Omega$.

The forces acting on the block are its weight mg, normal reaction N, applied force P and frictional force f.

Resolve mg along and normal to the inclined plane and apply Newton's second law to get

$0 = P + f - mg\sin \theta$,

which gives

$f = - P + mg\sin \theta$. ...... (1)

This is a straight line with slope $-$1. Substitute the values of P₁ and P₂ in equation (1) to get the frictional force at those points i.e.,

${f_1} = \mu mg\cos \theta$ and ${f_2} = - \mu mg\cos \theta$.

According to VSEPR, $\mathrm{BrF}_5$ has square pyramidal structure with axial plane containing a lone pair and fluorine. The other four fluorine are arranged in square planner configuration around central metal atom. Thus, $\mathrm{BrF}_5$ assumes square pyramidal shape where the valence electron pairs surrounding an atom tend to repel each other and will, therefore, adopt an arrangement that minimises this repulsion, thus, determining the molecule's geometry. All four planar bonds $(\mathrm{F}-\mathrm{Br}-\mathrm{F})$ will reduce from $90^{\circ}$ to $84.8^{\circ}$ after lone pair - bond pair repulsion.

So, there are no 90 -degree $\mathrm{F}-\mathrm{Br}-\mathrm{F}$ angles in $\mathrm{BrF}_5$.

To get appropriate bromoalkane an alkyne that were involved in the synthesis, we have to cleave the molecule and we know that sodium amide is involved, so the synthesis of 3-octyne using the alkyne and bromoalkane was $\mathrm{S}_{\mathrm{N}}2$ reaction. And as the sodium amide is involved, the alkyne involved must be terminal then only it could abstract the acidic hydrogen 3-octyne is a molecule that will have eight carbon atoms in their parent chain since the prefix-oct represents 8 and as in the name it is given that 3 octyne, ' 3 ' represents the position of triple bond. Thus suffix-one represents triple bond.

So, in the molecule triple bond will be at the third carbon. Structure of 3-octyne :

Now, cleave the molecule between $\mathrm{C}_2-\mathrm{C}_3$ or $\mathrm{C}_4$ - $C_5$.

Now, see the products formed if cleavage takes place between C₂-C₃, the reactants will be

If cleavage takes place between C₄-C₅, the reactants will be :

Then correct option is D.

EDTA (ethylenediaminetetracetic acid) is a hexadentate ligand. It is also called the amino poly-carboxylic acid.

It is structure consists of the ethane group which is attached to tertiary amine groups at the terminal positions forming an ethylene diamine. Tertiary amine group is situated by the acetic acid groups $\left(-\mathrm{CH}_2 \mathrm{COOH}\right)$.

Alkyl aryl ethers are cleaved at the alkyl-oxygen bond due to stronger aryl-oxygen bond. The reaction yields phenol and alkyl halide. Thus,

The reaction is

Iron obtained in form of its oxide is removed by slag formation with SiO$_2$.

FeO + SiO$_2$ $\to$ FeSiO$_ 3$ (slag)

In the structure of 2,2-dimethylbutane:

On C$_2$$-$C$_3$ bond axis X = CH$_3$, Y = CH$_3$.

On C$_1$$-$C$_2$ bond axis X = H, Y = C$_2$H$_5$.

The standard molar enthalpy of formation, denoted as $\Delta H_f^\circ$, is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K and 1 atm pressure). By definition, the standard molar enthalpy of formation of a pure element in its most stable form at 298 K is zero.

Looking at the options:

• Br₂ (g) - Bromine's standard state at room temperature is liquid (Br₂(l)), not gas.
• Cl₂ (g) - Chlorine's standard state at room temperature is a gas, which means the standard molar enthalpy of formation for Cl₂(g) is indeed zero.
• H₂O (g) - Water in gaseous form is not a basic element but a compound of hydrogen and oxygen, so its enthalpy of formation is not zero.
• CH₄ (g) - Methane, represented here, is also a compound composed of carbon and hydrogen, thus its enthalpy of formation is also not zero.

Therefore, the correct option is Option B, Cl₂ (g), as chlorine gas is a diatomic molecule and an element in its standard state at 298 K, and as such, its standard molar enthalpy of formation is zero.

To find the average titre value, first add up the three measurements provided and then divide by the number of measurements.

The sum of the measurements is:

Since there are three measurements, divide this sum by 3 to calculate the average:

When reporting the average, we must consider the significant figures of the original measurements. The number of significant figures is determined by the least precise measurement, which in this case is 25.0 mL with three significant figures. Therefore, we should report the average value to three significant figures as well.

The average value of 25.15 mL has four significant figures, so we need to round it to three significant figures. However, this is slightly tricky since 25.15 already appears to be rounded to four significant figures. We should consult the original measurements to decide on the best course of action.

