Atoms having the same number of neutrons, but different number of electrons or protons are called Isotones.

Nucleoids having the same atomic number, but a different mass number are known as Isotopes.

Atoms of different chemical elements that have the same number of nucleons are called Isobars.

Conductors $AOB$ and $COD$ are perpendicular to each other shown in figure.

At distance $a$ above $O$,

$B_1=\frac{{\mu }_0{i}_1}{2\pi a}$ 

And $B_2=\frac{{\mu }_0{i}_2}{2\pi a}$

$B_1$ is perpendicular to $B_2$.

Resultant of $B_1$ and $B_2$,

$B=\sqrt{B_1^2+B_2^2}$

$=\frac{{\mu }_0}{2\pi a}\sqrt{i_1^2+i_2^2}$

Due to friction between skin and cloths, electrostatic charge is built up on the skin. Hence, electrical discharge may occur when a man touches somebody else. This phenomenon is more significant in winters because due to low humidity, charge has a tendency to stay longer on the body.

The direction of the equipotential surface is from low potential to high potential is not the property of equipotential surfaces.

Any surface over which the electric potential is same everywhere is called an equipotential surface. No work is required to move a charge from one point to another on the equipotential surface. Properties of equipotential surface are:

• The electric field is always perpendicular to an equipotential surface.
• Two equipotential surfaces can never intersect.
• For a point charge, the equipotential surfaces are concentric spherical shells.
• For a uniform electric field, the equipotential surfaces are planes normal to the x-axis.
• The direction of the equipotential surface is from high potential to low potential.

Let $\mathrm{L}$ is the original length of the wire and $\mathrm{k}$ is force constant of wire.

Final length $=$ initial length $+$ elongation

${\mathrm{L}}^{\prime }=\mathrm{L}+\frac{\mathrm{F}}{\mathrm{k}}$

For first condition $\mathrm{a}=\mathrm{L}+\frac{4}{\mathrm{k}}$.....(i)

For second condition $\mathrm{b}=\mathrm{L}+\frac{5}{\mathrm{k}}$......(ii)

By solving Eqs. (i) and (ii), we get

$\mathrm{L}=5\mathrm{a}-4\mathrm{b}\text{ and }\mathrm{k}=\frac{1}{\mathrm{b}-\mathrm{a}}$

Now, when the longitudinal tension is $9\mathrm{N}$. length of the string.

$=\mathrm{L}+\frac{9}{\mathrm{k}}=5\mathrm{a}-4\mathrm{b}+9(\mathrm{b}-\mathrm{a})$

$=5\mathrm{b}-4\mathrm{a}$

The shear is experienced along the surface area of the punch. The surface of the punch is cylindrical with the diameter of 1 cm and a height of 0.3 cm which is the thickness of the sheet.

Therefore force needed to oppose the shear force is = force needed to punch the hole in the steel sheet.

$F=$ Area $\times$ Stress

$F=$ Shear $\times \pi dt$

$F=3.5\times {10}^8\times \pi \times 1\times {10}^{-2}\times 0.3\times {10}^{-2}$

$F=3.29\times {10}^{4}N\approx 3.3\times {10}^{4}N$

Given,

$N_p=4N_o$

$N_Q=N_o$

Also,

$T_P=1\min$

$T_Q=2min$

Now,

Amount left after time

$N_{P_t}=4N_O{\left(\frac{1}{2}\right)}^{\frac{t}{1}}$

And

$N_{Q_t}=N_o{\left(\frac{1}{2}\right)}^{\frac{t}{2}}$

Now,

According to question,

$N_{P_t}=N_{Q_t}$

Thus,

$4N_o{\left(\frac{1}{2}\right)}^t=N_o{\left(\frac{1}{2}\right)}^{\frac{t}{2}}$

Then, we get 

$4={\left(\frac{1}{2}\right)}^{\frac{-t}{2}}$

$\Rightarrow 4=2^{\frac{t}{2}}$

Further

$\frac{t}{2}=2$

Thus,

$t=4\min$

Thus,

For $R$

$N_R=(N_o-N_{P_t})+(N_o-N_{Q_t})$

Then,

$N_R=(N_o-\frac{N_o}{4})+(N_o-\frac{N_o}{4})$

Then, we get

$N_R=\frac{15N_o}{4}+\frac{3N_o}{4}$

Then,

$N_R=\frac{9N_o}{2}$

Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave- This statement is incorrect. 

