NEET-UG 2008

As we know,

$V=\frac{4}{3}\pi R^3$

$\therefore$ $\frac{dV}{V}=3\frac{dR}{R}$

Error in the dimension of the volume is = 3 $\times$ 2% = 6 %

Energy density = $\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L^3\right]}=\left[M{L}^{-1}{T}^{-2}\right]$

Young's Modulus(Y) = $\frac{F}{A}\times \frac{l}{\Delta l}$

$\Rightarrow$ [Y] = $\frac{\left[ML{T}^{-2}\right]}{\left[L^3\right]}\frac{\left[L\right]}{\left[L\right]}$ = $\left[M{L}^{-1}{T}^{-2}\right]$

v² = u² + 2$a$s

$\Rightarrow$ $a=\frac{v^2-u^2}{2s}$

         = $\frac{{\left(20\right)}^2-{\left(10\right)}^2}{2\times 135}$

         = $\frac{400-100}{270}$

         = $\frac{300}{270}$

         = $\frac{10}{9}$ m/s²

$v=u+at$

$\Rightarrow$ $t=\frac{v-u}{a}$

= $\frac{20-10}{\frac{10}{9}}$

= $\frac{10}{\frac{10}{9}}$ = 9 m/s

distance travelled in nth seconds

S(n th) = u + $\frac{a}{2}(2n-1)$

Here u = 0, $a$ = $\frac{4}{3}$m s^$-$2, n = 3

S(3 rd) = $\frac{4}{3\times 2}(2\times 3-1)$ = $\frac{10}{3}$ m

The slope of the graph $\frac{ds}{dt}$ is maximum at C and hence the instantaneous velocity is maximum at C.

The components of 1N and 2N forces along + x axis = 1 cos 60° + 2 sin 30°

$=1\times \frac{1}{2}+2\times \frac{1}{2}=\frac{1}{2}+1=\frac{3}{2}=1.5N$



The component of 4 N force along – x-axis

$=4\sin 30^o=4\times \frac{1}{2}=2N$

Therefore, if a force of 0.5N is applied along + x-axis, the resultant force along x-axis will become zero and the resultant force will be obtained only along y-axis.

For the riders to experience weightlessness at the top of the hill, the weight of the rider must be balanced by the centripetal force.



i.e., $mg=\frac{m{v}^2}{R}$

$\Rightarrow v=\sqrt{gR}=\sqrt{10\times 20}=14.1m{s}^{-1}$

Hence, the speed of the car should be between 14 ms^–1 and 15 ms^–1.

$F=\frac{d}{dt}\left(Mv\right)=v\frac{dM}{dt}+M\frac{dv}{dt}$

As v is a constant,

$F=v\frac{dM}{dt}$    

But $\frac{dM}{dt}=Mkg/s$

$\therefore$ To keep the conveyer belt moving at v m/s, force needed = vM newton.

It is seen that input power to turbine will be the potential energy due to water falling in a second which is equal to

mgh = 15×10×60

= 9000 watt

Now power generated by turbine is 10% less than above value equal to

(9000 – 900) watt = 8100 watt

= 8.1 kW

The magnitude of the resultant velocity at the point of projection and the landing point is same.

Clearly, change in momentum along horizontal (i.e along x-axis)

= mvcos$\theta$ – mvcos$\theta$ = 0

Change in momentum along vertical (i.e. along y–axis) = mv sin$\theta$ – (–mv sin$\theta$)

= 2 mvsin$\theta$ = 2mv × sin 45°

= $2mv\times \frac{1}{\sqrt{2}}=\sqrt{2}mv$

Hence, resultant change in momentum = $\sqrt{2}mv$

As per laws of conservation of momentum and energy, if v and V represents initial velocity of shell of mass m and recoil velocity of gun of mass M respectively, then

mv + MV = 0

$V=-\left(\frac{m}{M}\right)v$

$=-\left(\frac{0.2}{4}\right)v=-\frac{v}{20}m/s$

Here (–) sign shows that recoil velocity of gun is opposite to velocity of shell.

The energy of explosion is used to impart kinetic energy to shell and gun so that we have

$\frac{1}{2}$ (mv² + MV²) = 1.05 × 10³

$\frac{1}{2}$ [0.2v² + (4v² / 400)] = 1.05 ×10³

$\Rightarrow$ v² = 10⁴

$\Rightarrow$ v = 100 m/s

Total mass = M, total length = L
Moment of inertia of OA about O = Moment of inertia of OB about O.

