WBJEE 2019 Paper 2 (Physics & Chemistry) Question Paper with Solutions

Given,

Maximum error in the measurement of mass = 0.3% Maximum error in the measurement of length = 0.2%

We known that,

Error in density is given as,

Density, $\rho = {{mass(m)} \over {volume(v)}} = {m \over {{L^3}}}$

where, L = side of cube

Error in density is given as,

$\left( {{{\Delta \rho } \over \rho }} \right) = {{\Delta m} \over m} + {{3\Delta L} \over L}$

or $\left( {{{\Delta \rho } \over \rho }} \right) \times 100 = \left( {{{\Delta m} \over m} + {{3\Delta L} \over L}} \right) \times 100$

Substituting the given values, we get

${\left( {{{\Delta \rho } \over \rho }} \right)_{\max }} = (0.3\% + 3(0.2)\% ) = 0.3\% + 0.6\%$

$\therefore$ Maximum percentage error measurement of density ${\left( {{{\Delta \rho } \over \rho }} \right)_{\max }} = 0.9\%$

According to the Kepler's third law

T² $\propto$ r³

where, T = time period of revolution

r = radius

Now, ${\left( {{{{T_E}} \over {{T_P}}}} \right)^2} = {\left( {{{{r_E}} \over {{r_P}}}} \right)^3} = {{{r_E}} \over {{r_P}}} = {\left( {{{{T_E}} \over {{T_P}}}} \right)^{2/3}}$

${{{r_E}} \over {{r_P}}} = {\left( {{{2\pi /{\omega _E}} \over {2\pi /{\omega _P}}}} \right)^{2/3}}$ $\because$ $\left[ {T = {{2\pi } \over \omega }} \right]$

${{{r_E}} \over {{r_P}}} = {\left( {{{{\omega _P}} \over {{\omega _E}}}} \right)^{2/3}}$

According to the question, ${\omega _P} = 2{\omega _E}$ and ${r_E} = R$

$\Rightarrow {R \over {{r_P}}} = {\left( {{{2{\omega _E}} \over {{\omega _E}}}} \right)^{2/3}}$

${R \over {{r_P}}} = {(2)^{2/3}}$

${r_P} = {R \over {{{(2)}^{2/3}}}} = R{(2)^{ - 2/3}}$

According to the Bohr's atomic model,

Angular momentum, $L = mvr = {{nh} \over {2\pi }}$ .... (i)

Since, angular velocity $\omega = {v \over r}$

$\Rightarrow v = r\omega$

From Eq. (i), we get

$m(r\omega )r = {{nh} \over {2\pi }}$

$m\omega {r^2} = {{nh} \over {2\pi }}$

$\omega = {{nh} \over {2\pi m{r^2}}}$ .... (ii)

Since, the radius of the electron in n^th orbit of Bohr's atomic model is given as, $r = {{{n^2}{h^2}{\varepsilon _o}} \over {\pi mz{e^2}}}$ .... (iii)

Squaring the Eq. (iii) and substituting its value in Eq. (ii), we get,

$\omega = {{nh{{(\pi mze)}^2}} \over {2\pi m({n^4}{h^4}\varepsilon _0^2)}} \Rightarrow \omega \propto {1 \over {{n^3}}}$

According to the question,

Area of both bodies A and B = A

Temperature of body A = 27$^\circ$C = 27 + 273 K

Temperature of body B = 177$^\circ$C = 177 + 273 K

Now, by Stefan-Boltzmann law, thermal energy radiated per second by a body

Q = $\sigma$AT⁴,

where, A = Area

T = Temperature

and $\sigma$ = Stefan-Boltzmann's constant

So, the ratio of thermal energy radiated per second by A that by B is

${{{Q_1}} \over {{Q_2}}} = {{\sigma A} \over {\sigma A}}{\left( {{{{T_1}} \over {{T_2}}}} \right)^4}$

Now, ${{{Q_1}} \over {{Q_2}}} = {\left( {{{{T_1}} \over {{T_2}}}} \right)^4}$

${{{Q_1}} \over {{Q_2}}} = {\left( {{{273 + 27} \over {273 + 177}}} \right)^4}$

${{{Q_1}} \over {{Q_2}}} = {\left( {{{300} \over {450}}} \right)^4}$

${{{Q_1}} \over {{Q_2}}} = {\left( {{2 \over 3}} \right)^4} = \left( {{{16} \over {81}}} \right)$

Ratio of thermal energy radiated per second

${Q_1}:{Q_2} = 16:81$

According to the question, we can draw the following diagram


Hence, T is the tension in the string.

