MHT CET 2022 11th August Evening Shift (Physics & Chemistry) Question Paper with Solutions

Draw a line parallel to $\phi$ axis

$\because \phi=\mathrm{LI} \quad \therefore \mathrm{L}=\frac{\phi}{\mathrm{I}}=$ slope of $\phi-\mathrm{I}$ curve

$\therefore$ We find that the slope is maximum for A

$\therefore$ A has the maximum self inductance as OP is same for all.

The self inductance of a solenoid is given by

$\mathrm{L}=\frac{\mu_0 \mu_{\mathrm{r}} \cdot \mathrm{N}^2 \mathrm{~A}}{l}$

L depends upon $\mathrm{N}, \mathrm{A}$ and $l$. It does not depend upon the current flowing through it. Change in current does not affect $\mathrm{L}$.

The force of attraction between the plates of a parallel plate capacitor is

$$\begin{aligned} & \mathrm{F}=\frac{\sigma^2 \mathrm{~A}}{2 \varepsilon_0}=\frac{\mathrm{Q}^2}{\mathrm{~A}^2} \times \frac{\mathrm{A}}{2 \varepsilon_0}\left(\because \sigma=\frac{\mathrm{Q}}{\mathrm{A}}\right) \\ & \quad=\frac{\mathrm{Q}^2}{2 \varepsilon_0 \mathrm{~A}}=\frac{\mathrm{C}^2 \mathrm{~V}^2}{2 \varepsilon_0 \mathrm{~A}}(\because \mathrm{Q}=\mathrm{CV}) \\ & \therefore \mathrm{F}=\mathrm{C}\left[\frac{\mathrm{CV}^2}{2 \varepsilon_0 \mathrm{~A}}\right]=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \times \frac{\mathrm{CV}^2}{2 \varepsilon_0 \mathrm{~A}} \\ & \therefore \mathrm{F}=\frac{1}{2} \frac{\mathrm{CV}^2}{\mathrm{~d}} \end{aligned}$$

To answer this question, we need to consider the forces in action when the bucket is at the topmost point in its circular path. At that point, the centripetal force required to keep the water moving in a circular path must be greater than or equal to the gravitational force acting on the water, to prevent it from falling out of the bucket.

The centripetal force ($F_c$) can be described by the following equation, where $m$ is the mass of water, $v$ is the linear velocity of the bucket, and $r$ is the radius of the circle:

$F_c = \frac{m v^2}{r}$

At the minimum velocity needed to keep the water in the bucket, this centripetal force is provided entirely by the weight of the water, which is $mg$, where $g$ is the acceleration due to gravity. Therefore, we can set $F_c = mg$ and solve for the velocity:

$mg = \frac{m v^2}{r}$

$v^2 = rg$

Now, velocity can also be related to the frequency of revolution ($f$) and the circumference of the circle ($C$) using the relation:

$v = f \times C$

The circumference of the circle is given by:

$C = 2 \pi r$

Let's substitute this into the velocity equation and solve for frequency:

$\sqrt{rg} = f \times 2 \pi r$

$\frac{\sqrt{rg}}{2 \pi r} = f$

To isolate $f$, since $r$ is in a square root in the numerator and is not in a square root in the denominator, we can simplify:

$f = \frac{1}{2 \pi} \sqrt{\frac{g}{r}}$

So the correct answer to the minimum frequency required to prevent the water from falling out of the bucket is Option C:

$\frac{1}{2 \pi} \sqrt{\frac{g}{r}}$

For a parallel plate capacitor, $\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$

When the charging battery is disconnected and $\mathrm{d}$ is increased then

(a) Q remains constant

(b) $\mathrm{C} \propto \frac{1}{\mathrm{~d}}$ hence if $\mathrm{d}$ is increased, $\mathrm{C}$ is decreased.

(c) $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$ or $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$ if $\mathrm{C}$ is decreased, $\mathrm{V}$ will increase

(d) $\mathrm{E}=\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}$ if $\mathrm{C}$ is decreased, then $\mathrm{E}$ will increase

Thus (a), (b) and (d) are wrong. Only (c) is correct.

