NEET-UG 2010

$\frac{1}{2}{\epsilon }_0{E}^2$ = Energy density of an electric field E

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

x = $\frac{1}{t+5}$

$\therefore$ v = $\frac{dx}{dt}$ = $\frac{-1}{{\left(t+5\right)}^2}$ .......(i)

$\therefore$ $a$ = $\frac{dv}{dt}$ = $\frac{2}{{\left(t+5\right)}^3}$ = 2x³ .......(ii)

$\therefore$ $a$ $\propto$ (distance)³

From equation (i), we get

$v^{3/2}=\frac{-1}{{\left(t+5\right)}^3}$

Putting this in equation (ii), we get

$a$ = -2$v^{3/2}$

$\therefore$ $a$ $\propto$ (velocity)^3/2

From the question, we can say

distance moved by 1^st ball in 18 s = distance moved by 2^nd ball in 12 s.

So, distance moved by 1^st ball in 18 s

= $\frac{1}{2}\times 10\times 18^2$ = 1620 m

and distance moved by 2^nd ball in 12 s

= $vt+\frac{1}{2}g{t}^2$

$\therefore$ 1620 = $v\left(12\right)+\frac{1}{2}\times 10{\left(12\right)}^2$

$\Rightarrow$ $v$ = 75 m/s

$\overrightarrow{u}=3\hat{i}+4\hat{j}$,  $\overrightarrow{a}=0.4\hat{i}+0.3\hat{j}$

$\Rightarrow u_x=3units,u_y=4units$

$a_x=0.4units,a_y=0.3units$

$\therefore$ v_x = u_x + a_x $\times$ 10 = 3 + 4 = 7 units

and v_y = 4 + 0.3 $\times$ 10 = 4 + 3 = 7 units

$\therefore v=\sqrt{v_x^2+v_y^2}=7\sqrt{2}$ units

Using the law of vector addition, ($\overrightarrow{d}+\overrightarrow{e}$) is as shown in the fig.


$\therefore \overrightarrow{d}+\overrightarrow{e}=\overrightarrow{f}$

Forces acting on the block are as shown in the fig. Normal reaction N is provided by the force ma due to acceleration $\alpha$


$\therefore$ N = m$\alpha$

For the block not to fall, frictional force, $F_f\ge mg$

$\Rightarrow \mu N\ge mg$

$\Rightarrow \mu m\alpha \ge mg$

$\Rightarrow \alpha \ge g/\mu$

Here,

Mass per unit length of water, $\mu$ = 100 kg/m

Velocity of water, v = 2 m/s

Power of the engine, P = $\mu$v³ = (100 kg/m) (2 m/s)³

= 800 W

Here, m₁ = m, m₂ = 2m, u₁ = 2 m/s, u₂ = 0

Coefficient of restitution, e = 0.5
Let v₁ and v₂ be their respective velocities after collision.
Applying the law of conservation of linear momentum, we get

m₁u₁ + m₂u₂ = m₁ v₁ + m₂ v₂

$\therefore$ m × 2 + 2m × 0 = m × v₁ + 2m × v₂

or 2m = mv₁ + 2mv₂ or 2 = (v₁ + 2v₂) ...(i)

By definition of coefficient of restitution,

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

or e(u₁ – u₂) = v₂ – v₁ $\Rightarrow$ 0.5(2 – 0) = v₂ – v₁ ...(ii)

Solving equations (i) and (ii), we get
v₁ = 0 m/s, v₂ = 1 m/s

If no external force actson a system of particles, the centre of mass remains at rest. So, speed of centre of mass is zero.

The coin will revolve with the record, if Force of friction $\ge$ Centrifugal force

$\mu mg\ge mr{\omega }^2$

$\Rightarrow r\le \frac{\mu g}{{\omega }^2}$

As no external torque is applied to the system, the angular momentum of the system remains conserved.

