MHT CET 2019 2nd May Morning Shift (Physics & Chemistry) Question Paper with Solutions

Free body diagram (FBD) of a stone moving in a vertical circular path, which has tension force at point $𝐴$ and $𝐵$ as represented by ${𝑇}_{𝐴}$ and ${𝑇}_{𝐵}$ respectively as given below in the figure,

Given, mass of stone $\left(𝑚\right)=1\mathrm{kg}$,

length of the string $\left(𝑅\right)=2m$

and rotating linear speed $\left(𝑣\right)=40{\mathrm{ms}}^{-1}$

As, we know that the tension at position $𝐴$,

$\begin{array}{cc}& {𝑇}_{𝐴}=\frac{𝑚{𝑣}^2}{𝑅}+𝑚𝑔(\therefore {𝐹}_{𝑐}=\frac{𝑚{𝑣}^2}{𝑅})\\ & \Rightarrow {𝑇}_{𝐴}=\frac{1\times (40{)}^2}{2}+1\times 10=810N\end{array}$

Similarly, tension at position $𝐵$,

$\Rightarrow {𝑇}_{𝐵}=\frac{𝑚{𝑣}^2}{𝑅}-𝑚𝑔=\frac{1\times (40{)}^2}{2}-1\times 10=790N$

So, the ratio of ${𝑇}_{𝐵}$ and ${𝑇}_{𝐴}$ i.e.,

$\frac{{𝑇}_{𝐵}}{{𝑇}_{𝐴}}=\frac{790}{810}=\frac{79}{81}$

Hence, the ratio of tension at position $𝐵$ and tension at position $𝐴$ is $79:81$.

Given, mutual inductance in the coil, $𝑀=0.01H$ and current in coil, $𝑙=5\sin 200𝜋𝑡$.

Emf induced in secondary coil is given,

$𝑒=𝑀\frac{𝑑𝑙}{𝑑𝑡}$

putting the value in above relation,

$\begin{array}{cc}𝑒& =0.01\times \frac{𝑑}{𝑑𝑡}(5\sin 200𝜋𝑡)\\ & =0.01\times 5\times 200𝜋\cos 200𝜋𝑡=10𝜋\cos 200𝜋𝑡\end{array}$

Compairing the above equation with the general equation of induced emf, which is represented as,

$𝑒={𝑒}_0\cos 𝜔𝑡$

where, ${𝑒}_0$ is the maximum value of the emf.

$\therefore$ In the question, the maximum value of emf induced in the coil is; ${𝑒}_0=10𝜋V$.

Given, radius of earth, ${𝑅}_{𝑒}=6371\mathrm{km}$ and radius of orbit around the sun, ${𝑅}_0=149\times {10}^6\mathrm{km}$

Now, the ratio of the diameters of the earth and orbit around the sun can be expressed as

$\frac{{𝐷}_0}{{𝐷}_{𝜃}}=\frac{2{𝑅}_0}{2{𝑅}_{𝜃}}=\frac{2\times 14.9\times {10}^6}{2\times 6371}=2.33\times {10}^4$

From above ratio, it can be concluded that the order of magnitude of the diameter of the orbit is greater than that of earth by multiple of ${10}^4$.

The fundamental frequency of an open organ pipe of length $𝑙$ is given by the formula:

$𝑛=\frac{𝑣}{2𝑙}$

where $𝑣$ is the velocity of sound in the pipe.

For an open pipe with frequency ${𝑛}_1$ and length ${𝑙}_1$, we have:

${𝑛}_1=\frac{𝑣}{2{𝑙}_1}\Rightarrow {𝑙}_1=\frac{𝑣}{2{𝑛}_1}\text{(Equation 1)}$

Similarly, for an open pipe with frequency ${𝑛}_2$ and length ${𝑙}_2$:

${𝑛}_2=\frac{𝑣}{2{𝑙}_2}\Rightarrow {𝑙}_2=\frac{𝑣}{2{𝑛}_2}\text{(Equation 2)}$

When both pipes are joined to form a single pipe, the total length $𝐿$ is:

$𝐿={𝑙}_1+{𝑙}_2\text{(Equation 3)}$

Let the fundamental frequency of the joined pipe be ${𝑁}_0$. Therefore, we can write:

$𝐿=\frac{𝑣}{2{𝑁}_0}={𝑙}_1+{𝑙}_2$

Substituting from Equations 1 and 2, we have:

$\frac{𝑣}{2{𝑁}_0}=\frac{𝑣}{2{𝑛}_1}+\frac{𝑣}{2{𝑛}_2}$

By solving the equation above, we find:

${𝑁}_0=\frac{{𝑛}_1{𝑛}_2}{{𝑛}_1+{𝑛}_2}$

Given that the ratio of the molar specific heats at constant pressure (${𝐶}_{𝑝}$) and constant volume (${𝐶}_{𝑣}$) is $𝛾$, we need to find the expression for ${𝐶}_{𝑝}$. The universal gas constant is denoted by $𝑅$.

