NEET-UG 2002

Time required to reach the ground is dependent on the vertical motion of the particle. Vertical motion of both the particles A and B are exactly same. Although particle B has an initial velocity, but that is in horizontal direction and it has no component in vertical (component of a vector at a direction of 90^o = 0) direction. Hence they will reach the ground simultaneously.

m = 10 kg, R = mg

$\therefore$ Frictional force = f_k

= ${\mu }_{k}R={\mu }_{k}mg$

= 0.5 × 10 × 10

= 50 N [g = 10 m/sec²]

$\therefore$ Net force acting on the body = F = P – f_k
= 100 – 50 = 50 N.

$\therefore$ Acceleration of the block = a = F/m
= 50/10 = 5 m/sec² .

Mass (m) = 3 kg, force (F) = $6t^2\hat{i}+4t\hat{j}$

$\therefore$ acceleration
$a=F/m=\frac{6t^2\hat{i}+4t\hat{j}}{3}=2t^2\hat{i}+\frac{4}{3}t\hat{j}$

Now, $a=\frac{dv}{dt}=2t^2\hat{i}+\frac{4}{3}t\hat{j}$

$dv=\left(2t^2\hat{i}+\frac{4}{3}t\hat{j}\right)dt$

$\therefore v=\underset{0}{\overset{3}{\int }}\left(2t^2\hat{i}+\frac{4}{3}t\hat{j}\right)dt$

= ${\frac{2}{3}{t}^3\hat{i}+\frac{4}{6}{t}^2\hat{j}|}_0^3=18\hat{i}+6\hat{j}$

For a lift which is moving in upward direction with an acceleration a, the tension T developed in the string connected to the lift is given by

T = m (g + a).

Here m = 1000 kg, a = 1 m/s², g = 9.8 m/s²

$\therefore$ T = 1000(9.8 + 1) = 10,800 N.

Let us consider an elementary length dx at a distance x from one end.

It’s mass = k ·x ·dx

[k = proportionality constant]

Then centre of gravity of the rod x_c is given by

$x_c=\frac{\underset{0}{\overset{3}{\int }}kxdx.x}{\underset{0}{\overset{3}{\int }}kxdx}=\frac{\underset{0}{\overset{3}{\int }}{x}^{2}dx}{\underset{0}{\overset{3}{\int }}xdx}=\frac{{\frac{x^3}{3}|}_0^3}{{\frac{x^2}{2}|}_0^3}$

$\Rightarrow x_c=\frac{\frac{27}{3}}{\frac{9}{2}}=2$

$\therefore$ Centre of gravity of the rod will be at distance of 2 m from one end.

Let m be the mass of the body and v₁ and v₂ be the initial and final velocities of the body respectively.

$\therefore$ Initial kinetic energy = $\frac{1}{2}m{v}_1^2$

Final kinetic energy = $\frac{1}{2}m{v}_2^2$

Initial kinetic energy is increased 300% to get the final kinetic energy.

$\therefore$ $\frac{1}{2}m{v}_2^2=\frac{1}{2}\left(1+\frac{300}{100}\right)m{v}_1^2$

$\Rightarrow$ v₂ = 2v₁ or v₂/v₁ = 2 ... (i)

Initial momentum = p₁ = mv₁

Final momentum = p₂ = mv₂

$\therefore$$\frac{p_2}{p_1}=\frac{m{v}_2}{m{v}_1}=\frac{v_2}{v_1}=2$

$\therefore$ $p_2=2p_1=\left(1+\frac{100}{100}\right)p_1$

So momentum has increased 100%.

If external torque is zero, angular momentum remains conserved.
[External torque is zero because the weight of child acts downward]

L = I$\omega$ = constant

Since there is no friction at the contact surface (smooth horizontal surface) there will be no rolling. Hence, the acceleration of the centre of mass of the sphere will be independent of the position of the applied force F. Therefore, there is no relation between h and R.

In half rotation point P has moved horizontally.

$\frac{\pi d}{2}=\pi r=\pi \times 1m=\pi m$   $\left[\because radius=1m\right]$

In the same time, it has moved vertically a distance which is equal to its diameter = 2 m.