$\Rightarrow$ M.I total = $2\times \left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2.\frac{1}{3}=\frac{M{L}^2}{12}$

M.I. of a circular disc, $M{k}^2=\frac{M.R^2}{2}$

M.I. of a circular ring = MR².

$\therefore$ Ratio of their radius of gyration = $\frac{1}{\sqrt{2}}:1$

$\Rightarrow$ 1 : $\sqrt{2}$

W is the temperature on new scale corresponding to 39ºC on ºC scale.

So, (C – 0)/(100 – 0) = (W – 39)/(239 – 39)

$\Rightarrow$ C/100 = (W – 39)/200

$\Rightarrow$ W = (C/100) × 200 + 39

= (39/100) × 200 + 39

= 78 + 39 = 117

So, temperature on new scale is 117°W corresponding to 39°C.

Internal energy depends only on the initial and final states of temperature and not on the path. In a cyclic process, as initial and final states are the same, change in internal energy is zero. Hence E is $\Delta$U, the change in internal energy.

PV = nRT

$\Rightarrow PV=\frac{m}{M}RT$

$\Rightarrow \frac{PV}{m}=\frac{RT}{M}$

$\Rightarrow \frac{P}{\rho }=\frac{RT}{M}$

$\Rightarrow \frac{P}{\rho }\propto T$

$\therefore \frac{P}{\rho }\propto \frac{1}{T}$

$\frac{\frac{P_1}{{\rho }_1}}{\frac{P_2}{{\rho }_2}}=\frac{T_1}{T_2}=\frac{383}{283}$

$\Rightarrow \frac{x}{\frac{P_2}{P_2}}=\frac{383}{283}$

$\therefore \frac{P_2}{P_2}=\frac{383}{283}x$

Maximum acceleration of a particle in the simple harmonic motion is directly proportional to the square of angular frequency i.e. $a\propto {\omega }^2$

$\therefore$ $\frac{a_1}{a_2}=\frac{{\omega }_1^2}{{\omega }_2^2}=\frac{{\left(100\right)}^2}{{\left(1000\right)}^2}=\frac{1}{100}$

$\Rightarrow$ a₁ : a₂ = 1 : 10².

y = 0.25sin(10$\pi$x – 2$\pi$t)

$y_{\max }=0.25$

$k=\frac{2\pi }{\lambda }=10\pi \Rightarrow \lambda =0.2m$

$\omega =2\pi f=2\pi \Rightarrow f=1Hz$

The sign is negative inside the bracket. Therefore this wave travels in the positive x-direction.

Other factors such as $\omega$ and v remaining the same, I = A² constant K, or A = $\sqrt{\frac{I}{K}}$

On superposition

$A_{\max }=A_1+A_2$ and $A_{\min }=A_1-A_2$

$\therefore A_{\max }^2=A_1^2+A_2^2+2A_1{A}_2$

$\Rightarrow \frac{I_{\max }}{K}=\frac{I_1}{K}+\frac{I_2}{K}+\frac{2\sqrt{I_1{I}_2}}{K}$

$A_{\min }^2=A_1^2+A_2^2-2A_1{A}_2$

$\Rightarrow \frac{I_{\min }}{K}=\frac{I_1}{K}+\frac{I_2}{K}-\frac{2\sqrt{I_1{I}_2}}{K}$

$\therefore I_{\max }+I_{\min }=2I_1+2I_2$

We have $x=a\sin \left(\omega t+\frac{\pi }{6}\right)$

$\therefore$ Velocity, $v=\frac{dx}{dt}=a\omega \cos \left(\omega t+\frac{\pi }{6}\right)$

Maximum velocity = a$\omega$
According to question,

$\frac{a\omega }{2}=a\omega \cos \left(\omega t+\frac{\pi }{6}\right)$

$\Rightarrow \cos \left(\omega t+\frac{\pi }{6}\right)=\frac{1}{2}=\cos 60^o$ or $\cos \frac{\pi }{3}$

$\Rightarrow \omega t=\frac{\pi }{3}-\frac{\pi }{6}\Rightarrow \omega t=\frac{\pi }{6}$

$\Rightarrow \frac{2\pi }{T}.t=\frac{\pi }{6}\Rightarrow t=\frac{T}{12}$

By the symmetry of the figure, the electric fields at O due to the portions AC and BD are equal in magnitude and opposite in direction. So, they cancel each other.
Similarly, the field at O due to CD and AKB are equal in magnitude but opposite in direction. Therefore, the electric field at the centre due to the charge on the part ACDB is E along OK.

