NEET-UG 2018

Diameter of the ball

= MSR + CSR × (Least count) – Zero error

= 5 mm + 25 × 0.001 cm – (–0.004) cm

= 0.5 cm + 25 × 0.001 cm – (–0.004) cm = 0.529 cm.

We know,

F = $\mu$N

$\Rightarrow$ MLT^–2 = $\mu$MLT^–2

$\Rightarrow$ $\mu$ = M⁰L⁰T⁰

$\therefore$ Coefficient of sliding friction has no dimension.

At equilibrium

ma cos $\theta$ = mg sin $\theta$

$\Rightarrow$ a = g tan $\theta$

Since from the given figure, in order to complete the vertical circle, velocity at the bottom of vertical circle should be $\sqrt{5gR}$.

As body is at rest initially, i.e., speed = 0.
At point A, speed = v.
As track is frictionless, so total mechanical energy will remain constant.

$\therefore$ 0 + mgh = $\frac{1}{2}m{v}^2$ + 0

$\Rightarrow$ h = $\frac{v^2}{2g}$

For completing the vertical circle, v $\ge$ $\sqrt{5gR}$

$\therefore$ h = $\frac{5gR}{2g}$ = $\frac{5}{2}R$ = $\frac{5}{4}D$

From the law of conservation of linear momentum,

momentum before collision = momentum after collision, so

mv + 4m × 0 = m × 0 + 4mv′

mv = 4mv′ or v = 4v′

Coefficient of restitution,

e = $\frac{v^{\prime }}{v}=\frac{\frac{v}{4}}{v}$ = 0.25

In rolling motion,

Translational kinetic energy, K_t = $\frac{1}{2}m{v}^2$

Rotational kinetic energy, K_r = $\frac{1}{2}I{\omega }^2$

$\therefore$ K_t + K_r = $\frac{1}{2}m{v}^2$ + $\frac{1}{2}I{\omega }^2$

= $\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\left(\frac{v}{r}\right)}^2$

= $\frac{7}{10}m{v}^2$

$\therefore$ $\frac{K_t}{K_t+K_r}=\frac{\frac{1}{2}m{v}^2}{\frac{7}{10}m{v}^2}$ = $\frac{5}{7}$

Given, $\overrightarrow{F}=4\hat{i}+5\hat{j}-6\hat{k}$

We know,
Moment of force, $\overrightarrow{\tau }=\left(\overrightarrow{r}-{\overrightarrow{r}}_0\right)\times \overrightarrow{F}$

Where ${\overrightarrow{r}}_0=2\hat{i}-2\hat{j}-2\hat{k}$

and $\overrightarrow{r}=2\hat{i}+0\hat{j}-3\hat{k}$

$\overrightarrow{r}-{\overrightarrow{r}}_0$ = $\left(2\hat{i}+0\hat{j}-3\hat{k}\right)-\left(2\hat{i}-2\hat{j}-2\hat{k}\right)$

= $0\hat{i}+2\hat{j}-\hat{k}$

$\overrightarrow{\tau }=\left(\overrightarrow{r}-{\overrightarrow{r}}_0\right)\times \overrightarrow{F}$

= $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& -1\\ 4& 5& -6\end{array}\right|$

= (-12 + 5)$\hat{i}$ - (+4)$\hat{j}$ + (-8)$\hat{k}$

= $-7\hat{i}-4\hat{j}-8\hat{k}$

As sphere is in free space and no external torque is acting over it so its angular momentum will remain constant.

Work done required to bring them rest

$\Delta$W = $\Delta$K_E (work-energy theorem)

$\Delta$W = $\frac{1}{2}I{\omega }^2$

For same $\omega$, $\Delta$W $\infty$ I

For a solid sphere, I_A = $\frac{2}{5}M{R}^2$

For a thin circular disk, I_B = $\frac{1}{2}M{R}^2$

For a circular ring, I_C = MR²

$\therefore$ I_C > I_B > I_A

So, W_C > W_B > W_A

If universal gravitational constant becomes ten times, then G' = 10 G

Acceleration due to gravity, g = $\frac{GM}{R^2}$

So, acceleration due to gravity increases, hence ‘g’ on the Earth will not change is not correct.