$\therefore$ L_i = L_f

According to given problem,

$I_t{\omega }_i=\left(I_t+I_b\right){\omega }_f$

$\Rightarrow {\omega }_f=\frac{I_t{\omega }_i}{\left(I_t+I_b\right)}$    ...(i)

Initial energy, $E_i=\frac{1}{2}{I}_t{\omega }_i^2$   ...(ii)

Final energy, $E_f=\frac{1}{2}\left(I_t+I_b\right){\omega }_f^2$   ...(iii)

Substituting the value of ${\omega }_f$ from equation (i) in equation (iii), we get

Final energy, $E_f=\frac{1}{2}\left(I_t+I_b\right){\left(\frac{I_t{\omega }_i}{\left(I_t+I_b\right)}\right)}^2$

$=\frac{1}{2}\frac{I_t^2{\omega }_i^2}{I_t+I_b}$   ...(iv)

Loss of energy, $\Delta$E = E_i – E_f

$=\frac{1}{2}{I}_t{\omega }_i^2-\frac{1}{2}\frac{I_t^2{\omega }_i^2}{\left(I_t+I_b\right)}$     (Using (ii) and (iv))

$=\frac{{\omega }_i^2}{2}\left(I_t-\frac{I_t^2}{\left(I_t+I_b\right)}\right)=\frac{{\omega }_i^2}{2}\left(\frac{I_t^2-I_b{I}_t-I_t^2}{\left(I_t+I_b\right)}\right)$

$=\frac{1}{2}\frac{I_b{I}_t}{\left(I_t+I_b\right)}{\omega }_i^2$

Since the man is in gravity free space, force on man + stone system is zero.

Therefore centre of mass of the system remains at rest. Let the man goes x m above when the stone reaches the floor, then

M_man × x = M_stone × 10

$x=\frac{0.5}{50}\times 10$

x = 0.1 m

Therefore final height of man above floor = 10 + x = 10 + 0.1 = 10.1 m

Here, Mass of a particle = M
Mass of a spherical shell = M
Radius of a spherical shell = a
Let O be centre of a spherical shell.
Gravitational potential at point P due to particle at O is

$V_1=-\frac{GM}{a/2}$

Gravitational potential at point P due to spherical shell is

$V_2=-\frac{GM}{a}$

Hence, total gravitational potential at point P is V = V₁ + V₂

$=-\frac{GM}{a/2}+\left(-\frac{GM}{a}\right)=-\frac{2GM}{a}-\frac{GM}{a}=-\frac{3GM}{a}$

$\left|V\right|=\frac{3GM}{a}$

Orbital velocity of a satellite in a circular orbit of radius a is given by

$v=\sqrt{\frac{GM}{a}}$

$\Rightarrow v\propto \sqrt{\frac{1}{a}}$

$\Rightarrow \frac{v_2}{v_1}=\sqrt{\frac{a_1}{a_2}}$

$\therefore$ $v_2=v_1\sqrt{\frac{4R}{R}}$ = 2v₁ = 6V

The amount of heat flows in time t through a cylindrical metallic rod of length L and uniform area of cross-section $A\left(=\pi R^2\right)$ with its ends maintained at temperatures T₁ and T₂ (T₁ > T₂) is given by

$Q=\frac{KA\left(T_1-T_2\right)t}{L}$   ...(i)

where K is the thermal conductivity of the material of the rod.
Area of cross-section of new rod

$A^{\prime }=\pi {\left(\frac{R}{2}\right)}^2=\frac{\pi R^2}{4}=\frac{A}{4}$   ...(ii)

As the volume of the rod remains unchanged

$\therefore$ AL = A'L'

where L' is the length the new rod

$\Rightarrow L^{\prime }=L\frac{A}{A^{\prime }}$   ...(iii)

= 4L     (Using (ii))

Now, the amount of heat flows in same time t in the new rod with its ends maintained at the same temperatures T₁ and T₂ is given by

$Q^{\prime }=\frac{K{A}^{\prime }\left(T_1-T_2\right)t}{L^{\prime }}$   ...(iv)

Substituting the values of A' and L' from equations (ii) and (iii) in the above equation, we get

$Q^{\prime }=\frac{K(A/4)\left(T_1-T_2\right)t}{4L}=\frac{1}{16}\frac{KA\left(T_1-T_2\right)t}{L}=\frac{1}{16}Q$
   (Using (i))