$V=\frac{1}{4\pi {\epsilon }_0}\times \frac{Q}{R}$

= Q × 10¹¹ volt    …(i)

$E=\frac{1}{4\pi {\epsilon }_0}\times \frac{Q}{R^2}$

= V/R = Q × 10¹¹ × $4\pi {\epsilon }_0$ × 10¹¹    from ...(i)

= $4\pi {\epsilon }_0$ × Q × 10²² volt/m

As the P.D. between 4 $\Omega$ and 3 $\Omega$ (in parallel), are the same,

4 × 1 amp = 3 × i₁ $\Rightarrow i_1=\frac{4}{3}A$

Total resistance of 4 $\Omega$ and 3 $\Omega$ = 12/7 $\Omega$.

Current in MQP (upper one) = $1+\frac{4}{3}=\frac{7}{3}A$

$\therefore$ $P.D.=\frac{12}{7}\times \frac{7}{3}=4V$

Current in MNP = $\frac{4}{1.25}=\frac{4\times 4}{5}=\frac{16}{5}A$

$\therefore$ P.D. across $1\Omega =\frac{16}{5}A\times 1\Omega =\frac{16}{5}volt$

$\Rightarrow$ P.D. across 1$\Omega$ = 3.2 volt.

$\frac{\Delta l}{l}=0.1$  $\therefore$ $l=1.1$

but the area also decreases by 0.1.

mass = $\rho lA=V\rho$, $\ln l+\ln A=\ln$ mass.

$\therefore$ $\frac{\Delta l}{l}+\frac{\Delta A}{A}=0\Rightarrow \frac{\Delta l}{l}=\frac{-\Delta A}{A}$

Length increases by 0.1, resistance increases, area decreases by 0.1, then also resistance will increase. Total increase in resistance is approximately 1.2 times, due to increase in length and decrease in area. But specific resistance does not change.

Here $E>\frac{ER}{R+r}$, hence the lengths 110 cm and 100 cm are interchanged.

Without being short-circuited through R, only the battery E is balanced.

$E=\frac{V}{L}\times l_1=\frac{V}{L}\times 110$   ...(i)

When R is connected across E, $Ri=\frac{V}{L}\times l_2$

$\Rightarrow R\left(\frac{E}{R+r}\right)=\frac{V}{L}\times 100$   ...(ii)

Dividing (i) by (ii), we get

$\frac{R+r}{R}=\frac{110}{100}$

$\Rightarrow$ 100 R + 100 r = 110 R

$\Rightarrow$ 10 R = 100 r

$\therefore$ $r=\frac{10R}{100}=\frac{10\times 10}{100}$  ($\because$ R = 10$\Omega$)

$\Rightarrow$ r = 1$\Omega$.

Clearly, 2$\Omega$ , 4$\Omega$ and ( 1 + 5) $\Omega$ resistors are in parallel. Hence, potential difference is same across each of them.

$\therefore$ I₁ × 2 = I₂ × 4 = I₃ × 6


Given I₁ = 3A    $\therefore$ I₁ × 2 = I₃ × 6

Given I₁ = 3A.

$\therefore$ I₁ × 2 = I₃ × 6 provides

$I_3=\frac{I_1\times 2}{6}=\frac{3\times 2}{6}=1A$

Now, the potential across the 5$\Omega$ resistor is
V = I₃ × 5 = 1 × 5 = 5V.

$\therefore$ the power dissipated in the 5$\Omega$ resistor

$P=\frac{V^2}{R}=\frac{5^2}{5}=5watt.$

Power = 220 V × 4 A = 880 watts. = 880 J/s.
Heat needed to raise the temperature of 1 kg water through 80°C

= ms.$\Delta$T × 4.2 J/cal

= 1000 g × 1 cal/g × 80 × 4.2 J/cal.

$\therefore$ Time taken = $\frac{1000\times 1\times 80\times 4.2}{880}=\frac{336\times {10}^3}{880}$

= 382 s = 6.3 min.