The relationship between ${𝐶}_{𝑝}$ and ${𝐶}_{𝑣}$ for an ideal gas is given by Mayer's relation:

${𝐶}_{𝑝}-{𝐶}_{𝑣}=𝑅$

By substituting ${𝐶}_{𝑝}$ from the above relation into the expression for $𝛾$, we have:

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑝}-𝑅}$

To solve for ${𝐶}_{𝑝}$, multiply both sides by $({𝐶}_{𝑝}-𝑅)$:

$𝛾({𝐶}_{𝑝}-𝑅)={𝐶}_{𝑝}$

Expanding and rearranging terms, we get:

$𝛾{𝐶}_{𝑝}-{𝐶}_{𝑝}=𝛾𝑅$

Factor out ${𝐶}_{𝑝}$ on the left side:

${𝐶}_{𝑝}(𝛾-1)=𝛾𝑅$

Finally, solve for ${𝐶}_{𝑝}$:

${𝐶}_{𝑝}=\frac{𝛾𝑅}{𝛾-1}$

Thus, ${𝐶}_{𝑝}$ is equal to $\frac{𝛾𝑅}{𝛾-1}$.

Key Idea Impedance for series $\angle 𝐶𝑅$ circuit is given by $𝑍=\sqrt{\surd {𝑅}^2+{(𝜔𝐿-\frac{1}{𝜔𝐶})}^2}$

Given, resistance, $𝑅=300\Omega$, inductance, $𝐿=0.9H$, capacitance, $𝐶=2𝜇F=2\times {10}^{-6}F$ and angular frequency, $𝜔=1000\mathrm{rad}/s$.

Substituting the given values in the above equation, we get

$\begin{array}{cc}\Rightarrow & 𝑍=\sqrt{\surd {300}^2+{(1000\times 0.9-\frac{1}{1000\times 2\times {10}^{-6}})}^2}\\ \Rightarrow & 𝑍=\sqrt{\surd 90000+(900-500{)}^2}\\ \Rightarrow & 𝑍=\sqrt{\surd 250000}=500\Omega \end{array}$

Hence, the impedance of $𝐿𝐶𝑅$ circuit is $500\Omega$

For a particle executing simple harmonic motion (SHM),

Total energy $=$ kinetic energy + potential energy

$\begin{array}{cc}& ={𝐸}_{𝐾}+{𝐸}_{𝑃}=\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}𝑘{𝑥}^2\\ & =\frac{1}{𝑚}({𝐴}^2-{𝑥}^2)+\frac{1}{2}𝑚{𝜔}^2{𝐴}^2\\ & (\because 𝑣=\sqrt{\surd {𝐴}^2-{𝑥}^2}\text{ where, }𝐴\text{ is the }\\ & \text{ amplitude of the particle) }\\ & =\frac{1}{2}𝐾{𝐴}^2\end{array}$

It is independent of $𝑡,𝑥$ is remains constant i.e., it does not vary periodically.

However, the displacement of a particle executing SHM along $𝑥$-axis is given by

$𝑥=𝐴\sin (𝜔𝑡+𝜙)$

(where, $𝐴=$ amplitude, $𝜔=$ angular velocity $𝑡=$ time, $𝜙=$ initial phase of the particle)

i.e., displacement varies periodically with time $𝑡$. velocity of particles executing SHM,

$𝑣=\frac{𝑑𝑥}{𝑑𝑡}=𝜔\sqrt{\surd {𝐴}^2-{𝑥}^2}$

which varies with $𝑥$.

Acceleration of the particle,

$𝑎=\frac{𝑑𝑣}{𝑑𝑡}=-{𝜔}^2𝑥$

Which also varies with $𝑥$.

For $𝑛$ capacitors i.e., ${𝐶}_4,{𝐶}_2,\dots \dots \dots .{𝐶}_{𝑛}$ their equivalent capacitance, when connected in

(a) series is given as, $\frac{1}{{𝐶}_{𝑠}}=\frac{1}{{𝐶}_1}+\frac{1}{{𝐶}_2}+\dots \dots$

and

(b) parallel is given as, ${𝐶}_{𝑝}={𝐶}_1+{𝐶}_2+\dots \dots \dots$

In the given problem, we will check the net capacitance in each option and compare it with the desired capacitance mentioned in it.

(a) 3 capacitors are in parallel and 4 capacitors are connect in series, i.e.,

where, ${𝐶}_{𝐴𝐵}$ is the equivalent capacitance of parallel branches between $𝐴$ and $𝐵$ and ${𝐶}_{𝐵𝐷}$ is equivalent capacitance of series capacitors between $𝐵$ and $𝐷$. So, equivalent capacitance.