$\therefore$ Displacement of P = $\sqrt{{\pi }^2+2^2}=\sqrt{{\pi }^2+4}$ m

A circular disc may be divided into a large number of circular rings. Moment of inertia of the disc will be the summation of the moments of inertia of these rings about the geometrical axis.

Now, moment of inertia of a circular ring about its geometrical axis is MR², where M is the mass and R is the radius of the ring.

Since the density (mass per unit volume) for iron is more than that of aluminium, the proposed rings made of iron should be placed at a higher radius to get more value of MR². Hence to get maximum moment of inertia for the circular disc, aluminium should be placed at interior and iron at the outside.

Rate of heat loss in rod 1 = Q₁ $=\frac{K_1{A}_1\left(T_1-T_2\right)}{l_1}$

Rate of heat loss in rod 2 = Q₂ $=\frac{K_2{A}_2\left(T_1-T_2\right)}{l_2}$

By problem, Q1 = Q2.

$\therefore \frac{K_1{A}_1\left(T_1-T_2\right)}{l_1}=\frac{K_2{A}_2\left(T_1-T_2\right)}{l_2}$

$\therefore$ K₁A₁ = K₂A₂.   $\left[\because l_1=l_2\right]$

Radiating power of a black body

$=E_0=\sigma \left(T^4-T_0^4\right)A$

where $\sigma$ is known as the Stefan-Boltzmann constant, A is the surface area of a black body, T is the temperature of the black body and T₀ is the temperature of the surrounding.

$\therefore$ 60 = $\sigma$(1000⁴ – 500⁴)    ...(i)

[T = 727^oC = 727 + 273 = 1000 K, T0 = 227^oC = 500 K].

In the second case, T = 1227^oC = 1500 K and let E' be the radiating power.

$\therefore$ E' = $\sigma$(1500⁴ – 500⁴)    ...(ii)

From (i) and (ii) we have

$\frac{E^{\prime }}{60}=\frac{{1500}^4-{500}^4}{{1000}^4-{500}^4}=\frac{{15}^4-5^4}{{10}^4-5^4}=\frac{50000}{9375}$

$\therefore$ $E^{\prime }=\frac{50000}{9375}\times 60=320W$

Unit of Stefan’s constant is watt/m²K⁴.

An ideal black body is one which absorbs all the incident radiation without reflecting or transmitting any part of it. Black lamp absorbs approximately 96% of incident radiation.

An ideal black body can be realized in practice by a small hole in the wall of a hollow body (as shown in figure) which is at uniform temperature. Any radiation entering the hollow body through the holes suffers a number of reflections and ultimately gets completely absorbed. This can be facilitated by coating the interior surface with black so that about 96% of the radiation is absorbed at each reflection. The portion of the interior surface opposite to the hole is made conical to avoid the escape of the reflected ray after one reflection.

Wien’s displacement law states that the product of absolute temperature and the wavelength at which the emissive power is maximum is constant i.e. ${\lambda }_{max}$ T = constant. Therefore it expresses relation between wavelength corresponding to maximum energy and temperature.

Efficiency ($\eta$) of a carnot engine is given by $\eta =1-\frac{T_2}{T_1}$, where T₁ is the temperature of the source and T₂ is the temperature of the sink.

Here, T₂ = 500 K

$\therefore$ $0.5=1-\frac{500}{T_1}\Rightarrow T_1=1000K$

Now, $\eta =0.6=1-\frac{{T_2}^{\prime }}{1000}$ (T₂' is the new sink temperature)

$\Rightarrow {T_2}^{\prime }=400K$

The time period of a spring mass system as shown in figure 1 is given by $T=2\pi \sqrt{m/k}$, where k is the spring constant.