The period of an electron in the $\mathrm{n}^{\text {th }}$ orbit of a hydrogen atom is given by

$$\begin{aligned} & \mathrm{T}_{\mathrm{n}}=\frac{2 \pi \mathrm{r}_{\mathrm{n}}}{\mathrm{v}_{\mathrm{n}}} \\ & \text { and } \mathrm{r}_{\mathrm{n}}=\frac{\varepsilon_0 \mathrm{n}^2 \mathrm{~h}^2}{\pi \mathrm{me}^2} \\ & \text { and linear speed } \mathrm{v}_{\mathrm{n}}=\frac{\mathrm{e}^2}{2 \varepsilon_0 \mathrm{nh}} \\ & \therefore \mathrm{T}_{\mathrm{n}}=\frac{2 \pi \cdot \varepsilon_0 \mathrm{n}^2 \mathrm{~h}^2}{\pi \mathrm{me}^2} \times \frac{2 \varepsilon_0 \mathrm{nh}}{\mathrm{e}^2}=\frac{4 \varepsilon_0^2 \mathrm{n}^3 \mathrm{~h}^3}{\mathrm{me}^4} \end{aligned}$$

Three equal charges are kept at the corners A, B, C of a square $\mathrm{ABCD}$.

$\therefore$ The force between $\mathrm{q}_1$ and $\mathrm{q}_2$ is $\mathrm{F}_{12}$

$\text { and } \mathrm{F}_{12}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q} \times \mathrm{q}}{\mathrm{a}^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{\mathrm{a}^2}$

and the force between $\mathrm{q}_1$ and $\mathrm{q}_3$ at $\mathrm{A}$ and $\mathrm{C}$ is

$$\begin{aligned} & \mathrm{F}_{13}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{2 \mathrm{a}^2} \\ & \therefore \frac{\mathrm{F}_{12}}{\mathrm{~F}_{13}}=\frac{\mathrm{q}^2}{\mathrm{a}^2} \times \frac{2 \mathrm{a}^2}{\mathrm{q}^2}=2 \text { (in magnitude) } \end{aligned}$$

For body A, mass $=\mathrm{m}$, radius of the circle $=\mathrm{r}$

For body B, mass $=3 \mathrm{~m}$ and radius of the circle $=\frac{r}{3}$ and $\mathrm{v}$ and $\mathrm{v}$' are the tangential speeds

$\text { For A, C.P. force }=\frac{\mathrm{mv}^2}{\mathrm{r}}\quad \text{.... (1)}$

For B, C.P. force $=\frac{3 \mathrm{~m} \cdot \mathrm{v}^{\prime 2}}{\mathrm{r} / 3}=\frac{9 \mathrm{mv}^{\prime 2}}{\mathrm{r}}\quad \text{.... (2)}$

and it is given that $\mathrm{v}=\mathrm{nv}{ }^{\prime}$

Since the C.P. force is same for both

$$\begin{aligned} & \therefore \frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{9 m v^{\prime 2}}{\mathrm{r}} \\ & \therefore \mathrm{v}^2=9 \mathrm{v}^{\prime 2} \text { but } \mathrm{v}=\mathrm{nv^{ \prime }} \\ & \therefore \mathrm{n}^2 \mathrm{v}^{\prime 2}=9 \mathrm{v}^{\prime 2} \\ & \therefore \mathrm{n}^2=9 \quad \therefore \mathrm{n}=3 \end{aligned}$$

The first minimum in a Fraunhofer diffraction pattern occurs when the path difference between the light from the two edges of the slit is equal to the wavelength of the light, $\lambda$. The condition for the first minimum can be written as:

$d \sin \theta = m\lambda$

For the first minimum, $m = 1$ and we know that the angle is $30^{\circ}$. Using the given wavelength $\lambda = 5500 \mathop A\limits^o = 550 \times 10^{-9}$ meters (since $1 \mathop A\limits^o = 10^{-10}$ meters), we can write:

