NEET-UG 2007

We know,

$R=\frac{V}{I}$

$\therefore$ $\left[R\right]=\frac{\left[V\right]}{\left[I\right]}$ = $\frac{\left[M{L}^2{T}^{-3}{I}^{-1}\right]}{\left[I\right]}$ = $\left[M{L}^2{T}^{-3}{I}^{-2}\right]$

f = f₀$\left(1-\frac{t}{T}\right)$

When f = 0 then,

0 = f₀$\left(1-\frac{t}{T}\right)$

$\Rightarrow$ $\left(1-\frac{t}{T}\right)$ = 0

$\Rightarrow$ t = T

Also f = $\frac{dv}{dt}$

$\therefore$ $\underset{0}{\overset{v_x}{\int }}dv=\underset{t=0}{\overset{t=T}{\int }}fdt$

v_x = $\underset{t=0}{\overset{t=T}{\int }}{f}_0\left(1-\frac{t}{T}\right)dt$

= $f_0{\left[\left(t-\frac{t^2}{2T}\right)\right]}_0^T$

= $f_0\left(T-\frac{T^2}{2T}\right)$

= $f_0\left(T-\frac{T}{2}\right)$

= $\frac{1}{2}$ f₀T

Let total distance = 2S

Let car take t₁ time to cover first S distance from X to Y,

$\therefore$ t₁ = $\frac{S}{v_u}$

Let car take t₂ time to cover S distance,from Y to X,

$\therefore$ t₂ = $\frac{S}{v_d}$

$\therefore$ Average speed = $\frac{2S}{\frac{S}{v_u}+\frac{S}{v_d}}$ = $\frac{2v_u{v}_d}{v_u+v_d}$

Speed v = $\frac{dx}{dt}$ = $\frac{d}{dt}\left(9t^2-t^3\right)$ = 18t - 3t²

At maximum speed

$\frac{dv}{dt}$ = 0

$\Rightarrow$ 18 - 6t = 0

$\Rightarrow$ t = 3

$\therefore$ Position of the particle = 81 - 27 = 54 m

Let $\theta$ be the angle which the particle makes with an x-axis.

From figure,

$\tan \theta =\frac{3}{\sqrt{3}}=\sqrt{3}$

$\Rightarrow \theta ={\tan }^{-1}\left(\sqrt{3}\right)=60^{\circ }$

$\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{3}\left(\overrightarrow{A}.\overrightarrow{B}\right)$

$\therefore$ $AB\sin \theta =\sqrt{3}AB\cos \theta$

or $\tan \theta =\sqrt{3}$ or $\theta ={\tan }^{-1}\sqrt{3}=60^{\circ }$;

Given u = V, final velocity = 0.
Using v = u + at

$\therefore$ $0=V-at$

$\Rightarrow$ $-a=\frac{0-V}{t}=-\frac{V}{t}$

f = $\mu$R = $\mu$mg (f is the force of friction)

$\therefore$ Retardation, a = $\mu g$

So, $t=\frac{V}{a}=\frac{V}{\mu g}$

As seen, gravitational potential energy of the ball gets converted into elastic potential energy of the spring.

$mg\left(h+d\right)=\frac{1}{2}k{d}^2$

W_net = work done by gravity + work done by spring

$\Rightarrow$ W_net = mg(h + d) - $\frac{1}{2}k{d}^2$

Torque about A,

$\tau =mg\times \frac{l}{2}=\frac{mgl}{2}$

$\tau =I\alpha$

$\therefore$ Angular acceleration,
$\alpha =\frac{\tau }{I}=\frac{mgl/2}{m{l}^2/3}=\frac{3g}{2l}$

Angular momentum = Linear momentum × perpendicular distance between line of action of liner momentum from origin

Let d be the perpendicular distance.
p_A and p_B be the linear momentum at A and B

Angular momentum L_A = p_A × d
L_B = p_B × d

So, linear momentum will be equal, i.e., L_A = L_B.here, p_A and p_B are equal as have equal velocity.

L_A = L_B

Given: Angular acceleration, $\alpha$ = 3 rad/sec²
Initial angular velocity ${\omega }_i$ = 2 rad/sec

Time t = 2 sec

Using, $\theta ={\omega }_{i}t+\frac{1}{2}\alpha t^2$

$\therefore$ $\theta =2\times 2+\frac{1}{2}\times 3\times 4=4+6=10$ radian.