$\therefore$ $t_1=2\pi \sqrt{m/k_1}$   ...(i)
and $t_2=2\pi \sqrt{m/k_2}$   ...(ii)


Now, when they are connected in parallel as shown in figure 2(a), the system can be replaced by a single spring of spring constant, k_eff = k₁ + k₂.
[Since mg = k₁x + k₂x = k_effx]

$\therefore$ $t_0=2\pi \sqrt{m/k_{eff}}=2\pi \sqrt{m/\left(k_1+k_2\right)}$   ...(iii)

From (i), $\frac{1}{t_1^2}=\frac{1}{4{\pi }^2}\times \frac{k_1}{m}$   ...(iv)

From (ii), $\frac{1}{t_2^2}=\frac{1}{4{\pi }^2}\times \frac{k_2}{m}$   ...(v)

From (ii), $\frac{1}{t_0^2}=\frac{1}{4{\pi }^2}\times \frac{k_1+k_2}{m}$   ...(vi)

Now (iv) + (v)

$\frac{1}{t_1^2}+\frac{1}{t_2^2}=\frac{1}{4{\pi }^{2}m}\left(k_1+k_2\right)=\frac{1}{t_0^2}$

$\therefore$ $t_0^{-2}=t_1^{-2}+t_2^{-2}$

For a simple harmonic motion between A and B, with O as the mean position, maximum kinetic energy of the particle executing SHM will be at O and maximum potential energy will be at A and B.


$\therefore$ Displacement between maximum potential energy and maximum kinetic energy is $\pm a$.

This is a case of damped vibration as the amplitude of vibration is decreasing with time.
Amplitude of vibrations at any instant t is given by $a=a_0{e}^{–bt}$, where $a_0$ is the initial amplitude of vibrations and b is the damping constant.

Now, when t = 100T, $a=a_0/3$  [T is time period]

Let the amplitude be $a^{\prime }$ at t = 200T.
i.e. after completing 200 oscillations.

$\therefore$ $a=a_0/3=a0e^{–100Tb}$   ...(i)

and $a^{\prime }=a_0{e}^{–200Tb}$ ...(ii)

From (i), $\frac{1}{3}=e^{-100Tb}$
$\therefore$ $e^{-200Tb}=1/9$

From (ii), $a^{\prime }=a_0\times \frac{1}{9}=\frac{a_0}{9}$

$\therefore$ The amplitude will be reduced to 1/9 of initial value.

The equation of progressive wave travelling in positive x-direction is given by

$y=a\sin \frac{2\pi }{\lambda }\left(vt-x\right)$

Here a = 0.2 m, v = 360 m/sec, $\lambda$ = 60 m

$\therefore y=0.2\sin \frac{2\pi }{60}\left(360t-x\right)$

$=0.2sin\left[2\pi \left(6t-\frac{x}{60}\right)\right]$

The whistle is revolving in a circle of radius 50 cm. So the source (whistle) is moving and the observer is fixed.


The minimum frequency will be heard by the observer when the linear velocity of the whistle (source) will be in a direction as shown in the figure, i.e. when the source is receding.

The apparent frequency heard by the observer is then given by ${\upsilon }^{\prime }=\upsilon \left(\frac{v}{V+v}\right)$

where V and v are the velocities of sound and source respectively and $\upsilon$ is the actual frequency.

Now $v=r\omega =0.5\times 20=10m/s$

V = 340 m/s, $\upsilon$ = 385 Hz.

$\therefore$ ${\upsilon }^{\prime }=385\times \frac{340}{340+10}=374$ Hz

Electric field intensity E is zero within a conductor due to charge given to it.

Also, $E=-\frac{dV}{dx}\Rightarrow \frac{dV}{dx}=0$   (inside the conductor)

$\therefore$ V = constant. [V is potential]

So potential remains same throughout the conductor.

There are eight corners of a cube and in each corner there is a charge of (–q). At the centre of the corner there is a charge of (+q). Each corner is equidistant from the centres of the cube and the distance (d) is half of the diagonals of the cube.

