MHT CET 2023 11th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

The direction of angular moment remains constant as the angular momentum is a vector quantity, and its direction is perpendicular to the plane of motion. As the speed decreases, the linear momentum decreases, but the angular momentum remains constant due to the conservation of angular momentum.

Given: $N=100, I=1 \mathrm{~A}, \phi=2.5 \times 10^{-5} \mathrm{~Wb} /$ turn

Self-inductance of coil,

$$\begin{aligned} \mathrm{L} & =\frac{\mathrm{N} \phi}{\mathrm{I}} \\ \therefore \quad \mathrm{L} & =\frac{100 \times 2.5 \times 10^{-5}}{1}=2.5 \times 10^{-3} \mathrm{H}=2.5 \mathrm{~mH} \end{aligned}$$

Separation between consecutive nodes, $\frac{\lambda}{2}=5 \mathrm{~cm}$

$\therefore \quad \lambda=10 \mathrm{~cm}=0.1 \mathrm{~m}$

The frequency of vibration is given as:

$\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{2}{0.1}=20 \mathrm{~Hz}$

From lens formula,

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{~F}}$

For a convex lens, $\mathrm{u}$ is negative, $\mathrm{v}$ and $\mathrm{F}$ are positive.

$$\begin{aligned} & \therefore \quad \frac{1}{\mathrm{v}}-\frac{1}{-\mathrm{u}}=\frac{1}{\mathrm{~F}} \\ & \therefore \quad \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{~F}}-\frac{1}{\mathrm{u}} \end{aligned}$$

Multiplying by $\mathrm{v}$,

$$\begin{aligned} 1 & =\frac{\mathrm{v}}{\mathrm{F}}-\frac{\mathrm{v}}{\mathrm{u}} \\ \therefore \quad 1 & =\frac{\mathrm{v}}{\mathrm{F}}-\mathrm{m} \quad ...\left(\because \frac{v}{u}=m\right)\\ \therefore \quad \frac{\mathrm{v}}{\mathrm{F}} & =1+\mathrm{m} \\ \therefore \quad \mathrm{v} & =\mathrm{F}(1+\mathrm{m}) \end{aligned}$$

Let the radius of the disc be $R$

$$\begin{aligned} & \therefore \quad \mathrm{K}_{\mathrm{c}}=\frac{\mathrm{R}}{\sqrt{2}} \\ & \therefore \quad\mathrm{K}_{\mathrm{d}}=\frac{\mathrm{R}}{2} \end{aligned}$$

Taking the ratio,

$\therefore \quad \frac{\mathrm{K}_{\mathrm{c}}}{\mathrm{K}_{\mathrm{d}}}=\frac{\frac{\mathrm{R}}{\sqrt{2}}}{\frac{\mathrm{R}}{2}}=\frac{\sqrt{2}}{1}$

The tension in the string can be resolved in two components along the perpendicular axis. The gravitational force is acting downwards and the centrifugal force is acting in $-x$ direction $T \sin \theta=m r \omega^2$

$\therefore \quad \omega^2=\frac{T \sin \theta}{\mathrm{mr}}$

$\therefore \quad \omega=\sqrt{\frac{T \sin \theta}{\mathrm{mr}}}$

From figure, $\sin \theta=\frac{r}{l}$

$$\begin{aligned} & \therefore \quad \omega=\sqrt{\frac{\mathrm{Tr}}{\mathrm{mr} l}} \\ & \therefore \quad \omega=\sqrt{\frac{\mathrm{T}}{\mathrm{m} l}} \end{aligned}$$

Using dimensional analysis, dimensions of only option (D) matches with the dimensions of angular velocity.

Kinetic energy of the photoelectrons

$\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi$

$\therefore \quad$ For wavelength $\lambda_1$,

$\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_1}-\phi$ ..... (i)

$\therefore \quad$ For wavelength $\lambda_1$,

$\mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\phi$ ..... (ii)

Subtracting equation (i) from equation (ii),

$$\begin{aligned} & \mathrm{K}_2-\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_2}-\phi-\frac{\mathrm{hc}}{\lambda_1}+\phi \\ & \mathrm{K}_2-\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_2}-\frac{\mathrm{hc}}{3 \lambda_2} \\ & \mathrm{~K}_2-\mathrm{K}_1=\frac{2}{3} \frac{\mathrm{hc}}{\lambda_2} \\ & \frac{\mathrm{hc}}{\lambda_2}=\frac{3}{2}\left(\mathrm{~K}_2-\mathrm{K}_1\right) \quad \text{..... (iii)} \end{aligned}$$