Since orbital velocity of satellite is $v=\sqrt{\frac{GM}{r}}$

it does not depend upon the mass of the satellite.

Therefore, both satellites will move with same speed.

Power radiated by the sun at t°C

$=\sigma {\left(t+273\right)}^{4}4\pi r^2$

Power received by a unit surface

$=\frac{\sigma {\left(t+273\right)}^{4}4\pi r^2}{4\pi R^2}=\frac{r^2\sigma {\left(t+273\right)}^4}{R^2}$

According to Stefan’s law, rate of energy radiated $E\propto T^4$

where T is the absolute temperature of a black body.

$\therefore E\propto {\left(727+273\right)}^4\Rightarrow E\propto {\left[1000\right]}^4$

Since efficiency of engine is $\eta =1-\frac{T_2}{T_1}$

According to problem,

$\frac{1}{6}=1-\frac{T_2}{T_1}$   ...(i)

When the temperature of the sink is reduced by 62°C, its efficiency is doubled

$2\left(\frac{1}{6}\right)=1-\frac{T_2-62}{T_1}$   ...(ii)

Solving (i) and (ii)
T₂ = 372 K

T₁ = 99°C = Temperature of source.

Kinetic energy + potential energy = total energy
When kinetic energy is maximum, potential energy is zero and vice versa.

$\therefore$ Maximum potential energy = total energy.

0 + K₀ = K₀ (K.E. + P.E. = total energy).

Let y = Asin$\omega$t

$\frac{dy}{dt}=A\omega \cos \omega t=A\omega \sin \left(\omega t+\frac{\pi }{2}\right)$

Acceleration = $-A{\omega }^2\sin \omega t$

The phase difference between acceleration and velocity is $\pi /2$.

Mass gets detached at the upper extreme position when pan returns to its mean position. At that point, $R=mg-m{\omega }^{2}a=0$,
i.e. $g={\omega }^{2}a$

$\Rightarrow a=g/{\omega }^2=mg/k$

$\Rightarrow a=\frac{2\times 10}{200}$   $\left[\because As={\omega }^2=\frac{k}{m}\right]$

$\Rightarrow$ a = 1/10 m = 10 cm

Displacement from the mean position

$y=a\sin \left(\frac{2\pi }{T}\right)t$

According to problem y = a/2

$a/2=a\sin \left(\frac{2\pi }{T}\right)t$

$\Rightarrow \frac{\pi }{6}=\left(\frac{2\pi }{T}\right)t\Rightarrow t=T/12$

This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.

Let ${\varphi }_A$, ${\varphi }_B$ and ${\varphi }_C$ are the electric flux linked with A, B and C.

According to Gauss theorem,

${\varphi }_A+{\varphi }_B+{\varphi }_C$ = $\frac{q}{{\epsilon }_0}$

Since ${\varphi }_A$ = ${\varphi }_C$

$\Rightarrow 2{\varphi }_A=\frac{q}{{\epsilon }_0}-{\varphi }_B$

$\Rightarrow 2{\varphi }_A=\frac{q}{{\epsilon }_0}-\varphi$   $\left(Given{\varphi }_B=\varphi \right)$

$\therefore$ ${\varphi }_A=\frac{1}{2}\left(\frac{q}{{\epsilon }_0}-\varphi \right)$

Potential at C = V_C = 0
Potential at D = V_D

$=k\left(\frac{-q}{L}\right)+\frac{kq}{3L}=-\frac{2}{3}\frac{kq}{L}$

Potential difference
V_D – V_C $=-\frac{2}{3}\frac{kq}{L}=\frac{1}{4\pi {\epsilon }_0}\left(-\frac{2}{3}.\frac{q}{L}\right)$

$\Rightarrow$ Work done = Q (V_D – V_C)

$=-\frac{2}{3}\times \frac{1}{4\pi {\epsilon }_0}\frac{qQ}{L}=\frac{-qQ}{6\pi {\epsilon }_{0}L}$

This consists of two dipoles, –q and +q with dipole moment along with the +y-direction and –q and +q along the x-direction.


$\therefore$ The resultant moment = $\sqrt{q^2{a}^2+q^2{a}^2}=\sqrt{2}qa$

Along the direction 45° that is along OP where P is (+a, +a, 0).