Diagonal of the cube = $\sqrt{b^2+b^2+b^2}=\sqrt{3}b$

$\therefore$ $d=\sqrt{3}b/2$

Now, electric potential energy of the charge (+q) due to a charge (–q) at one corner

U $=\frac{q_1{q}_2}{4\pi {\epsilon }_{0}r}=\frac{\left(+q\right)\times \left(-q\right)}{4\pi {\epsilon }_0\left(\sqrt{3}b/2\right)}=-\frac{q^2}{2\pi {\epsilon }_0\left(\sqrt{3}b\right)}$

$\therefore$ Total electric potential energy due to all the eight identical charges
= $8U=-\frac{8q^2}{2\pi {\epsilon }_0\sqrt{3}b}=\frac{-4q^2}{\sqrt{3}\pi {\epsilon }_{0}b}$

Resistance of a conductor is given by R = $\rho \frac{l}{A}$, where $\rho$ is the specific resistance, $l$ is the length and A is the cross-sectional area of the conductor. Now, when $l$ = 1 and A = 1, R = $\rho$.

So specific resistance or resistivity of a material may be defined as the resistance of a specimen of the material having unit length and unit cross-section. Hence, specific resistance is a property of a material and it will increase with the increase of temperature, but will not vary with the dimensions (length, crosssection) of the conductor.

Terminal potential difference is 2.2 V when circuit is open.

$\therefore$ e.m.f. of the cell = E = 2.2 volt

Now, when the cell is connected to the external resistance, circuit current I is given by

$I=\frac{E}{R+r}=\frac{2.2}{5+r}$ ampere, where r is the internal resistance of the cell.

Potential difference across the cell = IR

$=\frac{2.2}{5+r}\times 5=1.8$

$\therefore$ 5 + r = 11/1.8.

$\therefore r=\frac{11}{1.8}-5=\frac{110-90}{18}=\frac{10}{9}\Omega$

Charge Q = C₁V
Total capacity of combination (parallel)

C = C₁+ C₂

$P.D.=\frac{Q}{C}=\frac{C_{1}V}{C_1+C_2}$

B₁ = B = $\frac{{\mu }_{0}I}{2R}$

Further, B₂ = $\frac{{\mu }_0\left(2I\right)}{2r}$

Now 2 × 2$\pi$r = 2$\pi$R or r = R/2

Hence B₂ = $4\times \frac{{\mu }_{0}I}{2R}=4B$

Lorentz forece

$\overrightarrow{F_L}=\overrightarrow{F_e}+\overrightarrow{F_m}$

= $q\overrightarrow{E}+q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

For converting galvanometer to voltmeter, a high resistance should be connected in series with galvanometer.

Time period, T = $2\pi \sqrt{\frac{I}{MB}}$

$\therefore$ T $\propto$ $\frac{1}{\sqrt{M}}$

M₁ =2M + M = 3M  

M₂ = 2M – M = M

$\frac{T_1}{T_2}=\sqrt{\frac{M}{3M}}=\frac{1}{\sqrt{3}}$

$\Rightarrow$ T₂ = $\sqrt{3}T$

$\therefore$ T₁ $<$ T₂

The impedance Z of a series LCR circuit is given by,

Z = $\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

At resonance, X_L = X_C, hence Z = R.

Let, supply voltage = V_R = V

$\therefore$ R.M.S. current, I = $\frac{V}{R}$

$\therefore$ Power loss = VI = I²R

Since, the direction of propagation of electromagnetic waves is perpendicular to Electric field and Magnetic field. Also electric and magnetic fields perpendicular so, direction of the wave will be parallel to $\overrightarrow{E}\times \overrightarrow{B}$.

Infrared ray is the cause of Green House effect. The glass transmits visible light while infrared rays are absorbed by plants. Then it emits long infrared rays, which are reflected back by glass.

Applying Snell’s law of refraction at A,

$\mu =\frac{\sin 45^{\circ }}{\sin r}$

$\Rightarrow$ r = ${\sin }^{-1}\left(\frac{1}{\sqrt{2}\mu }\right)$ ....(1)

Applying the condition of total internal reflection at B,

i_c = ${\sin }^{-1}\left(\frac{1}{\mu }\right)$ ..........(2)

where i_c is the critical angle.

