The scientific principle that forms the basis of tokamak technology is magnetic confinement of plasma. This involves using strong magnetic fields to confine hot plasma within a toroidal chamber, allowing for controlled nuclear fusion reactions. While superconductivity is often employed to generate the necessary magnetic fields efficiently, the fundamental principle underlying tokamaks is magnetic confinement rather than superconductivity itself. Controlled nuclear fission is a different process used in nuclear reactors, and the motion of charged particles in electromagnetic fields is a general principle relevant to many areas of physics but not specific to tokamak technology.

Thus, the scientific principle that forms the basis of tokamak technology is magnetic confinement of plasma.

The dimensional formula for  \(E=E_oe^{\frac{-t}{t_o}}\) as \(e^{\frac{-t}{t_o}}\) is a dimensionless quantity.

In the question, \(E\) and \(E_o\) represents the energies. \(E\) is the time-averaged energy of the oscillator which is dimensionally represented as \(ML^2T^{-2}\) and \(E_o\) is the total energy of an undamped oscillator which is represented by \(E_o=\frac12mA^2w_o^2\), where \(m\) is the mass, \(A\) is the amplitude, and \(w\) is the angular velocity.

The dimensional representation of the total energy of an undamped oscillator is

\(E_o=ML^2T^{-1^2}=ML^2T^{-2}\)

Thus, the dimensional formula is \(E=E_oe^{\frac{-t}{t_o}}\).

We have given that the bridge is 45 m above the water. When the ball is dropped such that it directly falls in the boat, the boat has moved a distance of 12 m by the time the ball reaches the boat. Since the boat is moving, we can observe the ball is dropped into the boat even when the boat is not directly at the bottom of the bridge.

We can calculate the time taken by the ball to fall into the moving boat using kinematic equation,

\(s=ut+\frac12gt^2\)

Here, s is the distance covered by the ball, u is the initial velocity of the ball, g is the acceleration due to gravity and t is the time.

Since the ball is dropped from that height, we can see the ball has zero initial velocity. Therefore, we can write the above equation as follows,

\(s=\frac12gt^2\)

\(\Rightarrow \;t=\sqrt{\frac{2s}g}\)

We substitute 45 m for s and \(10\;m/s^2\)  for g in the above equation.

\(t=\sqrt{\frac{2(45)}{10}}\)

\(t=3\;s\)

Now, we have the relation between distance, velocity and time,

\(v=\frac dt\)

We substitute 12 m for d and 3 s for t in the above equation.

\(v=\frac{12}3=4\;m/s\)

Thus \(v=4\;m/s\)

A ball is dropped from a height \(H\) from rest and the ball travels \(\frac H2\) in last \(1\;s\).

We are required to find the total time taken by the ball to hit the ground.For that we shall use equation of motions for the ball along its entire journey and journey along \(\frac H2\) height.

Let the time taken by the ball to cover entire height be \(t\) sec.Now for entire journey the equation of motion is-

\(-H=0+\frac12(-g)t^2\)

\(H=\frac12gt^2\)              (1)

Similarly,for covering half of the height the equation is

\(-\frac H2=0+\frac12(-g)(t-1)^2\)

\(H=g(t-1)^2\)                       (2)

Comaring (1) and (2) we get

\(\frac{t^2}2=(t-1)^2\)

\(t^2=2(t^2-2t+1)\Rightarrow \;t^2-4t+2=0\)

\(t=\frac{4\pm\sqrt{16-8}}2=2\pm\sqrt2\)

\(t=2+\sqrt2=3.41\;s\)

Thus \(t=3.41\;s\)

Given the planet completes 2 revolutions in 360 days.

One rotation of the planet moves  \(2\pi\) radians angle.

Thus, angular frequency  \(\omega=\frac\theta t\) where \(\theta\) represents the angle moved by theobject and \(t\) represents the time taken by the object to travel through \(\theta\).

\(\omega=\frac{2\times2\pi}{360}=0.0348=3.5\times10^{-2}\;rad/day\)

If the vector sum of two forces is perpendicular to their vector difference, it implies that the two forces are equal to each other in magnitude. This is a property of vectors known as the parallelogram law of vector addition. When the resultant (sum) of two vectors is perpendicular to their difference, it means the vectors are of equal magnitude. Therefore, the correct option is: "are equal to each other in magnitude."

A mass \(m\) is attached by a massless string that is wounded around a uniform hollow cylinder. Now, according to Newton's second law of motion, the force acting on a body is equal to the product of mass and acceleration.

\(F=ma\)

Now, if the tension is produced in the string, than the force is given by

\(F=mg-T\)

Now, putting the value of  \(F\)  in the above equation, we get

\(ma=mg-T\)

The linear acceleration  \(a\) will rise tangentially with the angular acceleration of the string, therefore, the acceleration is given by

\(a=R\alpha\)

Now, the torque produced by the angular movement of the string is the result of the angular acceleration of the cylinder. Therefore, the torque is given by

\(τ =I\times\alpha\)

Here, \(I\) is the rotational inertia in the cylinder and is given by

\(I=mR^2\)

\(τ=mR^2\alpha=mRa\)

Which is the rotational torque.