$d \sin 30^{\circ} = 1 \times 550 \times 10^{-9} \text{ m}$

$d \times \frac{1}{2} = 550 \times 10^{-9} \text{ m}$

$d = 2 \times 550 \times 10^{-9} \text{ m}$

$d = 1100 \times 10^{-9} \text{ m}$

$d = 1100 \mathop A\limits^o$

The secondary maxima occur in between the primary minima. The first secondary maxima (also known as the first 'bright' fringe other than the central maximum) occurs when the path difference is $3/2$ times the wavelength (this is the condition for the maximum that lies between the first and second minima, $m = 1$ and $m = 2$, respectively). This results in the following condition:

$d \sin \theta = \left(m + \frac{1}{2}\right) \lambda$

For the first secondary maxima $m = 1$:

$d \sin \theta = \left(1 + \frac{1}{2}\right) 550 \times 10^{-9} \text{ m}$

$d \sin \theta = \frac{3}{2} \times 550 \times 10^{-9} \text{ m}$

$1100 \sin \theta = 3 \times 550 \times 10^{-9} \text{ m}$

$\sin \theta = \frac{3}{2} \times \frac{550 \times 10^{-9}}{1100 \times 10^{-9}}$

$\sin \theta = \frac{3}{2} \times \frac{1}{2}$

$\sin \theta = \frac{3}{4}$

Therefore, the correct answer is:

Option B

$\sin ^{-1}\left(\frac{3}{4}\right)$

The intensity of the electric field (E) between two plane parallel sheets of equal and opposite charges is given by $\mathrm{E}=\frac{\sigma}{\varepsilon_0}$

$\therefore \sigma=\mathrm{E} \varepsilon_0$ where $\sigma=$ surface density of charge $=\frac{\mathrm{Q}}{\mathrm{A}}$

$\therefore$ Charge on either plate of the capacitor is $\mathrm{Q}=\sigma \mathrm{A}=\varepsilon_0 \mathrm{EA}$ and $\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$

$\therefore$ The energy stored in the capacitor is

$$\begin{aligned} & \mathrm{U}=\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}=\frac{\left(\varepsilon_0 \mathrm{EA}\right)^2}{2 \cdot \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}}=\frac{\varepsilon_0^2 \mathrm{~A}^2 \mathrm{E}^2 \times \mathrm{d}}{2 \varepsilon_0 \mathrm{~A}} \\ & \therefore \mathbf{U}=\frac{\mathbf{1}}{\mathbf{2}} \varepsilon_0 \mathbf{E}^{\mathbf{2}} \mathbf{A d} \end{aligned}$$

The value of $g$ at a depth $h$ below the surface of the earth of radius $\mathrm{R}$ is

$$\begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right] \\ & \therefore \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=1-\frac{\mathrm{d}}{\mathrm{R}} \quad \text{..... (1)} \end{aligned}$$

It is given that $\mathrm{mg}=500 \mathrm{~N}$ and $\mathrm{mg}^{\prime}=250 \mathrm{~N}$

$\therefore \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{250}{500}=\frac{1}{2}\quad \text{..... (2)}$

$\therefore$ From (1) and (2), $\frac{1}{2}=1-\frac{\mathrm{d}}{\mathrm{R}}$

$\therefore \frac{\mathrm{d}}{\mathrm{R}}=\frac{1}{2}$

$\therefore \mathrm{d}=\frac{\mathrm{R}}{2}=\frac{6400}{2}=3200 \mathrm{~km}$

The excess pressure $(\mathrm{P})$ in a soap bubble of radius $2 \mathrm{~cm}$ is $50$ dyne/$\mathrm{cm}^2$.