From figure, r + i_c = $\frac{\pi }{2}$

$\Rightarrow$ ${\sin }^{-1}\left(\frac{1}{\sqrt{2}\mu }\right)=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{1}{\mu }\right)$

$\Rightarrow$ ${\sin }^{-1}\left(\frac{1}{\sqrt{2}\mu }\right)={\cos }^{-1}\left(\frac{1}{\mu }\right)$ = ${\sin }^{-1}\left(\frac{\sqrt{{\mu }^2-1}}{\mu }\right)$

$\Rightarrow$ $\frac{1}{\sqrt{2}\mu }=\frac{\sqrt{{\mu }^2-1}}{\mu }$

$\Rightarrow$ ${\mu }^2=1+\frac{1}{2}$

$\Rightarrow$ $\mu =\sqrt{\frac{3}{2}}$

For the aparture, limit of resolution :

$\frac{y}{D}\ge \frac{\lambda }{d}$

$\Rightarrow$ y $\ge$ $\frac{\lambda D}{d}$

$\Rightarrow$ y $\ge \frac{5\times {10}^{-7}\times 50}{2\times {10}^{-3}}$ $\ge$ 1.25 cm

Using the lens formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Given v = d, for equal size image |v| = |u| = d

By sign convention u = -d

$\therefore$ $\frac{1}{f}=\frac{1}{d}-\frac{1}{-d}$

$\Rightarrow$ f = $\frac{d}{2}$

₈O¹⁶ + ₁H² $$\to$$ ₇N¹⁴ + ₂He⁴

So when a deuteron is bombarded on ₈O¹⁶ nucleus then an $\alpha$-particle (₂He⁴) is emitted and the product nucleus is ₇N¹⁴.

N = 4 $\times$ 10¹⁶${\left(\frac{1}{2}\right)}^{\frac{30}{10}}$ = $\frac{1}{2}$ $\times$ 10¹⁶

$\therefore$ Atoms decayed = 4 $\times$ 10¹⁶ - $\frac{1}{2}$ $\times$ 10¹⁶

= 3.5 $\times$ 10¹⁶

The nuclei of light elements have a lower binding energy than that for the elements of intermediate mass. They are therefore less stable; consequently the fusion of the light elements results in more stable nucleus.

The value of Planck’s constant is 6.63 $\times$ 10^$-$34 J-sec.

It occurs by incidence of X-rays.

de Broglie wavelength for a particle is given by

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

Since all the particles are moving with same velocity, the particle with least mass will have maximum de-Broglie wavelength.

Here $\beta$-particles(electrons) has the lowest mass and therefore it has maximum wavelength.

Cathode rays are basically negatively charged particles (electrons). If the cathode rays are allowed to pass between two plates kept at a difference of potential, the rays are found to be deflected from the rectilinear path. The direction of deflection shows that the rays carry negative charges.

In forward biasing, the resistance of p-n junction diode is very low to the flow of current. So practically all the voltage will be dropped across the resistance R, i.e. voltage across R will be V.

In reverse biasing, the resistance of p-n junction diode is very high. So the voltage drop across R is zero.

This truth table represents NAND gate.

For conduction, p-n junction must be forward biased. For this p-side should be connected to higher potential and n-side to lower potential.

$\frac{I_C}{I_E}$ = 0.96 = $\alpha$

$\therefore$ Current gain, $\beta$ = $\frac{I_C}{I_B}$ = $\frac{\alpha }{1-\alpha }$ = 24

In body-centred cubic (b.c.c.) lattice there are eight atoms at the corners of the cube and one at the centre.

Therefore number of atom per unit cell

= $\frac{1}{8}\times 8$ + 1 = 2

From molarity equation

M₁V₁ + M₂V₂ = MV

$\Rightarrow$ 1× 2.5 + 0.5 × 3 = M × 5.5

$\Rightarrow$ M = $\frac{4}{5.5}$ = 0.73M

	Element	%	Atomic mass	Relative no. of atoms	Simplest ratio of atoms
1	C	40	12	$\frac{40}{12}$ = 3.33	$\frac{3.33}{3.3}$ = 1
2	H	13.3	1	$\frac{13.3}{1}$ = 13.3	$\frac{13.3}{3.3}$ = 4
2	N	46.7	14	$\frac{46.7}{14}$ = 3.3	$\frac{3.3}{3.3}$ = 1

$\therefore$ The empirical formula is CH₄N.