Electromagnetic waves or EM waves: The waves that are formed as a result of vibrations between an electric field and a magnetic field and are perpendicular to each other and to the direction of the wave is called an electromagnetic wave.

Electromagnetic waves do not require any matter to propagate from one place to another as it consists of photons. They can move in a vacuum.

Properties of electromagnetic waves:

• Not have any charge or we can say that they are neutral.
• Propagate as a transverse wave.
• They move with the velocity the same as that of light i.e., $3\times {10}^8\mathrm{m}/\mathrm{s}$.
• It contains energy and they also contain momentum.
• They can travel in a vacuum also.

From above, it is clear that, electromagnetic waves do not require any matter to propagate from one place to another as it consists of photons. Therefore, statement in option (A) is correct.

In an electromagnetic wave, the electric field and magnetic field vary continuously with maxima and minima at the same place and same time. Therefore, statement in option (B) is correct.

The energy in an electromagnetic wave is divided equally between electric and magnetic fields. Therefore, statement in option (C) is correct. 

An electromagnetic wave is a perpendicular variation in both the electric field (E) and Magnetic field (B). Therefore, statement in option (D) is incorrect.

In this figure $Q$ and $P$ are at the same phase. Therefore, at $P$ point the path difference between ray $BP$ and reflected ray $OP$.

We can say, angles of $QO$ and $OP$ are the same. 

In triangle $POR,OP=\frac{PR}{cos\theta }=\frac{d}{cos\theta }$

In triangle $QOP,QO=OP\sin ({90}^{\circ }-2\theta )=OP\cos 2\theta$

$\mathrm{\Delta }=OP\cos 2\theta +OP$

$=OP(\cos 2\theta +1)$

$=2OP{\cos }^2\theta$

$=2\times \frac{d}{\cos \theta }\times {\cos }^2\theta$

$=2d\cos \theta$

Now, path difference is $\frac{\lambda }{2}$

Due to reflection at point $P$ 

$\mathrm{\Delta }=\frac{\lambda }{2},\frac{3\lambda }{2}\dots \dots \dots \dots \dots \dots$

$2d\cos \theta =\frac{\lambda }{2},\frac{3\lambda }{2}\dots \dots \dots$

$\cos \theta =\frac{\lambda }{4d},\frac{3\lambda }{4d}\dots \dots \dots \dots$

A pressure change at any point in the fluid is transmitted throughout the fluid such that the same change occurs everywhere – Pascal’s Law.

For every action, there is an equal and opposite reaction – Newton’s Third Law

Force is the time rate of change of momentum – Newton’s Second Law

For an ideal gas, the pressure is directly proportional to temperature and constant volume and mass – Ideal Gas Law

Given,

$V=220\mathrm{V}$

$\mathrm{f}=50\mathrm{Hz}$

$\omega =2\pi \mathrm{f}=2\pi \times 50=100\pi \mathrm{rad}/\mathrm{s}$

$\mathrm{R}=20\mathrm{\Omega }$

$\mathrm{L}=0.4\mathrm{H}$

$\Rightarrow {\mathrm{X}}_{\mathrm{L}}=\omega \mathrm{L}=100\pi \times 0.4=40\mathrm{\Omega }$

In the given circuit, capacitor is absent.

$\therefore \tan \theta =\frac{X_L-X_C}{R}=\frac{X_L}{R}=\frac{40\pi }{20}=2\pi$

Therefore, $\theta ={\tan }^{-1}(2\pi )$

Here, $H=1500\mathrm{A}{\mathrm{m}}^{-1},\varphi =2.4\times {10}^{-5}$ weber

$A=0.5{\mathrm{cm}}^2=0.5\times {10}^{-4}{\mathrm{m}}^2$

$\therefore B=\frac{\varphi }{A}=\frac{2.4\times {10}^{-5}}{0.5\times {10}^{-4}}=4.8\times {10}^{-1}\mathrm{T}$

and $\mu =\frac{B}{H}=\frac{4.8\times {10}^{-1}}{1500}=3.2\times {10}^{-4}$

So relative permeability,

${\mu }_r=\frac{\mu }{{\mu }_0}=\frac{3.2\times {10}^{-4}}{4\pi \times {10}^{-7}}=0.255\times {10}^3=255$

The unit of light year, wavelength and displacement is meter.