This torque also produces a linear torque and is given by

\(τ=TR\)

Now, equating the equations of  \(τ\), we get

\(mRa=TR\)

\(ma=T\)

Now, the tension in the string is given by

\(T=mg-ma\)

\(2ma=mg\)

Thus, \(a=\frac g2\).

According to the equation for the coefficient of restitution

\(e=\frac{velocity\;of\;seperation\;after\;collision}{velocity\;of\;approach\;before\;collision}\)

\(e=\frac{v_2-v_1}{u_1-u_2}\) along the line of impact

From the given information, \(e=\frac{v_2-0}{u_1cos\theta-0}=1\)

\(u_1cos\theta=v_2\)

According to the conservation of momentum along the line of impact

\(m_1u_1cos\theta=m_2v_2\)

\(m_1=m_2\)

Thus, \(\frac{m_2}{m_1}=1\)

Given

Mass of the block \(100\;g\)

Initial velocity of the block \(u=2\;m/s^2\)

Final velocity of the block \(v=\frac u2=1\;m/s^2\)

As non-conservative forces are not acting on the system we can conserve energy.

Thus, loss in kinetic energy of block  \(=\)  Gain in potential energy of spring

\(TE_i=TE_f\)

\(\frac12mu^2=\frac12mv^2+\frac12kx^2\)

\(\frac12\times0.1\times2^2=\frac12\times0.1\times1^2+\frac12k\times0.02^2\)

Thus, \(k=750\;N/m\).

Given: Mass of bullet \(m\), the velocity of bullet \(v_0\) and mass of the wooden block \(M\).

Let the velocity of the bullet after the collision with the wooden block be  \(v\)

Applying the law of conservation of momentum, we obtain.

\(mv_o=M+mv\)

\(v=\frac{mv_o}{M+m}\)

Now applying the law of conservation of mechanical energy, we obtain

\(\frac12mv_o^2=mgh\)

Putting the value of  \(v\)  in the above equation, we get 

\(\frac12M+m(\frac{mv_o}{M+m})^2=M+mgh\)

Maximum height of the block-bullet system 

Thus,\(h=\frac12mv_o^2(\frac{M}{m+M})\).

Given: Masses of the balls  \(m_1\) and \(m_2\) are \(100\;g\) and \(200\;g\) respectively

Applying the equation of motion  \(s=ut+\frac12gt^2\)  for two cases.

As the bast is at rest initially, \(u=0\) thus distance travelled by the ball will be \(s=\frac12gt^2\)

Now, for the first ball \(s_1=\frac12gt_1^2\) and for the second ball \(s_2=\frac 12gt_2^2\) 

Here \(t_1\) and \(t_2\)  are the time of fall of both balls.

 The distance between the center of mass of two balls is given by

\(s_{cm}=\frac{m_1s_1+m_2s_2}{m_1+m_2}\)

\(s_{cm}=\frac{100\times0.4^2\times\frac{10}2+200\times0.2^2\times\frac{10}2}{100+200}=0.4\;m\)

Kinetic energy of the centre of mass \(KE=\frac43\;joule\)

The velocity of centre of mass  \(v_{cm}\) is equal to the summation of the momentum of each ball to the total mass of the system.

\(v_{cm}=\frac{2\times M+1\times1}{M+1}=\frac{2M+1}{M+1}\)

Now kinetic energy of the centre of mass is  \(KE=\frac12mv^2\)

\(KE=\frac12(M+1)(\frac{2M+1}{M+1})^2=\frac43\)

\((2M+1)^2=\frac{8(M+1)}3\)

\(3(2M+1)^2=8(M+1)\)

\(12M^2+4M-5=0\)

\(M=\frac12=0.5\;kg\).

Frequency of the simple harmonic oscillator \(f=1\;Hz\), and phase \(\theta=1\;radian\)

The expression of the phase of a simple harmonic oscillator is stated as  \(\theta=\omega t\) where \(\omega\) is the angular velocity \(t\) is the time.

\(\omega=\frac\theta t\)

we know that \(\omega=2\pi f=2\pi\)

Thus time, \(t=\frac\theta\omega=\frac1{2\pi}\)

Thus, shift in time should be \(-\frac1{2\pi}\;\;\,s\).

Given the orbit of the planet around the Sun, the time taken to travel different segments of the orbit depends on the relative positions and speeds of the planet and the Sun.

If we consider DAB and BCD as segments of the orbit

DAB: This segment represents the part of the orbit where the planet is moving from point D to point B. As the planet moves closer to the Sun, it speeds up due to gravitational acceleration.

BCD: This segment represents the part of the orbit where the planet is moving from point B to point C. As the planet moves away from the Sun, it slows down due to gravitational deceleration.

Therefore, the time taken to travel DAB is generally less than that for BCD because the planet moves faster when closer to the Sun.

In fact, gases are highly compressible compared to solids and liquids. The compressibility of a substance refers to its ability to be squeezed into a smaller volume under pressure. Gases have much greater compressibility compared to liquids and solids because their particles are more widely spaced and have weaker intermolecular forces. Therefore, the statement "Gases are least compressible" is incorrect.

The relation of viscous force between different layers of fluid is given as

\(F=\eta A\frac{dv}{dx}=\eta A\frac VH\)

Here \(\eta\) is the viscosity of water, \(A\) is the area, and \(\frac{dv}{dx}\) is the rate of change of velocity with the change in position.