According to the Stefan Boltzmann law, the power radiated by the star whose outer surface radiates as a black body at temperature T K is given by

$P=\sigma 4\pi r^2{T}^4$


where, r = radius of the star
$\sigma$ = Stefan’s constant

The radiant power per unit area received at a distance R from the centre of a star is

$S=\frac{P}{4\pi R^2}=\frac{\sigma 4\pi r^2{T}^4}{4\pi R^2}=\frac{\sigma r^2{T}^4}{R^2}$

Power radiated by the sun at t°C

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

Power received by a unit surface

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

By first law of thermodynamics,

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

In adiabatic process, $\Delta Q=0$

$\therefore$ $\Delta U=-\Delta W$

In isothermal process, $\Delta U=0$

$\therefore$ $\Delta Q=\Delta W$

$T=2\pi \sqrt{\frac{m}{K}}$

$\therefore \frac{T_1}{T_2}=\sqrt{\frac{M_1}{M_2}}$

$\therefore$ $T_2=T_1\sqrt{\frac{M_2}{M_1}}=T_1\sqrt{\frac{2M}{M}}$

$T_2=T_1\sqrt{2}=\sqrt{2}T$     (where T₁ =T)

$x=a{\sin }^2\omega t=a\left(\frac{1-\cos 2\omega t}{2}\right)$

   $\left(\because \cos 2\theta =1-2{\sin }^2\theta \right)$

$=\frac{a}{2}-\frac{a\cos 2\omega t}{2}$

$\therefore$ Velocity, $v=\frac{dx}{dt}=\frac{2\omega a\sin 2\omega t}{2}=\omega \alpha \sin 2\omega t$

Acceleration, $a=\frac{dv}{dt}=2{\omega }^{2}a\cos 2\omega t$

For the given displacement x = $\alpha \sin 2\omega t$,

$a\propto -x$ is not satisfied.

Hence, the motion of the particle is non simple harmonic motion.

Note : The given motion is a periodic motion with a time period

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

y = A sin ($\omega$t–kx)

Particle velocity,

$v_p=\frac{dy}{dt}=A\omega \cos \left(\omega t-kx\right)$

$\therefore v_{p\max }=A\omega$

wave velocity = $\frac{\omega }{k}$

$\therefore$ $A\omega =\frac{\omega }{k}$

i.e, $A=\frac{1}{k}$ But $k=\frac{2\pi }{\lambda }$

$\therefore \lambda =2\pi A$

Let the frequencies of tuning fork and piano string be ${\upsilon }_1$ and ${\upsilon }_2$ respectively.

$\therefore$ ${\upsilon }_2={\upsilon }_1\pm 4=512Hz\pm 4$

= 516 Hz or 508 Hz



Increase in the tension of a piano string increases its frequency.

If ${\upsilon }_2$ = 516 Hz, further increase in ${\upsilon }_2$, resulted in an increase in the beat frequency. But this is not given in the question.

If ${\upsilon }_2$ = 508 Hz, further increase in ${\upsilon }_2$ resulted in decrease in the beat frequency. This is given in the question. When the beat frequency decreases to 2 beats per second. Therefore, the frequency of the piano string before increasing the tension was 508 Hz.

According to Coulomb’s law, the force of repulsion between the two positive ions each of charge q, separated by a distance d is given by

$F=\frac{1}{4\pi {\epsilon }_0}\frac{\left(q\right)\left(q\right)}{d^2}$

$F=\frac{q^2}{4\pi {\epsilon }_0{d}^2}$

$q^2=4\pi {\epsilon }_{0}F{d}^2$

$q=\sqrt{4\pi {\epsilon }_{0}F{d}^2}$   ...(i)

Since, q = ne
where, n = number of electrons missing from each ion
e = magnitude of charge on electron

$\therefore n=\frac{q}{e}$

$n=\frac{\sqrt{4\pi {\epsilon }_{0}F{d}^2}}{e}$   (Using (i))

= $\sqrt{\frac{4\pi {\epsilon }_{0}F{d}^2}{e^2}}$

Electric flux, $\varphi$ = EA cos $\theta$ , where $\theta$ = angle between E and normal to the surface.