Speed of the planet will be maximum when its distance from the sun is minimum as mvr = constant.

Point A is perihelion and C is aphelion.

Clearly, V_A > V_B > V_C

So, K_A > K_B > K_C

Power = rate of production of heat = F.V

= 6$\pi$$\eta$rV_T.V_T = = 6$\pi$$\eta$rV_T²

[ As F = 6$\pi$$\eta$rV_T from stoke’s formula ]

$\because$ Terminal velocity V_T = $\frac{2r^2\left(\rho -\sigma \right)g}{9\eta }$

$\therefore$ V_T $\propto$ r²

$\therefore$ Power $\propto$ 6$\pi$$\eta$r$\left[\frac{4r^4{\left(\rho -\sigma \right)}^2{g}^2}{81{\eta }^2}\right]$

$\Rightarrow$ Power $\propto$ r⁵

Let the wire 1 is of length = l and wire 2 of length = $\frac{l}{3}$

Now area of wire 1 = A while area of wire 2 = 3A

For wire 1,
$\Delta$l = $\left(\frac{F}{AY}\right)l$ ....(i)

For wire 2, let F' force is applied

$\Delta$l = $\left(\frac{F^{\prime }}{3AY}\right)\frac{l}{3}$ ....(2)

From equation (i) & (ii)

$\left(\frac{F}{AY}\right)l$ = $\left(\frac{F^{\prime }}{3AY}\right)\frac{l}{3}$

$\Rightarrow$ F' = 9F

From Wien's law, $\lambda$_maxT = constant

$\therefore$ $\lambda$_max1T₁ = $\lambda$_max2T₂

$\Rightarrow$ $\lambda$₀T₁ = $\frac{3{\lambda }_0}{4}$T₂

$\Rightarrow$ $\frac{T_2}{T_1}=\frac{4}{3}$

According to Stefan-Boltzmann law, energy emitted
unit time by a black body is Ae$\sigma$T⁴,

$\therefore$ P $\propto$ T⁴

So $\frac{P_2}{P_1}={\left(\frac{T_2}{T_1}\right)}^4$ = ${\left(\frac{4}{3}\right)}^4$

Here, P₂ = nP and P₁ = P

So, $\frac{nP}{P}=n={\left(\frac{T_2}{T_1}\right)}^4={\left(\frac{4}{3}\right)}^4=\frac{256}{81}$

Efficiency of ideal heat engine,

$\eta =\left(1-\frac{T_2}{T_1}\right)$

Freezing point of water = 0^oC = 273 K

Boiling point of water = 100^oC = (100 + 273) K = 373 K

$\therefore$ Sink temperature, T₂ = 0^oC = 0 + 273 = 273 K

$\therefore$ Source temperature, T₁ = 100^oC = 100 + 273 = 373 K

Percentage efficiency, %$\eta =\left(1-\frac{T_2}{T_1}\right)\times 100$

= $\left(1-\frac{273}{373}\right)\times 100$

= $\left(\frac{100}{373}\right)\times 100$ = 26.8%

Using first law of thermodynamics,

$\Delta$Q = $\Delta$U + $\Delta$W

$\Rightarrow$ 54 × 4.18 = $\Delta$U + 1.013 × 10⁵(167.1 × 10^–6 – 0)

$\Rightarrow$ $\Delta$U = 208.7 J

Gas is monatomic, so C_p = $\frac{5}{2}R$

Given process is isobaric.

$\therefore$ dQ = nC_pdT

or dQ = n$\left(\frac{5}{2}R\right)$dT

Also, dW = PdV = nRdT ($\because$ PV = nRT)

$\therefore$ Required ratio = $\frac{dW}{dQ}$ = $\frac{nRdT}{n\left(\frac{5}{2}R\right)dT}$ = $\frac{2}{5}$

Escape velocity from the Earth’s surface is vescape = 11200 m s^–1

Let at temperature T, rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere.