Shearing stress \(=\frac FA=\eta\frac VH\).

Length of steel rod at \(25^oC\) is \(1\,m\).

The actual length measured by another metal scale at \(25^oC\) is \(L=1+\alpha_m\Delta T\)

\(L=1+25\times20\times10^{-6}=1.0005\;m\)

This is the actual length measured by the other metal scale.

Now, the steel rod has the length \(1.0005\;m\) at the temperature of \(25^oC\)

Let \(L_o\) be the length of the steel rod at \(0^oC\),

\(L=L_o+\alpha_s\Delta T\)

Putting the values in the above equation, we get

\(1.0005=L_o+12\times10^{-6}\times25\)

\(L_o=\frac{1.0005}{1.0003}=1.00019\approx1.0002\;m\).

Given Latent heat of vaporisation of water \(=22.6\times10^5\;J/kg\)

Mass of water \(=100\;kg\)

We know that to condense water from vapour we need \(L_v\times m\) of energy

Thus, \(Q=22.6\times10^5\times100=22.6\times10^7\;J\) is needed.

According to the first law of thermodynamics

\(Q=\Delta U+W\), where Q is the heat transfer to the system, \(\Delta U\) is the change in the internal energy, and \(W\) is the work done by the system.

Applying for the path 1, puttingthe value from the given figure

\(5PV=\frac f2nR\Delta T+W\)

\(5PV=\frac f2 2PV+2PV\)

\(f=3\)

Now applying for path 2,

\(Q=\Delta U+W \)

\(Q=\frac32\frac{PV}2+\frac32\frac{3PV}2+\frac{PV}2+2PV\)

Thus, \(Q=\frac{11}2PV\).

Abdus Salam, a Pakistani theoretical physicist, was known for his work on the unification of weak and electromagnetic interactions. This work, which laid the foundation for the electroweak theory, was a significant contribution to our understanding of fundamental particle interactions. Salam shared the 1979 Nobel Prize in Physics with Sheldon Glashow and Steven Weinberg for this achievement.

A light year is a unit of length, not time. It represents the distance that light travels in one year in a vacuum. Since it's a measure of distance, its dimensions are the same as those of length. Therefore, the dimension of a light year is simply [L], which stands for length.

Distance travelled in \(t\;s\)

\(x=a+bt^2\)

when object covers 16 m in 2 sec

\(16=a+b\times4\)    (1)

Distance travelled between

\(t=3\) and \(t=5\)

\(s'=(a+25b)-(a+9b)=16b\)

Average velocity \(=\frac{s'}{\Delta t}=\frac{16b}2=28\)

\(b=3.5\,m/s^2\)

Substituting in (1)

\(16=a+3.5\times4\)

\(a=16-14=2\;m\).

Thus, \(a=2\,m, \;b=3.5\,m/s^2\).

Area under velocity-time graph will give the value of displacement

Area from \(t=0\) to \(t=5\) 

\(A= \frac12\times5\times12=30\)

The slop of graph will give the value of acceleration, from \(t=5\) to \(t=10\),

Acceleration \(a=\frac{-12}5=-2.4\;m/s^2\)

Distance travelled from \(t=5\) to \(t=6\)

\(s=ut+\frac12at^2\)

\(s=12-\frac12\times2.4\times1=10.8\)

Total distance \(=30+10.8=40.8\;m\)

Given

\(\vec a=\hat i+2\hat j+2\hat k\)

\(\vec b =\hat i-\hat j+\hat k\)

\(\vec a+\vec b=(1+1)\hat i+(2-1)\hat j+(2+1)\hat k=2\hat i+1\hat j+3\hat k\)

\(|\vec a+\vec b|=\sqrt{2^2+1^2+3^2}=\sqrt{14}\)

 unit vector can be given as \(\frac{\vec a+\vec b}{|\vec a+\vec b|}=\frac{2\hat i+\hat j+3\hat k}{\sqrt{14}}\)

Range \((R)\;\;\,=\frac{u^2sin2\theta}g\)

When \(\theta=15^o\)

\(R-10=\frac{u^2}{2g}\)                       (1)

When \(\theta=45^o\)

\(R+15=\frac{u^2}g\)                       (2)

dividing eq(1) by eq(2)

We get,

\(\frac{R-10}{R+15}=\frac12\)

\(R=35\)

\(R=\frac{u^2sin2\theta}g=50sin\)

\(sin2\theta=\frac{35}{50}=\frac7{10}\)

\(\theta=\frac12sin^{-1}\frac7{10}\)

Given

Mass of ball A = 50 kg

Mass of ball B = 20 kg

Initial velocity of ball A = 9 km/h

Initial velocity of ball B = 18 km/h

Final velocity of ball B = 27 km/h

Let final velocity of ball A be 'v'

According to law of conservation of momentum

\(m_1u_1+m_2u_2=m_1v_1+m_2v_2\)

Thus, \(50\times9+20\times18=50\times v+20\times(-27)\)         [ here \(v\) is negative due to the opposite direction of motion]

\(v=27\)

Given

Radius of turn \(R=10\;m\)

Speed of car \(=10\;m/s\)

Let the angle made by the string with the vertical be \(\theta\)

Let the mass of the bob be \(m\)

Let \(T\) be the tension in stringof bob

Bob moves in a horizontal circle of radius  \(R\), Because the length of string is very small compared to \(R\)

Thus, No net force acts in vertical direction.