Here, $\theta =\frac{\pi }{2}$

$\Rightarrow \varphi$ = 0

(i) When key between the terminals 1 and 2 is plugged in,

P.D. across R = IR = $k{l}_1$

$\Rightarrow$ R =$k{l}_1$ as I = 1A

(ii) When key between terminals 1 and 3 is plugged in,

P.D. across (X + R) = I(X + R) = $k{l}_2$

$\Rightarrow$ X + R = $k{l}_2$

$\therefore$ X = $k(l_2–l_1)$

$\therefore$ R = $k{l}_1$ and X = $k(l_2–l_1)$

Kirchhoff’s junction law or Kirchhoff’s first law is based on the conservation of charge.

Kirchhoff’s loop law or Kirchhoff’s second law is based on the conservation of energy.

Hence both statements (A) and (B) are correct.

A series combination of n1 capacitors each of capacitance C₁ are connected to 4V source.

Total capacitance of the series combination of the capacitors is

1/C_s = 1/C₁ + 1/C₁ + 1/C₁ .... upto n₁ terms = n₁/C₁

$\Rightarrow$ C_s = C₁/n₁

Total energy stored in a series combination of the capacitors is

U_s = (1/2) C_s (4V)² = (1/2) (C₁/n₁) (4V)²    …(1)

Now a parallel combination of n₂ capacitors each of capacitance C₂ are connected to V source

Total capacitance of the parallel combination of capacitors is

C_p = C₁ + C₂ + ... + upto n₂ terms = n₂C₂

$\Rightarrow$ C_p = n₂C₂

Total energy stored in a parallel combination of capacitors is

U_p = (1/2)C_pV² = (1/2)(n₂C₂ )(V)²    ...(ii)

According to the given problem, U_s = U_p

(1/2) (C₁/n₁)(4V)² = (1/2)(n²C²)(V)²

C₁ × 16/n₁ = n₂C₂

C₂ = 16 × C₁ / (n₁ × n₂)

Let the resistance to be added be R, then

30 = I_g (r + R)

$\therefore$ R = $\frac{30}{I_g}-r$

= $\frac{30}{30\times {10}^{-3}}-100$

= 1000 - 100 = 900 $\Omega$

When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by $\tau =NBiA\sin \theta$. Torque is maximum when $\theta$ = 90^o,i.e., the plane of the coil is parallel to the field ${\tau }_{\max }=NBiA$

Forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ acting on the coil are equal in magnitude and opposite in direction. As the forces ${\overrightarrow{F}}_2$ and ${\overrightarrow{F}}_2$ have the same line of action their resultant effect on the coil is zero. The two forces ${\overrightarrow{F}}_3$ and ${\overrightarrow{F}}_4$ are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque. Thus, if the force on one arc of the loop is $\overrightarrow{F}$ , the net force on the remaining three arms of the loop is - F.

Here,
Resistance of galvanometer, G = 100 $\Omega$

Current for full scale deflection, I_g = 30 mA = 30 × 10^–3 A

Range of voltmeter, V = 30 V

To convert the galvanometer into an voltmeter of a given range, a resistance R is connected in series with it as shown in the figure.


From figure, V = I_g(G + R)

$\Rightarrow R=\frac{V}{I_g}-G=\frac{30}{30\times {10}^{-3}}-100\Omega$

= 1000 – 100 = 900 $\Omega$

When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is

I = $\frac{q}{T}$ = qf

Magnetic field at the centre of the ring is

B = $\frac{{\mu }_{0}I}{2R}$ = $\frac{{\mu }_{0}qf}{2R}$

Electromagnet are made of soft iron because  soft iron has low retentivity and low coercive force  or low coercivity.