1 mole = 6.023 $\times$ 10²³ number of molecules.

1 g mole of O₂ = 32 g of O₂

$\Rightarrow$ 16 g of O₂ = 0.5 g mole of O₂

1 g mole of N₂ = 28 g of N₂

$\Rightarrow$ 7 g of N₂ = 0.25 g mole of N₂

1 g mole of H₂ = 2 g of H₂

$\Rightarrow$ 2 g of H₂ = 1 g mole of H₂

1 g of NO₂ = 14 + 16 $\times$ 2 = 46

$\Rightarrow$ 16 g of NO₂ = 0.35 mole NO₂

K.E = 1/2 mv²

= ${\left(\frac{\pi e^2}{nh}\right)}^2\times 2m$ [ $\because$ v = $\frac{2\pi e^2}{nh}$]

$\therefore$ Total energy = E_n = $\frac{-2{\pi }^{2}m{e}^4}{n^2{h}^2}$

= -${\left(\frac{\pi e^2}{nh}\right)}^2$ $\times$ 2m = -K.E

$\therefore$ K.E = - E_n

Energy of first excited state is -3.4 eV

$\therefore$ Kinetic energy of the same orbit (n = 2) will be +3.4 ev

At higher temperature and low pressure real gas acts as an ideal gas.

For the reaction

BaO_2(g) $\rightleftharpoons$ BaO_(s) + O_2(g); H = +ve.

At equilibrium K_p = P_O2

[For solid and liquids concentration term is taken as unity]

Hence, the value of equilibrium constant depends only upon partial pressure of O₂. Further on increasing temperature formation of O₂ increases as this is an endothermic reaction.

Na₂CO₃ is a salt of weak acid H₂CO₃ and strong base NaOH, therefore, its aqueous solution will be basic hence has pH more than 7.

MX2	⇌	Ag2+	+	2X-
s		s		2s

K_sp = [M²⁺] [X^–]² = (S)(2S)² = 4S³

$\Rightarrow$ K_sp = 4(0.5 × 10^–4)³ = 5 × 10^–13

NH₄OH and NH₄Cl constitute to form a basic buffer.

pOH = pK_b + log $\frac{\left[Salt\right]}{\left[Base\right]}$

We know, pOH+ pH = 14 or pOH = 14 – pH

$\therefore$ 14 - pH - $\log \frac{\left[Salt\right]}{\left[Base\right]}$ = pK_b

$\Rightarrow$ 14 - 9.25 - $\log \frac{0.1}{0.1}$ = pK_b

$\Rightarrow$ 14 – 9.25 – 0 = pK_b

$\Rightarrow$ pK_b = 4.75

For dilute solution, the van’t Hoff equation is

$\pi =\frac{n}{V}RT$

$\Rightarrow$ $\pi V=nRT$

$\Rightarrow$ $\pi V=\frac{m_2}{M}RT$

$\Rightarrow$ $M_2=\left(\frac{m_2}{V}\right)\frac{RT}{\pi }$

These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B.

Entropy is the change in heat per degree. Thus, its unit is as follows.

Entropy = $\frac{Jmo{l}^{-1}}{K}$ = J mol^-1 K^-1

2KBr + I₂ $\to$ 2KI + Br₂

reaction is not possible because Br^– ion is not oxidised in Br₂ with I₂ due to higher electrode (oxidation) potential of I₂ than bromine.

When in a closed insulated container, a liquid is stirred, it increases the temperature of liquid thus it started behaving as adiabatic process resulting no change in heat and only temperature increases. Therefore, q = 0 for this process.

Also, $\Delta$E = q + W

$\Delta$E = W (As, q = 0)

As, there is rise in temperature thus, there must be change in internal energy, i.e., $\Delta$E = W ≠ 0.