${𝐶}_{\text{eq }}=\frac{{𝐶}_{𝐴𝐵}\cdot {𝐶}_{𝐵𝐷}}{{𝐶}_{𝐴𝐵}+{𝐶}_{𝐵𝐷}}=\frac{6\times \frac{2}{4}}{6+\frac{2}{4}}=\frac{6}{13}𝜇F$

Similarly, (b) 2 capacitors are in parallel and 5 capacitors are connect in series i.e.,

So, equivalent capacitance,

${𝐶}_{\text{eq }}=\frac{4\times \frac{2}{5}}{4+\frac{2}{5}}=\frac{4}{11}𝜇F$

(c) 4 capacitors are in parallel and 3 capacitors are connect in series i.e.,

So, equivalent capacitance,

${𝐶}_{\text{eq }}=\frac{8\times \frac{2}{3}}{8+\frac{2}{3}}=\frac{16}{26}=\frac{8}{13}𝜇F$

(d) 5 capacitors are in parallel and 2 capacitors are connect in series i.e.,

So, equivalent capacitance,

${𝐶}_{𝑒𝑞}=\frac{10\times 1}{10+1}=\frac{10}{11}𝜇F$

So, the correct option is (d).

For a particle moving in a magnetic field, then applied two forces are equal.

centripetal force $\left({𝐹}_{𝑐}\right)=$ magnetic force $\left({𝐹}_{𝑚}\right)$

$\begin{array}{cc}& \Rightarrow \frac{𝑚{𝑣}^2}{𝑟}=𝑞𝑣𝐵\\ & \Rightarrow 𝑚{𝑣}^2=𝑞𝐵\left(𝑣𝑟\right)\text{.... (i)}\end{array}$

Also, from Bohr's model,

$\begin{array}{cc}𝑚𝑣𝑟& =\frac{𝑛ℎ}{2𝜋}\\ \therefore 𝑣𝑟& =\frac{𝑛ℎ}{2𝜋𝑚}\text{.... (ii)}\end{array}$

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

$𝑚{𝑣}^2=\frac{𝑛ℎ}{2𝜋𝑚}\cdot 𝑞𝐵\text{... (iii)}$

Energy of the electron moving in $𝑛$th orbit,

$\begin{array}{cc}𝐸& =\frac{1}{2}\cdot 𝑚{𝑣}^2=\frac{1}{2}\cdot \frac{𝑛ℎ𝑞𝐵}{2𝜋𝑚}\text{(using Eq. (iii)}\\ \Rightarrow & 𝐸=\frac{𝑛ℎ𝑞𝐵}{4𝜋𝑚}\end{array}$

Hence, the energy of the charged particle in the $𝑛$th level will be $\frac{𝑛ℎ𝑞𝐵}{4𝜋𝑚}$.

The horseshoe magnet has cylindrically concave pole-pieces. Due to its shape the magnet produces radial magnetic field, so that when the coil rotates in any position its plane is always parallel to the direction of the magnetic field. This has been shown in the figure given below.

To analyze the problem, we can use the relation that describes the extension of a wire when a force is applied. For a wire of initial length ${𝐿}_0$, cross-sectional area $𝐴$, and Young's modulus $𝑌$, the extension $\Delta 𝐿$ is given by:

$\Delta 𝐿=\frac{𝐹{𝐿}_0}{𝐴𝑌}$

Here's a step-by-step breakdown:

Since both wires are identical in dimensions and are subjected to the same force $𝐹$, the other factors (${𝐿}_0$ and $𝐴$) remain the same for both.

Therefore, the extension $\Delta 𝐿$ depends only on the modulus of elasticity $𝑌$. The relationship shows that:

$\Delta 𝐿\propto \frac{1}{𝑌}$

Given that the elongation of substance $𝑄$ is more than that of $𝑃$, it means:

$\frac{1}{{𝑌}_{𝑄}}>\frac{1}{{𝑌}_{𝑃}}$

or equivalently,

${𝑌}_{𝑄}<{𝑌}_{𝑃}$

A higher modulus of elasticity indicates that the material is stiffer, or in other words, more elastic in the sense that it is less easily deformed.

Thus, since wire $𝑃$ has a higher Young's modulus than wire $𝑄$, it is more elastic (stiffer) compared to wire $𝑄$.

Therefore, the correct choice is:

Option B: $𝑃$ is more elastic than $𝑄$.

Gravitational force is a conservative frorce. As we know that the work done by a conservative force is independent of the path followed and depends only on the end points of the motion. Since in the given pattern ${𝑤}_1,{𝑤}_2$ and ${𝑤}_3$ have the same end point $𝐴$ and $𝐵$ as shown in the figure below,

So, the correct relation between is ${𝑤}_1={𝑤}_2={𝑤}_3$

In the given circuit, the diode is forward biased, hence the current will flow through it

Now, applying KVL in the above circuit,

$\begin{array}{cc}+3-100𝑖-(+1)& =0\\ \Rightarrow 3-1& =100/\\ \Rightarrow 𝑖=\frac{2}{100}& =0.02A\\ & 𝑖=20\mathrm{mA}\end{array}$

Hence, the current flow in the circuit is 20 mA .