Also, V_rms = V_escape = $\sqrt{\frac{3k_{B}T}{m_{O_2}}}$

$\Rightarrow$ 11200 = $\sqrt{\frac{3\times 1.38\times {10}^{-23}\times T}{2.76\times {10}^{-26}}}$

$\Rightarrow$ T = 8.360 × 10⁴ K

Given, acceleration, a = 20 m/s²,
and displacement, y = 5m

Magnitude of acceleration of a particle
moving in a SHM is, |a| = ${\omega }^2$y; where y is amplitude.

$\Rightarrow$ 20 = ${\omega }^2$(5)

$\Rightarrow$ $\omega =2$ rad/s

Time period of pendulum,

T = $\frac{2\pi }{\omega }=\frac{2\pi }{2}$ = $\pi$ s

Two successive resonance are produced at 20 cm and 73 cm of column length

$\therefore$ $\frac{\lambda }{2}$ = (73 - 20) $\times$ 10^-2 m

$\Rightarrow$ $\lambda$ = 2$\times$ (73 - 20) $\times$ 10^-2 m

Velocity of sound, v = n$\lambda$

= 2 $\times$ 320 $\times$(73 - 20) $\times$ 10^-2 m

= 339 m/s

For closed organ pipe, third harmonic

n = $\frac{\left(2N-1\right)V}{4l}=\frac{3V}{4l}$ ($\because$ N = 2 )

For open organ pipe, fundamental frequenty

n = $\frac{NV}{2l}=\frac{V}{2l^{\prime }}$ ($\because$ N = 1 )

Given, third harmonic for closed organ pipe = fundamental frequency for open organ pipe.

$\therefore$ $\frac{3V}{4l}=\frac{V}{2l^{\prime }}$

$\Rightarrow$ l' = $\frac{4l}{3\times 2}$ = $\frac{2l}{3}$ = $\frac{2\times 20}{3}$ = 13.33 cm

Acceleration, a = $\frac{v-u}{t}$ = $\frac{6-0}{1}$ = 6 ms^-2

For t = 0 to t = 1 s,

S₁ = $\frac{1}{2}\times 6{\left(1\right)}^2$ = 3 m ...(i)

For t = 1 s = to t = 2 s,

S₂ = 6.1 - $\frac{1}{2}\times 6{\left(1\right)}^2$ = 3 m ...(ii)

For t = 2 s to t = 3 s,

S₃ = 0 - $\frac{1}{2}\times 6{\left(1\right)}^2$ = -3 m ...(ii)

Total displacement S = S₁ + S₂ + S₃ = 3 m

Average velocity = $\frac{3}{3}$ = 1 ms^-1

Total distance travelled = 9 m

Average speed = $\frac{9}{3}$ = 3 ms^-1

Force experienced by a charged particle in an electric field, F = qE

As F = ma

$\therefore$ ma = qE

$\Rightarrow$ a = $\frac{qE}{m}$

As electron and proton both fall from same height at rest. Then initial velocity = 0

From the formula, s = ut + $\frac{1}{2}a{t}^2$

$\therefore$ h = $\frac{1}{2}a{t}^2$

$\Rightarrow$ h = $\frac{1}{2}\frac{qE}{m}{t}^2$

$\therefore$ t = $\sqrt{\frac{2hm}{qE}}$

$\Rightarrow$ t $\propto$ $\sqrt{m}$ as ‘q’ is same for electron and proton.

Since, electron has smaller mass so it will take smaller time.

Colour code for carbon resistor

0 - Black
1 - Brown
2 - Red
3 - Orange
4 - Yellow
5 - Green
6 - Blue
7 - Violet
8 - Grey
9 - White

Tolerance :
± 5% Gold
± 10% Silver
± 20% No colour

(47 $\pm$ 4.7) k$\Omega$ = 47 $\times$ 10³ $\pm$ 10% $\Omega$

Which shows the colour code Yellow – Violet – Orange – Silver.