In equilibrium condition,

For mass m₁, T $-$ m₁g = m₁a ... (i)

For mass m₂, m₂g $-$ T = m₂a ....(ii)

After adding Eqs. (i) and (ii), we get

\eqalign{ & {m_2}g - {m_1}g = {m_1}a + {m_2}a \cr \cr \Rightarrow& ({m_2} - {m_1})g = ({m_1} + {m_2})a \cr \cr \Rightarrow& a = {{({m_2} - {m_1})} \over {({m_1} + {m_2})}}g \cr}

The clock diagram is as shown below


In the above diagram, charge +Q is not placed at 10 h position.

So, net electric field strength at centre C,

${E_{net}} = {E_1} + {E_2} + {E_3} + {E_4} + {E_5} + {E_6} + {E_7} + {E_8} + {E_9} + {E_{11}} + {E_{12}}$

$\Rightarrow {E_{net}} = {E_1} + {E_2} + {E_3} + {E_4} + {E_5} + {E_6} + ( - {E_1} - {E_2} - {E_3} - {E_5} - {E_6})$

$\Rightarrow {E_{net}} = {E_4} = {Q \over {4\pi {\varepsilon _0}{r^2}}}$ from centre towards the mark 10

Given, Power (P) = constant

Kinetic Energy (KE) = ${1 \over 2}m{v^2}$

We know that, $P = {{KE} \over {\Delta t}} \Rightarrow P = {{m{v^2}} \over 2}$

$\because$ P = constant,

Hence, velocity of the body $v \propto \sqrt t$ .... (i)

As, velocity $v = {{ds} \over {dt}}$ ...(ii)

From Eqs. (i) and (ii), we get

So, ${{ds} \over {dt}} \propto \sqrt t$

integrating the above equation w.r.t. time (t),

$\int {{{ds} \over {dt}}} \propto \int {\sqrt t }$

we get, displacement of the body $s \propto {t^{3/2}}$

$\because$ Displacement $s = a{t^{3/2}}$, where a is constant.

As we know that de-Broglie wavelength is given as $\lambda = {h \over p} = {h \over {mv}}$ ....(i)

where, h = Plank constant

p = momentum of particle

v = velocity of particle

and m = mass of the particle.

Eq. (i) can be written as,

$\lambda = {h \over {2m(KE)}} = {h \over {\sqrt {2mqv} }}$ [$\because$ KE = qv]

where, KE = Kinetic energy of particle

Hence, $\lambda \propto {1 \over {\sqrt m }}$

Now, ${{{\lambda _{proton}}} \over {{\lambda _{electron}}}} = \sqrt {{{{m_{electron}}} \over {{m_{proton}}}}}$

Given, mass of proton, ${m_{proton}} = 2000{m_{electron}}$

${{{\lambda _{proton}}} \over {{\lambda _{electron}}}} = \sqrt {{1 \over {2000}}}$

${{{\lambda _{proton}}} \over {{\lambda _{electron}}}} = {1 \over {20\sqrt 5 }} \Rightarrow {\lambda _p} = {{{\lambda _e}} \over {20\sqrt 5 }}$

Given, electric current in circular wire = I

Radius of wire = r

Charge on particle = q

Speed of the particle = v

The given loop can be as shown in the figure below.


$\because$ Net magnetic field at the centre = 0

So, magnetic force F_magnetic = q(v $\times$ B) = 0

So, the magnetic force experienced by the particle when it passes through the centre is F = 0

According to the question,


We know that incident ray, reflected ray and normal lie in the same plane.

and angle of incidence = angle of reflection

Therefore, $\widehat n$ will be along the angle bisector of $\widehat e$ and $-$$\widehat e$₀,

i.e. $\widehat n = {{\widehat e + ( - {{\widehat e}_0})} \over {|\widehat e - {{\widehat e}_0}|}}$ ... (i)

[$\because$ Bisector will along a vector dividing in same ratio as the ratio of sides forming that angle]

