Understanding the impact of Earth's rotation on gravity.

The effective acceleration due to gravity, taking into account Earth's rotation, is given by: \[ g_{eff} = g - \omega^2 R \cos^2 \theta \]
where \( \theta \) is the latitude angle.

At the poles, where \( \theta = 90^\circ \), the effect of rotation on gravity is zero because \( \cos(90^\circ) = 0 \). Hence, there is no change in gravity at the poles.
At the equator, where \( \theta = 0^\circ \), the effect is maximal: \[ g_{eff} = g - \omega^2 R \]
Therefore, the maximum reduction in gravity occurs at the equator, and it is zero at the poles. This contradicts Statement II, but Statement I remains correct. Quick Tip: When dealing with questions related to rotation, always consider the impact of centrifugal force and how it varies with latitude.

Step 1: Using the formula for intensity in a Young's double slit experiment.
The intensity at any point in Young's double slit experiment is given by: \[ I_{res} = 4I_0 \cos^2 \left( \frac{\theta}{2} \right) \]
Where \( I_0 \) is the maximum intensity, and \( \theta \) is the phase difference.

For the first point \( P \), where \( \theta = \frac{\pi}{3} \): \[ I_1 = 4I_0 \cos^2 \left( \frac{\pi}{6} \right) = 4I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = 4I_0 \times \frac{3}{4} = 3I_0 \]

For the second point \( Q \), where \( \theta = \frac{\pi}{2} \): \[ I_2 = 4I_0 \cos^2 \left( \frac{\pi}{4} \right) = 4I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = 4I_0 \times \frac{1}{2} = 2I_0 \]

Thus, the ratio of intensities is: \[ \frac{I_1}{I_2} = \frac{3I_0}{2I_0} = \frac{3}{2} \] Quick Tip: For intensity problems in interference, use the formula \( I = 4I_0 \cos^2 \left( \frac{\theta}{2} \right) \), where \( \theta \) is the phase difference between the waves.

Step 1: Apply the time period formula for a satellite: \[ T^2 = \frac{4\pi^2 r^3}{GM} \]

Step 2: Substituting \( r = 2R \): \[ T^2 = \frac{4\pi^2 (2R)^3}{GM} = \frac{4 \times 8 \times \pi^2 R^3}{GM} = \frac{4 \times 8 \times g \times R^3}{GR^2} \]

Step 3: Simplify the expression:
At the Earth's surface, \( g = \frac{GM}{R^2} \).
Substitute \( GM = gR^2 \) and \( \pi^2 = g \), yielding: \[ T^2 = 32R \quad \Rightarrow \quad T = \sqrt{32R} \] Quick Tip: When calculating the time period of a satellite, use the appropriate formula and ensure you account for the satellite's height and the distance from Earth.

In a metallic conductor, free electrons are in constant random motion due to thermal energy. When an electric field is applied, these electrons experience a force that acts in the direction opposite to the field (since electrons are negatively charged).


The velocity of the electrons may make any angle with the acceleration vector during successive collisions. As a result, their motion is not along a straight line, but instead follows curved paths. This curving of the path is caused by the continuous collisions between electrons and atoms in the conductor, which constantly alter the direction and velocity of the electrons.


The overall motion of the electrons is not uniform, but they gradually drift towards the positive end (higher potential) of the conductor due to the applied electric field. This drift is characterized by the "drift velocity," which represents the average velocity of electrons moving toward the positive terminal under the influence of the electric field.


Thus, the electrons follow curved trajectories as they move from lower potential to higher potential. This happens because the direction of their velocity and acceleration vectors are not aligned, owing to the random collisions between electrons and atoms in the conductor.


Step 1: Random motion of electrons:

In the absence of an electric field, electrons in a conductor move randomly in all directions due to their thermal energy.

Step 2: Effect of the electric field:

When an external electric field is applied, it exerts a force on the electrons, causing them to drift in the opposite direction of the field. However, due to frequent collisions, the velocity of the electrons continuously changes, and they follow curved paths.