Current drawn from the cell is

I = $\frac{n\epsilon }{nr}=\frac{\epsilon }{r}$


So, I is independent of n and I is constant.

n series grouping equivalent resistance

R_series = nR

In parallel grouping equivalent resistance

R_parallel = $\frac{R}{n}$

Current drawn from a battery when n resistors are connected in series is

I = $\frac{E}{nR+R}$ ......(i)

Current drawn from same battery when n resistors are connected in parallel is

10I = $\frac{E}{\frac{R}{n}+R}$ ......(ii)

On dividing eqn. (ii) by (i),

10 = $\frac{\left(n+1\right)R}{\left(\frac{1}{n}+1\right)R}$

Solving we get, n = 10

In an isolated capacitor Q is constant, so an electrostatic force between metal plates is given as:

F = QE = Q $\times$ $\frac{\sigma }{2{\epsilon }_0}$ = $\frac{Q^2}{2A{\epsilon }_0}$ [$\because$ $\sigma =\frac{Q}{A}$]

F is independent of the distance between plates.

Given Mass per unit length of a metallic rod is
$\frac{m}{l}$ = 0.5 kg m^-1

From figure, for equilibrium

mg sin 30^o = I$l$B cos 30^o

$\Rightarrow$ I = $\frac{mg}{lB}\tan 30^{\circ }$

= $\frac{0.5\times 9.8}{0.25\times \sqrt{3}}$ = 11.32 A

Current sensitivity of moving coil galvanometer

I_s = $\frac{NBA}{C}$ .....(i)

Voltage sensitivity of moving coil galvanometer,

V_s = $\frac{NBA}{C{R}_G}$

Dividing eqn. (i) by (ii)

Resistance of galvanometer

R_G = $\frac{I_s}{V_S}=\frac{5\times 1}{20\times {10}^{-3}}=\frac{5000}{20}=250\Omega$

Energy of current source will be converted into gravitational potential energy of the rod.

Average power in impedance

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

where X_C = capacitive reactance and X_L = inductive reactance.

Also X_C = $\frac{1}{\omega C}$ and X_L = $\omega L$

$\therefore$ Z = $\sqrt{{\left(50\right)}^2+{\left(\frac{1}{314\times 100\times {10}^{-6}}-314\times 20\times {10}^{-3}\right)}^2}$

$\Rightarrow$ Z = 56.15 $\Omega$

I_rms = $\frac{V_{rms}}{Z}$ = $\frac{V_m}{\sqrt{2}\times Z}=\frac{10}{\sqrt{2}\times 56}$

Hence power loss in the circuit

= ${\left(\frac{10}{\sqrt{2}\times 56}\right)}^2\times 50$ = 0.79 W

From question energy stored in inductor, U = 25 × 10^–3 J

Current, I = 60 mA

Also, energy stored in inductor, U = $\frac{1}{2}L{I}^2$

25 × 10^–3 = $\frac{1}{2}\times L\times {\left(60\times {10}^{-3}\right)}^2$

$\Rightarrow$ L = $\frac{25\times 2\times {10}^6\times {10}^{-3}}{3600}$ = 13.89 H

As we know,

$\overrightarrow{E}\times \overrightarrow{B}=\overrightarrow{V}$

$\Rightarrow$ $E\hat{j}\times \overrightarrow{B}=v\hat{i}$

[ As given Electric field vector is along +y axis and $\overrightarrow{v}=v\hat{i}$ ]

$\hat{i}=\hat{j}\times \hat{k}$ so $\overrightarrow{B}=B\hat{k}$.

$\therefore$ Direction of magnetic field vector is along +z direction.

Using mirror formula,

$\frac{1}{f}=\frac{1}{v_1}+\frac{1}{u_1}$

$\frac{1}{-15}=\frac{1}{v_1}-\frac{1}{40}$

$\Rightarrow$ v₁ = –24 cm

When object is displaced by 20 cm towards mirror.

Here u₂ = –20

So, $\frac{1}{f}=\frac{1}{v_2}+\frac{1}{u_2}$

$\frac{1}{-15}=\frac{1}{v_2}-\frac{1}{20}$

$\Rightarrow$ v₂ = -60 cm

So, the image will be shift away from mirror by (60 – 24) cm = 36 cm.

For telescope, angular magnification = $\frac{f_0}{f_e}$, which shows that focal length of objective lens to be large.