\(Tcos\theta=mg\)                          (1)

Horizontal force on the bob,

\(Tsin\theta=F_c\)                          (2)

Where \(F_c\) is centripetal force

\(F_C=\frac{mv^2}R\)                             (3)

From equations (1),(2) and(3)

\(tan\theta=\frac{v^2}{Rg}\)

\(\theta=45^o\)

Given

Spring force constant \(K\)

compression of the spring \(x\)

Energy stored in the spring \(\frac12Kx^2\)

Energy after reaching height h is \(mgh\)

Thus according to law of conservation of energy 

\(\frac12Kx^2=mgh\)

\(\Rightarrow\;x=\sqrt{\frac{2mgh}K}\)

Given 

Both masses of 1g and 4g are having equal kinetic energy.

Let \(p_1\) and \(p_2\) are momentums of 1g and 4g respectively.

Thus,

\(\frac{\frac12m_1v_1^2}{\frac12m_2v_2^2}=\frac11\)

\(\Rightarrow\;\frac{m_1v_1^2}{m_2v_2^2}=\frac11\)

Now multiplying \(\frac{m_1}{m_2}\) on both sides

\(\Rightarrow\;\frac{(m_1v_1)^2}{(m_2v_2)^2}=\frac{m_1}{m_2}\)

\(\Rightarrow\;\frac{p_1^2}{p_2^2}=\frac{m_1}{m_2}\)

Thus,\(\frac{p_1}{p_2}=\frac12\)

\(\therefore \,p_1:p_2=1:2\)

Given 

\(\omega=at\hat i+bt^2\hat j\)

\(a=1\;rad/s^2\) , \(b=0.5\;rad/s^3\)

angular accleration \(\alpha=\frac{d\omega}{dt}\)

\(\Rightarrow\;\alpha=at\hat i+2bt\hat i\)

Now \(\vec{\alpha}.\vec \omega=|\alpha ||\omega |cos \theta\)

\(cos\theta=\frac{\vec \alpha\vec\omega}{|\alpha||\omega|}\)

\(cos\theta=\frac{(a\hat i+2bt \hat j).(at\hat i+bt^2\hat j)}{\sqrt{(at)^2+(bt^2)^2}\sqrt{a^2+(2bt)^2}}\)

Now substituting value of a abd b and \(t=1\;s\)

\(cos\theta=\frac3{\sqrt{10}}\)

thus, \(\theta=cos^{-1}(\frac3{\sqrt{10}})\)

The uniform wire is bent to form a circle. The length of the wire equals the circumference of the circle. As the circumference of the circle is the value of its boundary. In simple words, if we reform the circle into a straight wire, then, the length of the wire will be equal to that of the circumference of the circle.

 The circumference of the circle is given by the formula as \(C=2\pi R\)

Where R is the radius of the circle.

From given, we have the length of the wire to be equal to “L”.

So, we have, the circumference of the circle is equal to the length of the wire.

\(L=C=2\pi R\)

\(R=\frac L{2\pi}\)

We have found the equation in terms of the radius, because the centre of the mass shifts by a value equal to that of the radius of the circle.

At the value of the shift in the centre of the mass equals \(\frac L{2\pi}\).

Given, the function \((\;sin\omega t-cos\omega t\;)\) representing the S.H.M

let \(x=sin\omega t-cos\omega t\)

\(x=2(\frac12sin\omega t-\frac12 cos\omega t)\)

\(x=2sin(\omega t+\frac{\pi}2)\)

On comparing with the SHM equation \(x=Asin(\omega t+\phi)\)

Thus, \(T=\frac{2\pi}{\omega}\).

The potential energy stores per unit volume is given as

\(\rho=\frac12\times stress\times strain\)

we know that, \(Y=\frac{stress}{strain}\)

\(\rho=\frac12\times Y\times strain^2=\frac12\times16\times10^{10}\times\frac{5\times10^{-3^2}}{10}\)

\(\rho=2\times10^4\; J/m^3\)

When the cylinder displaced slightly by some distance  \(x\) , the net restoring force acting on the cylinder due to upthrust

\(F=V\rho g=\pi r^2\rho g x =m\omega^2 x\)

\(\omega=\sqrt{\frac{\pi r^2\rho g}m}=2\pi n\)

\(n=\frac1{2\pi}\sqrt{\frac{\pi r^2\rho g}m}\)

Now substituting the values

we get,

\(n=\frac1{20}\sqrt{\frac\rho\pi}\).

Amount of heat energy required to convert  \(50\;g\) of water at \(80^oC\) to water at \(0^oC\).

\(Q=ms\Delta T=50\times1\times80=4000\;cal\)

The heat energy required to melt \(50\;g\) ice

\(Q=mL=50\times80=4000\;cal\)

So the equilibrium temperature \(0^oC\).