The equation for a simple harmonic progressive wave is given by $𝑌=𝑎\sin 2𝜋(𝑏𝑡-𝑐𝑥)$.

To find the condition under which the maximum particle velocity is twice the wave velocity, we start with the wave equation $𝑦=𝑎\sin 2𝜋(𝑏𝑡-𝑐𝑥)$.

Comparing this with the general equation of a progressive wave, $𝑦={𝐴}_0\sin 2𝜋(𝑓𝑡-\frac{𝑥}{𝜆})$, we identify the components:

Frequency, $𝑓=𝑏$

Wavelength, $𝜆=\frac{1}{𝑐}$

Amplitude, ${𝐴}_0=𝑎$

The maximum particle velocity (${𝑣}_{max}$) can be expressed as:

${𝑣}_{max}={𝐴}_0𝜔=𝑎\times 2𝜋𝑏\text{.... (i)}$

The wave velocity (${𝑣}_{\text{wave}}$) is given by:

${𝑣}_{\text{wave}}=𝑓𝜆=\frac{𝑏}{𝑐}\text{.... (ii)}$

Given that the maximum particle velocity is twice the wave velocity, we have:

${𝑣}_{max}=2{𝑣}_{\text{wave}}$

Substituting values from equations (i) and (ii) into this relation, we obtain:

$𝑎\times 2𝜋𝑏=2\times \frac{𝑏}{𝑐}$

From this equation, we derive:

$𝑐=\frac{1}{𝑎𝜋}$

In the case of a fundamental mode, the time required for a sound wave to travel to the closed end of a pipe filled with air is denoted as ' $𝑡$ ' seconds. The frequency of vibration of the air column can be calculated as follows:

For a closed organ pipe, the fundamental frequency is expressed by the formula:

${𝑓}_0=\frac{𝑣}{4𝐿}\text{.... (i)}$

where:

$𝑣$ is the velocity of the sound wave inside the pipe.

$𝐿$ is the length of the pipe.

Given that the time taken for the wave to reach the other end is $𝑡$ seconds, we can relate this to the length of the pipe with the equation:

$𝑡=\frac{𝐿}{𝑣}\text{.... (ii)}$

Substitute equation (ii) into equation (i):

${𝑓}_0=\frac{1}{4}\cdot \frac{1}{𝑡}=(4𝑡{)}^{-1}$

Thus, the frequency of vibration of the air column is $(4𝑡{)}^{-1}$.

Given that, the two small drops of mercury and radius $𝑅$ of each drop coalesce to form a large drop of radius $𝑟$, so the net volume remains constant i.e,

$\begin{array}{cc}\text{ initial volume }& =\text{ final volume }\\ {𝑉}_{\text{initial }}& =V_{\text{final }}\\ 2\times \frac{4}{3}𝜋{𝑅}^3& =1\cdot \frac{4}{3}𝜋{𝑟}^3\Rightarrow 𝑟=2^{1/3}𝑅\text{.... (i)}\end{array}$

As, surface tension is constant for both the drops.

So, the surface energy of two small drops,

${𝐸}_1=2\times 𝑇{𝐴}_1=2\times 4𝜋{𝑅}^2\times 𝑇$

Surface energy of one big drop,

${𝐸}_2=𝑇{𝐴}_2=4𝜋{𝑟}^2\times 𝑇=2^{1/3}.4𝜋{𝑅}^2𝑇\text{ (using Eq. (i)) }$

Ratio the total surface energy before and after the change is given as,

$\frac{{𝐸}_1}{{𝐸}_2}=\frac{8𝜋}{2^{2/3}}\frac{{𝑅}^2𝑇}{4𝜋{𝑅}^2𝑇}=\frac{2^{1-\frac{2}{3}}}{1}=2^{\frac{1}{3}}:1$

Hence, the ratio of surface energy of the each drop is $2^{\frac{1}{3}}:1$.

Key Idea Moment of inertia of a solid sphere about its diameter is given as, $𝐼=\frac{2}{5}𝑀{𝑅}^2$, where $𝑀$ and $𝑅$ are the mass and radius of the sphere, respectively.

Moment of inertia of a solid sphere about its diameter is $𝐼=\frac{2}{5}𝑀{𝑅}^2\text{.... (i)}$

As the radius of a solid sphere is doubled then the new moment of inertia of the sphere will be

${𝐼}^{\prime }=\frac{2}{5}𝑀(2𝑅{)}^2=\frac{8}{5}𝑀{𝑅}^2\text{.... (ii)}$

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

$\frac{𝐼}{{𝐼}^{\prime }}=\frac{\frac{2}{5}𝑀{𝑅}^2}{\frac{8}{5}𝑀{𝑅}^2}=\frac{1}{4}=1:4$

The ratio of their moment of inertia about any of its diameter is $1:4$.