C_(s) + 2H_2(g) $\to$ CH_4(g), $\Delta$H^o = ?

C_(s) + O_2(g) → CO_2(g), $\Delta$H = – 94 kcal/mol ....(1)

H_2(g) + $\frac{1}{2}$O_2(g) $\to$ H₂O_(g), $\Delta$H = – 68 kcal/mol ....(2)

CH_4(g) + 2O_2(g) $\to$ CO_2(g) + H₂O_(l), $\Delta$H = – 213 kcal/mol ....(3)

Performing (1) + 2 × (2) – (3)

$\Delta$H^o = [– 94 – 2(68)] – (–213) kcal/mol

= [– 94 – 136] + 213 k cal/mol

= – 230 + 213 = – 17 kcal/mol

For isothermal reversible expansion

w = q = nRT $\times$ 2.303 $\log \frac{v_2}{v_1}$

= 2RT $\times$ 2.303 $\log \frac{20}{2}$

= 2 × 2 × T × 2.303 × 1 = 9.2 T

Entropy change, $\Delta$S = $\frac{q}{T}$ = $\frac{9.2T}{T}$ = 9.2 cal/mol K

In electrolysis of NaCl when Pt electrode is taken then H₂ liberated at cathode while with Hg cathode it forms sodium amalgam because more voltage is required to reduce H⁺ at Hg than Pt.

2A $\to$ B + C

If it is zero order reaction r = k [A]^o.

i.e the rate remains same at any concentration of 'A'. i.e independent upon concentration of A.

${}_{92}^{235}$U + ${}_0^1$ n $\to$ ${}_{54}^{139}$Xe + ${}_{38}^{94}$Sr + 3${}_0^1$ n

Micelles are formed by the association of colloids. They are formed by lyophilic and lyophobic groups. As the concentration increases the lyophobic parts receding away from the solvent approach each other are forms a cluster, the lyophobic ends are in the interior, lyophilic groups projecting outwards in contact with the solvent.

Along the period, I.P. generally increases but not regularly. Be and B are exceptions. First I.P. increases in moving from left to right in a period, but I.P. of B is lower than Be.

As in case of boron, 2p¹ electron have to be removed to get B⁺(1s² 2s²) from B(1s² 2s² 2p¹), while in case of Be (1s²2s²) electron have to be removed to get Be⁺ (1s²2s¹). p electron can be removed more easily than s electron so the energy required to remove electron will be less in case of boron.

$O^{-}—\stackrel{\stackrel{O}{\uparrow }}{N}=O$

In NO₃^- ion, nitrogen has 4 bond pair of electrons and no lone pair of electrons.

Electronic structure of S atom in excited state


In 'S' unhybride d- orbital is present, which will involved in bond formation with oxygen atom.

In oxygen two unpaired p- orbital is present in these one is involved in $\sigma$ bond formation while other is used in $\pi$ bond formation.

So, in SO₃^2$-$, p$\pi$ $-$ d$\pi$ bonding present.

In CO, the number of electrons

$\Rightarrow$ 6 + 8 = 14 [ Z of C = 6 and O = 8]

Electronic configuration of molecular orbital of CO :

${\left(\sigma 1s\right)}^2{\left({\sigma }^{\ast }1s\right)}^2{\left(\sigma 2s\right)}^2{\left({\sigma }^{\ast }2s\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2{\left(\pi 2p_z\right)}^2$

CN^- also get (6 + 7 + 1) 14 electrons and the configuration is similar to that CO.

So, CN and CO is isoelectronic.

In Castner’s process, for production of sodium metal, sodium hydroxide (NaOH) is electrolysed at temperature 330ºC.

Fluorine is more reactive than chlorine. So, bond energy of chlorine is greater than fluorine.

Silicon exhibits 6 coordination numbers.

In aqueous state Mn(II) is more stable.

Mn ⇌ Mn²⁺ + 2e^-

Zinc is on the top position of hydrogen in electrochemical series. So Zn displaces H₂ from dilute H₂SO₄ and HCl with liberation of H₂ .