Substituting equation (iii) in equation (ii),

$\mathrm{K}_2=\frac{3}{2}\left(\mathrm{~K}_2-\mathrm{K}_1\right)-\phi$

$$\begin{aligned} & 2 \mathrm{~K}_2=3 \mathrm{~K}_2-3 \mathrm{~K}_1-2 \phi \\ & \mathrm{K}_2-3 \mathrm{~K}_1=2 \phi \\ & \mathrm{As}, \phi>0 \\ & \mathrm{~K}_2-3 \mathrm{~K}_1>0 \\ & \mathrm{~K}_1<\frac{\mathrm{K}_2}{3} \end{aligned}$$

Potential energy: $U=\frac{1}{2} m \omega^2 x^2$ and

Kinetic energy: $K=\frac{1}{2} m \omega^2\left(A^2-x^2\right)$

Given: Potential energy $=8 \times$ Kinetic energy

$$\begin{aligned} & \therefore \quad \mathrm{U}=8 \mathrm{~K} \\ & \frac{1}{2} m \omega^2 x^2=8 \times \frac{1}{2} m \omega^2\left(A^2-x^2\right) \\ & x^2=8 A^2-8 x^2 \\ & 9 x^2=8 A^2 \\ & x^2=\frac{8 A^2}{9} \\ & x=\frac{2 \sqrt{2} A}{3} \\ \end{aligned}$$

The ratio of force when the medium is changed will be:

$\frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{1}{\mathrm{k}}$

Here, $\mathrm{k}=5$,

Therefore, the force experience by each will be $\frac{F}{5}$.

With increase in frequency current increases initially, becomes maximum and then decreases with further increase in frequency.

Option (B) is photodiode and option (D) is LED with reverse and forward biasing respectively. Therefore both will operate properly.

Work done, $\mathrm{W}=\mathrm{S} \times \Delta \mathrm{A}$

Where, $\Delta \mathrm{A}$ is a change in surface area.

Since, the radius of the big drop be $\mathrm{R}$

$\mathrm{A}_{\text {initial }}=4 \pi \mathrm{R}^2$

Let the radius of the small drops be $r$

$\mathrm{A}_{\text {final }}=8 \times 4 \pi \mathrm{r}^2$

The volume is constant

$$\begin{array}{ll} \therefore & \frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3 \\ \therefore & R^3=8 r^3 \\ \therefore & R=2 r \end{array}$$

Substituting the values in the formula.

$$\begin{aligned} & \mathrm{W}=\mathrm{S} \times \Delta \mathrm{A} \\ & \mathrm{W}=\mathrm{S}\left(\mathrm{A}_{\text {final }}-\mathrm{A}_{\text {initial }}\right) \end{aligned}$$

$$\begin{aligned} & \mathrm{W}=\mathrm{S}\left(8 \times 4 \pi \mathrm{r}^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=\mathrm{S}\left(32 \pi\left(\frac{\mathrm{R}}{2}\right)^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=\mathrm{S}\left(8 \pi \mathrm{R}^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=4 \pi \mathrm{R}^2 \mathrm{~S} \end{aligned}$$

Power Radiated by black body, $\mathrm{P}=\sigma \mathrm{AT}^4$

$\therefore \quad$ For first black body:

$\mathrm{P}_1=\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4$

$\therefore \quad$ For second black body:

$\mathrm{P}_2=\sigma 4 \pi \mathrm{r}_2{ }^2 \mathrm{~T}_2{ }^4$

$\therefore \quad$ The ratio will be:

$$\begin{aligned} & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4}{\sigma 4 \pi \mathrm{r}_2^2 \mathrm{~T}_2^4} \\ & \frac{1}{2}=\frac{\mathrm{r}_1^2 \mathrm{~T}_1^4}{\mathrm{r}_2^2 \mathrm{~T}_2^4} \\ & \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{1}{2} \frac{\mathrm{T}_2^4}{\mathrm{~T}_1^4} \\ & \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2 \end{aligned}$$

Potential energy of '$n$' point charges,

\mathrm{U}=\frac{1}{4 \pi \varepsilon_0} \sum_\limits{\text {all pairs }} \frac{\mathrm{q}_{\mathrm{j}} \mathrm{q}_{\mathrm{k}}}{\mathrm{r}_{\mathrm{jk}}}