But $\widehat n$ is a unit vector where $|\widehat e - {\widehat e_0}| = OC$

$= 2OP = 2|\widehat e|$

$\cos \theta = 2\cos \theta$

Substituting this value in Eq. (i) , we get

$\widehat n = {{\widehat e - {{\widehat e}_0}} \over {2\cos \theta }}$

$\widehat e = {\widehat e_0} + (2\cos \theta )\widehat n$

$\widehat e = {\widehat e_0} - 2(\widehat n\,.\,{\widehat e_0})\widehat n$ [$\because$ $\widehat n\,.\,{\widehat e_0} = - \cos \theta$]

We know that, energy stored in the capacitor

= ${1 \over 2}C{E^2}$

and energy supplied by the source of emf = CE²

$\therefore$ Energy dissipated in resistance R

= Energy supplied by the source of emf E $-$ Energy stored in the capacitor

= $C{E^2} - {1 \over 2}C{E^2}$

$= \left( {1 - {1 \over 2}} \right)C{E^2} = {1 \over 2}C{E^2}$

If velocity of particle, v is perpendicular to magnetic field, B i.e. $\theta$ = 90$^\circ$, then particle will experience maximum magnetic force, i.e. F_max = qvB. This force acts in a direction perpendicular to the motion of charged particle. Therefore the trajectory of the particle is a circle. In this case path of charged particle is circular and magnetic force provides the necessary centripetal force,

i.e. $qvB = {{m{v^2}} \over r}$

$\Rightarrow$ Radius of path, $r = {{mv} \over {qB}}$

$r = {p \over {qB}}$ [$\because$ p = mv]

where, p = momentum of the particle

$\therefore$ r $\propto$ momentum

Hence, option (d) is correct.

The circuit diagram is as shown below,


According to question, when given pair of point in options are connected through a resistance R, then equivalent resistance between point A and B (R_AB) remains unchanged. It is only possible when current does not flows through resistance R and circuit become Wheatstone bridge as shown in the figure below,


At the balanced condition, ${P \over Q} = {R \over S}$

The current flow in CD branch will be zero. Now by checking each option,

from option (a), circuit is,


Now, $\because$ ${P \over Q} = {R \over S} \Rightarrow {{2r} \over {4r}} = {r \over {2r}} \Rightarrow {1 \over 2} = {1 \over 2}$

Hence, option (a) is correct.

From option (b), circuit is


Now, ${P \over Q} = {R \over S} \Rightarrow {{3r} \over {3r}} \ne {r \over {2r}}$

So, option (b) is also incorrect.

From option (c) circuit is,


Now, ${P \over Q} = {R \over S} \Rightarrow {{4r} \over {2r}} = {{2r} \over r} \Rightarrow {2 \over 1} = {2 \over 1}$

So, option (c) is also correct.

From option (d), circuit is


Now, ${P \over Q} = {R \over S} \Rightarrow {{4r} \over {2r}} \ne {r \over {2r}}$

So, option (d) is also incorrect.

Hence, option (a) and (c) are correct.

According to the question, we can draw the following diagram


Given, u = $-$0.2 m and f = 0.05 m

As we know that,

${1 \over f} = {1 \over v} - {1 \over u}$ ...(i)

${1 \over v} = {1 \over f} + {1 \over u}$

${1 \over v} = {1 \over {0.05}} - {1 \over {0.2}}$

${1 \over v} = {{100} \over 5} - {{10} \over 2}$

${1 \over v} = 20 - 5 \Rightarrow v = {1 \over {15}}$ m

Now, differentiating eq. (i), we get

$- {{dv} \over {{v^2}}} = - {{du} \over {{u^2}}}dv = du{{{v^2}} \over {{u^2}}}$

${A_{\max }} = A \times {\left( {{1 \over {15}}} \right)^2} \times {\left( {{1 \over {( - 0.2)}}} \right)^2}$

${A_{\max }} = A \times {1 \over {225}} \times 25{A_{\max }} = {A \over 9}$

Here A is in cm.