Power dissipiated = P

$=\frac{V^2}{R}=\frac{{\left(18\right)}^2}{6}=54W$

A balanced wheatstone bridge simply requires

$\frac{P}{Q}=\frac{R}{S}\Rightarrow \frac{2}{2}=\frac{2}{S}$

Therefore, S should be 2$\Omega$.
A resistance of 6$\Omega$ is connected in parallel.
In parallel combination,

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{2}=\frac{1}{6}+\frac{1}{S}$

$\Rightarrow S=3\Omega$

As the capacitors are connected in parallel, therefore potential difference across both the condensors remains the same.

$\therefore$ $Q_1=CV$

$Q_2=\frac{C}{2}V$

Aslo, $Q=Q_1+Q_2$

$=CV+Q_2=\frac{3}{2}CV$

Work done in charging fully both the condensors is given by

$W=\frac{1}{2}QV=\frac{1}{2}\times \left(\frac{3}{2}CV\right)V=\frac{3}{4}C{V}^2$

We know

$\frac{I}{I_S}=1+\frac{G}{S}$

$\frac{750}{100}=1+\frac{13}{S}$

$\Rightarrow S=2\Omega$

For the circular motion in a cyclotron,

$qvB=\frac{m{v}^2}{r}\Rightarrow qB=m\omega =\frac{m\times 2\pi }{T}$

$\therefore T=\frac{2\pi m}{qB}$ is independent of v and r.

Magnetic moment $\mu$ = IA

T = $\frac{2\pi R}{v}$ and I = $\frac{q}{T}=\frac{qv}{2\pi R}$

$\therefore$ $\mu$ = $\left(\frac{qv}{2\pi R}\right)\left(\pi R^2\right)$ = $\frac{qvR}{2}$

Above Curie temperature, ferromagnetic material become paramagnetic.

Efficiency of the transformer

= $\frac{P_{output}}{P_{input}}\times 100$

= $\frac{100}{220\times 0.5}\times 100$ = 90 %

Voltage across the primary,

V_p = $\frac{d\varphi }{dt}=\frac{d}{dt}\left({\varphi }_0+4t\right)=$ 4 volt

Also $\frac{V_s}{V_p}=\frac{N_s}{N_p}$

Where, Ns = No. of turns across primary coil = 50

N_p = No. of turns across secondary coil = 1500

$\therefore$ V_s = $\frac{1500}{50}\times 4$ = 120 V

Condition for which current is maximum in a series LCR circuit is,

$\omega =\frac{1}{\sqrt{LC}}$

$\Rightarrow$ 1000 = $\frac{1}{\sqrt{L\left(10\times {10}^6\right)}}$

$\Rightarrow$ L = 100 mH

In electromagnetic waves, E and B are in same phase and perpendicular to each other.

As $\frac{1}{\mu }=\frac{\sin i}{\sin 90^{\circ }}$

$\Rightarrow$ $\mu$ = $\frac{1}{\sin i}$ = $\frac{5}{3}$

Speed, v = $\frac{c}{\mu }$ = $\frac{3\times {10}^8}{5/3}$ = 1.8 $\times$ 10⁸ m/s

By using v = n$\lambda$

$\therefore$ v = 2 $\times$ 10¹⁴ $\times$ 5000 $\times$ 10^-10 = 10⁸ m/s

Refractive index of the material,

$\mu$ = $\frac{c}{v}$ = $\frac{3\times {10}^8}{{10}^8}$ = 3

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

E₁ = -13.6 eV

E₂ = $-\frac{13.6}{2^2}$ = -3.4 eV

Kinetic energy of an electron in the first excited state is

K = –E₂ = 3.4 eV.

A nucleus of mass number A has radius

R = R₀ A^1/3,

where R₀ = 1.2 × 10^–15 m and A = mass number

For ${}_{13}^{27}Al$, R₁ = R₀ (27)^1/3 = 3R₀

For ${}_{32}^{125}Te$, R₂ = R₀ (125)^1/3 = 5R₀

$\therefore$ $\frac{R_2}{R_1}=\frac{5R_0}{3R_0}$

$\Rightarrow$ $R_2=\frac{5}{3}\times R_1$ = $\frac{5}{3}\times 3.6$ = 6.0 fm

The difference in mass of a nucleus and its constituents, $\Delta$M, is called the mass defect and is given by

$\Delta$M = [ZM_p + (A $-$ Z)M_n] - M(A, Z)

binding energy = $\Delta$Mc²

= [ZM_p + (A $-$ Z)M_n $-$ M(A, Z)]c²

In mass spectrometer, when ions are accelerated through potential V,

$\frac{1}{2}m{v}^2=qV$ ....(1)

As the magnetic field curves the path of the ions in a semicircular orbit.