Step 3: Drift of electrons:

The combined motion of the electrons, resulting from both their random motion and the drift caused by the electric field, leads to a net drift from lower potential to higher potential along curved paths.

Conclusion:

Thus, the free electrons move along curved paths from lower potential to higher potential. Quick Tip: In metallic conductors, while electrons undergo random motion, an applied electric field causes them to drift towards the positive terminal, moving along curved paths due to continuous collisions.

Step 1: Understanding bandwidth in amplitude modulation.

In Amplitude Modulation (AM), the bandwidth of the modulated signal depends on the frequency of the message signal. The formula for the bandwidth of an AM signal is:
\[ AM signal bandwidth = 2 \times f_{message signal} \]

where \( f_{message signal} \) is the frequency of the message signal that modulates the carrier wave.

Step 2: Using the given values.

From the problem, we are provided with the message signal frequency: \( f_{message signal} = 3 \, kHz \).

Now, applying the bandwidth formula: \[ AM signal bandwidth = 2 \times 3 \, kHz = 6 \, kHz \] Quick Tip: In Amplitude Modulation, the bandwidth is directly proportional to the frequency of the message signal. The bandwidth is twice the frequency of the message signal.

Step 1: Determining the zero error and the actual measurement.
The zero error of the Vernier scale is calculated as: \[ Zero error = 6 \times 0.1 \, mm = 0.6 \, mm \, (positive zero error) \]
Next, the diameter measured using the Vernier scale is: \[ Diameter = Main Scale Reading + Vernier Scale Reading \times Least Count \] \[ Diameter = 3.2 \, cm + 0.1 \, mm \times 4 = 3.2 \, cm + 0.4 \, mm = 3.24 \, cm \]
To find the actual diameter, we subtract the zero error: \[ Actual diameter = 3.24 \, cm - Zero error = 3.24 \, cm - 0.06 \, cm = 3.18 \, cm \] Quick Tip: When using a Vernier scale, always account for the zero error before finalizing the measurement.

In projectile motion, the maximum height \( H \) reached by a projectile is given by the formula: \[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

For \( \theta = 30^\circ \), the maximum height \( H_1 \) is: \[ H_1 = \frac{u^2 \sin^2 30^\circ}{2g} = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2}{8g} \]

For \( \theta = 60^\circ \), the maximum height \( H_2 \) is: \[ H_2 = \frac{u^2 \sin^2 60^\circ}{2g} = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3u^2}{8g} \]

Now, we find the ratio of the maximum heights: \[ \frac{H_1}{H_2} = \frac{\frac{u^2}{8g}}{\frac{3u^2}{8g}} = \frac{1}{3} \]
Thus, the ratio of the maximum heights is \( 1 : 3 \). Quick Tip: In projectile motion, the maximum height is proportional to the square of the sine of the angle of projection. A larger angle results in a greater height.

Assertion A is correct because an electric fan continues to rotate for a short time after the current is turned off. This happens due to the inertia of motion, which is the tendency of an object to resist changes in its state of motion. Even when the current is no longer supplied, the fan blades keep spinning for a while.


Reason R is also correct, as it explains that the fan continues to rotate because of inertia of motion. Inertia is the property of matter that causes an object to keep moving unless acted upon by an external force.


Therefore, both Assertion A and Reason R are correct, and Reason R provides the correct explanation for Assertion A.
Quick Tip: Inertia of motion is a property of matter that enables an object to keep moving after the applied force is removed. In the case of a fan, the blades continue to spin due to inertia, even after the power is switched off.

The capacitance \( C \) of a parallel plate capacitor is given by the formula: \[ C = \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between the plates.

For the initial configuration, the capacitance is \( C_1 \) with air as the dielectric, so: \[ C_1 = \frac{\epsilon_0 A}{d} \]

When a metal sheet of thickness \( \frac{2d}{3} \) is introduced between the plates, the effective distance between the plates becomes \( d - \frac{2d}{3} = \frac{d}{3} \), and the capacitance becomes \( C_2 \).