Hence, ${A_{\max }} = {A \over 9} \times {10^{ - 2}}$ m

According to the question,


Total resistance of the given circuit

${R_{eq}} = 2R$

Now, circuit


$\therefore$ $I = {{V/2R} \over 8}$ ($\because$ ${I_1} = V/2R$)

So, current I in the circuit $I = {V \over {16R}}$ $\because$ $\left( {I = {{{I_1}} \over 8}} \right)$

Given, decrement in the length = 1%

Poisson's ratio for material of the rod, $\sigma$ = 0.2

As we know that,

Volume = v = $\pi$r²l, [where, l is the length of the rod]

${{\Delta V} \over V} = {{2\Delta r} \over r} + {{\Delta l} \over l}$ ...(i)

Since, Poisson's ratio, $\sigma = {{ - \Delta D/D} \over {\Delta L/L}}$

$= {{ - \Delta r/r} \over {\Delta L/L}}$

So, Eq. (i) can be written as,

${{\Delta V} \over V} = {{\Delta l} \over l}(1 - 2\sigma )$

${{\Delta V} \over V} \times 100 = \left( {{{\Delta l} \over l} \times 100} \right) \times (1 - 2\sigma )$

$= - 1 \times [1 - 2 \times (0.2)]$

$= - 1 \times [1 - 0.4]$

$= - 0.6\%$

Here, negative sign shows the decrement in the volume.

In given, Series R - L - C circuit

Resistance R = 100$\Omega$

Inductance of Inductor, L = 20 mH = 20 $\times$ 10^$-$3 H

Resonance frequency f = ${{1250} \over \pi }$ Hz

Source voltage V_DC = 25V

According to resonant frequency,

${\omega _0} = 2\pi {f_0} = {1 \over {\sqrt {LC} }}$

or ${(2\pi {f_0})^2} = {1 \over {LC}}$

or $4{\pi ^2}{{1250 \times 1250} \over {\pi \times \pi }} = {1 \over {LC}}$

or $C = {{1000} \over {1250 \times 1250 \times 4 \times 20}}$ ($\because$ By substituting L = 20 $\times$ 10^$-$3)

or $C = 8 \times {10^{ - 6}}$F

We know that, Charge ${Q_s} = CV$

${Q_s} = 8 \times {10^{ - 6}} \times 25 = 0.2$ mC

So, the amount of charge stored in each plate of capacitor is 0.2 mC.

According to the question,


Input voltage V_s = 10V

Source resistance R_s = 1k$\Omega$

Zener diode voltage V_z = 6V

$\because$ Breakdown voltage of zener diode is = 6V, and the potential difference across the zener diode 5V.

$\therefore$ Current flow in zener diode I_z = 0

As we know that,

Impulse, I = force (F) $\times$ small time interval

I = ma $\times$ t [$\because$ F = ma]

I = [M] [LT^$-$2] $\times$ [T]

I = [MLT^$-$1]

Hence, the correct dimensional formula for impulse is given by [MLT^$-$1].

The equation of state for an ideal gas undergoing adiabatic process is given as,

TV^$\gamma$$-$1 = constant

Let the temperature after adiabatic compression as given in the question be T then,

${T_0}{V^{\gamma - 1}} = T{\left( {{{{V_0}} \over 2}} \right)^{\gamma - 1}}$

$\Rightarrow T = {T_0}{2^{\gamma - 1}}$

Now, heat released at volume ${{{V_0}} \over 2}$ to achieve temperature ${T_0}$.

The net heat released can be determined by the equation,

$\Delta Q = \mu {C_V}\Delta T$

$= \mu \times {R \over {\gamma - 1}}$ $({T_0}{2^{\gamma - 1}} - {T_0})$

$= {{\mu R{T_0}} \over {\gamma - 1}}({2^{\gamma - 1}} = 1)$ $\because$ $(\mu {T_0}R = {p_0}{V_0})$

$\therefore$ Heat released = ${{{p_0}V} \over {\gamma - 1}}({2^{\gamma - 1}} - 1)$

According to the question,

X\buildrel {\alpha - particle} \over \longrightarrow Y\buildrel {\beta - particle} \over \longrightarrow Z(say)

According to Rutherford Soddy law of radioactive decay, the rate of decay of radioactive along at any instant is proportional to the number of atoms present at that instant

${{dN} \over {dt}} = - \lambda N$

Since, $\lambda$ = constant (decay constant) and N = number of atoms

Rate of decay of Y particle is given as,

${\lambda _X}{N_X} - {\lambda _Y}{N_Y} = {{d{N_Y}} \over {dt}} = 0$

[$\because$ Decay rate for $\beta$-particle become constant after some time]

Given, rate of emission of $\beta$-particle = ${10^7}/h$

$\therefore$ ${\lambda _X}{N_X} - {\lambda _Y}{N_Y} = {10^7}/h$

According to the question,

Charge on particle is = q

Velocity of particle = v

Due to uniform electric field,

electric force on particle

F_electric = qE ....(i)

Due to a uniform magnetic field,

,magnetic force on particle is given by,

F_magnetic = q(v $\times$ B) ....(ii)

When v is perpendicular to E and B, which are mutually perpendicular to each other,

F_electric = F_magnetic

From Eqs. (i) and (ii)

qE = qvB

$\therefore$ $v = {E \over B}$

So, $v = {E \over B}$ is the condition for which the particle moves undeflected in its original trajectory.