$\therefore$ Bqv = $\frac{m{v}^2}{R}$

$\Rightarrow$ v = $\frac{BqR}{m}$ .....(2)

Now by substituting (ii) in (i),

$\frac{1}{2}m{\left(\frac{BqR}{m}\right)}^2=q$


$\Rightarrow$ $\frac{q}{m}=\frac{2V}{B^2{R}^2}$

As V and B are constants,

$\therefore$ $\frac{q}{m}$ $\propto$ $\frac{1}{R^2}$

In beta minus decay ($\beta$^–), a neutron is transformed into a proton, and an electron is emitted from the nucleus along with antineutrino.

n = p + e^$$-$$ + $\overline{\nu }$

N_A = N₀e^{–5$\lambda$t}
N_B = N₀e^{–$\lambda$t}

Given $\frac{N_A}{N_B}$ = ${\left(\frac{1}{e}\right)}^2$

$\Rightarrow$ $\frac{N_0{e}^{-5\lambda t}}{N_0{e}^{-\lambda t}}$ = ${\left(\frac{1}{e}\right)}^2$

$\Rightarrow$ e^{–4$\lambda$t} = e^-2

$\Rightarrow$ t = $\frac{1}{2\lambda }$

For a light source of power P watt, the intensity at a distance d is given by

I = $\frac{P}{4\pi d^2}$

$\therefore$ No. of photoelectrons or intensity $\propto$ $\frac{1}{{\left(dis\tan ce\right)}^2}$

In magnetic field a charged particle moves in a circular orbit.

Since p = nh$\nu$

$\Rightarrow$ n = $\frac{p}{h\nu }$ = $\frac{2\times {10}^{-3}}{6.6\times {10}^{34}\times 6\times {10}^{14}}$

= 5 $\times 10^{15}$

A diode is said to be forward biased if p-side is at higher potential than n-side of p-n junction.

In a cubic crystal structure
a = b = c  and  $\alpha$ = $\beta$ = $\gamma$ = 90^o

For a p-type semiconductor, the acceptor energy level, as shown in the diagram, is slightly above the top E_v of the valence band. With very small supply of energy an electron from the valence band can jump to the level E_C and ionise acceptor negatively.

Y' = $\overline{A+B}$

Y = $\overline{Y^{\prime }}$

Y = $\overline{\overline{A+B}}$

Y = A + B which is OR-gate.

Average isotope mass of X

= $\frac{200\times 90+199\times 8+202\times 2}{90+8+2}$

= $\frac{18000+1592+404}{100}$

= 199.96 a.m.u $\approx$ 200 a.m.u

Normality = $\frac{98\times 1.8\times 10}{49}$ = 36 N

N₂ = 0.1 $\times$ 2 = 0.2 N

N₂V₂ = N₁V₁

$\Rightarrow$ 36 $\times$ V = 0.2 $\times$ 1000

$\Rightarrow$ V = $\frac{0.2\times 1000}{36}$ = 5.55 mL

(i) represents an electron in 3s orbital

(ii) is not possible as value of m varies from 0, 1, .... ($n$ -1)

(iii) represents an electron in 4f orbital

(iv) is not possible as value of m varies from -$l$ ... +$l$

(v) is not possible as value of m varies from -$l$ ... +$l$, it can never be grater than $l$

2NH₃ $\rightleftharpoons$ N₂ + 3H₂;     $\frac{1}{K_1}$

N₂ + O₂ $\rightleftharpoons$ 2NO;     K₂

3H₂ + $\frac{3}{2}$O₂ $\rightleftharpoons$ 3H₂O;     (K₃)³

By adding all equations, we get

2NH₃ + $\frac{5}{2}$ O₂ $\rightleftharpoons$ 2NO + 3H₂O

$\therefore$ K = $\frac{K_2\times {\left(K_3\right)}^3}{K_1}$

For weak acid degree of dissociation,

$\alpha$ = $\sqrt{\frac{K_a}{C}}$

= $\sqrt{\frac{1\times {10}^{-5}}{0.1}}=10^{-2}$ = 1.00 %

Given, [H₃O⁺.] = 1 $\times$ 10^$-$10

$\Rightarrow$ pH = 10

Also we know, pH + pOH = 14

$\Rightarrow$ pOH = 14 - pH = 14 - 10 = 4

	HX	⇌	H+	+	X-
Initially	1		0		0
At equilibrium	1 - $\alpha$		$$\alpha$$		$$\alpha$$