The formula for \( C_2 \) becomes: \[ C_2 = \frac{\epsilon_0 A}{d - t + \frac{t}{K}} \]
where \( t = \frac{2d}{3} \) and \( K = \infty \) for metals. Substituting these values: \[ C_2 = \frac{\epsilon_0 A}{\frac{d}{3}} = 3 \times \frac{\epsilon_0 A}{d} = 3 C_1 \]

Thus, the ratio \( \frac{C_2}{C_1} = 3 : 1 \). Quick Tip: When a metal sheet is inserted between the plates of a capacitor, it effectively reduces the distance between the plates, which increases the capacitance.

In an electromagnetic wave, the amplitude of the electric field \( E_0 \) is connected to the amplitude of the magnetic field \( B_0 \) by the equation: \[ E_0 = c B_0 \]
where \( c \) represents the speed of light (\( 3 \times 10^8 \, m/s \)).

Given that the amplitude of the magnetic field is \( B_0 = 6.0 \times 10^{-7} \, T \), we can calculate \( E_0 \) as follows: \[ E_0 = (6.0 \times 10^{-7}) \times (3 \times 10^8) = 18 \times 10^1 = 180 \, Vm^{-1} \]

Therefore, the maximum value of the electric field is \( 180 \, Vm^{-1} \). Quick Tip: In electromagnetic waves, the relationship between the electric and magnetic fields is given by \( E_0 = c B_0 \), where \( c \) is the speed of light. Use this equation to calculate the missing amplitude of the field.

Analyzing the circuit.

In the given circuit, we observe that Diodes \( D_1 \) and \( D_3 \) are conducting, while \( D_2 \) is reverse biased. As a result, the current \( I_1 \) is divided between \( I_3 \) and \( I_4 \), and we have \( I_2 = 0 \).

Applying Kirchhoff's Current Law (KCL), we get: \[ I_1 = I_2 + I_4 + I_3 = 2I_2 \]
Therefore, the ratio \( \frac{I_1}{I_2} \) is: \[ \frac{I_1}{I_2} = 2 \] Quick Tip: When analyzing circuits with diodes, always identify which diodes are conducting and which are reverse biased. This will allow you to split the currents correctly using KCL.

Step 1: Calculating the internal energy of each gas.

The internal energy of \( O_2 \) is: \[ U_{O_2} = \frac{5}{2}nRT = \frac{5}{2} \times 2 \times RT = 5RT \]

The internal energy of \( Ne \) is: \[ U_{Ne} = \frac{3}{2}nRT = \frac{3}{2} \times 4 \times RT = 6RT \]

Step 2: Determining the total internal energy.
The total internal energy of the system is the sum of the individual energies: \[ U_{total} = 5RT + 6RT = 11RT \] Quick Tip: To find the total internal energy of a gas mixture, simply add the internal energy of each gas, considering their moles and specific heat capacities.

The equation of motion for simple harmonic motion (SHM) is expressed as: \[ y = A \sin(\omega t) + B \cos(\omega t) \]
The amplitude of the motion is determined by the formula: \[ Amplitude = \sqrt{A^2 + B^2} \]
In the equation \( y = \sin(\omega t) + \cos(\omega t) \), we have \( A = 1 \) and \( B = 1 \).

Therefore, the amplitude is: \[ Amplitude = \sqrt{(1)^2 + (1)^2} = \sqrt{2} \]

Thus, the amplitude is \( \sqrt{2} \). Quick Tip: For a periodic motion expressed as the sum of sine and cosine terms, the amplitude is found by taking the square root of the sum of the squares of the coefficients of the sine and cosine terms.