For a gas at temperature T, the internal energy,

$U = {f \over 2}\mu RT$

where, f = degree of freedom

$\Rightarrow$ Change in energy, $\Delta U = {f \over 2}\mu R\Delta T$ .... (i)

Also, as we know for any gas heat supplied at constant volume

$(\Delta {Q_v}) = \mu {C_v}\Delta T = \Delta U$ .... (ii)

where, C_v = molar specific heat at constant volume

From Eqs. (i) and (ii), we get

${C_v} = {1 \over 2}fR$

For diatomic gas, degree of freedom, f = 5

${C_v} = {5 \over 2}R$

Given, inductance of inductor, L = 60 mH

= 60 $\times$ 10^$-$3 H

Phase difference between voltage and current in L-R circuit, $\theta$₁ = 60$^\circ$

Capacitance of capacitor, C = 0.5 $\mu$F = 0.5 $\times$ 10^$-$6F

Phase difference between voltage and current in R-C circuit, $\theta$₂ = 30$^\circ$

For L-R circuit,

$\tan {\theta _1} = {{{X_L}} \over R}$

$\tan {\theta _1} = {{\omega L} \over R}$ .... (i) [$\because$ ${X_L} = \omega L$]

Similarly, for R-C circuit,

$\tan {\theta _2} = {{{X_C}} \over R}$

$\tan {\theta _2} = {{{1 \over {\omega C}}} \over R}$ ...(ii) $\because$ $\left[ {{X_C} = {1 \over {\omega C}}} \right]$

From Eqs. (i) and (ii), we get

${{\tan {\theta _1}} \over {\tan {\theta _2}}} = {{\omega L} \over {R \times {1 \over {\omega CR}}}} = {\omega ^2}LC$

${{\tan 60^\circ } \over {\tan 30^\circ }} = {\omega ^2}LC$

${{\sqrt 3 } \over {1/\sqrt 3 }} = {\omega ^2}LC$

${\omega ^2}LC = 3$

${\omega ^2} = {3 \over {LC}}$${\omega ^2} = {3 \over {60 \times {{10}^{ - 3}} \times 0.5 \times {{10}^{ - 6}}}}$

${\omega ^2} = {3 \over {30 \times {{10}^{ - 9}}}}$

$\omega = \sqrt {{{10}^8}}$

$\omega = {10^4}$

As we know that,

$\omega = 2\pi f$

$f = {\omega \over {2\pi }} = {{{{10}^4}} \over {2\pi }}Hz = {1 \over {2\pi }} \times {10^4}Hz$

Given, initial velocity of particle, u = 10 ms^$-$1

Range, R = 5 m

Gravitational acceleration, g = 10 ms^$-$2

As we known that,

Range of projectile, $R = {{{u^2}\sin 2\theta } \over g}$

$5 = {{{{(10)}^2}\sin 2\alpha } \over {10}}$

$\sin 2\alpha = {5 \over {10}}$

$\sin 2\alpha = {1 \over 2}$

$\alpha = 15^\circ$

or $(90^\circ - 15^\circ = 75^\circ )$

Given, cross sectional area of nozzle $A = {5 \over {\sqrt {21} }} \times {10^{ - 3}}{m^2}$

and rate of heat transfer $Q = {1 \over {10}} = {10^{ - 1}}{m^3}/s$

$\because$ Heat transfer Q = AV

$\therefore$ $V = {Q \over A} = {{{{10}^{ - 1}}{m^3}/s} \over {{5 \over {\sqrt {21} }} \times {{10}^{ - 3}}{m^2}}} = 2\sqrt {21} \times 10m/s$

When in hits a rigid wall then maximum possible increase in temperature of water can be expressed as

${1 \over 2}m{v^2} = ms\Delta T$ ...(i)

where, m = mass, s = specific heat of water = 1 cal / g = 4.2 $\times$ 10³ J / kg and $\Delta T$ = increase in temperature