The unit of light year is meter. A light year is the distance traveled by light in one year. and wavelength is the distance between two consecutive vertices or descents. The unit of wavelength is also the meter. The minimum distance covered by an object in a certain direction with respect to a reference point is called displacement. The unit of displacement is also the meter.

When we rub a glass rod with silk, then the charge on the glass rod will be positive. When we rub a glass rod with silk, some of the electrons from the rod are transferred to the silk cloth. Thus the rod gets positively charged and the silk gets negatively charged.

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

Work done, $(W)=\mathrm{\Delta }K=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

Where $m$ is the mass of the object, $v$ is the final velocity of the object and $u$ is the initial velocity of the object.

Work-energy theorem for a variable force:

Kinetic energy, $K=\frac{1}{2}m{v}^2$

$\frac{dK}{dt}=\frac{d\left(\frac{1}{2}m{v}^2\right)}{dt}$

$\Rightarrow \frac{dK}{dt}=m\frac{dv}{dt}v$

$\Rightarrow \frac{dK}{dt}=mav$

$\Rightarrow \frac{dK}{dt}=Fv$

$\Rightarrow \frac{dK}{dt}=F\frac{dx}{dt}$

$\Rightarrow dK=Fdx$

On integrating, we get

${\int }_{K_i}^{K_f}dK={\int }_{x_i}^{x_f}Fdx$

$\Rightarrow \mathrm{\Delta }K={\int }_{x_i}^{x_f}Fdx$

From the work-energy theorem, a change in kinetic energy equals the work done.

$\mathrm{\Delta }K={\int }_{x_i}^{x_f}Fdx$

$\Rightarrow \mathrm{\Delta }K={\int }_{-a}^a(P{y}^2+Qy+R)dy$

$\Rightarrow \mathrm{\Delta }K={[\frac{P{y}^3}{3}+\frac{Q{y}^2}{2}+Ry]}_{-a}^a$

$\Rightarrow \mathrm{\Delta }K=[(\frac{P{a}^3}{3}+\frac{Q{a}^2}{2}+Ra)-(\frac{-P{a}^3}{3}+\frac{Q{a}^2}{2}-Ra)]$

$\Rightarrow \mathrm{\Delta }K=\frac{2P{a}^3}{3}+2Ra$

Given,

 $n=800$

$\mathrm{A}=2.5\times {10}^{-4}{\mathrm{m}}^2$

$\mathrm{I}=3.0\mathrm{A}$

A magnetic field develops along the axis of the solenoid. Therefore current-carrying solenoid acts like a bar magnet.

Associated magnetic moment,

$\mathrm{m}=\mathrm{nIA}$

$=800\times 3\times 2.5\times {10}^{-4}$

$=0.6{\mathrm{JT}}^{-1}$

Maxwell Boltzmann distribution for velocity is given by:

$n(E)dE=\frac{2\pi N}{V(\pi kT{)}^{\frac{3}{2}}}{E}^{\frac{1}{2}}{e}^{-\frac{E}{kT}}dE$

Where, $n=$ number of molecules, $T=$ temperature, $k=$ Boltzmann constant, and $E=$ energy

From the above equation, it is clear that the fraction of gas molecules having an energy between $E$ and $E+dE$ is proportional to $E^{\frac{1}{2}}\exp (-\frac{E}{kT})$.

Given,

Mass $(m)=5kg$

Radius $(R)=1m$

Velocity $(v)=300rpm=\frac{300}{60}=5rpm$

As we know,

The angular speed is given by,

$\omega =2\pi v$

$\therefore \omega =2\pi \times 5=10\pi rad{s}^{-1}$

$v=\omega R$

$\therefore v=10\pi \times 1=10\pi m{s}^{-1}$

Kinetic energy, $(KE)=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2{R}^2$

$=\frac{1}{2}\times 5\times (10\pi {)}^2$

$=250{\pi }^2$

The loudness and pitch of a sound depends on intensity and frequency.

Loudness is a sensation of how strong a sound wave is at a place. It is always a relative term. It is a dimensionless quantity. Its unit is decibel (dB).