Angular resolution = $\frac{D}{1.22\lambda }$ which is large, hence objective should have large focal length and larger diameter.

For retracing the path, light ray should be normally incident on silvered face.



Applying Snell’s law at point M,

$\frac{\sin i}{\sin 30^{\circ }}=\frac{\sqrt{2}}{1}$

$\Rightarrow$ sin i = $\sqrt{2}\times \frac{1}{2}$ = $\frac{1}{\sqrt{2}}$

$\Rightarrow$ i = 45^o

Angular width = $\frac{\lambda }{d}$

$\therefore$ 0.20^o = $\frac{\lambda }{2}$

$\Rightarrow$ $\lambda$ = 0.2 $\times$ 2

Again, 0.21^o = $\frac{\lambda }{d}$

So using value of λ, we have

d = $\frac{0.20\times 2}{0.21}$ = 1.9 mm

If reflected and refracted light rays are perpendicular, reflected light gets polarised with electric field vector perpendicular to the plane of incidence.

Also, tan i = $\mu$ (i = Brewster's angle)

Given N₀ = 600, N₁ = 450, T = 10 min

Number of nuclei remaining, N = 600 – 450 = 150 after time ‘t’

$\frac{N}{N_0}=\frac{150}{600}=\frac{1}{4}$

According to the law of radioactive decay,

N = N₀e^-$\lambda$t

$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}$

$\therefore$ $\frac{150}{600}={\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}$

$\Rightarrow$ ${\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}$

$\Rightarrow$ t = 2T_1/2 = 2 $\times$ 10 minutes = 20 minutes

In a Bohr orbit of the hydrogen atom,

Kinetic energy = – (Total energy)

So, Kinetic energy : Total energy = 1 : –1

Given, Initial velocity v = v₀ $\hat{i}$

Electric field E = – E₀$\hat{i}$

$\therefore$Initial de-Brogile wavelength,

$\lambda$₀ = $\frac{h}{m{v}_0}$

Now force due to electric field on electrons,

$\overrightarrow{F}=\left(-e\right)\left(-E_0\hat{i}\right)=e{E}_0\hat{i}$

Acceleration produced in the electron,

$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{e{E}_0}{m}\hat{i}$

Now, velocity of electron after time t,

$\overrightarrow{v_t}=\overrightarrow{v}+\overrightarrow{a}t$ = $\left(v_0+\frac{e{E}_{0}t}{m}\right)\hat{i}$

$\Rightarrow$ $\left|\overrightarrow{v_t}\right|=v_0+\frac{e{E}_{0}t}{m}$

Now, $\lambda$_t = $\frac{h}{m{v}_t}$

= $\frac{h}{m\left(v_0+\frac{e{E}_{0}t}{m}\right)}$

= $\frac{h}{m{v}_0\left(1+\frac{e{E}_{0}t}{m{v}_0}\right)}$

= $\frac{{\lambda }_0}{\left(1+\frac{e{E}_{0}t}{m{v}_0}\right)}$ [As ${\lambda }_0=\frac{h}{m{v}_0}$]

From Einstein’s equation

$\frac{1}{2}m{v}_1^2=2h{\nu }_0-h{\nu }_0=h{\nu }_0$

Again, $\frac{1}{2}m{v}_2^2=5h{\nu }_0-h{\nu }_0=4h{\nu }_0$

$\therefore$ $\frac{v_1}{v_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

From question, V_BE = 0, V_i = 20 V

V_CE = 0

V_b = 0 (earthed)

We need to find values of base and collector currents I_B and I_C and amplification factor $\beta$, so

V_CC = I_CR_C + V_CE

I_C = V_CC – V_CE/R_C = 20 V – 0 V/4000 = 5 mA

Base current I_B = V_i /R_B = $\frac{20}{500\times {10}^3}$ = 40 μA

Amplification factor $\beta$ = $\frac{I_C}{I_B}$ = $\frac{25\times {10}^{-3}}{40\times {10}^{-6}}$ = 125

As a result of heating, temperature increases which generates large number of electron-hole pairs which lead to increase in conductivity. As
current increases I = I₀($e^{-\frac{qv}{KT}}$), overall resistance of diode changes which affects forward and reversed biasing.