The time period T of oscillation of a magnet is given by

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

As I, B remains the same here

So T $\propto$ $\frac{1}{\sqrt{B}}$

or $\frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}}$

$\Rightarrow$ T₂ = $T_1\sqrt{\frac{B_1}{B_2}}$

= $2\sqrt{\frac{24\times {10}^{-6}}{6\times {10}^{-6}}}$ = 4 s

Constant rate at which radius of the loop shrinks

$\frac{dr}{dt}=1\times 10^{-3}$ m s^-1

Magnetic flux linked with the loop is

$\varphi$ = BAcos$\varphi$ = B($\pi$r²)cos0^o = B$\pi$r²

The magnitude of the induced emf is

|e| = $\frac{d\varphi }{dt}=\frac{d}{dt}\left(B\pi r^2\right)=B\pi 2r\frac{dr}{dt}$

= 0.025 $\times$ $\pi$ $\times$ 2 $\times$ 2 $\times$ 10^-2 $\times$ 1 $\times$ 10^-3

= $\pi$ $\times$ 10^-6 V

= $\pi \mu V$

Efficiency of the transformer,
$\eta$ =

Output power

Input power

$\Rightarrow$ $\frac{80}{100}=\frac{V_s{I}_s}{V_p{I}_p}$

$\Rightarrow$ I_p = $\frac{440\times 2}{\frac{80}{100}\times 220}$ = 5 A

As V_L = V_C = 300 V, resonance will take place

$\therefore$ V_R = 220 V

Current, I = $\frac{220}{100}$ = 2.2 A

$\therefore$ Hence, the reading of the voltmeter V₃ is 220 V and the reading of ammeter A is 2.2 A.

In an electromagnetic wave it is observed that electric and magnetic vectors are perpendicular to each other and are perpendicular to the direction of propagation of wave.

For total internal reflection, sini $>$ sinC

where, i = angle of incidence, C = critical angle

But, sinC = $\frac{1}{\mu }$

$\therefore$ sini $>$ $\frac{1}{\mu }$

$\Rightarrow$ $\mu$ $>$ $\frac{1}{\sin i}$

$\therefore$ $\mu$ $>$ $\frac{1}{\sin 45^{\circ }}$

$\Rightarrow$ $\mu$ $>$ $\sqrt{2}$

By covering aperture, focal length does not change. But intensity is reduced by $\frac{1}{4}$ times, as aperture diameter $\frac{d}{2}$ is covered.

$\therefore$ I' = I - $\frac{I}{4}$ = $\frac{3I}{4}$

$\therefore$ New focal length = f

and intensity = $\frac{3I}{4}$

Now N = N₀e^{–$\lambda$t}

$\Rightarrow$ $\frac{N_0}{e}$ = N₀e^–5t

$\Rightarrow$ $\lambda$ = $\frac{1}{5}$ per minute

Further, $\frac{N_0}{2}$ = N₀e^–λ(t)

t = $\frac{1}{\lambda }\ln 2$ = 5log_e 2

Energy of a H-like atom in it's n^th state is given by,

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

For, first excited state of He⁺, n = 2, Z = 2

$\therefore$ E_He⁺ = $-\frac{13.6}{2^2}\times 2^2$ = -13.6 eV

For ${}_3^{7}Li$ nucleus,

Mass defect, $\Delta$M = 0.042 u

$\therefore$ 1 u = 931.5 MeV/c²

$\therefore$ $\Delta$M = 0.042 × 931.5 MeV/c² = 39.1 MeV/c²

Number of nucleons in ${}_3^{7}Li$ is 7.

Binding energy per nucleon,

E_bn = $\frac{39.1}{7}$ = 5.6 MeV

Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\times \frac{\left(Ze\right)\times \left(2e\right)}{d_{in}}$

$\therefore$ d_min $\propto$ $\frac{1}{m}$

According to Einstein’s photoelectric equation

eV_s = h$\nu$ - $\varphi$

$\Rightarrow$ eV_s = $\frac{hc}{\lambda }$ - $\varphi$

= $\frac{1240}{200}$ - 5.01

= 6.2 eV – 5.01 eV = 1.2 eV

Stopping potential, V_s = 1.2 V

The potential difference that must be applied to stop photoelectrons = – V_s = – 1.2 V