If $T$ is the S.T., then $\mathrm{P=\frac{4 T}{R}}$

$\therefore \mathrm{T}=\frac{\mathrm{PR}}{4}=\frac{50 \times 2}{4}$

$\therefore \mathrm{T}=25$ dyne/$\mathrm{cm}$

The relative permeability $\mu_{\mathrm{R}}=1+\chi$

$\therefore$ Absolute permeability

$$\begin{aligned} \mu & =\mu_{\mathrm{R}} \mu_0=\mu_0(1+\chi) \\ & =4 \pi \times 10^{-7}[1+349] \\ & =350 \times 4 \times 3.142 \times 10^{-7} \\ & \doteqdot 4400 \times 10^{-7} \text { SI units } \end{aligned}$$

The force produced by the magnetic field on a moving charged particle is $\mathrm{F}=\mathrm{qvB}$ and this gives a C.P. force

$$\begin{aligned} & \frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \therefore \frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{qvB} \\ & \therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{p}}{\mathrm{qB}} \text { where } \mathrm{p}=\text { momentum } \end{aligned}$$

It is given that both the particles (electron and proton) have the same momenta.

Similarly they have the same charge in magnitude (e and $-$e) and they move in the same field (B).

$$\begin{aligned} & \therefore \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\mathrm{p}_1}{\mathrm{q}_1 \mathrm{~B}} \times \frac{\mathrm{q}_2 \mathrm{~B}}{\mathrm{p}_2}=1 \\ & \because \mathrm{p}_1=\mathrm{p}_2 \text { and } \mathrm{q}_1=\mathrm{q}_2 \\ & \therefore \mathrm{r}_1=\mathrm{r}_2 \end{aligned}$$

$\therefore$ They will describe the same curved path.

[One will move clockwise and the other anticlockwise.]

The heat given to a body, which raises its temperature by 1ºC (or 1 K), is known as the thermal capacity or sometimes referred to as the heat capacity of the body. The correct answer is Option B: thermal capacity.

The thermal capacity (C) of a body is defined as the amount of heat energy (Q) required to raise the temperature of the entire body by one degree Celsius (or one Kelvin). The formula for thermal capacity is given by:

where Q is the heat energy supplied to the body and ΔT is the change in temperature.

To elaborate on the other options provided:

Option A: Specific heat (sometimes called specific heat capacity) is the amount of heat required to raise the temperature of one kilogram of the substance by one degree Celsius (or one Kelvin). It is an intrinsic property of the substance and is expressed in units such as joules per kilogram Kelvin (J/kg·K). The formula for specific heat (c) is given by:

where m is the mass of the substance.

Option C: Water equivalent is a somewhat outdated term used to describe a quantity of a substance that would absorb the same amount of heat as a given mass of water. It is based on the high specific heat capacity of water, which has historically been used as a benchmark. It's not precisely a term for the heat to raise the temperature but rather a comparative metric.

Option D: Temperature gradient refers to the rate of change of temperature with respect to distance in a particular direction. It is a vector quantity that illustrates how temperature changes from one point to another and is not related to the amount of heat energy that is supplied to a body. Therefore, it does not describe the amount of heat needed to increase the temperature of a body by 1ºC.

Given $\mathrm{e}=220 \sin (50 \mathrm{t})$

Comparing with $\mathrm{e}=\mathrm{e}_0 \sin \omega \mathrm{t}$, we get

$$\begin{aligned} & \mathrm{e}_0=220 \mathrm{~V}, \text { peak voltage } \\ & \mathrm{C}=50 \mu \mathrm{F}=50 \times 10^{-6} \mathrm{~F} \\ & \therefore \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{50 \times 50} \times 10^6=400 \Omega \end{aligned}$$

$\therefore \mathrm{I}_0=\frac{\mathrm{e}_0}{\mathrm{X}_{\mathrm{C}}}=\frac{220}{400}=0.55 \mathrm{~A}$

Peak current $=0.55 \mathrm{~A}$

The thermodynamic process, in which no work is done on or by the system is isochoric process.

In an isochoric process, $\mathrm{V}=\mathrm{constant} \quad \therefore \mathrm{dV}=0$

$\therefore$ Work done $(\mathrm{dW})=\mathrm{PdV}=0$

Note : $\mathrm{dQ}=0$ in adiabatic, $\mathrm{dT}=0$ in isothermal.

In isochoric, $\mathrm{dV}=0$ and in isobaric, $\mathrm{dP}=0$

If the radius of the spherical Gaussian surface is increased, then the electric flux due to a point charge enclosed by the surface remains constant.

Flux depends only on the enclosed charge. It does not depend upon the size or shape of the Gaussian surface.