Capacitance of a parallel plate capacitor is

$C=\frac{{\epsilon }_{0}A}{d}$   ...(i)

Potential difference between the plates is

V = Ed    ...(ii)

The energy stored in the capacitor is

$U=\frac{1}{2}C{V}^2=\frac{1}{2}\left(\frac{{\epsilon }_{0}A}{d}\right){\left(Ed\right)}^2$   (Using (i) and (ii))

$=\frac{1}{2}{\epsilon }_0{E}^{2}Ad$

Total initial resistance

= R_G + R₁ = (50 + 2950) $\Omega$ = 3000 $\Omega$

$\epsilon =3V$

$\therefore$ Current = $\frac{3V}{3000\Omega }=1\times 10^{-3}mA$

If the deflection has to be reduced to 20 divisions, current i = 1 mA $\times$ $\frac{2}{3}$ as the full deflection scale for 1 mA = 30 divisions.

$3V=3000\Omega \times 1mA=x\Omega \times \frac{2}{3}mA$

$\Rightarrow x=3000\times 1\times \frac{3}{2}=4500\Omega$

But the galvanometer resistance = 50 $\Omega$

Therefore the resistance to be added

= (4500 – 50)$\Omega$ = 4450 $\Omega$.

According to the figure the magnitude of force on the segment QM is F₃ – F₁ and PM is F₂.


Therefore, the magnitude of the force on

segment PQ is $\sqrt{{\left(F_3-F_1\right)}^2+F_2^2}$

When a charged particle having a given K.E., T enters in a field of magnetic induction, which is perpendicular to its velocity, it takes a circular trajectory. It does not increase in energy, therefore T is the K.E.

Curie temperature is the temperature above which ferromagnetic material becomes paramagnetic material.

Total number of turns in the solenoid,

N = 500 Current, I = 2A.

Magnetic flux linked with each turn = 4 × 10^–3 Wb

Self inductance of coil :

L = $\frac{N\varphi }{I}=\frac{500\times 4\times {10}^3}{2}$ = 1 H

Given B = $\frac{1}{\pi }\left(\frac{Wb}{m^2}\right)$

$\theta$ = 60^o

Area normal to the plane of the disc

= $\pi$r² cos60^o = $\frac{\pi r^2}{2}$

Flux = B × normal area

= $\frac{0.02\times 0.02}{2}$ = 0.02 Wb

Average power = $\frac{E_0{I}_0}{2}\cos \varphi$

The velocity of electromagnetic radiation in vacuum is = $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

$\therefore$ Power P = $\frac{f_1+f_2}{f_1{f}_2}$

$$\left|\frac{v}{u}\right|=$$

size of image

size of object

Size of image = $\left|\frac{v}{u}\right|$ $\times$ Size of object

= $\left(\frac{{10}^{-1}}{1.5\times {10}^{11}}\right)\times \left(1.39\times {10}^9\right)$

= 9.2 ×10^–4 m

$\therefore$ Diameter of the sun’s image = 9.2 ×10^–4 m

Mass defect = ZM_p + (A –Z)M_n – M(A, Z)

$\Rightarrow$ $\frac{B.E}{c^2}$ = ZM_p + (A –Z)M_n – M(A, Z)

$\therefore$ M(A, Z) = ZM_p + (A $-$ Z)M_n $-$ $\frac{B.E}{c^2}$

X₁ = N₀e^{–5$\lambda$t}
X₂ = N₀e^{–$\lambda$t}

Given $\frac{X_1}{X_2}$ = $\frac{1}{e}$

$\Rightarrow$ $\frac{N_0{e}^{-5\lambda t}}{N_0{e}^{-\lambda t}}$ = $\frac{1}{e}$

$\Rightarrow$ e^{–4$\lambda$t} = e^-1

$\Rightarrow$ t = $\frac{1}{4\lambda }$

E_n = $-\frac{13.6}{n^2}$

E₁ = -13.6 eV

E₂ = $-\frac{13.6}{2^2}$ = -3.4 eV

$\therefore$ $\Delta$E = ( – 3.4) – ( – 13.6) = 10.2 eV.

A₁ : A₂ = 1 : 3

Their radii will be in the ratio

$R_0{A}_1^{1/3}$ : $R_0{A}_2^{1/3}$ = 1 : 3^1/3

$\therefore$ ${\rho }_{A_1}$ : ${\rho }_{A_2}$ = $\frac{A_1}{\frac{4}{3}\pi R_1^3}$ : $\frac{A_2}{\frac{4}{3}\pi R_2^3}$

= $\frac{1}{1^3}:\frac{3}{{\left(3^{1/3}\right)}^3}$

= 1 : 1

$\varphi$ = 6.2 eV.