Rayleigh-Jeans formula is an approximation of the Planck's formula for the spectral radiance of blackbody radiation in the limit of low frequencies or long wavelengths. It is given by

\(B(ν,T)=\frac{8\pi ν^2 kT}{c^3}\)

In the limit of low frequencies or long wavelengths, Planck's formula reduces to the Rayleigh-Jeans formula. This limit is known as the Rayleigh-Jeans limit. It's important to note that the Rayleigh-Jeans formula diverges at high frequencies, which is known as the ultraviolet catastrophe. Planck's formula, on the other hand, correctly describes the observed blackbody radiation spectrum over all frequencies

Thus, at long wavelength region Plank's formula reduces to Rayleigh Jean's formula.

Coefficient of performance of air conditioner \((\alpha)=\frac{heat\;removed}{energy\;supplied}=\frac{1200}{400}=3\)

Coefficient of performance of refrigerator \((\beta)=30\)

if the efficiency of the heat engine is   \(\eta\)

\(\beta=\frac{1-\eta}\eta\)

\(30=\frac{1-\eta}\eta\)

\(\eta=\frac1{31}\)

\(\eta=1-\frac{T_i}{T_o}\)

\(\frac1{31}=1-\frac{T_i}{310}\)

\(T_i=300\;K\)

Thus, \(T_i=27^oC\)

Given, Pressure of gas A is three times the pressure of the gas B 

\(P_A=3P_B\)

Let the densities of gasses be \(\rho_A\) and \(\rho_B\) then as per the given condition of the question

\(\rho_A=2\rho_B\)

According to the ideal gas equation  \(PV=RT\)

\(P=\frac{\rho RT}M\)

\(M=\frac{\rho RT}M\)

Now taking relation

\(\frac{M_A}{M_B}=\frac{\rho_A}{P_A}\times\frac{P_B}{\rho_B}\)

Now substituting the given values

\(\frac{M_A}{M_B}=\frac{2\times1}{1\times3}=\frac23\)

Phase difference between two coherent plane wave \(\phi=60^o\)

The resultant intensity of two coherent waves is given as,  \(I_R=I_1+I_2+2\sqrt{I_1I_2}cos\phi\)

Where \(I_1\) and \(I_2\) are the intensities of two coherent waves.

Since the intensities of both waves are identical \(I_1=I_2=I\)

Thus \(I_R=3I\)

Let us assume that the focal length is denoted by \(f\), the object distance is denoted by \(u\), and the image distance is \(v\) from the lens.

The lens formula to be used in the given question to find the position of the image is as below:

\(f_1=+10\;cm\)

\(u_1=(-30)\;cm\)

Using the lens formula,

\(\frac1{v_1}-\frac1{u_1}=\frac1{f_1}\)

On rearranging the above lens formula, we get

\(\frac1{v_1}=\frac1{f_1}+\frac1{u_1}\)

\(\Rightarrow\;v_1=15\;cm\)

Now, considering the case of the second lens.

\(f_2=(-10)\;cm\)

\(u_2=15-5=10\;cm\)

Again substituting the values in the above given equation we get,

\(\frac1{v_2}=\frac1{f_2}+\frac1{u_2}\)

\(\Rightarrow\;v_2=\infty\)

The real image thus formed by the second lens at infinite distance. This image will act as an object for the lens that is placed in front of this current image that is for the third lens.

Now calculating the distance of the image that is formed by the third lens is

\(f_3=30\;cm\)

\(u_3=\infty\)

\(\frac1{v_3}=\frac1{f_3}+\frac1{u_3}\)

\(v_3=30\;cm\)

So, the final position of the image formed on this combination is  \(30\;cm\) to the right of the third lens.

Given the object is placed at a distance of 40 cm in front of a concave mirror with a focal length of 20 cm, we can use the mirror equation to determine the characteristics of the image formed.

The mirror equation is

\(\frac1{f}=\frac1v+\frac1u\)

Given \(f=-20\;cm\) and \(u=-40\;cm\) 

\(\frac1{-20}=\frac1{-40}+\frac1v\)

\(v=40\;cm\)

The image distance is positive, indicating that the image is formed on the same side as the object. This means the image is real.

Since the image is formed on the same side as the object and the image distance is equal to the object distance, the image is of the same size as the object. However, it is inverted due to the nature of reflection by a concave mirror.

Therefore, the correct option is: "real, inverted, and of the same size."

Given: Intensities of two beams of monochromatic light

\(I_1=64\;mW\) and \(I_2=4\;mW\)

Applying the formula of resultant intensity,

\(I=I_1+I_2+2\sqrt I_1 \sqrt I_2cos\theta\)

Putting the values from the question,

\(84=64+4+2\times8\times2\times cos\theta\)

\(cos\theta=\frac12\)

\(\theta=60^o\)

As we can see on the left hand side of A and B two capacitors of capicitance \(2C\) each are in series so 

\(\frac1C_{eq1}=\frac1{2C}+\frac1{2C}=\frac1C\)

\(\Rightarrow C_{eq1}=C\)

On the Right hand side of A and B two capacitors of capicitance \(C\) are in parallel and the resultant is series with the capacitors of capicitance \(2C\) so

\(\frac1C_{eq2}=\frac1{C+C}+\frac1{2C}=\frac1C\)

\(\Rightarrow C_{eq2}=C\)

Now All \(C_{eq1},C_{eq2}\) and capicitor of capicitance of \(C\) between A and B are in parallel.