The value of the required shunt $(\mathrm{S})$ is calculated by using

$$\begin{aligned} & \frac{I_{\mathrm{g}}}{\mathrm{I}}=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}} \\ & \therefore \frac{3}{100}=\frac{\mathrm{S}}{\mathrm{S}+200} \\ & \therefore 100 \mathrm{~S}=3 \mathrm{~S}+600 \quad \therefore 97 \mathrm{~S}=600 \\ & \therefore \mathrm{S}=\frac{600}{97} \doteqdot 6 \Omega \end{aligned}$$

[Note for $\mathrm{S}=6,97 \mathrm{~S}=582$ and for $\mathrm{S}=7,97 \mathrm{~S}=679$ ]

$$\begin{aligned} & \text { } \mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}=\mathrm{A} \sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right) \text { where } \mathrm{x}=\mathrm{A} / 2 \\ & \therefore \frac{\mathrm{A}}{2}=\mathrm{A} \sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right) \\ & \therefore \frac{1}{2}=\sin \frac{\pi}{6}=\sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right) \\ & \therefore \frac{2 \pi \mathrm{t}}{\mathrm{T}}=\frac{\pi}{6} \\ & \therefore \mathrm{t}=\frac{\mathrm{T}}{12} \mathrm{~s} \end{aligned}$$

To find the minimum distance between an object and its real image formed by a convex lens, we need to take into account the lens formula, which is given by:

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

where:

• $f$ is the focal length of the lens,
• $d_o$ is the distance of the object from the lens, and
• $d_i$ is the distance of the image from the lens.

To find the minimum distance between the object and its image, we should consider the case where the object is at a distance of $2f$ from the lens. In this case, according to the properties of a convex lens, the image will also be formed at a distance $2f$ on the other side of the lens. This situation corresponds to the object and the image being at the same distance from the lens and both being twice the focal length. The separation between the object and the image would then be the sum of these distances. Thus, the object distance $d_o = 2f$ and the image distance $d_i = 2f$.

To find the minimum total distance between the object and image ($D$), we simply add the object distance ($d_o$) and the image distance ($d_i$):

$D = d_o + d_i = 2f + 2f = 4f$

Therefore, the minimum distance between an object and its real image formed by a convex lens of focal length $f$ is $4f$, which corresponds to Option B.

• Surface tension (T): A property of liquids that causes their surface to behave like a stretched membrane, minimizing surface area. It's represented by the symbol $T$ and has units of force per unit length.


• Work done in forming a bubble: When you blow a bubble, you increase its surface area. This requires work to be done against the surface tension forces trying to minimize the area.

The work done (W) in increasing the surface area of a liquid is:

$W = T \Delta A$

where:

• $T$ is the surface tension


• $\Delta A$ is the change in surface area

1. Bubble 1:

• Radius = $R$


• Surface Area = $4\pi R^2$ (Since a soap bubble has two surfaces, inner and outer)


• Work Done = $W_1$

1. Bubble 2:

• Radius = $2R$


• Surface Area = $4\pi (2R)^2 = 16\pi R^2$


• Work Done = $W_2$

• Surface tension of liquids generally decreases with an increase in temperature.

Since the surface area of bubble 2 is four times that of bubble 1, you might think $W_2 = 4W_1$. However, the surface tension is lower for bubble 2 due to the heated soap solution.

Because the surface tension decreases with temperature:

• Work done ($W_2$) to form bubble 2 will be less than four times the work ($W_1$) needed to form bubble 1.

A body moving in a circular path at constant speed has constant kinetic energy. The directions of momentum, velocity and acceleration change from point to points. Hence they do not remain constant. K.E. is a scalar. Others are vectors.