K_max = 5 eV;

$\therefore$ h$\nu$ = 11.2 eV = E

$\therefore$ $\lambda$ = $\frac{hc}{E}$ = $\frac{12400}{11.2}$ = 1107 $\stackrel{o}{A}$

de-Broglie wavelength associated with electron moving with velocity ν,

$\lambda$ = $\frac{h}{mv}$

So, $\lambda$_e = $\frac{h}{9.1\times {10}^{-31}\times 3\times {10}^6}$

Wavelength of particle of mass 1 mg moving with velocity ν.

$\lambda$_p = $\frac{h}{{10}^{-6}\times v}$

As given, $\lambda$_e = $\lambda$_p

$\Rightarrow$ $\frac{h}{9.1\times {10}^{-31}\times 3\times {10}^6}$ = $\frac{h}{{10}^{-6}\times v}$

$\Rightarrow$ v = $\frac{27.3\times {10}^{-25}}{{10}^{-6}}$ = $2.7\times 10^{-18}m{s}^{-1}$

In the phenomenon of electric discharge tube through gases at low pressure, the coloured glow in the tube appears as a result of collisions between the charged particles emitted from cathode and the atoms of the gas.

The atomic radius in a f.c.c. crystal is = $\frac{a}{2\sqrt{2}}$

$\therefore$ Atomic radius = $\frac{3.6}{2\sqrt{2}}$ = 1.27 $\stackrel{\circ }{A}$

Negative feedback is applied to reduce the output voltage of an amplifier. If there is no negative feedback, the value of output voltage could be very high. In the options given, the maximum value of voltage gain is 100. Hence it is the correct option.

Energy gap E_g = 2.0 eV

E_g = 2.0 × 1.6 × 10^–19 J

E_g = hf

f = frequency of radiation

$\therefore$ f = $\frac{E_g}{h}$

= $\frac{3.2\times {10}^{-19}}{6.625\times {10}^{-34}}$

= 5 $\times$ 10¹⁴ Hz

Element	%	Atomic mass	mole ratio	simple ratio
C	38.71	12	$$\frac{38.71}{12}=3.22$$	$$\frac{3.22}{3.22}=1$$
H	9.67	1	$$\frac{9.67}{1}=9.67$$	$$\frac{9.67}{3.22}=1$$
O	51.62	16	$$\frac{51.62}{16}=3.22$$	$$\frac{3.22}{3.22}=1$$

Hence empirical formula of the compound would be CH₃O

PbO	+	2HCl	$$\to$$	PbCl2	+	H2O
n mol		2n mol		n mol
$\frac{6.5}{224}$ mol		$\frac{3.2}{36.5}$ mol
0.029 mol		0.087 mol

Formation of moles of lead(II) chloride depends upon the no. of moles of PbO which acts as a limiting reagent here. So no of moles of PbCl₂ formed will be equal to the no of PbO (i.e 0.029)

C3H8	+	5O2	$$\to$$	3CO2	+	4H2O	(balanced equation)
1 vol.		5 vol.		3 vol.		4 vol.

According to the above equation 1 vol. or 1 litre of propane requires to 5 vol. or 5 litre of O₂ to burn completely.

$\Delta v$ = $\frac{1\times {10}^{-18}}{9\times {10}^{-28}}=1.1\times 10^{9}cm{s}^{-1}$

According to Heisenberg uncertainty principle

$\Delta p.\Delta x\ge \frac{h}{4\pi }$ or m$\Delta v.\Delta x\ge \frac{h}{4\pi }$

$\Rightarrow$ (m$\Delta v$)² $\ge \frac{h}{4\pi }$ ($\because$ $\Delta x$ = $\Delta p$)

$\Rightarrow$ $\Delta v\ge$ $\frac{1}{2m}\sqrt{\frac{h}{\pi }}$

The balanced ionic equation for oxidation of ferrous oxalate by MnO₄^– in acidic medium is as follows :

3MnO₄^- + 5FeC₂O₄ + 24H⁺ $\to$

3Mn²⁺ + 10CO₂ + 12H₂O + 5Fe³⁺

Thus, 5 moles of FeC₂O₄ require 3 moles of MnO₄^-.