	HCOOH	$$\frac{\text{buildrel}{H}_{2}S{O}_4}{⟶}$$	H2O	+	CO
Initial	$$\frac{2.3}{46}=\frac{1}{20}$$		0		0
Final	0		$$\frac{1}{20}$$		$$\frac{1}{20}$$

	H2C2O4	$$\frac{\text{buildrel}{H}_{2}S{O}_4}{⟶}$$	H2O	+	CO	+	CO2
Initial	$$\frac{4.5}{90}=\frac{1}{20}$$		0		0		0
Final	0		$$\frac{1}{20}$$		$$\frac{1}{20}$$		$$\frac{1}{20}$$

KOH will absorb CO₂ and conc. H₂SO₄ will absorb obtain water so in final solution only CO left.

Moles of CO formed from both reactions,
= $\frac{1}{20}$ + $\frac{1}{20}$ = $\frac{1}{10}$

$\therefore$ Mass of CO = moles × molar mass = $\frac{1}{10}\times 28$ = 2.8 g

(a) 18 mL water :
Since, density of H₂O = 1 g/mL
Mass of water = V × d = 18 × 1 = 18 g
No. of molecules of H₂O = $\frac{18}{18}\times$ N_A = 1 $\times$ N_A

(b) 0.18 g of water :
No. of molecules of H₂O = $\frac{0.18}{18}\times$ N_A = 0.01$\times$ N_A

(c) Volume of H₂O(g) STP = 0.00224 L
No. of molecules of H₂O = $\frac{V}{22.4}$ $\times$ N_A
= $\frac{0.00224}{22.4}$ $\times$ N_A = 0.0001 $\times$ N_A

(d) No of Molecules of water = mole × N_A = 10^–3 N_A

In degnerate orbital all unpaired electrons show same spin. So the correct configuration of N atom is

The correct balanced equation is
2MnO₄^– + 5C₂O₄ ^2– + 16H⁺ $\to$ 2Mn²⁺ + 10CO₂ + 8H₂O

In Ideal gases ‘a’ account for the intermolecular attractive forces between gas molecules.

van der Waals’ constant ‘a’ signifies the intermolecular forces of attraction between the particle of gas. So, higher the value of ‘a’, easier will be the liquefaction of gas.

On increasing the pressure and decreasing the temperature, equilibrium will shift in forward direction.

Given, Solubility of BaSO₄ = 2.42 × 10^–3 g L^–1

Convert solubility in mol/lit.