The magnetic field at $\mathrm{O}$, due to current in $\mathrm{P}$ is

$\mathrm{B}_1=\frac{\mu_0}{4 \pi}\left(\frac{2 \mathrm{I}_1}{\mathrm{~d} / 2}\right)=\frac{\mu_0}{4 \pi} \times \frac{4 \mathrm{I}_1}{\mathrm{~d}}=\frac{\mu_0 \mathrm{I}_1}{\pi \mathrm{d}}$

and the magnetic field at $\mathrm{O}$ due to current $\mathrm{I}_2$ in the wire $\mathrm{Q}$ is

$\mathrm{B}_2=\frac{\mu_0}{4 \pi}\left(\frac{2 \mathrm{I}_2}{\mathrm{~d} / 2}\right)=\frac{\mu_0}{4 \pi} \times \frac{4 \mathrm{I}_2}{\mathrm{~d}}=\frac{\mu_0 \mathrm{I}_2}{\pi \mathrm{d}}$

The currents in the wires are in opposite directions.

Hence the magnetic fields will be added.

$\therefore$ Resultant field $B=B_1+B_2=\frac{\mu_0}{\pi d}\left(I_1+I_2\right)$

For a spring $\mathrm{F}=\mathrm{Kx}$

$$\begin{aligned} & \therefore \mathrm{x}=\frac{\mathrm{F}}{\mathrm{K}} \\ & \therefore \mathrm{x}_1=\frac{\mathrm{F}}{\mathrm{K}_1} \text { and } \mathrm{x}_2=\frac{\mathrm{F}}{\mathrm{K}_2} \\ & \text { or } \mathrm{e}_1=\frac{\mathrm{F}}{\mathrm{K}_1} \text { and } \mathrm{e}_2=\frac{\mathrm{F}}{\mathrm{K}_2} \end{aligned}$$

$\therefore \mathrm{e}_1+\mathrm{e}_2=\mathrm{F}\left(\frac{1}{\mathrm{~K}_1}+\frac{1}{\mathrm{~K}_2}\right)$

Note: The springs are connected in series.

$\therefore$ The effective spring constant $\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}$

$\therefore$ Total extension

$\mathrm{e}=\frac{\mathrm{F}}{\mathrm{K}}=\frac{\mathrm{F}\left(\mathrm{K}_1+\mathrm{K}_2\right)}{\mathrm{K}_1 \mathrm{~K}_2}=\frac{\mathrm{F}}{\mathrm{K}_2}+\frac{\mathrm{F}}{\mathrm{K}_1}$

To open or close the door, a force of $1 \mathrm{~N}$ is applied at a distance of $1.2 \mathrm{~m}$ from the hinges.

Moment of the force $=\mathrm{F} \times \mathrm{d}=1 \times 1.2=1.2 \mathrm{~N}-\mathrm{m}$

When the force is applied at $\mathrm{P}$ at a distance of $0.2 \mathrm{~m}$ from $\mathrm{O}$, then the force required to have the same moment is given by

$1.2=\mathrm{F} \times 0.2 \quad \therefore \mathrm{F}=\frac{1.2}{0.2}=6 \mathrm{~N}$

Diode $\mathrm{D}_1$ is forward biased and diode $\mathrm{D}_2$ is reverse biased. Hence no current till flow in the branch of $D_2$.

$\therefore$ The total effective resistance in the circuit is

$2+3=5 \Omega$

$\therefore$ Current $\mathrm{I}=\frac{10}{5}=2 \mathrm{~A}$

$\therefore$ Current through $\mathrm{R}_1=2 \mathrm{~A}$

The correct answer to this question is Option D: unidirectional but not steady current.

In a full-wave rectifier circuit, the objective is to convert alternating current (AC) into direct current (DC). The fundamental characteristic of a full-wave rectifier is that it inverts the negative half-cycles of the AC input voltage so that on the output side, the current flows in only one direction for both half-cycles. This makes the output current unidirectional. However, without a filter to smooth out the ripples, the output current is not a constant direct current but a pulsating direct current that still retains the frequency of the original AC signal.

The output current of a full-wave rectifier without a filter can be represented as an absolute value of a sinusoidal wave, meaning it maintains the sinusoidal shape but with all the values above the zero line since the negative values are inverted. This waveform is not constant or steady; instead, it rises and falls with the frequency of the AC input, typically at twice the frequency in the case of a full-wave rectifier because it rectifies both halves of the input sine wave.