Zn + H₂SO₄ $\to$ ZnSO₄ + H₂

On the other hand HNO₃ is an oxidising agent. Hydrogen obtained in this reaction is converted into H₂O.

Zn + 2HNO₃ $\to$ Zn(NO₃ )₂ + 2H

2HNO₃ $\to$ H₂O + 2NO₂ + O

2H + O $\to$ H₂O.

Given electronic configuration of atom 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s².

In the configuration, the last electron of the atom is filled in d sub-shell as 3d³. Thus this element belongs to d-block of the periodic table with group no. V.

Mn : [Ar] 3d⁵4s²

Shows +2, +3, +4, +5, +6 & +7 oxidation states

General electronic configuration of lanthanides is

$\left(n-2\right)f^{1-14}\left(n-1\right)s^2{p}^6{d}^{0-1}n{s}^2$ .

Copper being more electropositive readily precipitate silver from their salt (Ag⁺) solution.

Cu + 2AgNO₃ $\to$ Cu(NO₃)₂ + Ag

In K[Ag(CN)₂] solution a complex anion [Ag(CN)₂]^– is formed so Ag⁺ ions are less available in the solution and Cu cannot displace Ag from this complex ion.

Copper sulphate react with KCN to give white ppt of CuCN and cyanogen gas. The insoluble copper cyanide dissolve in excess of KCN and give soluble potassium cuprocyanide

2CuSO₄ + 4KCN $\to$ 2CuCN + (CN)₂ + 2K₂SO₄

CuCN + 3KCN $\to$ K₃[Cu(CN)₄]

The hypothetical complex "chloro diaquatriammine cobalt(III) chloride" can be broken down as follows :

• "chloro" indicates the presence of a chloride ion (Cl⁻) as a ligand.


• "diaqua" indicates the presence of two water molecules (H₂O) as ligands.


• "triammine" indicates the presence of three ammonia molecules (NH₃) as ligands.


• "cobalt(III)" indicates that cobalt is in the +3 oxidation state.


• The additional "chloride" at the end suggests the presence of chloride ions outside the coordination sphere, likely as counterions.

Given this, let's analyze the options :

• Option A - [CoCl(NH₃)₃(H₂O)₂]Cl₂ : This matches the description. It has one chloro ligand, three ammonia ligands, two water ligands, and two chloride ions as counterions, with cobalt in the +3 oxidation state.


• Option B - [Co(NH₃)₃(H₂O)Cl₃] : Incorrect, as it suggests three chloride ligands inside the coordination sphere, which is not specified in the name.


• Option C - [Co(NH₂)₃(H₂O)₂Cl] : Incorrect, as NH₂ is not the same as NH₃ (ammonia), and it also suggests a different ligand arrangement.


• Option D - [Co(NH₃)₃(H₂O)₃]Cl₃ : Incorrect, as it has three water molecules instead of two and lacks the chloro ligand inside the coordination sphere.

Therefore, the correct representation is Option A - [CoCl(NH₃)₃(H₂O)₂]Cl₂.

Atoms, ions or molecules containing unpaired electrons are paramagnetic.

In [Cr(NH₃)₆]³⁺

Cr⁺³= [Ar]3d³4s⁰

NH₃ is weak field ligand so pairing of d-electrons take place odes not happen.

Here, Number of unpaired electrons = 3

So it is paramagnetic.

In other cases pairing of d-electrons take place in presence of strong field ligands such as CO or CN^–.

Geometrical isomers have same molecular formula but differ in the position of atoms or groups in space due to hindered rotation about a double bond.

IUPAC Name is 1-hexene-5-yne.

atomic radii order :
I > Br > Cl > F

F, Cl, Br, I belongs to the same group orderly. Atomic radii go on increasing as the nuclear charge increases from top to bottom in a group. The decreasing order of bond length C – I > C – Br > C – Cl > C – F. The order of bond dissociation energy
R – F > R – Cl > R – Br > R – I. During dehydrohalogenation C – I bond breaks more easily than C – F bond.So reactivity order of halides

$R-\mathrm{I}>R-Br>R-Cl>R-F$