Let the person travel the first distance \( x \) with velocity \( v_1 \), and the next distance \( x \) with velocity \( v_2 \). The time taken for the first part of the journey is: \[ t_1 = \frac{x}{v_1} \]
The time taken for the second part of the journey is: \[ t_2 = \frac{x}{v_2} \]
The total displacement is \( x + x = 2x \), and the total time is: \[ t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2} = x \left( \frac{1}{v_1} + \frac{1}{v_2} \right) \]
The average velocity is given by: \[ v = \frac{Total displacement}{Total time} = \frac{2x}{t_1 + t_2} = \frac{2x}{x \left( \frac{1}{v_1} + \frac{1}{v_2} \right)} = \frac{2}{\left( \frac{1}{v_1} + \frac{1}{v_2} \right)} \]
Thus, the relationship is: \[ \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2} \] Quick Tip: When a person covers the same distance with two different velocities, the average velocity is the harmonic mean of the two velocities.

Step 1: Understanding the decay process.
Radioactive decay is governed by the equation: \[ N = N_0 \times \left( \frac{1}{2} \right)^n \]
where \( N_0 \) is the initial number of nuclei, \( N \) is the number of nuclei remaining after \( n \) half-lives, and \( n \) is the number of half-lives that have passed.

If \( \frac{7}{8} \) of the substance has decayed, then \( \frac{1}{8} \) of the substance remains. This implies that the remaining number of radioactive nuclei is \( \frac{N_0}{8} \).

Step 2: Determining the number of half-lives.
Using the decay equation: \[ \frac{N_0}{2^n} = \frac{N_0}{8} \]
Simplifying: \[ 2^n = 8 \] \[ n = 3 \]
Therefore, 3 half-lives have passed.

Step 3: Calculating the time.
Given that the half-life is \( T \), the total time for the disintegration of \( \frac{7}{8} \) of the substance is: \[ Time = 3 \times T = 3T \] Quick Tip: To solve decay problems, use the equation \( 2^n = \frac{N_0}{N} \), where \( n \) represents the number of half-lives, and \( N_0 \) is the initial number of nuclei.

Step 1: Understanding the adiabatic process.

In an adiabatic process, no heat is exchanged between the system and its surroundings, i.e., \( \Delta Q = 0 \). According to the first law of thermodynamics: \[ \Delta Q = \Delta U + W \]
where \( \Delta U \) is the change in internal energy, and \( W \) is the work done by the system. Since there is no heat exchange in an adiabatic process, we get: \[ 0 = \Delta U + W \quad \Rightarrow \quad \Delta U = -W \]
This implies that the change in internal energy is equal to the negative of the work done by the system.

Step 2: Conclusion.

Hence, the internal energy of the gas does change during adiabatic compression. Therefore, the statement "There is no change in the internal energy" is incorrect. Quick Tip: In an adiabatic process, the change in internal energy is equal to the work done on or by the gas. No heat is exchanged.

Statement I is accurate: For diamagnetic substances, the magnetic susceptibility \( X \) is in the range \( -1 \leq X < 0 \), as diamagnetic materials are repelled by magnetic fields and exhibit negative susceptibility.


Statement II is also correct: Diamagnetic substances move from regions of stronger magnetic fields to weaker ones due to their negative susceptibility. This occurs because diamagnetic materials generate an opposing magnetic field that repels the external field, causing them to move towards areas with weaker magnetic fields.


Hence, both statements are correct. Quick Tip: Diamagnetic materials possess negative magnetic susceptibility and are repelled by magnetic fields. They tend to migrate from areas of high magnetic field strength to regions with weaker magnetic fields.

The formula for the elongation \( \Delta l \) of a wire under a load \( W \) is given by: \[ \frac{W}{A} = Y \cdot \frac{\Delta l}{l} \]
where \( W \) is the applied load, \( A \) is the area of cross section, \( Y \) is Young's modulus, \( \Delta l \) is the elongation, and \( l \) is the length of the wire.