Total moles = 1 – $\alpha$ + $\alpha$ + $\alpha$ = 1 + $\alpha$

$\therefore$ i = $\frac{1+\alpha }{1}$

Given, $\alpha$ = 20% = 0.2

$\therefore$ i = 1 + $\alpha$ = 1 + 0.2 = 1.2

$\Delta$T_f = ik_f m = 1.2 × 1.86 × 0.5 = 1.12 K

Chemical equation for the formation of H₂O(l) is

H_2(g) + 1/2O_2(g) = H₂O_(l)

Because enthalpy of formation of a compound is the heat absorbed or released when one mole of this substance is formed from its constituent elements.

Thus, enthalpy of formation of H₂O is –X₂ kJ mol^–1 where negative sign shows that the reaction is exothermic.

Equation (i) represents nutralisation reaction.

Equation (iii) represents hydrogenation reaction.

Equation (iv) represents combustion reaction.

$\frac{1}{2}$H₂ + $\frac{1}{2}$Cl₂ $\to$ HCl

$\Delta$H_f = –90 kJ mol^–1

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

$\Rightarrow$ -90 = [ $\frac{1}{2}$(430)_H2 + $\frac{1}{2}$(240)_Cl2] - (B.E)_HCl

$\Rightarrow$ -90 = [215 + 120] - (B.E)_HCl

$\Rightarrow$ (B.E)_HCl = 425 kJ mol^–1

Efficiency of a fuel cell ($\varphi$) = $\frac{\Delta G}{\Delta H}\times 100$

Generally, fuel cells are expected to have an efficiency of 100 percent.

RT ln K = nFE°

ln K = $\frac{nF{E}^{\circ }}{RT}$

= $\frac{2\times 0.46}{0.0591}$

$\Rightarrow$ K = 4 $\times$ 10¹⁵

The slow step is the rate determining step and it involves 1 molecule of H₂(g) and 1 molecule of ICl(g) . Hence the rate will be,

r = k[H₂(g)] [ICl(g)]

$\therefore$ The reaction is 1^st order with respect to H₂(g) and ICl(g).

For first order reaction

k = $\frac{2.303}{t}\log \frac{a}{a-x}$

at $t_{1/2}$, x = $\frac{a}{2}$

$t_{1/2}$ = $\frac{2.303}{k}\log \frac{a}{a-\frac{a}{2}}$

= $\frac{\ln 2}{k}$

For a first order reaction,

k = $\frac{2.303}{t}\log \frac{a}{a-x}$

k = $\frac{2.303}{60}\log \frac{100}{40}$

= $\frac{2.303}{60}\log 2.5$

= 0.0153

Also, $t_{1/2}$ = $\frac{2.303}{k}\log \frac{100}{50}$

= $\frac{2.303}{0.0153}\log 2$

= 45 min.

As each Sr²⁺ ion introduces one cation vacancy, therefore concentration of cation vacancies = mole % of SrCl₂ added.

$\therefore$ Concentration of cation vacancies

= 10^–4 mole %

= $\frac{{10}^{-4}}{100}\times 6.023\times 10^{23}$

= $6.023\times 10^{23}$ $\times$ 10^-6

= $6.023\times 10^{17}$

The maximum properties of the available volume which may be filled by hard sphere in simple cubic arrangement is $\frac{\pi }{6}$ or 0.52.

Langmuir adsorption isotherm is based on the assumption that every adsorption site is equivalent and that the ability of a particle to bind there is independent of whether nearby sites are occupied or not occupied.

The larger the atomic size, smaller is the value of the ionisation enthalpy. Again higher the screening effect, lesser is the value of ionisation potential, So option(b) has lowest ionisation enthalpy.