Power of S₂

Power of S₁

= $\frac{n_2\left[\frac{hc}{{\lambda }_2}\right]}{n_1\left[\frac{hc}{{\lambda }_1}\right]}$ = $\frac{n_2{\lambda }_1}{n_1{\lambda }_2}$

= $\frac{1.02\times {10}^{15}\times 5000}{{10}^{15}\times 5100}$ = 1

Force on electron due to = Force on electron due magnetic field to electric field

Bev = eE

$\Rightarrow$ v = $\frac{E}{B}$ .......(1)

If V is the potential difference between the anode and the cathode, then

$\frac{1}{2}m{v}^2$ = eV

$\Rightarrow$ $\frac{e}{m}=\frac{v^2}{2V}$ .....(2)

Substituting the value of v from equation (1) in equation (2), we get

$\frac{e}{m}=\frac{E^2}{2V{B}^2}$

The Boolen expression for the given combination is

output Y = (A + B) · C

So it is clear that Y = 1, when A = 1, B = 0 and C = 1

Voltage gain= $\beta$ × Impedance gain

$\Rightarrow$50 = β × (200/100)

$\Rightarrow$ $\beta$ = 25

Power gain = $\beta$ × Voltage gain

= 25 × 50 = 1250

In case of metal, conductivity decreases with increase in temperature and metal has high melting point.

Integrated circuit can act as a complete electronic circuit.

In a n-type semiconductors, electrons are majority carriers and holes are minority carriers.

No. of atoms

= N_A $\times$ No. of moles $\times$ 3

= 6.023 $\times$ 10²³ $\times$ 0.1 $\times$ 3

= 1.806 $\times$ 10²³

Given in the question that molar mass of Na₂CO₃ = 106 g

$\therefore$ Molarity of solution = $\frac{25.3\times 1000}{106\times 250}$

= 0.9547 M = 0.955 M

Na₂CO₃ $\to$ 2Na⁺ + $C{O}_3^{2-}$

[Na⁺] = 2[Na₂CO₃] = 2 $\times$ 0.955 = 1.910 M

[$C{O}_3^{2-}$] = [Na₂CO₃] = 0.955 M

As we know, K_p = K_c × (RT)^$\Delta$n_g

So, for reaction having same number of gaseous moles on reactants and products side will have same value of Kc and Kp otherwise their values are different.

For reaction, 2NO_(g) $\rightleftharpoons$ N_2(g) + O_2(g)

$\Delta$n_g = 2 – 2 = 0

$\Rightarrow$ K_p = K_c × (RT)⁰

$\therefore$ K_p = K_c

For reaction, SO_2(g) + NO_2(g) $\rightleftharpoons$ SO_3(g) + NO_(g)

$\Delta$n_g = 2 – 2 = 0

$\Rightarrow$ K_p = K_c × (RT)⁰

$\therefore$ K_p = K_c

For reaction, H_2(g) + I_2(g) $\rightleftharpoons$ 2HI_(g)

$\Delta$n_g = 2 – 2 = 0

$\Rightarrow$ K_p = K_c × (RT)⁰

$\therefore$ K_p = K_c

For reaction, H_2(g) + I_2(g) $\rightleftharpoons$ 2HI_(g)

$\Delta$n_g = 2 – 2 = 0

$\Rightarrow$ K_p = K_c × (RT)⁰

$\therefore$ K_p = K_c

For reaction, 2C_(s) + O_2(g) $\rightleftharpoons$ 2CO_2(g)

$\Delta$n_g = 2 – 3 = -1

$\Rightarrow$ K_p = K_c × (RT)^-1

$\therefore$ K_p $\ne$ K_c

	Ba(OH)2	⇌	Ba2+	+	2OH
At equilibrium			x		2x

pH = – log[H⁺]

12 = – log [H⁺]

$\Rightarrow$ [H⁺] = 10^–12

As, [H⁺][OH^– ] = 10^–14

10^–12 [OH^– ] = 10^–14

$\Rightarrow$ [OH^– ] = 10^–2

As [OH^– ] = 2x = 10^–2 then x = 5.0 × 10^–3

Now, K_sp = [Ba²⁺][OH^– ]²

K_sp = (5 × 10^–3) (10^–2)² = 5.0 × 10^–7

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

$\Rightarrow$ pOH = - log K_b + log $\frac{\left[Salt\right]}{\left[Acid\right]}$

$\Rightarrow$ pOH = –log10^–10 + log 1

[As conc. of HB and B^– are same]

$\Rightarrow$ pOH = 10

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

CH₃COOH and CH₃COONa constitute to form an acidic buffer.