For wire A: \[ \Delta l_1 = \frac{W l}{A Y} \]
For wire B: \[ \Delta l_2 = \frac{W l}{A' Y'} \]
Given that \( \frac{Y_A}{Y_B} = \frac{1}{4} \) and \( \frac{A_A}{A_B} = \frac{1}{3} \), we substitute these ratios into the equation for elongation.

For wire A: \[ \Delta l_1 = \frac{W l}{A Y} \]
For wire B: \[ \Delta l_2 = \frac{W l}{3A \cdot \frac{Y}{4}} = \frac{4W l}{3A Y} \]

Thus, the ratio of elongation is: \[ \frac{\Delta l_1}{\Delta l_2} = \frac{\frac{W l}{A Y}}{\frac{4W l}{3 A Y}} = \frac{3}{4} = 12 : 1 \]

Thus, the ratio of the elongations is \( 12 : 1 \). Quick Tip: When dealing with elongation problems, remember that elongation is inversely proportional to Young's modulus and directly proportional to the area of the cross section.

Step 1: Understanding the photoelectric equation.

In the photoelectric effect, the stopping potential (\( V_0 \)) is related to the frequency (\( v \)) of the incident light by the equation:
\[ eV_0 = h(v - v_{th}) \]
Where:

\( V_0 \) is the stopping potential (in volts).

\( v \) is the frequency of the incident light (in Hz).

\( v_{th} \) is the threshold frequency (below which no photoelectric emission occurs).

\( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, C \)).

\( h \) is Planck's constant (\( 6.6 \times 10^{-34} \, J \cdot s \)).


Step 2: Identifying the threshold frequency.

From the graph, we can observe that the stopping potential becomes non-zero at a frequency of approximately \( 5 \times 10^{14} \, Hz \). This is the threshold frequency \( v_{th} \).


Step 3: Calculating the work function.

At the threshold frequency, the stopping potential is zero. We use the equation: \[ \phi = h v_{th} \]
Substituting the values: \[ \phi = (6.6 \times 10^{-34}) \times (5 \times 10^{14}) = 33 \times 10^{-20} \, J \] \[ \phi = 3.3 \times 10^{-19} \, J \]

To convert this to eV, divide by the charge of the electron: \[ \phi = \frac{3.3 \times 10^{-19}}{1.6 \times 10^{-19}} \, eV = 2.07 \, eV \]

Thus, the work function is \( \phi = 2.07 \, eV \).
Quick Tip: In the photoelectric effect, the work function \( \phi \) is related to the threshold frequency \( v_{th} \) by \( \phi = h v_{th} \). The stopping potential can be used to find this frequency.

Step 1: Understanding Lenz's Law.

Lenz's Law states that the induced current in the copper tube due to the motion of the bar magnet will generate a magnetic field that opposes the motion of the magnet. This means the magnet will experience a resistive force as it falls through the tube.


Step 2: Analyzing the forces.

When the bar magnet is initially released, it accelerates under the influence of gravity (\( g \)). However, as the magnet moves, the changing magnetic flux through the conducting tube induces an electromotive force (EMF), which drives a current in the tube. This current creates a magnetic force that resists the motion of the magnet, in accordance with Lenz's Law.


Step 3: Reaching terminal velocity.

As the magnet’s velocity increases, the opposing magnetic force grows stronger, eventually balancing the downward force due to gravity. Once these forces are equal, the net force on the magnet becomes zero, and the magnet moves with a constant speed, known as terminal velocity. Quick Tip: When a magnet falls through a conducting tube, the induced current creates a magnetic force that opposes the magnet’s motion, resulting in the magnet reaching a constant speed after a while.

For the Lyman series, the shortest wavelength corresponds to the transition from \( n = \infty \) to \( n = 1 \), with a wavelength of 917 Å.


For the Balmer series, the shortest wavelength corresponds to the transition from \( n = \infty \) to \( n = 2 \).