Looking at the individual measurements (25.2, 25.25, and 25.0), we should consider the lowest decimal place which they all have in common, which is the first decimal place. The third measurement has no second decimal place, indicating its level of precision. Thus, the number of significant figures for the average titre value should be in line with this level of precision. Since the average calculated is 25.15, when we adjust to the first decimal place for consistent significant figures, the average is 25.1 mL with three significant figures.

Therefore, the number of significant figures in the average titre value is three: 25.1 mL.

Partial roasting of chalcopyrite gives

2CuFeS$_2$ + O$_2$ $\to$ Cu$_2$S + 2FeS + SO$_2$

2CuFeS$_2$ + 4O$_2$ $\to$ Cu$_2$S + 2FeO + 3SO$_2$

Since iron is more electropositive than copper, so its sulphide is oxidised in preference. Cu$_2$S remains mostly unaffected and the small amount of Cu$_2$O formed on oxidation, reacts with FeS to give back Cu$_2$S.

2Cu$_2$S + 3O$_2$ $\to$ 2Cu$_2$O + 2SO$_2$

Cu$_2$O + FeS $\to$ Cu$_2$S + FeO

To understand how the concentration of ions affects the cell potential, we will first use the Nernst equation. The Nernst equation gives us a way to calculate the potential of a cell under non-standard conditions and is represented as follows:

$E = E^0 - \frac{RT}{nF} \ln \frac{a_{\text{Red}}}{a_{\text{Ox}}}$

In our case, we're analyzing a concentration cell where the metal M is the same in both the anode and the cathode but with different concentrations of M⁺. Here, the standard potential $E^0$ is zero because the same substance is used as both the anode and cathode.

The cell reaction becomes:

$\text{M}(s) | \text{M}^{+}(0.05 \text{ M at anode}) \Longrightarrow \text{M}(s) | \text{M}^{+}(1 \text{ M at cathode})$

Since $E^0 = 0$, the Nernst equation simplifies to:

$E = - \frac{RT}{nF} \ln \frac{[\text{M}^{+} \text{ (cathode)}]}{[\text{M}^{+} \text{ (anode)}]}$

Which turns into:

$E = - \frac{RT}{nF} \ln \frac{1}{0.05} ,$

and we can further simplify using $\ln(\frac{1}{0.05}) = -\ln(0.05)$. Assume the reaction involves the transfer of 1 mole of electrons (n=1), the value of F (Faraday constant) is approximately 96485 C/mol, and the temperature T is 298K (room temperature). The gas constant R is 8.314 J/(mol·K). Plugging in these values:

$E \approx - \frac{(8.314 \, \text{J/mol·K})(298 \, \text{K})}{(1)(96485 \, \text{C/mol})} \ln(0.05)$

This equation can be used to find the potential in volts when the concentration at anode was 0.05 M and was resulting in 70 mV.

Now, switching the anode concentration to 0.0025 M, we reapply the Nernst equation:

$E = - \frac{RT}{nF} \ln \frac{1}{0.0025}$

We can see the ratio of concentrations has changed, moving from 1/0.05 to 1/0.0025. Thus, the concentration difference across the membrane has increased, which should increase the voltage according to the Nernst equation. Specifically, the ratio changed by a factor of 0.05/0.0025 = 20 times.

Given that $\ln(\frac{1}{x})$ is proportional to the voltage, when the ratio of concentration changes by twenty-fold, and because $\ln(0.0025) = -\ln(400)$ which is twice the $\ln(20)$, the potential will double. Since the initial 70 mV doubles, the new cell potential will be:

$E \approx 2 \times 70 \, \text{mV} = 140 \, \text{mV}.$

Therefore, the correct answer is:

Option C: 140 mV

Concept:
Temporary hardness in water is due to the presence of dissolved bicarbonates of calcium and magnesium, namely $\mathrm{Ca(HCO_3)_2}$ and $\mathrm{Mg(HCO_3)_2}$.

1. Clark's Method (using slaked lime, $\mathrm{Ca(OH)_2}$):
When a calculated amount of calcium hydroxide is added, it reacts with the soluble bicarbonates to form insoluble carbonates and hydroxides that precipitate out:

Hence, (B) is correct.

2. Using Washing Soda ($\mathrm{Na_2CO_3}$):
Sodium carbonate precipitates the calcium ions from $\mathrm{Ca(HCO_3)_2}$ as insoluble calcium carbonate:

Hence, (C) is correct.

$\mathrm{Ca_3(PO_4)_2}$ is an insoluble salt not used for softening, and $\mathrm{NaOCl}$ (sodium hypochlorite) is a bleaching and disinfecting agent, not a water softener.

Correct options: (B) and (C)