For 3 point charges,

$$\begin{aligned} \therefore \quad U & =-\frac{q(2 q)}{4 \pi \varepsilon_0 r}-\frac{q(2 q)}{4 \pi \varepsilon_0 r}+\frac{q(q)}{4 \pi \varepsilon_0(2 r)} \\ U & =-\frac{2 q^2}{4 \pi \varepsilon_0 r}-\frac{2 q^2}{4 \pi \varepsilon_0 r}+\frac{q^2}{4 \pi \varepsilon_0(2 r)} \\ U & =-\frac{4 q^2}{8 \pi \varepsilon_0 r}-\frac{4 q^2}{8 \pi \varepsilon_0 r}+\frac{q^2}{8 \pi \varepsilon_0 r} \\ U & =-\frac{7 q^2}{8 \pi \varepsilon_0 r} \end{aligned}$$

Given: rate of flow of heat (conduction rate) $\mathrm{P}_{\text {cond }}=1600 \mathrm{~cal} / \mathrm{s}$

Thermal resistance,

$$\begin{aligned} & \mathrm{R}_{\mathrm{T}}=\frac{\Delta \mathrm{T}}{\mathrm{P}_{\text {cond }}} \\ & \mathrm{R}_{\mathrm{T}}=\frac{40}{1600} \\ & \mathrm{R}_{\mathrm{T}}=0.025^{\circ} \mathrm{Cs} / \mathrm{cal} \end{aligned}$$

Magnetic field, $B=\frac{\mu_0 I}{2 R}$

Flux passing through the coil,

$$\begin{aligned} \phi & =\mathrm{B} \times \pi \mathrm{r}^2 \\ \therefore \quad \phi & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \times \pi \mathrm{r}^2 \end{aligned}$$

Mutual Inductance, $M=\frac{\phi}{I}$

$\therefore \quad M=\frac{\frac{\mu_0 I}{2 R} \times \pi r^2}{I}=\frac{\mu_0 \pi r^2}{2 R}$

Given: Density ratio: $\frac{\rho_1}{\rho_2}=\frac{4}{3}$

and surface tension ratio: $\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{6}{5}$

$\therefore \quad$ Rise of liquid in a capillary:

$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$

For constant $\theta, \mathrm{r}$ and $\mathrm{g}$,

$$\begin{aligned} \mathrm{h} & \propto \frac{\mathrm{T}}{\rho} \\ \therefore \quad & \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{\mathrm{T}_1 \rho_2}{\mathrm{~T}_2 \rho_1}=\frac{6 \times 3}{5 \times 4}=0.9 \end{aligned}$$

When an electron is accelerated through a voltage its kinetic energy is converted into electric potential energy:

$$\begin{aligned} & \mathrm{K}=\mathrm{U} \\ & \frac{1}{2} \mathrm{mv}^2=\mathrm{eV} \\ \therefore \quad \mathrm{V} & =\frac{\mathrm{mv}^2}{2 \mathrm{e}} \\ & \mathrm{V}=\frac{\left(9 \times 10^{-31}\right)\left(1.6 \times 10^7\right)^2}{2 \times 1.6 \times 10^{-19}}=7.2 \times 10^2 \mathrm{~V} \end{aligned}$$

Emf induced due to mutual inductance per coil,

$\mathrm{e}=\frac{\mathrm{M} \Delta \mathrm{I}}{\Delta \mathrm{t}}$

Here, the current changes from I to zero in time $\mathrm{t}$ seconds.

$\therefore \quad e=\frac{M I}{t}$

The magnetic force experienced by the electron, $\mathrm{F}=\mathrm{qvB} \sin \theta$

As electron is projected along the axis, $\theta=0$

$\therefore \quad \mathrm{F}=0$

For potentiometer,

$$\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1+l_2}{l_1-l_2} \\ \therefore \quad & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{64+32}{64-32} \quad=\frac{96}{32}=\frac{3}{1}=3: 1 \end{aligned}$$

Time period of a simple pendulum is $\mathrm{T}=2 \pi \sqrt{\left(\frac{l}{\mathrm{a}}\right)}$

In stationary lift, the value of acceleration is $\mathrm{a}=\mathrm{g}$

$\mathrm{T}=2 \pi \sqrt{\left(\frac{l}{\mathrm{~g}}\right)}$

When the lift is accelerating in an upward direction, there is a pseudo force acting in a downward direction.

$$\begin{aligned} & \therefore \quad \mathrm{ma}=\mathrm{mg}+\frac{\mathrm{mg}}{3} \\ & \therefore \quad \mathrm{a}=\mathrm{g}+\frac{\mathrm{g}}{3} \\ & \therefore \quad \mathrm{a}=\frac{4 \mathrm{~g}}{3} \end{aligned}$$