So, 1 mole of FeC₂O₄ requires

= $\frac{3}{5}$ = 0.6 moles of MnO₄^-

1 mole of water contains 6.023 × 10²³ molecules of water

6.023 × 10²³ molecules of water weigh = 18 g

So, 1 molecule of water weighs = $\frac{18}{6.023\times {10}^{23}}$

Now, volume of 1 molecule of water

=

Mass of 1 molecule of water

Density of water

.

= $\frac{18}{6.023\times {10}^{23}}\times \frac{1}{1gc{m}^{-3}}$

= 3 $\times$ 10^-23 cm³

Kinetic energy of a gas is expressed as

K.E = $\frac{3}{2}nRT$

Thus, on expansion of fixed amount of gas at constant temperature the kinetic energy remains constant.

Given

X $\rightleftharpoons$ Y + Z      . . . .(i)

A $\rightleftharpoons$ 2B       . . . .(ii)

Let the total pressure for reaction (i) and (ii) be P₁ and P₂ respectively, then

$\frac{K_{P_1}}{K_{P_2}}=\frac{9}{1}$

	X	⇌	Y	+	Z
Initial mole	1		0		0
At equilibrium	1 - $\alpha$		$$\alpha$$		$$\alpha$$

Total number of moles at equilibrium

= 1 - $\alpha$ + $\alpha$ + $\alpha$ = 1 + $\alpha$

$\therefore$ K_P1 = $\frac{P_Y\times P_Z}{P_X}$ = $\frac{\frac{\alpha }{1+\alpha }\times P_1\times \frac{\alpha }{1+\alpha }\times P_1}{\frac{1-\alpha }{1+\alpha }\times P_1}$

	A	⇌	2B
Initial mole	1		0
At equilibrium	1 - $\alpha$		2$\alpha$

Total number of moles at equilibrium

= 1 - $\alpha$ + 2$\alpha$ = 1 + $\alpha$

$\therefore$ K_P2 = $\frac{{\left(P_B\right)}^2}{P_A}$ = $\frac{{\left(\frac{2\alpha }{1+\alpha }\times P_2\right)}^2}{\frac{1-\alpha }{1+\alpha }\times P_2}$

$\therefore$ $\frac{K_{P_1}}{K_{P_2}}=\frac{P_1}{4P_2}$

$\Rightarrow$ $\frac{P_1}{4P_2}=\frac{9}{1}$

$\Rightarrow$ $\frac{P_1}{P_2}=\frac{36}{1}$

HI_(g) $\rightleftharpoons$ $\frac{1}{2}$H_2(g) + $\frac{1}{2}$I_2(g),

For this reaction, K = 8.0

Reversing the equation,

$\therefore$$\frac{1}{2}$H_2(g) + $\frac{1}{2}$I_2(g) $\rightleftharpoons$ HI_(g) ......(1)

For this reaction, K₁ = $\frac{1}{8}$

H_2(g) + I_2(g) $\rightleftharpoons$ 2HI_(g)

We get this equation by multiplying equation (1) by 2,

$\therefore$ For this reaction, K₂ = ${\left(\frac{1}{8}\right)}^2$ = $\frac{1}{64}$

Fe(OH)_3(s) $\rightleftharpoons$ Fe³⁺_(aq) + 3OH^$$-$$_(aq)

Equilibrium constant, K_c = $\frac{\left[F{e}^{3+}\right]{\left[O{H}^{-}\right]}^3}{\left[Fe{\left(OH\right)}_3\right]}$

We know, equilibrium constant remains same at constant temperature.

Now, let the increase in concentration of Fe³⁺ be x times.

$\therefore$ K_c = $\frac{\left[x\times F{e}^{3+}\right]{\left[\frac{1}{4}\times O{H}^{-}\right]}^3}{\left[Fe{\left(OH\right)}_3\right]}$

= $\frac{x}{64}\frac{\left[F{e}^{3+}\right]{\left[O{H}^{-}\right]}^3}{\left[Fe{\left(OH\right)}_3\right]}$

$\Rightarrow$ K_c = $\frac{x}{64}$K_c

$\Rightarrow$ $\frac{x}{64}$ = 1

$\Rightarrow$ x = 64

	2AB2(g)	⇌	2AB(g)	+	B2(g)
Initial mole	2		0		0
At equilibrium	2(1 - x)		2x		x

Amount of moles at equilibrium = 2(1 – x) + 2x + x = 2 + x

$K_p=\frac{{\left[p_{AB}\right]}^2\left[p_{B_2}\right]}{{\left[p_{A{B}_2}\right]}^2}$