Pitch is the characteristic of sound by which an acute (or shrill) note can be distinguished from a grave or a flat note. The term 'pitch' is often used in music. It depends on the frequency of the sound wave. A note of higher frequency is at a higher pitch than a note of lower frequency. It is a qualitative term and cannot be quantified.

The following figure shows the cyclic process of gas. If an object returns to its initial position after one or more processes it went through.

$ABCDA$ is the cycle. The pressure at the points both $D$ and $C$ remain the same, which is $2P$. The volume at the point $D$ is $V$ and at the point $C$ is $3V$. So, the work done by the gas from point $D$ to point $C,W_{DC}=2P(3V-V)=4PV$

The pressures at the points $C$ and $B$ are $2P$ and $P$ respectively. The volume at the points both $C$ and $B$ remain the same, which is $3V$. So, the work done by the gas from point $C$ to point $B,$

$W_{CB}=P(3V-3V)=0$

The pressure at the points both $B$ and $A$ remain the same, which is $P$. The volume at the point $B$ is $3V$ and at the point $A$ is $V$,

So, the work done by the gas from point $B$ to point $A,W_{BA}=P(V-3V)=-2PV$

The pressures at the points $A$ and $D$ are $P$ and $2P$ respectively. The volume at the points both $A$ and $D$ remain the same, which is $V$.

So, the work done by the gas from point $A$ to the point $D,W_{AD}=P(V-V)=0$

Hence the total work done in the whole cycle,

$W=4PV-2PV=2PV$

We know the heat rejected from the cycle is equal to the amount of total work done by the gas, so $Q=W$

$Q=2PV$

Work energy theorem.

$W=\frac{1}{2}I({\omega }_f^2-{\omega }_i^2)$

Here, $\theta =2\pi$ revolution

$=2\pi \times 2\pi =4{\pi }^{2}rad$

$W_i=3\times \frac{2\pi }{60}rad/s$

$\Rightarrow -\tau \theta =\frac{1}{2}\times \frac{1}{2}m{r}^2(0^2-{\omega }_1^2)$

$\Rightarrow -\tau =\frac{\frac{1}{2}\times \frac{1}{2}\times 2\times (4\times {10}^{-2}){(-3\times \frac{2\pi }{60})}^2}{4{\pi }^2}$

$\Rightarrow \tau =2\times {10}^{-6}Nm$

The work to be done in the process is given by $dW=PdV$

$=1\times {10}^{5}Pa\times (1.671-0.001)m^3$

$=1.670\times {10}^{5}J$

The change in heat energy during the vaporisation process can be calculated as follows-

$\mathrm{\Delta }{Q}_{\text{supplied }}=2257\times 1\times {10}^{3}J$

$=22.57\times {10}^{5}J$

Hence, the change in internal energy in the process is given by

$\mathrm{\Delta }U=\mathrm{\Delta }{Q}_{\text{supplied }}-\mathrm{\Delta }W$

$=(22.57-1.67)\times {10}^{5}J$

$=20.9\times {10}^{5}J$

$=2090kJ$

Threshold frequency is given as:

${\mathrm{v}}_0=\frac{W_0}{h}$

Here $W_0=3.3\times {10}^{-19}\mathrm{J}$

$h=6.6\times {10}^{-34}\mathrm{Js}$

${\mathrm{v}}_0=\frac{(3.3\times {10}^{-19})}{(6.6\times {10}^{-34})}$

${\mathrm{v}}_0=5\times {10}^{14}\mathrm{Hz}$

The velocity of sound in air is given by:

${\mathrm{C}}_{\mathrm{s}}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{M}}}$.....(1)

where $\gamma$ is the degree of freedom

$\mathrm{T}=\text{ temperature }$

$\mathrm{M}=\text{ mass }$

According to the kinetic theory of gases, The rms velocity of sound is given by:

$\mathrm{C}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}}$.....(2)

Dividing equation (1) and (2) we get:

$\frac{{\mathrm{C}}_{\mathrm{s}}}{\mathrm{C}}=\sqrt{\frac{\gamma }{3}}$

$\Rightarrow {\mathrm{C}}_{\mathrm{s}}=\mathrm{C}\sqrt{\frac{\gamma }{3}}$

Equation is given as

$\sqrt{x}=2t+3$

Taking square both side

$(\sqrt{x}{)}^2=(2t+3{)}^2$

$x=(2t+3{)}^2$

$x=(2t{)}^2+(3{)}^2+2(2t)(3)$

$x=4t^2+12t+9$

Taking derivative both side relative to ${}^{\prime }{t}^{\prime }$

$\frac{dx}{dt}=2(4t)+12$

$\frac{dx}{dt}=8t+12$

We know that,

$v=\frac{dx}{dt}$

Hence, $v=8t+12$

at $t=0$

$v=8(0)+12$

$v=12m/s$

Then, the initial velocity is $12m/s$.