$\therefore \quad$ Period for a pendulum in accelerating lift is

$$\begin{aligned} \quad \mathrm{T}^{\prime} & =2 \pi \sqrt{\frac{l}{\mathrm{a}}}=2 \pi \sqrt{\frac{3 l}{4 \mathrm{~g}}} \\ \therefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2}\left(2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\right) \\ \therefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2} \mathrm{~T} \end{aligned}$$

After isothermal expansion:

$$\begin{aligned} & P_1 V_1=P_2 V_2 \\ & P_2=P_1 \frac{V_1}{V_2} \\ & P_2=P_1 \frac{V}{2 V} \\ & P_2=\frac{P}{2} \end{aligned}$$

After adiabatic expansion:

$$\begin{aligned} & \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma \\ & \mathrm{P}_3=\mathrm{P}_2\left(\frac{\mathrm{V}_2}{\mathrm{~V}_3}\right)^\gamma \\ & \mathrm{P}_3=\frac{\mathrm{P}}{2}\left(\frac{2 \mathrm{~V}}{16 \mathrm{~V}}\right)^{5 / 3} \\ & \mathrm{P}_3=\frac{\mathrm{P}}{2}\left(\frac{1}{8}\right)^{5 / 3} \\ & \mathrm{P}_3=\frac{\mathrm{P}}{64} \end{aligned}$$

When the velocity of the bullet changes from V to $\frac{\mathrm{V}}{2}$ the distance travelled by the bullet is $30 \mathrm{~cm}$.

Using $3^{\text {rd }}$ equation of motion,

$$\begin{aligned} & \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\ & \left(\frac{\mathrm{V}}{2}\right)^2=\mathrm{V}^2+2 \mathrm{a}(30) \\ & \frac{\mathrm{V}^2}{4}=\mathrm{V}^2+60 \mathrm{a} \\ & \frac{-3 \mathrm{~V}^2}{4}=60 \mathrm{a} \\ & \mathrm{a}=\frac{-\mathrm{V}^2}{80} \end{aligned}$$

Further, when a bullet penetrates it comes to rest. So, the final velocity of the bullet becomes zero.

Using the relation,

$$\begin{aligned} & \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\ & 0=\left(\frac{\mathrm{V}}{2}\right)^2+2\left(-\frac{\mathrm{V}^2}{80}\right) \mathrm{s} \\ & \frac{\mathrm{V}^2}{4}=\left(\frac{\mathrm{V}^2}{40}\right) \mathrm{s} \\ & \mathrm{s}=\frac{40}{4} \\ & \mathrm{~s}=10 \mathrm{~cm} \end{aligned}$$

Initial energy of the combined system

$\mathrm{U}_1=\frac{1}{2} \mathrm{CV}_1^2+\frac{1}{2} \mathrm{CV}_2^2=\frac{\mathrm{C}}{2}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2\right)$

On joining the two condensers in parallel, common potential,

$\mathrm{V}=\frac{\mathrm{V}_1+\mathrm{V}_2}{2}$

$\therefore \quad$ Final energy of the combined system:

$\mathrm{U}_2=\frac{1}{2}(\mathrm{C}+\mathrm{C})\left(\frac{\mathrm{V}_1+\mathrm{V}_2}{2}\right)^2$

$\therefore \quad$ Decrease in energy will be:

$$\begin{aligned} \Delta \mathrm{U} & =\mathrm{U}_1-\mathrm{U}_2 \\ & =\frac{\mathrm{C}}{2}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2\right)-\frac{1}{2}(\mathrm{C}+\mathrm{C})\left(\frac{\mathrm{V}_1+\mathrm{V}_2}{2}\right)^2 \\ & =\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_2\right)^2 \end{aligned}$$

The length of closed pipe is denoted using L.

Let $l$ be the length of open pipe and $\mathrm{v}$ be the velocity.