From Eq. (i), $\Delta T = {1 \over 2}{{{v^2}} \over s}$

$= {{{{(2\sqrt {21} \times {{10}^1})}^2}} \over {2 \times 4.2 \times {{10}^3}}}$

$= {{84 \times {{10}^2}} \over {8.4 \times {{10}^3}}}$

= $1^\circ$C

Given, initial frequency of light,

${f_1} = 4 \times {10^{14}}{s^{ - 1}}$

Final frequency of light, ${f_2} = 5 \times {10^{14}}{s^{ - 1}}$

Change in wavelength, $\Delta \lambda = {\lambda _1} - {\lambda _2}$

or $\Delta \lambda = {c \over {{f_1}}} - {c \over {{f_2}}}$

$= {{3 \times {{10}^8}} \over {4 \times {{10}^{14}}}} - {{3 \times {{10}^8}} \over {5 \times {{10}^{14}}}}$

$= {{3 \times {{10}^8}} \over {{{10}^{14}}}}\left( {{1 \over 4} - {1 \over 5}} \right)$

$= {{3 \times {{10}^8}} \over {{{10}^{14}}}} \times {1 \over {20}}$

$= 1.5 \times {10^{ - 7}}$ m

Now, we know that,

Angular width of central maxima,

$\theta = {{2\lambda } \over d}$ or $\Delta \theta = {{2\Delta \lambda } \over d}$

$d = {{2\Delta \lambda } \over {\Delta \theta }}$

Here, $\Delta \theta$ = 0.6 radian

$d = {{2 \times 1.5 \times {{10}^{ - 7}}} \over {0.6}} = 5 \times {10^{ - 7}}$ m

According to the question,


In two dimension, the position vectors r₁ and r₂ represented as

${r_1} = vt\widehat i - {1 \over 2}g{t^2}\widehat j$ ...(i)

${r_2} = vt( - \widehat i) - {1 \over 2}g{t^2}(\widehat j)$ ... (ii)

$\because$ we know that, when the two vectors are mutually perpendicular, i.e.

$\theta = 90^\circ$

So, ${r_1}\,.\,{r_2} = {r_1}{r_2}\cos 90^\circ$

${r_1}\,.\,{r_2} = 0$

Substituting the values r₁ and r₂ in the above relation, we get

$((vt)\widehat i - {1 \over 2}g{t^2}\widehat j)\,.\,(vt( - \widehat i) - {1 \over 2}g{t^2}(\widehat j)) = 0$

$- {v^2}{t^2} + {1 \over 4}4{g^2}{t^4} = 0$ (where, $\widehat i\,.\,\widehat i = \widehat j\,.\,\widehat j = \widehat k\,.\,\widehat k = 1$)

${v^2}{t^2} = {1 \over 4}{g^2}{t^4}$

$\Rightarrow {v^2} = {1 \over 4}{g^2}{t^2}$

$\therefore$ Magnitude of velocity of the particles, $v = {1 \over 2}gt$

We know that, separation distance between particles at a time t

$\Delta x = 2vt$

$\Delta x = 2 \times v \times {{2v} \over g} \Rightarrow \Delta x = {{4{v^2}} \over g}$

According to the balance condition of Wheatstone bridge,

In the first case,

${P \over Q} = {R \over S}$

${P \over Q} = {5 \over S}$ ....(i)

In the second case

${P \over Q} = {7 \over {S + 3}}$ ... (ii)

From Eqs. (i) and (ii), we get

${5 \over S} = {7 \over {S + 3}}$

$5(S + 3) = 7S$

$5S + 15 = 7S$

$2S = 15$

$S = 7.5\Omega$

The given circuit shows a balanced Wheatstone bridge. Now, the circuit becomes


In the branch AB, both capacitor are arranged in the series combination. Hence, its equivalent capacitance is given by

${1 \over {{C_{AB}}}} = {1 \over {20}} + {1 \over {20}} = {2 \over {20}} = {1 \over {10}}$

${C_{AB}} = 10\mu F$

Similarly, in branch CD,

${C_{CD}} = 10\mu F$

Now, C_AB and C_CD are connected in parallel combination. Hence, the equivalent capacitance of the circuit,

${C_{eq}} = 10 + 10 = 20\mu F$

We know that, charge $Q = {C_{eq}}V$

$Q = 20 \times {10^{ - 6}} \times 150V$

Amount of charge stored $Q = 3 \times {10^{ - 3}}C$