s = $\frac{2.42\times {10}^{-3}}{233}=1.04\times 10^{-5}$ mol L^-1

BaSO4(s)	⇌	Ba2+(aq)	+	SO42-(aq)
		s		s

K_sp = [Ba²⁺][SO₄^2-] = s²

= ${\left(1.04\times {10}^{-5}\right)}^2$

= 1.08 $\times$ 10^-10 mol² L^–2

(A) 60 mL $\frac{M}{10}$ HCl + 40 mL $\frac{M}{10}$ NaOH

Mili. moles of HCl = 60 $\times$ $\frac{1}{10}$ = 6

Mili. moles of NaOH = 40 $\times$ $\frac{1}{10}$ = 4

Mili. moles of HCl remaining = 6 - 4 = 2

Total volume will be 60 + 40 = 100 mL

Concentration of [H⁺] = $\frac{2}{100}$ = 2 $\times$ 10^-2

$\therefore$ pH = 2 - log2 = 1.7

(B) 55 mL $\frac{M}{10}$ HCl + 45 mL $\frac{M}{10}$ NaOH

Mili. moles of HCl = 55 $\times$ $\frac{1}{10}$ = 5.5

Mili. moles of NaOH = 45 $\times$ $\frac{1}{10}$ = 4.5

Mili. moles of HCl remaining = 5.5 - 4.5 = 1

Concentration of [H⁺] = $\frac{1}{100}$ = 10^-2

$\therefore$ pH = 2

(C) 75 mL $\frac{M}{5}$ HCl + 25 mL $\frac{M}{5}$ NaOH

Mili. moles of HCl = 75 $\times$ $\frac{1}{5}$ = 15

Mili. moles of NaOH = 25 $\times$ $\frac{1}{5}$ = 5

Mili. moles of HCl remaining = 15 - 5 = 10

Total volume will be 75 + 25 = 100 mL

Concentration of [H⁺] = $\frac{10}{100}$ = 10^-1

$\therefore$ pH = 1

(D) 100 mL $\frac{M}{10}$ HCl + 100 mL $\frac{M}{10}$ NaOH

Mili. moles of HCl = 100 $\times$ $\frac{1}{10}$ = 10

Mili. moles of NaOH = 100 $\times$ $\frac{1}{10}$ = 10

Mili. moles of HCl remaining = 10 - 10 = 0

So, it is neutral solution.

$\therefore$ pH = 7

Let bond dissociation energies of X₂ , Y₂ and XY are x kJ mol^–1 , 0.5x kJ mol^–1 and x kJ mol^–1 respectively.

$\frac{1}{2}$X₂ + $\frac{1}{2}$Y₂ $\to$ XY; $\Delta$H = -200 kJ mol^-1

$\Delta$H = Σ(B.E)_Reactant - Σ(B.E)_Product

$\Rightarrow$ -200 = $\left[\frac{1}{2}\times x+\frac{1}{2}\times 0.5x\right]$ - $\left[1\times x\right]$

On solving, x = 800 kJ mol^–1

Calculate E^o_cell corresponding to each compound undergoing disproportionation reaction. The reaction for which E^o_cell comes out + ve is spontaneous.

HBrO $\to$ Br₂, E^o_HBrO/Br2 = 1.595 V

HBrO $\to$ BrO₃^–, E^o_BrO3^-/HBrO = 1.595 V

E^o_cell for the disproportionation of HBrO,

E^o_cell = E^o_HBrO/Br2 - E^o_BrO3^-/HBrO

= 1.595 - 1.5

= 0.095 V = + ve

So, E^o_cell > 0

So, $\Delta$G^o < 0 (Spontaneous)

Half life of zero order
t_1/2 = $\frac{{\left[A\right]}_0}{2k}$

$\therefore$ t_1/2 $\infty$ [A]₀

$\therefore$ If [A]₀ = doubled then, t_1/2 = doubled.

For the first order reaction, t_1/2 = $\frac{0.693}{k}$
which is independent of initial concentration [A]₀.

For second order reaction, t_1/2 = $\frac{1}{k{\left[A\right]}_0}$
which depends on initial concentration [A]₀.

The density of a substance is given by the formula:

$$\rho =\frac{Z\times M_w}{N_A\times V}$$

Here, $Z$ is the number of atoms in one unit cell, $M_w$ is the molar mass, $N_A$ is Avogadro’s number, and $V$ is the volume of the unit cell.

1. For body-centered cubic (bcc) structure:

In bcc, $Z=2$ and the relation between atomic radius $r$ and lattice parameter $a$ is: $r=\frac{\sqrt{3}a}{4}$

Volume of one unit cell, $V=a^3$

2. For face-centered cubic (fcc) structure:

In fcc, $Z=4$ and the relation between atomic radius $r$ and lattice parameter $a$ is: $r=\frac{a}{2\sqrt{2}}$

Volume of one unit cell, $V=a^3$

3. Ratio of densities:

Since molar mass and atomic radius (hence atomic mass per atom) remain constant, the ratio of densities between bcc and fcc forms can be written as:

$$\frac{{\rho }_1}{{\rho }_2}=\frac{Z_1\times V_2}{Z_2\times V_1}$$

Substituting the values: $\frac{{\rho }_1}{{\rho }_2}=\frac{2\times \frac{\sqrt{3}\pi a^3}{16}}{4\times \frac{\pi a^3}{12\sqrt{2}}}$

Simplifying, $\frac{{\rho }_1}{{\rho }_2}=\frac{2\times \sqrt{3}\times \sqrt{2}\times 12}{4\times 16}$

Therefore, $\frac{{\rho }_1}{{\rho }_2}=\frac{3\sqrt{3}}{4\sqrt{2}}$

Hence, the ratio of densities of iron in bcc form (at room temperature) to fcc form (above 900°C) is $\frac{3\sqrt{3}}{4\sqrt{2}}.$

According to Hardy Schulze rule :
1. Coagulation of colloidal solution by using an electrolyte depends on the charge present (positive or negative) on colloidal particles as well as on its size.
2. Coagulating power of an electrolyte depends on the magnitude of charge present on effective ion of electrolyte.