Thus, \(C_{eq}=C+C+C=3C\)

Given: Current density  \(J=2\times10^{10}r^2\;Am^{-2}\), potential \(V=50\;V\), time \(t=100\;s\)

Energy is given as  \(E=VIt\)

Current \(I=\int^{2\times10^{-3}}_0JdA=\int^{2\times10^{-3}}_0 2\times10^{10}r^2\times2\pi rdr\)

\(I=2\times10^{10}2\pi\int^{2\times10^{-3}}_0r^3dr=4\pi\times10^{10}\frac{r^4}4^{2\times10^{-3}}\)

\(E=VIt=50\times16\pi=800\pi\)

Case 1

Net magnetic field at center is

\(B_1=|B_{upper\;part}+B_{lower\;part}|\)

\(=|\frac{\mu_oI}{4\times3r}+\frac{\mu_oI}{4\times r}|=|\frac{4\mu_oI}{12r}|\)

\(B_1=\frac{\mu_oI}{3r}\)

Case 2

\(B_2=|B_{upper\;part}-B_{lower\;part}|\)

\(=|\frac{\mu_oI}{4\times3r}-\frac{\mu_oI}{4\times r}|=|\frac{-2\mu_oI}{12r}|\)

\(B_2=|\frac{-\mu_oI}{6r}|=\frac{\mu_oI}{6r}\)

Case 3 

\(B_3=|B_{upper\;part}+B_{lower\;part}+B_{smallest\;part}|\)

\(=|\frac{\mu_oI}{4\times3r}-\frac{\mu_oI}{8\times 2r}+\frac{\mu_oI}{8r}|=|\frac{13\mu_oI}{48r}|\)

\(B_3=\frac{13\mu_oI}{48r}\)

Thus, \(B_1>B_3>B_2\)

When a charged particle moves through a magnetic field perpendicular to its direction, it experiences a force called the magnetic Lorentz force, which acts perpendicular to both the velocity of the particle and the magnetic field direction. This force causes the particle's path to curve, leading to circular motion.

In such a scenario

The magnitude of the velocity of the particle remains constant (assuming no other forces act on it).

The direction of the velocity of the particle changes continuously as it undergoes circular motion.

Since kinetic energy is given by \(KE=\frac12mv^2\) and the velocity v is constant, the kinetic energy remains constant.

However, since the direction of the velocity changes continuously, the momentum of the particle changes. Momentum is a vector quantity, and when the direction of motion changes, momentum changes too.

Therefore, the correct statement is: "the momentum changes, but the kinetic energy is constant."

The time period of oscillations of magnet

\(T=2\pi\sqrt{\frac I{MB_H}}\)     (1)

Where \(I=\) moment of inertia of magnet

\(I=\frac{mL^2}{12}\)

Now magnet is cut into three equal parts.

When the three equal parts of magnet are placed on one another with their like poles together, then

\(I'=\frac1{12}\times\frac m3\times\frac{L^2}3\times3=\frac I9\)

and \(M'=pole\;strength \times\frac l3\times3=M\)

Thus,

\(T'=\frac13\times T\)

\(T'=\frac23\;s\)

Given: Velocity of conducting bar  \(v=5\;m/s\)

Magnetic field \(B=0.1\;T\) pointing out the page.

For the given time interval distance travelled by the bar \(d=v\times t=4\times5=20\;m\)

In a time \(t=4\;s\), the bar will travel a distance equal to \(20\;m\) as seen from the diagram below.

Now induced emf is given as  \(e=Bvl\)

\(e=0.1\times5\times40=20\;V\)

Given

Current \(I=6\;A\)

Primary voltage\(V_p=220\;V\)
Secondary voltage \(V_s=2200\;V\)

\(60\%\) power is loss thus efficiency \(\eta=40\%\)

As efficiency is defined as the ratio of output power and input power

So,

\(40=\frac{1100\times I_s}{220\times6}\times100\)

\(I_s=0.48\;A\)

Given \(y_1=s\;sin(\omega t-kx)\)

\(y_2=b\;cos(\omega t-kx+\frac{\pi}3)\)

Now rearranging \(y_2\)

\(y_2=b\;sin(\omega t-kx+\frac\pi3+\frac\pi2)=b\;sin(\omega t-kx+\frac{5\pi}6)\)

Thus, the phase difference between \(y_1\) and \(y_2\) is \(\frac{5\pi}6\)

Maximum velocity of emitted photoelectrons is

\(v_{max}=1.6\times10^6m/s\)

Maximum kinetic energy of emitted photoelectrons is

\(K_{max}=\frac12mv^2_{max}\)

\(=\frac12\times9.1\times10^{-31}\times(1.6\times10^6)^2\;J\)

\(=\frac{9.1\times10^{-31}\times(1.6\times10^6)^2}{2\times1.6\times10^{-19}}\;=7.2\;eV\)

So, work function = incident energy −maximum kinetic energy of electron

\(=10.5-7.2=3.3\;eV\)

Given \(E_1\) is the energy for the transition from \(n=2\rightarrow n=1\) and \(E_2\) is the energy for the transition from \(n=3\rightarrow n=2\)