So, to summarize:

• Option A (an eddy current) is incorrect since eddy currents are localized currents induced in conductors when they are exposed to changing magnetic fields, which is not the output of a rectifier circuit.


• Option B (a constant direct current) is incorrect because without a filter (such as a capacitor or an inductor), the output is not constant but pulsating with ripples.


• Option C (a sinusoidal current) is incorrect because the resulting current is not sinusoidal; it's the absolute value of a sinusoidal wave, representing both halves of the wave above zero volts.


• Option D (unidirectional but not steady current) is correct as the output is indeed unidirectional due to the inversion of the negative half-cycles, but not steady due to the absence of a filter to smooth out the ripples.

For the NAND gate, the truth table is

Thus $\mathrm{C}=1, \mathrm{D}=1, \mathrm{E}=1$ and $\mathrm{F}=0$

For a NAND gate, there is a high output for a low input and a low output for high input.

$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ where $\mathrm{E}$ is the K.E.

$$\begin{aligned} & \therefore \lambda \propto \frac{1}{\sqrt{\mathrm{E}}} \quad \therefore \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{\mathrm{E}_1}{\mathrm{E}_2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \\ & \therefore \lambda_2=\frac{\lambda_1}{\sqrt{2}} \end{aligned}$$

Thus $\lambda$ charges by a factor $\frac{1}{\sqrt{2}}$

$y=10 \sin \left(\frac{\pi x}{4}\right) \cos (20 \pi t)$

Comparing with $\mathrm{y}=2 \mathrm{~A} \cos \left(\frac{2 \pi \mathrm{x}}{\lambda}\right) \sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right)$, we get

$$\begin{aligned} & \frac{2 \pi \mathrm{x}}{\lambda}=\frac{\pi \mathrm{x}}{\mathrm{T}} \quad \therefore \frac{2}{\lambda}=\frac{1}{4} \\ & \therefore \lambda=8 \mathrm{~cm} \end{aligned}$$

$\therefore$ The distance between two consecutive nodes

$=\frac{\lambda}{2}=\frac{8}{2}=4 \mathrm{~cm}$

The particle starts from the mean position.

$$\begin{aligned} \mathrm{x}=\mathrm{A} \sin \omega \mathrm{t} & =\mathrm{A} \sin \left(\frac{2 \pi}{\mathrm{T}}\right) \times \mathrm{t} \\ & =A \sin \left(\frac{2 \pi}{\mathrm{T}} \times \frac{\mathrm{T}}{12}\right) \\ & =A \sin \left(\frac{\pi}{6}\right) \end{aligned}$$

$\therefore \mathrm{x}=\mathrm{A} \sin 30^{\circ}=\frac{\mathrm{A}}{2} \quad \therefore \mathrm{x}^2=\frac{\mathrm{A}^2}{4}$

$\therefore$ The particle is at a distance $\mathrm{A} / 2$ from the mean position.

At this point its P.E. $=\frac{1}{2} K x^2=\frac{1}{2} m \omega^2 x^2\quad \text{...... (1)}$

and its K.E. $=\frac{1}{2} \mathrm{mv}^2$

$\therefore \mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right)\quad \text{..... (2)}$

$$\begin{aligned} & \therefore \frac{\text { P.E. }(\mathrm{U})}{\text { K.E. }(\mathrm{K})}=\frac{\mathrm{x}^2}{\mathrm{~A}^2-\mathrm{x}^2}=\frac{\frac{\mathrm{A}^2}{4}}{\mathrm{~A}^2-\frac{\mathrm{A}^2}{4}}=\frac{\frac{\mathrm{A}^2}{4}}{\frac{3 \mathrm{~A}^2}{4}} \\ & \therefore \frac{\mathrm{U}}{\mathrm{K}}=\frac{1}{3} \end{aligned}$$

The fundamental frequency of the closed pipe $(n)=\frac{V}{4 L}$

$\mathrm{n}=\frac{320}{4 \times 0.8}=\frac{320}{3.2}=100 \mathrm{~Hz}$

For the vibrating wire, fundamental frequency (n) is

$\mathrm{n}^{\prime}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$