$\Rightarrow$ pH = pKa + log$\frac{\left[Salt\right]}{\left[Acid\right]}$

pH = –log(1.8 × 10^–5) + log $\frac{\left(0.20\right)}{\left(0.10\right)}$

= 4.74 + log 2

= 4.74 + 0.3010 = 5.041

Now, pH = – log[H⁺]

$\Rightarrow$ 5.041 = – log[H⁺]

$\Rightarrow$ [H⁺] = 10^–5.041 = 9.0 × 10⁶ mol L^–1

Depression in freezing point,

$\Delta$T_f = K_f $\times$ m

m = $\frac{w_B}{M_B}\times \frac{1000}{W_A}$

= $\frac{68.5\times 1000}{342\times 1000}$

$\Delta$T_f = 1.86 $\times$ $\frac{68.5}{342}$

= 0.372 ^oC

$\therefore$ T_f = 0 - 0.372 ^oC = - 0.372 ^oC

Addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles resulting in decreased surface area. However, addition of solvent, i.e., dilution increases the surface area of the liquid surface, thus results in increased vapour pressure. Hence, addition of water to the aqueous solution of (1 molal) KI results in increased vapour pressure.

Here,

E_a = activation energy of forward reaction

E’_a = activation energy of backward reaction

$\Delta$H = enthalpy of the reaction
From the given diagram it is clear that

E_a = E’_a + $\Delta$H

$\therefore$ E_a > $\Delta$H

Given reaction is :

$\frac{1}{2}$X₂ + $\frac{3}{2}$Y₂ ⇌ XY₃

We know,

$\Delta$S^o = $\sum S_{products}^o-\sum S_{reac\tan ts}^o$

= 50 - (30 + 60) = -40 J K^-1 mol^-1

At equilibrium $\Delta$G^o = 0

$\Delta$H^o = T$\Delta$S^o

$\therefore$ $T=\frac{\Delta H^o}{\Delta S^o}$ = $\frac{-30\times {10}^3}{-40}$ = 750 K

The cell reaction

Cu + 2Ag⁺ $\to$ Cu²⁺ + 2Ag

We know, $\Delta$G° = – nFE°_cell

= – 2 × 96500 × 0.46 = – 88780 J

= – 88.780 kJ = – 89 kJ

Dilution of strong electrolytes increases ionisation, hence ionic mobility of ions increases which in turn increases equivalent conductance of the solution.

Rate = k[A]^x [B]^y

For the given situations

(I) rate = k(0.1)^x (0.1)^y = 6.0$\times$10^$-$3

(II) rate = k(0.2)^x (0.3)^y = 7.2$\times$10^$-$2

(III) rate = k(0.3)^x (0.4)^y = 2.88$\times$10^$-$1

(IV) rate = k(0.4)^x (0.1)^y = 2.40$\times$10^$-$2

Dividing eq. (I) by eq. (IV) we get

${\left(\frac{0.1}{0.4}\right)}^x{\left(\frac{0.1}{0.1}\right)}^y=\frac{6.0\times {10}^{-3}}{2.4\times {10}^{-2}}$

$\Rightarrow$ ${\left(\frac{1}{4}\right)}^x={\left(\frac{1}{4}\right)}^1$

$\Rightarrow$ x = 1

On dividing eq. (II) by eq. (III) we get

${\left(\frac{0.3}{0.3}\right)}^x{\left(\frac{0.2}{0.4}\right)}^y=\frac{7.2\times {10}^{-2}}{2.88\times {10}^{-1}}$

$\Rightarrow$ ${\left(\frac{1}{2}\right)}^y=\frac{1}{4}$

$\Rightarrow$ y = 2

$\therefore$ Rate = k[A][B]²