According to Ohm's law, the ratio of voltage to the corresponding current is represented as resistance, which can be expressed by the formula:

$\frac{𝑉}{𝐼}=𝑅\left(\text{resistance}\right)$

For a metallic wire, as the temperature increases, the resistance increases as well. This is due to the increased thermal motion of electrons within the wire, which causes more frequent collisions and impedes the flow of current.

The force per unit area is pressure Pressure inside a soap bubble of radius $𝑅$. is given by

$𝑃=\frac{4𝑇}{𝑅}$

where, $𝑇=$ surface tension and $𝑅=$ radius of the drop

Pressure inside a soap bubble, $𝑃=\frac{4𝑇}{𝑅}$.

If a bubble is break into 27 small soap bubbles then the volume of single bubble of radius $𝑅$ and the combined volume of 27 bubbles of radius $𝑟$ would be constant.

$27\times$ volume of small bubbles $=$ volume of larger bubble

$\begin{array}{cccc}\Rightarrow & & 27\left(\frac{4}{3}𝜋{𝑟}^3\right)& =\frac{4}{3}𝜋{𝑅}^3\\ \Rightarrow & & 27{𝑟}^3& ={𝑅}^3\\ \Rightarrow & & 𝑟& =\frac{𝑅}{3}\text{.... (i)}\end{array}$

Now, the pressure inside smaller soap bubble,

${𝑃}_{\text{small }}=\frac{4𝑇}{𝑟}=\frac{12𝑇}{𝑅}\text{ (using the relation) }$

and similarly ${𝑃}_{\text{large }}=\frac{4𝑡}{𝑅}$

$\therefore$ Ratio of pressure of the smaller and larger soap bubble is given as,

$\begin{array}{cc}& \frac{{𝑃}_{\text{larger }}}{{𝑃}_{\text{small }}}=\frac{4𝑇}{𝑅}\times \frac{𝑅}{12𝑇}=\frac{1}{3}\\ {𝑃}_{\text{larger }}:{𝑃}_{\text{small }}& =1:3\end{array}$

Hence, the ratio of mechanical force acting per unit area of big soap bubble to that of a small bubble is $1:3$.

Two wires 1 and 2 carrying currents ${𝑖}_1$ and ${𝑖}_2$ respectively are shown in the figure given below,

Force on wire 2 due to ${𝐵}_1$ (field of wire 1 ) is given by

${𝐹}_2={𝑖}_2\mid {𝐵}_1$

Using right hand screw rule, direction of ${𝐹}_2$ would be towards wire 1.

Similarly, force on wire due to ${𝐵}_2$ (field of wire 2)

${𝐹}_1={𝑖}_1/{𝐵}_2$

Again, by using right hand screw rule, direction of ${𝐹}_1$ would be towards wire 2.

If two parallel conductors carry (equal or unequal) current in the same direction, they will exerted force towards each other i.e., they will be attract each other.

The atmospheric $𝐸$ region or $𝐸$ layer is above the $𝐷$-region and below the F-region. It exists at altitude between 100 to 125 km above the earth surface $𝐸$-layer assists in long distance radio communication.

In this layer, gases like oxygen and nitrogen are ionized due to photo-ionization by energy released from the sun during day time which facilitates in radio communication so, this layer does not work during in the night time.

Given that, the linear mass density,

$𝑚={10}^{-3}\mathrm{kg}/m$ and equation of the wave

$𝑦=0.05\sin (𝑥+15𝑡)\text{.... (i)}$

Since, the general equation of wave,

$𝑦=𝑎\sin (𝑘𝑥+𝜔𝑡)\text{... (ii)}$

Now, compairing the Eqs. (i) and (ii) we get,

$𝑘=1,𝜆=2𝜋(\because 𝑘=\frac{2𝜋}{𝜆})$

and $𝜔=15\Rightarrow 𝑓=\frac{15}{2𝜋}(\because 𝜔=2𝜋𝑓)$

Velocity of the wave, $𝑣=𝑓𝜆=2𝜋\times \frac{15}{2𝜋}=15m/s$

As, we know, the tension force in the string,

$𝑇={𝑣}^2𝑚(\because 𝑣=\sqrt{\surd \frac{𝑇}{𝑚}})$

So, by substituting the values in the above relation, we get

$𝑇=(15{)}^2\times {10}^{-3}=0.225N$

Hence, the tension force in the string is 0.225 N.