Ga is slightly smaller than Al due poor shielding of d e^– so Z_eff increasing.

So, Atomic size : B < Ga < Al < In < Tl

In metals, on moving down the group, metallic character increases, so basic nature increases hence most acidic will be BeO. In fact, BeO is amphoteric oxide while other given oxides are basic oxides.

Hence, Cl has 2 lone pairs of electrons.

Molecular orbital configuration of NO (15 electrons) is

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^1}^{\ast }={\pi }_{2p_y^o}^{\ast }$

$\therefore$ N_b = 10

N_a = 5

$\therefore$ BO = $\frac{1}{2}\left[10-5\right]$ = 2.5

Moleculer orbital configuration of CN^– (14 electrons) is

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\sigma }_{2p_z^2}$

$\therefore$ N_b = 10

N_a = 4

$\therefore$ BO = $\frac{1}{2}\left[10-4\right]$ = 3

Moleculer orbital configuration of CN (13 electrons) is

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\sigma }_{2p_z^1}$

$\therefore$ N_b = 9

N_a = 4

$\therefore$ BO = $\frac{1}{2}\left[9-4\right]$ = 2.5

Moleculer orbital configuration of CN⁺ (12 electrons) is

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}$

$\therefore$ N_b = 8

N_a = 4

$\therefore$ BO = $\frac{1}{2}\left[8-4\right]$ = 2

Hence, CN^– has highest bond order.

Mg has more – $\Delta$G value than alumina. So it will be in the lower part of Ellingham diagram. Metals which have more – $\Delta$G value can reduce those metal oxides which have less – $\Delta$G value.

Smaller the size of cation, more will be its polarising power. Hence BeH2 will be least ionic.

BeH₂ < CaH₂ < BaH₂

Electronic configuration of element (X) is
1s²2s²2p³

So, valency of X will be 3.
Since, valency of Mg is 2.
So, formula of compound formed by Mg and X will be Mg₃X₂.

All halogens show both positive and negative oxidation states while fluorine shows only negative oxidation state except due to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluoric (I) acid. Oxidation number of F is +1 in HOF.

Boron does not have vacant d-orbitals in its valence shell, so it cannot extend its covalency beyond 4. i.e., ‘B’ cannot form the ions like MF₆^3–.

In CrO₄^2-       Cr⁺⁶ (n = 0)       diamagnetic

In Cr₂O₇^2–       Cr⁺⁶ (n = 0)       diamagnetic

In MnO₄^-       Mn⁺⁷ (n = 0)       diamagnetic

In MnO₄^2-       Mn⁺⁶ (n = 1)       paramagnetic

In MnO₄^2-, one unpaired electron(n) is present in d-orbital so, d-d transition is possible.

Co³⁺ = [Ar]3d⁶ , unpaired e^–(n) = 4

Spin magnetic moment ($\mu$) = $\sqrt{4\left(4+2\right)}=\sqrt{24}$ B.M

Cr³⁺ = [Ar]3d³ , unpaired e^–(n) = 3

Spin magnetic moment ($\mu$) = $\sqrt{3\left(3+2\right)}=\sqrt{15}$ B.M.

Fe³⁺ = [Ar]3d⁵ , unpaired e^–(n) = 5

Spin magnetic moment ($\mu$) = $\sqrt{5\left(5+2\right)}=\sqrt{35}$ B.M.

Ni²⁺ = [Ar]3d⁸ , unpaired e^–(n) = 2

Spin magnetic moment ($\mu$) = $\sqrt{2\left(2+2\right)}$ = $\sqrt{8}$ B.M.