Now the excitation energy is given as, 

\(\Delta E=-13.6(\frac1{n_1^2}-\frac1{n_2^2})\)

For the first case

\(E_1=-13.6(\frac1{1^2}-\frac1{2^2})=-13.6(\frac34)\)

For the second case

\(E_2=-13.6(\frac1{2^2}-\frac1{3^2})=-13.6(\frac5{36})\)

Thus, the ratio is \(\frac{E_2}{E_1}=\frac5{27}\)

Given mass of \(^{29}_{14}Si=28.976495\;u\)

Now mass of 14 protons \(=14\times\)mass of 1 proton

\(=14\times1.007276=14.101864\;u\)

Also, mass of 15 neutrons \(=15\times\) mass of 1 neutron

\(=15\times1.008664=15.12996\)

Total mass of constituents of Si atom\(=14.101864+15.12996=29.231824\;u\)

Mass defect = mass of constituents - mass of atom

\(=29.231824-28.976495=0.255329\;u\)

Binding energy = Energy equivalent of mass defect

\(=\Delta M\times 931.5=0.255329\times931.5=237.84\;MeV\)

In n-type semiconductors, the majority carriers are electrons, not holes. N-type semiconductors are doped with pentavalent impurities, which introduce extra electrons into the crystal lattice, making electrons the majority charge carriers. Therefore, the statement that holes are the majority carriers in n-type semiconductors is incorrect.

The behavior of the resistance of a material with temperature depends on its temperature coefficient of resistance.

For most conductors like copper, the resistance increases with increasing temperature, as they have a positive temperature coefficient of resistance. This is due to increased lattice vibrations that impede the movement of electrons, resulting in higher resistance.

On the other hand, for intrinsic semiconductors like germanium, the resistance decreases with decreasing temperature. This is because, at lower temperatures, fewer electrons are thermally excited from the valence band to the conduction band, leading to fewer charge carriers and therefore lower resistance.

So, the correct statement is: "The resistance of copper decreases, and germanium increases."

Given: For carrier signal

Frequency \(v_1\) and peak voltage \(V_1\)

For message signal

Frequency \(v_2\) and peak voltage \(V_2\)

The modulation index  \(m\) is defined as the ratio of modulation voltage to the carrier voltage.

Thus, \(m=\frac{V_2}{V_1}\)

The frequency of the carrier signal is given as,  \(v_1=\frac{v_++v_-}2\)

From ideal gas equation

\(P=\frac{nRT}V=\frac{mRT}{MV}\) where, \(n=\frac mM=\) number of moles.

So, at constant volume, pressure-temperature graph is a straight line passing through origin with slope  \(\frac{mR}{MV}\)

As the mass is doubled and volume is halved, slope becomes four times. Therefore, pressure versus temperature graph will be shown by the line B

A standing wave can be produced by combining two sinusoidal traveling waves of identical frequency traveling in opposite directions.

When these waves interfere constructively and destructively, they create stationary points called nodes and antinodes, resulting in the formation of a standing wave pattern.

This phenomenon occurs in various systems, including strings, pipes, and electromagnetic fields.

Thus, a standing wave can be produced by combining two sinusoidal traveling waves of identical frequency traveling in opposite directions.

Among the options given, violet color has the maximum dispersion.

When white light passes through a prism, it separates into its constituent colors due to the phenomenon of dispersion, where different colors of light bend by different amounts as they pass through the prism.

Violet light bends the most, followed by blue, green, yellow, orange, and red.

Therefore, violet color has the maximum dispersion.

\(|\frac{f_2}{f_1}|=\frac53 \)

\(\frac1{f_1}-\frac1{f_2}=\frac1{45}\)

\(f_2=\frac{5f_1}3\)

\(\frac1{f_1}-\frac1{\frac{5f_1}3}=\frac1{45}\)

\(\frac{5-3}{5f_1}=\frac1{45}\)

\(f_1=18\)

\(\frac1{18}-\frac1{f_2}=\frac1{45}\)

\(f_2=30\)

since second lens is convex lens \(f_2=-30\;cm\)

When a thin transparent sheet of thickness t is placed in front of a Young's double slit setup, it introduces an additional path difference between the light waves passing through the slits. This path difference causes a shift in the position of the fringes.

The expression for the fringe width \(\beta\)  in Young's double slit experiment without the thin transparent sheet is given by:

\(\beta=\frac{\lambda D}d\)

When the thin transparent sheet is introduced, the fringe width increases because the additional path difference caused by the sheet effectively increases the effective wavelength of light.

Therefore, the correct answer is increase

The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.

Where \(AB=BC=CD=AD=10\;cm\)

\(AC=BD=10\sqrt2\;cm\)

\(AO=OC=DO=OB=5\sqrt2\;cm\)

A charge of amount  \(1\mu C\) is placed at point O.

Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other.

Therefore, net force caused by the four charges placed at the corner of the square on 

To find the net electrostatic potential at the center of the circle due to the charges placed on its circumference, we need to consider the contribution from each individual charge.