As we know that the kinetic energy of a revolving satellite at a height $ℎ$ above the earth's surface is given as

$\begin{array}{cc}𝐾𝐸=\frac{1}{2}𝑚{𝑣}^2& =\frac{𝐺𝑀𝑚}{2𝑟}(\because 𝑣=\frac{\sqrt{\surd 𝐺𝑀}}{𝑟}\text{ for circular orbit })\\ & =\frac{𝐺𝑀𝑚}{2(𝑅+ℎ)}\end{array}$

where, R is the radius of the earth.

Given, height of a satellite from the earth surface $ℎ=3𝑅$ and mass of the satellite $=𝑚$

So,

$\begin{array}{cc}& 𝐾𝐸=\frac{𝐺𝑀𝑚}{2(𝑅+3𝑅)}=\frac{𝐺𝑀𝑚}{2\left(4𝑅\right)}\\ & 𝐾𝐸=\frac{𝐺𝑀𝑚}{8𝑅}\times \frac{𝑅}{𝑅}\\ \therefore & 𝐺𝑀𝑚\frac{𝑅}{8{𝑅}^2}=\frac{𝑔𝑚𝑅}{8}(\therefore 𝑔=\frac{𝐺𝑀}{{𝑅}^2})\end{array}$

So, the kinetic energy, $𝐾𝐸=\frac{𝑔𝑚𝑅}{8}$.

Key Idea One time period means the total time in which the particle returns to the same point and in the same direction from where it was released.

Now, in simple harmonic motion (SHM) total distance travelled is 4A.

This is because the particle covers $𝐴$ in $\frac{𝑇}{4}$ time period i.e., when it moves from $𝐴$ to $𝐵$, So in complete $𝑇$ time it covers 4 times the amplitude distance i.e., $4𝐴$.

A galvanometer is converted into a voltmeter by connecting a resistance in series with the galvanometer as shown in the circuit diagram,

where, $𝑅$ is resistance of the resistor connected in series.

Given, galvanometer resistance, ${𝑅}_{𝐺}=100\Omega$

Voltmeter range, ${𝑉}_{max}=50V$ and full deflection current ${𝐼}_{𝑓}=10\mathrm{mA}$

So, by applying the $𝐾𝑉𝐿$ in above circuit diagram,

$\begin{array}{cc}& {𝑉}_{𝐴𝐵}=100{𝐼}_{𝑛}+𝑅{𝐼}_{𝑛}\\ & \Rightarrow {𝑉}_{\text{AB }}=(100+𝑅{)}_{\text{I }}\end{array}$

$\therefore$ For a 50 V voltmeter range there must be,

${𝑉}_{𝐴𝐵}=50V\text{ and }{𝐼}_{𝑓}=10\mathrm{mA}$

Now, substituting the values of ${𝑉}_{𝐴𝐵}$ and ${𝐼}_{𝑓𝑙}$ in Eq. (i) we get,

$\begin{array}{cc}& 50=(100+𝑅)10\times {10}^{-3}\\ & \therefore 𝑅=4900\Omega \end{array}$

Hence, the resistance to be connected in series, to get a voltmeter of range 50 V is $4900\Omega$.

The relation between electron's angular momentum $𝐿$ and magnetic moment $𝜇$ given, $𝜇=-\frac{𝑒}{2{𝑚}_{𝑒}}𝐿$, where $𝑒$ and ${𝑚}_{𝑒}$ are the charges and mass of the electron.

Here, negative sign shows that the angular momentum and magentic moment are in opposite direction to each other i.e., if $𝐿$ in $+𝑧$-direction then $𝜇$ in-z-direction

So, the angle between $𝐿$ and $𝜇$ is ${180}^{\circ }$.

Condition for balanced wheatstone-bridge,

$\begin{array}{cc}& \frac{𝑃}{𝑄}=\frac{𝑅}{𝑆}\\ \Rightarrow & \frac{5}{1}=\frac{50}{10}\\ \Rightarrow & \frac{5}{1}=\frac{5}{1}\end{array}$

So, it is a balanced wheatstone-bridge and hence no current will flow through the arm $𝐺$.

The, circuit can be redrawn as given below,

As, we know that in a series branch, same current flows.

Hence, from the KVL loop rule,

${𝑉}_{𝑃𝑄}={𝑙}_1(5+1)=6{𝑙}_1\text{.... (i)}$

and ${𝑉}_{𝑅𝑆}=(50+10)(1.1-{𝑙}_1)\text{... (ii)}$

Since, for the parallel branches, the voltage drop across them remains same,

$\begin{array}{cc}& \text{ So, }\\ & {𝑉}_{𝑃𝑄}={𝑉}_{𝑅𝑆}\\ & \Rightarrow 6{𝐼}_1=60(1.1-{𝐼}_1)\text{ (using Eqs. (i) and (ii)) }\\ & \Rightarrow {𝐼}_1(60+6)=66\\ & \Rightarrow {𝐼}_1=1A\end{array}$ Hence, current of 1 A passes through $1\Omega$ resistance.