= $\frac{{\left[\frac{2x}{2+x}\times P\right]}^2\left[\frac{x}{2+x}\times P\right]}{{\left[\frac{2\left(1-x\right)}{2+x}\times P\right]}^2}$

= $\frac{\left[\frac{4x^3}{2+x}\times P\right]}{4{\left(1-x\right)}^2}$

$\Rightarrow$ K_p = $\frac{4x^3\times P}{2}\times \frac{1}{4}$

($\because$ 1 – x ≈ 1 and 2 + x ≈ 2)

$\Rightarrow$ x = ${\left(\frac{8K_p}{4P}\right)}^{\frac{1}{3}}$

$\Rightarrow$ x = ${\left(\frac{2K_p}{P}\right)}^{\frac{1}{3}}$

We know, pH = – log[H⁺]

For pH = 3, 3 = – log [H⁺]

$\Rightarrow$ [H⁺] = 10^–3 M

For pH = 4 , 4 = – log [H⁺]

$\Rightarrow$ [H⁺] = 10^–4 M

For pH = 5, 5 = – log [H⁺]

$\Rightarrow$ [H⁺] = 10^–5 M

Total concentration of [H⁺]

M = $\frac{{10}^{-3}\left(1+0.1+0.01\right)}{3}$

M(V₁ +V₂ + V₃) = M₁V₁ + M₂V₂ + M₃V₃

As V₁ = V₂ = V₃ = V

$\Rightarrow$M(3V) = (M₁ + M₂ + M₃)V

$\Rightarrow$ 3M = (10^–3 + 10^–4 + 10^–5)

= $\frac{1.11\times {10}^{-3}}{3}$

= 3.7 $\times$ 10^$-$4 M

$\frac{1}{2}$H₂ + $\frac{1}{2}$Cl₂ $\to$ HCl, $\Delta$H_f = ?

$\Delta$H_reaction = [ $\frac{1}{2}$(B.E)_H2 + $\frac{1}{2}$(B.E)_Cl2] - (B.E)_HCl

= [217 + 121] – 431

= 338 – 431 = – 93 kJ mol^–1

$\Delta$H = $\Delta$E + $\Delta$n_gRT

$\Delta$n_g = (1 + 1) – (1) = 1

$\Rightarrow$ $\Delta$H = $\Delta$E + RT

Thus, $\Delta$H is a positive quantity i.e., ∆H > 0. Now one mole of gaseous reactant dissociate into two moles of gaseous products thus, entropy increases i.e., $\Delta$S > 0.

Enthalpy (H = q + W) and Gibbs energy, (G = H –TS) are state functions which depend only on the initial and final states of system.

While, heat (q) and work done (W) are the path function which depends on the path followed in bringing the change between two states of the system.

Kohlrausch’s law states that the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductance of the component ions.

$\lambda$_$$\infty$$ = $\lambda$_a + $\lambda$_c

where, $\lambda$_a = equivalent conductance of the anion

$\lambda$_c = equivalent conductance of the cation.

Substances which have higher reduction potential are stronger oxidizing agent.

[Fe(CN)₆]^4$-$ $\to$ [Fe(CN)₆]^3$-$ + e^$$-$$;  E^o = $-$0.35 V

Fe²⁺ $\to$ Fe³⁺ + e^$$-$$;  E^o = $-$0.77 V

Higher the +ve reduction potential, stronger will be the oxidising agent. Oxidising agent oxidises other compounds and gets itself reduced easily.

At Anode:

C₅H₁₂ + 10H₂O $\to$ 5CO₂ + 32H⁺ + 32e^-

At Cathode:

8O₂ + 32H⁺ + 32e^- $\to$ 16H₂O
-------------------------------------------------
C₅H₁₂(g) + 8O₂(g) $\to$ 5CO₂(g) + 6H₂O(l)

$\Delta$G = 5×$\Delta$G_CO2 + 6 $\Delta$G_(H2O) – [$\Delta$G_(C5H12) +8 × $\Delta$G_O2 ]

= 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0)

= – 3387 kJ mol^–1

= – 3387 × 10³ J mol^–1

$\Delta$G = - nFE^o_cell

From the overall equation we find n = 32

– 3387 × 10³ = -32 $\times$ 96500 $\times$ E^o_cell

E^o_cell = $\frac{-3387\times {10}^3}{-32\times 96500}$ = 1.0968 V