The electrostatic potential \(V\) at a point due to a point charge \(Q\) at a distance \(r\) from the charge is given by Colulomb's law

\(V=\frac1{4\pi\epsilon_o}\frac{Q}r\)

For charges placed on the circumference of the circle, the distance from each charge to the center of the circle is \(R\), the radius of the circle. Therefore, the electrostatic potential at the center of the circle due to each charge is:

\(V_i=\frac1{4\pi\epsilon_o} \frac QR\)

Since all charges are identical and placed at equal distances from the center, the potential at the center due to each charge is the same.

The net electrostatic potential \(V_{net}\) at the center of the circle is the sum of the potentials due to each individual charge:

\(V_{net}=\frac1{4\pi\epsilon_o}\frac{NQ}R\)

\(\rho=\rho_oa+bx\)

Resistance \(R=\frac{\rho l}A\)

resistance of small element of length  \(dx,\;dR=\frac{\rho dx}A\)

\(dR=\frac{\rho_oa+bxdx}A\)

Total resistance \(R=\int^L_0\frac{\rho_oa+bxdx}A=\frac{\rho_oax+b(\frac{x^2}2)^L_o}A\)

\(R=\frac{\rho_o}A[a L+b\frac{L^2}2]\)

The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements on the loop. Applying Fleming’s left-hand rule on element AB , the direction of force will be leftwards and the magnitude will be

\(dF=IdlBsin90^o=IdlB\)

On element CD, the direction of force will be rightwards on the plane of the paper and the magnitude will be 

\(dF=IdlB\)

Hence, the net force on the loop due to current elements will be zero, which implies that the total force on the loop will be zero.

Given \(B_1=\frac{\mu_oI}{2R}\)

\(B_2=\frac{\mu_oIR^2}{2(R^2+x^2)^{\frac32}}\)

\(=\frac{\mu_oIR^2}{2(R^2+9R^2)^{\frac32}}\)

\(=\frac{\mu_oI}{2R\times10\sqrt{10}}\)

Thus, \(\frac{B_1}{B_2}=10\sqrt{10}\)

A magnetic needle kept in a non-uniform magnetic field experiences both a force and a torque.

Force: The magnetic needle experiences a force due to the non-uniform magnetic field acting on its magnetic moment. This force depends on the gradient of the magnetic field strength and the magnetic moment of the needle.

Torque: In addition to the force, the magnetic needle experiences a torque because the magnetic force tends to align the magnetic moment of the needle with the local direction of the magnetic field. This results in a torque that tends to rotate the needle until it aligns with the field lines.

So, the correct option is: "both force and torque."

Induced emf is equal to the rate of change of flu

\(e=-\frac{d\phi}{dt}=-\frac{B\times A\times N\times(cos0-cos45)}t\)

\(=-\frac{\sqrt2\times0.2\times1000}2(1-\frac1{\sqrt2})\)

Thus, \(e=41.4\;V\)

The current amplitude as a function of frequency for different types of circuits varies depending on the impedance characteristics of the circuit elements.

For an inductive circuit, the impedance of the inductor (L) is given by 

\(Z_L=jwL\)

The magnitude of the impedance of the inductor is directly proportional to the frequency of the applied alternating current. As the frequency increases, the impedance of the inductor increases.

Therefore, in an inductive circuit, the current amplitude decreases with increasing frequency due to the increasing impedance of the inductor.

Thus correct curve is 'a'.

Given 

Frequency \(=30MHz\)

Electric field \(\epsilon=30\;v/m\,\) (along \(-\hat i\))

Now we know that

\(c=\frac{\epsilon}{B_o}\)

\(B=\frac{30}{3\times10^8}=10^{-7}\;T\)

Now we have to find direction of magnetic field

We know that wave propagates in the direction perpendicular to oscillating electric field and magnetic field.

\(c=\vec\epsilon\times\vec B\) (only for direction)

\(c=\) along z axis

\(\epsilon=\) along - x axis

So B should be along -y axis

Thus, \(B=10^{-7}\) along \(-\hat j\)  direction.

\(E=\phi+KE_{max}\)

\(hv=\phi+eV_0\)

\(eV_0=hv-\phi\)

\(V_0=\frac{hv}e-\frac{\phi}e\)

Thus, Slope \(=\frac he\)

The series corresponding to the lowest wavelength transition in a hydrogen atom is the Lyman series.

In the Lyman series, electrons transition from higher energy levels (n > 1) to the lowest energy level (n = 1), which results in the emission of photons with the highest energy and shortest wavelength in the hydrogen spectrum.

Therefore, the correct option is "Lyman Series."

The universal gates are NOR and NAND.

A gate is considered universal if it can be used to implement any other logic gate.

NOR and NAND gates have this property because they can be used to construct all other basic logic gates, including AND, OR, NOT, and XOR gates.

Given the equation of modulating signal as \(=15sin(100\pi t)\)

we know the standard equation of modulating signal as \(=A_m cos(2\pi f_mt)\)

By comparing 

\(A_m=15 \;volts\)

Given, the equation of carrier signal is \(=45sin(3\times10^6\pi t)\)

The standard equation of carrier signal is \(=A_ccos(2\pi f_c t)\)

By comparing \(A_c=45\;volts\)

We know the formula for modulation index as

Thus, \(\mu=\frac{A_m}{A_c}=\frac{15}{45}=\frac13\)