Wavelength in Balmer series of hydrogen spectrum is given by relation,

$\frac{1}{𝜆}=𝑅[\frac{1}{2^2}-\frac{1}{{𝑛}^2}]\text{ where, }𝑛=3,4,5$

So, wavelength of first line,

$\begin{array}{cc}\frac{1}{{𝜆}_1}& =𝑅[\frac{1}{4}-\frac{1}{3^2}]\\ & =\frac{5𝑅}{36}\\ \Rightarrow 𝑅& =\frac{36}{5𝜆}& (\therefore {𝜆}_1=𝜆,\text{ given) ..... (i)})\end{array}$

Similarly wavelength of second line,

$\begin{array}{cc}& \frac{1}{{𝜆}_2}=𝑅[\frac{1}{4}-\frac{1}{4^2}]=\frac{3𝑅}{16}\\ & \Rightarrow {𝜆}_2=\frac{16}{3𝑅}\text{.... (ii)}\end{array}$

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

${𝜆}_2=\frac{20}{27}𝜆\text{. }$

Hence, the wavelength of the second line in the same series is $\frac{20}{27}𝜆$.

Given that, work done, $𝑊=2J$

As, we have two wires of same material and they are elongated to the same length.

So, this means, their Young's modulus $\left(𝑌\right)$ is same and $\Delta {𝐼}_1=\Delta {𝐼}_2=\Delta 𝐼$ (say)

The work done in stretching a wire,

$𝑊=\frac{1}{2}𝐹\Delta 𝐼$

Now, for both the wires we have

$\begin{array}{cc}& \frac{{𝑊}_1}{{𝑊}_2}=\frac{\frac{1}{2}{𝐹}_1\Delta 𝑙}{\frac{1}{2}{𝐹}_2\Delta 𝑙}\\ \Rightarrow & {𝑊}_2=2\times \frac{{𝐹}_2}{{𝐹}_1}\text{.... (i)}\end{array}$

The forces by which the wires are stretched can be expressed as the ratio,

$\begin{array}{c}\frac{{𝐹}_1}{{𝐹}_2}=\frac{𝑌{𝐴}_1\frac{\Delta 𝐼}{{𝐿}_1}}{𝑌{𝐴}_2\frac{\Delta 𝐼}{{𝐿}_2}}\\ =\frac{{𝑟}_2^1{𝐿}_2}{{𝑟}_2^2{𝐿}^2}\text{... (ii)}\end{array}$

Given, ${𝑟}_2=2{𝑟}_1$ and ${𝐿}_2=\frac{{𝐿}_4}{2}$

Now, substituting the given values in Eq. (ii), we get

$\frac{{𝐹}_1}{{𝐹}_2}=\frac{{𝑟}_2^1{𝐿}_1}{4{𝑟}_2^{1}2{𝐿}_2}=\frac{1}{8}\text{.... (iii)}$

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

$\begin{array}{cc}{𝑊}_2& =2\times 8\\ & =16J\end{array}$

Given that $𝐑$ is resultant of $𝐏$ and $𝐐$ as shown in the figure below,

$\mathrm{BC}=P,\mathrm{CA}=𝑄$ and $\mathrm{BA}=𝑅$

Given, $𝐵𝐴$ and $𝐵𝐶$ are perpendicular and equal in magnitude.

So, from property of triangle,

$\angle 𝐴𝐶𝐵={45}^{\circ }$

Now, BC has to be extended up to $𝐷$ so, that $𝐶𝐷=𝑃$

Now, CD and CA have the initial point $𝐶$, so the angle between CD and CA;

$={180}^{\circ }-{45}^{\circ }={135}^{\circ }=\frac{3𝜋}{4}$

So, angle between $𝑃$ and $𝑄$ is $3\frac{𝜋}{4}$.

Resolving power of a telescope is given,

$𝑥=\frac{𝑎}{1.22𝜆}\Rightarrow 𝑥\propto 𝑎$

where, $𝑎$ = diameter/aperture of objective a telescope and $𝜆=$ wavelength of light.

So, when the aperture (diameter) of objective is large, then its resolving power is high.

A rod of length 6L and mass 8 m is given in figure,

When two masses strike the rod, then angular momentum imparted to rod,

$\begin{array}{cc}{𝐿}_1+{𝐿}_2& =2𝑚𝑣\left(𝐿\right)+𝑚\left(2𝑣\right)\left(2𝐿\right)\\ & =6𝑚𝑣𝐿\end{array}$

Now, after striking of masses to rod the angular momentum of complete rod about the centre $𝐶$,

${𝐿}_{\text{od }}=10\text{... (i)}$

where, $𝜔=$ angular velocity of the rod

and $𝐼=$ moment of inertia of the rod.