Frequency of second overtone of an open organ pipe is $\mathrm{n}_{\mathrm{o}}=\frac{3 \mathrm{v}}{2 l}$

Frequency of first overtone of a closed pipe is $\mathrm{n}_{\mathrm{c}}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}$

Given: $\mathrm{n}_0=\mathrm{n}_{\mathrm{c}}$

$$\begin{aligned} \frac{3 \mathrm{v}}{2 l} & =\frac{3 \mathrm{v}}{4 \mathrm{~L}} \\ \mathrm{~L} & =\frac{l}{2} \\ \therefore \quad l & =2 \mathrm{~L} \end{aligned}$$

At equator, for the weight to be zero, the gravitational force must be equal to centrifugal force.

$$\begin{aligned} & \mathrm{mR} \omega^2=\mathrm{mg} \\ & \omega^2=\frac{\mathrm{g}}{\mathrm{R}} \\ & \omega=\sqrt{\frac{\mathrm{g}}{\mathrm{R}}} \\ & \omega=\sqrt{\frac{10}{6.4 \times 10^6}} \\ & \omega=\frac{1}{800} \frac{\mathrm{rad}}{\mathrm{s}} . \end{aligned}$$

The fifth bright band will be:

$\mathrm{y}_5=\frac{5 \lambda_1 \mathrm{D}}{\mathrm{d}}$

$\therefore \quad$ The sixth dark band will be:

$$\begin{aligned} & \mathrm{y}_6{ }^{\prime}=\frac{11 \lambda_2 \mathrm{D}}{2 \mathrm{~d}} \\ & \text { Given: } \mathrm{y}_5=\mathrm{y}_6^{\prime} \\ & \therefore \quad \frac{5 \lambda_1 \mathrm{D}}{\mathrm{d}}=\frac{11 \lambda_2 \mathrm{D}}{2 \mathrm{~d}} \\ & 5 \lambda_1= \frac{11 \lambda_2}{2} \\ & \frac{\lambda_2}{\lambda_1}=\frac{10}{11} \end{aligned}$$

In an LCR circuit, the inductor (L) and capacitor (C) introduce phase shifts in the current compared to the source voltage due to their reactive properties. The inductor causes a phase shift of +90 degrees, leading the voltage, while the capacitor causes a phase shift of $-$90 degrees, lagging the voltage. Thus, voltage across $\mathrm{L}$ and $\mathrm{C}$ are not in phase with the source voltage.

Inductance of solenoid, $\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$ .....(i)

where, $l=$ length of solenoid.

$\mathrm{A}=\pi \mathrm{r}^2=\text { area of solenoid }$

Let '$x$' be length of wire required.

$$\begin{aligned} \therefore \quad \mathrm{x} & =\text { circumference of solenoid } \times \text { no. of turns } \\ & =2 \pi \mathrm{rN} . \\ \therefore \quad \mathrm{N} & =\frac{\mathrm{x}}{2 \pi \mathrm{r}} \end{aligned}$$

Substituting in equation (i),

$$\begin{aligned} & \mathrm{L}=\frac{\mu_0\left(\frac{\mathrm{x}^2}{4 \pi^2 \mathrm{r}^2}\right) \times \pi \mathrm{r}^2}{l} \\ & \therefore \quad \mathrm{L}=\frac{\mu_0 \mathrm{x}^2}{4 \pi l} \\ & \therefore \quad \mathrm{x}^2=\frac{4 \pi \mathrm{L} l}{\mu_0} \\ & \therefore \quad \mathrm{x}=\sqrt{\frac{4 \pi \mathrm{L} l}{\mu_0}} \\ \end{aligned}$$

Substituting the values,

$$\begin{aligned} & x=\sqrt{\frac{4 \times \pi \times 10^{-3} \times 1}{4 \pi \times 10^{-7}}} \\ & x=\sqrt{10^4} \mathrm{~m} \\ & x=0.10 \mathrm{~km} \end{aligned}$$

Electric force, $F_{\text {electric }}=q E$

The effective force,

$$\begin{aligned} & \mathrm{mg}_{\text {eff }}=\mathrm{mg}-\mathrm{F}_{\text {electric }} \\ & \mathrm{g}_{\text {eff }}=\mathrm{g}-\frac{\mathrm{qE}}{\mathrm{m}} \end{aligned}$$

The period of oscillation for a simple pendulum,

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

Time period when pendulum is suspended between the plates,

$$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}_{\mathrm{eff}}}} \\ & \mathrm{T}=2 \pi\left[\frac{\mathrm{L}}{\mathrm{g}-\frac{\mathrm{qE}}{\mathrm{m}}}\right]^{\frac{1}{2}} \end{aligned}$$

$$\begin{aligned} & \lambda_{\text {liquid }}=\frac{\lambda_{\text {air }}}{\mu} \\ & \lambda_{\text {liquid }}=\frac{6300 \times 10^{-10}}{1.33} \end{aligned}$$

Fringe width,

$\mathrm{W}=\frac{\lambda_{\text {liquid }} \times \mathrm{D}}{\mathrm{d}}=\frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 0.001}=6.3 \times 10^{-4} \mathrm{~m}$