Going from left to right across a period in the periodic table, the basicity decreases as the electronegativity of the atom possessing the lone pair of electron increases. Hence basicity of $N{H}_2^{-}$ is higher than F^-. In moving down a group, as the atomic mass increases, basicity decreases. Hence F^- is more basic than I^- and HO^- is more basic than HS-. Hence among the ionic species, $N{H}_2^{-}$ has maximum proton affinity.

Among isoelectronic ions, ionic radius of anions is more than that of cations. Further size of the anion increases with increase in negative charge and size of the cation decreases with increase in positive charge.

All these structures exhibits resonance and can be represented by the following resonating structures.

More single bond character in resonance hybrid, more is the bond length. Hence the increasing bond length is

CO $<$ CO₂ $<$ CO${}_3^{2-}$

Hybridisation of Br in BrO₃^– :

= $\frac{1}{2}\left(7+0-0+1\right)$ = 4

Four hybrid orbital means sp³ hybridisation.

Hybridisation of Xe in XeO₃ :

= $\frac{1}{2}\left(8+0-0+0\right)$ = 4

Four hybrid orbital means sp³ hybridisation.

Thus, both BrO₃^– and XeO₃ are sp³ hybridised with three bond pairs of electrons and one lone pair of electrons and results in trigonal pyramidal shape.

Carbon and hydrogen are not suitable reducing agents for metal sulphides.

Galena (PbS), copper pyrites (CuFeS₂) and argentite (Ag₂S) are concentrated by froth floatation process but sphalerite (ZnS) is concentrated by chemical leaching.

As the basicity of metal hydroxides increases down the group from Be to Ba, the thermal stability of their carbonates also increases in the same order. Further group 1 compounds are more thermally stable than group 2 because their hydroxide are much basic than group 2 hydroxides therefore, the order of thermal stability is

BeCO₃ < MgCO₃ < CaCO₃ < K₂CO₃

Hydration energy of sulphate decreases on moving down the group-II. Mg²⁺ is smaller than other ions of group II, so Mg²⁺ is readily hydrated and hence MgSO₄ has higher hydration energy than the lattice energy.

[SiO₃^2–]_n and [Si₄O₁₁]^6– have chain structure of silicates.

The tendency to from +2 ionic state increases on moving down the group due to inert pair effect. Most characteristic oxidation state for lead and tin are +2, +4 respectively.

As the oxidation number of the central atom increases, strength of acid also increases.

In each vertical column of transition elements, the elements of second and third transition series resemble each other more closely than the elements of first and second transition series on account of lanthanide contraction. Hence the properties of elements of 4d series of the transition elements resemble with the properties of the elements of 5d series of the transition elements.

The exactly half-filled and completely filled d-orbital have extra stability and stability is directly proportional to number of unpaired electrons.

Mn⁺³ (22) = 3d⁴4s⁰ (4 unpaired electrons)

So, Mn³⁺ ion is most stable in aqueous solution. Rest all have less number of unpaired electrons.

Either a pair of crystals, molecules or compounds that are mirror images of each other but are not identical, and that rotate the plane of polarised light equally, but in opposite directions are called as enantiomorphs.

Cr²⁺ = 3d⁴ (4 unpaired electrons)

Mn²⁺ = 3d⁵ (5 unpaired electrons)

Fe²⁺ = 3d⁶ (4 unpaired electrons)

Ni²⁺ = 3d⁸ (2 unpaired electrons)

Thus, [Ni(H₂O)₆]²⁺ exhibit minimum paramagnetic behaviour.

Greater the number of unpaired electrons, higher is the paramagnetism. Hence Ni²⁺ will exhibit the minimum paramagnetic behaviour.

Compounds which do not show optical activity inspite of the presence of chiral carbon atoms are called meso-compounds. The absence of optical activity in these compound is due to the presence of a plane of symmetry in their molecules. e.g. meso-tartaric acid is optically inactive.

Benzene having any activating group, i.e., OH, R etc. undergoes electrophilic substitution very easily as compared to benzene itself. Thus, toluene (C₆H₅CH₃) and phenol (C₆H₅OH) undergo electrophilic substitution very readily than benzene. Chlorine with +E and +M effect deactivates the ring due to strong –I effect. So, it is difficult to carry out the substitution in chlorobenzene than in benzene, so correct order is

(iv) > (ii) > (i) > (iii)

Nucleophilicity increases down the periodic table :

I^– > Br^– > Cl^– > F^–

So, 2-methyl-2-butene yields acetone on ozonolysis.