$\begin{array}{rl}& R_1=\frac{R_2}{2}\\ & R_0{\left(A_1\right)}^{1/3}=\frac{R_0}{2}{\left(A_2\right)}^{1/3}\\ & A_1=\frac{1}{8}{A}_2\\ & A_1=\frac{192}{8}=24\end{array}$

The inverter converts DC into AC with the help of inductors and capacitors. 

The inverter is a static device that can convert one form of electrical power into another form of electrical power. The device can be used for back power supply in homes. 

Specific heat:  $\left[{\mathrm{L}}^2{\mathrm{T}}^{-2}{\mathrm{K}}^{-1}\right]$

Electrical potential: $\left[M^1{L}^2{T}^{-3}{A}^{-1}\right]$

Electrical resistance: $\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-2}\right]$

Magnetic flux: $\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}{\mathrm{A}}^{-1}\right]$

From the above information, it is clear that the specific heat does not have mass in its dimension. 

We know that,

$\text{ Escape velocity }(v_e)=\sqrt{\frac{2GM}{R}}\dots \dots ..(1)$

Where $G$ is the Universal Gravitational Constant and $M$ is the mass of the Earth.

Velocity of the object.

$v=\frac{v_e}{3}=\sqrt{\frac{2GM}{9R}}$..... (2)

The initial kinetic energy of the object.

$K{E}_1=\frac{1}{2}m{v}^2=\frac{1}{2}m\frac{2GM}{9R}\dots$... (3)

The initial potential energy of the object.

$P{E}_1=-\frac{GMm}{R}\dots \dots .$ (4)

Here $m$ is the mass of the object.

Let $h$ be the maximum height reached by the object. when the object reaches the maximum height, its velocity becomes zero which is why its final kinetic energy $=K{E}_2=0\dots \dots \dots$ (5)

At the maximum height, the potential energy of the object $\left(P{E}_2\right)=-\frac{GMm}{R+h}\dots \dots .(6)$

According to the law of conservation of energy,

$K{E}_1+P{E}_1=K{E}_2+P{E}_2$

(From $3,4,5$, and 6)

$\frac{1}{2}m\frac{2GM}{9R}-\frac{GMm}{R}=0-\frac{GMm}{R+h}$

$\Rightarrow \frac{1}{9R}-\frac{1}{R}=-\frac{1}{R+h}$

$\Rightarrow -\frac{8}{9R}=-\frac{1}{R+h}$

$\Rightarrow 8R+8h=9R$

$\Rightarrow h=\frac{R}{8}$

Thus, the maximum height reached will be $\frac{R}{8}$.

Dimension of pressure, stress & Young's modulus is $\left[{\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^{-2}\right]$.

While dimension of power $=\frac{\text{ Energy }}{\text{ Time }}$

$=\frac{\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-2}}{\mathrm{T}}$

$=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}\right]$

The rms speed of any homogeneous gas sample is given by:

$V_{rms}=\sqrt{\frac{3RT}{M}}$ $\cdots (1)$

Where, $R=$ universal gas constant, $T=$ temperature and $M=$ Molecular mass

Here, $M$ and $R$ is constant,

On increasing the value of T by 3 in equation (1) we get,

$V_{rms}\propto \sqrt{3T}$

If the temperature is increased to $3$ times, then $V_{rms}$ is increased by $\sqrt{3}$ times.

The direction of EM waves is found from the cross product of the electric field and magnetic field. 

Since both electric and magnetic fields are vectors, the direction of propagation of EM waves is obtained from the right-hand rule.

Let the electric field be denoted by $\overrightarrow{E}$ and magnetic field be denoted by $\overrightarrow{B}$.

According to the rule - If the fingers of the right hand are curled so that they follow a rotation from $\overrightarrow{E}$ to $\overrightarrow{B}$, then the thumb will point in the direction of the vector product i.e., the direction of EM waves.