The moment of inertia of rod

$𝐼=\frac{𝑀{𝐼}^2}{12}=\frac{8𝑚(6𝐿{)}^2}{12}=24𝑚{𝐿}^2$

$\therefore 𝑀=8m$ and $𝐼=6𝐿$ (given)

Now, moment of inertia of two masses after the striking to $𝐼$ rod

${𝐼}_1=2𝑚(𝐿{)}^2=2𝑚{𝐿}^2$

and ${𝐼}_2=𝑚(2𝐿{)}^2=4𝑚{𝐿}^2$

$\therefore$ The net moment of the inertia about the centre of rod,

$\begin{array}{cc}& 𝐼=24𝑚{𝐿}^2+2𝑚{𝐿}^2+4𝑚{𝐿}^2\\ \Rightarrow & 𝐼=30{\mathrm{mL}}^2\end{array}$

By putting this value in Eq. (i), we get

${𝐿}_{\text{rod }}=30𝑀{𝐿}^2𝜔$

From the law of conservation of angular momentum,

$\begin{array}{cccc}& & {𝐿}_1+{𝐿}_2& ={𝐿}_{\text{rod }}\\ \Rightarrow & & 6𝑚𝑣𝐿& =30𝑚{𝐿}^2𝜔\\ \Rightarrow & & 𝜔& =\frac{𝑣}{5𝐿}\end{array}$

Hence, the anguler velocity of the rod is $\frac{𝑣}{5𝐿}$.

As we know that the magnitude of the resultant of two vectors $X$ and $𝐘$,

${𝑅}^2={𝑋}^2+{𝑌}^2+2𝑋𝑌\cos 𝜃\text{.... (i)}$

where, $𝜃$ is the angle between $𝑋$ and $𝑌$.

Putting,

$\begin{array}{cc}& 𝑋=(𝐴+𝐵)\\ & 𝑌=(𝐴-𝐵)\end{array}$

$\begin{array}{cc}& \text{ and }𝑅=\sqrt{\surd {𝐴}^2+{𝐵}^2}\text{ in Eq. (i), we get }\\ & \begin{array}{cc}& \begin{array}{cc}& {𝐴}^2+{𝐵}^2=(𝐴+𝐵{)}^2+(𝐴-𝐵{)}^2+2(𝐴+𝐵)(𝐴-𝐵)\cos 𝜃\\ & \Rightarrow {𝐴}^2+{𝐵}^2={𝐴}^2+{𝐵}^2+2𝐴𝐵+{𝐴}^2\\ & +{𝐵}^2-2𝐴𝐵+2({𝐴}^2-{𝐵}^2)\cos 𝜃\end{array}\end{array}\\ & \begin{array}{cc}& \Rightarrow \frac{-({𝐴}^2+{𝐵}^2)}{2({𝐴}^2-{𝐵}^2)}=\cos 𝜃\\ & \text{ we get, }𝜃={\cos }^{-1}[-\frac{({𝐴}^2+{𝐵}^2)}{2({𝐴}^2-{𝐵}^2)}]\end{array}\end{array}$

Key Idea Moment of inertia of a thin circular coil about its diameter,

$𝐼=\frac{𝑀{𝑅}^2}{2}$

Moment of inertia of a thin circular coil,

$𝐼=\frac{𝑀{𝑅}^2}{2}$

Now, moment of inertia of a ring about axis $𝑋{𝑋}^{\prime }$ as in figure below,

${𝐼}_{𝑋{𝑥}^{\prime }}=\frac{𝑀{𝑅}^2}{2}+𝑀{𝑅}^2=\frac{3}{2}𝑀{𝑅}^2\text{.... (i)}$

(Using theorem of parallel axis)

Given, $𝐿=$ length of wire of ring

and $𝜌=$ linear mass density

Then, mass of the ring $=$ linear density $\times$ length

$\begin{array}{cc}\Rightarrow & 𝑀=𝜌𝐿\text{... (ii)}\\ \text{ and }& 𝐿=2𝜋𝑅\\ \Rightarrow & 𝑅=\frac{𝐿}{2𝜋}\text{... (iii)}\end{array}$

Now, putting the value from Eqs. (ii) and (iii) in (i), we get

${𝐼}_{𝑥{𝑥}^{\prime }}=\frac{3}{2}\left(𝜌𝐿\right)\frac{{𝐿}^2}{4{𝜋}^2}\Rightarrow \frac{3𝜌{𝐿}^3}{8{𝜋}^2}$

Hence, option a is correct.

Torque is expressed as,

$𝜏=$ force $\left(𝐹\right)\times$ perpendicular distance $\left(𝑟\right)$

So,

$\begin{array}{cc}\left[𝜏\right]& =\left[𝐹\right]\times \left[𝑟\right]=\left[{\mathrm{MLT}}^{-2}\right]\left[L\right]\\ & =\left[{\mathrm{ML}}^2{T}^{-2}\right]\end{array}$