The energy \( E_0 \) for the Lyman series is related to the wavelength \( \lambda_0 \) by the formula: \[ E_0 = \frac{hc}{\lambda_0} \]
For the Lyman series, with \( \lambda_0 = 917 \, Å \), we have: \[ E_0 = \frac{hc}{917 \, Å} \]

For the Balmer series, the energy is related by: \[ \frac{E_0}{4} = \frac{hc}{\lambda} \]
where \( \lambda \) is the wavelength of the Balmer series. Substituting the energy from the Lyman series: \[ \frac{hc}{4 \times 917 \, Å} = \frac{hc}{\lambda} \]

Thus, the wavelength for the Balmer series is: \[ \lambda = 917 \times 4 = 3668 \, Å \]

Therefore, the shortest wavelength of the Balmer series is 3668 Å. Quick Tip: To calculate the shortest wavelength in the Balmer series, note that the energy is divided by 4, corresponding to the transition from \( n = \infty \) to \( n = 2 \).

Step 1: Understanding the induced emf in a loop inside a solenoid.

The induced emf in the loop is given by the equation: \[ emf = B A W N \sin(\omega t) \]
Where:

- \( B \) is the magnetic field in the solenoid,

- \( A \) is the area of the loop,

- \( W \) is the angular frequency of the solenoid's current,

- \( N \) is the number of turns per unit length.


Step 2: Calculating the magnetic field \( B \) in the solenoid.

The magnetic field \( B \) inside a solenoid is given by: \[ B = \mu_0 n I \]
Where:

- \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \),

- \( n = 5000 \, turns/m \),

- \( I = 2.5 \, A \).


So, the magnetic field \( B \) is: \[ B = 4\pi \times 10^{-7} \times 5000 \times 2.5 = 5\pi \times 10^{-3} \, T \]

Step 3: Area of the loop.

The area of the loop is: \[ A = 2 \, cm \times 2 \, cm = 4 \, cm^2 = 4 \times 10^{-4} \, m^2 \]

Step 4: Calculating the induced emf.

Substitute the values into the equation for emf: \[ emf = (5 \times 10^{-3}) \times (4 \times 10^{-4}) \times 700 \times 1 \] \[ emf = 5 \times \frac{22}{7} \times 4 \times 700 \times 10^{-7} = 44 \times 10^{-4} \, V \] Quick Tip: For an induced emf in a loop inside a solenoid, use the equation \( emf = B A W N \sin(\omega t) \), where the magnetic field is calculated using \( B = \mu_0 n I \).

Step 1: Understanding the formula for resistance.

The resistance \( R \) between two opposite faces of a rectangular parallelepiped is given by the formula: \[ R = \rho \frac{l}{A} \]
Where:

\( \rho = 3 \times 10^{-7} \, \Omega \)-cm is the specific resistance,

\( l = 1 \, cm \) is the length of the parallelepiped,

\( A = 1 \, cm \times 100 \, cm = 100 \, cm^2 \) is the cross-sectional area.


Step 2: Calculating the resistance.

Substituting the given values:
\[ R = \frac{3 \times 10^{-7} \times 1}{100 \, cm^2} = 3 \times 10^{-7} \times \frac{1}{100 \times 10^{-4}} = 3 \times 10^{-5} \, \Omega \]

Thus, the resistance between the two opposite faces is \( 3 \times 10^{-5} \, \Omega \). Quick Tip: The resistance between opposite faces of a rectangular parallelepiped can be calculated using \( R = \rho \frac{l}{A} \), where \( \rho \) is the specific resistance, \( l \) is the length, and \( A \) is the cross-sectional area.

The torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force \( \vec{F} \): \[ \vec{\tau} = \vec{r} \times \vec{F} \]
The position vector \( \vec{r} \) from the origin to the point \( (2, -3) \) is: \[ \vec{r} = (0 - 2) \hat{i} + (0 - (-3)) \hat{j} = -2 \hat{i} + 3 \hat{j} \]
The force is given by: \[ \vec{F} = -p \hat{k} \]
Now, calculating the cross product: \[ \vec{\tau} = (-2 \hat{i} + 3 \hat{j}) \times (-p \hat{k}) \]
Using the cross product identity: \[ \vec{\tau} = (-2 \hat{i} + 3 \hat{j}) \times (-p \hat{k}) = -p(3 \hat{i} + 2 \hat{j}) \]
Given that the torque \( \vec{\tau} \) is \( P(a \hat{i} + b \hat{j}) \), comparing the coefficients, we get: \[ a = 3 \quad and \quad b = 2 \]
Thus, the ratio \( \frac{a}{b} = \frac{3}{2} \).

We are given that the ratio is \( \frac{x}{2} \), so: \[ \frac{3}{2} = \frac{x}{2} \]
Solving for \( x \), we get: \[ x = 3 \]

Thus, the value of \( x \) is \( 3 \). Quick Tip: To calculate torque, use the cross product of the position vector and force vector. Pay attention to the signs and direction of the vectors when performing the cross product.

The magnetic field produced by an arc is given by the formula: \(\)B = \frac{\mu_0 I \theta{4 \pi r\(\)
where:

\( B \) is the magnetic field,
\( \mu_0 = 4\pi \times 10^{-7} \, T m/A \),
\( I = 14 \, A \),
\( \theta = \pi \) radians (for a semicircle),
\( r = 2.2 \, cm = 0.022 \, m \).

Substituting the given values: \(\)B = \frac{(4\pi \times 10^{-7 \, T m/A) (14 \, \text{A) (\pi){4 \pi (0.022 \, \text{m)\(\) \(\)B = \frac{14 \pi \times 10^{-7{0.022\(\) \(\)B = \frac{14 \pi{2.2 \times 10^{-5\(\) \(\)B \approx 20 \times 10^{-5 \, \text{T\(\) \(\)B = 2 \times 10^{-4 \, \text{T\(\)



Answer:
The magnetic field is \( 2 \times 10^{-4 \, T \). Thus, the answer is 2. Quick Tip: For a current carrying semicircular arc, the magnetic field at the center is given by \( \frac{\mu_0 I}{4R} \), which depends on the current and the radius of the arc.

By the equation of continuity: \[ A_1 V_1 = A_2 V_2 \]
Where \( A_1 \) and \( A_2 \) are the areas of cross-section at the piston and the outlet, and \( V_1 \) and \( V_2 \) are the velocities at the piston and the outlet respectively.

Given: \[ A_1 = 2 \, cm^2, \quad V_1 = 4 \, cm/s^{-1}, \quad A_2 = 10 \, mm^2 = 10 \times 10^{-2} \, cm^2 \]
Substituting in the continuity equation: \[ 2 \times 4 = (10 \times 10^{-2}) \times V_2 \]
Solving for \( V_2 \): \[ V_2 = \frac{2 \times 4}{10 \times 10^{-2}} = 80 \, cm/s^{-1} \]

Thus, the speed of the outgoing fluid is 80 cm/s\(^-1\). Quick Tip: The equation of continuity is used to relate the velocity of fluid at different cross-sections of a pipe. The product of cross-sectional area and velocity remains constant along the pipe.

Step 1: Calculate the angular frequency \( \omega \).
The angular frequency for simple harmonic motion is given by: \[ \omega = \frac{2\pi}{T} \]
Substituting the time period \( T = 3.14 \, s \): \[ \omega = \frac{2 \times \frac{22}{7}}{3.14} = 2 \, rad/s \]

Step 2: Calculate the maximum acceleration \( a_{max} \).
The maximum acceleration in simple harmonic motion is given by: \[ a_{max} = \omega^2 A \]
Substituting \( \omega = 2 \, rad/s \) and \( A = 1 \, m \): \[ a_{max} = 2^2 \times 1 = 4 \, m/s^2 \]

Step 3: Calculate the maximum force \( F_{max} \).
The maximum force is given by: \[ F_{max} = m a_{max} \]
Substituting \( m = 5 \, kg \) and \( a_{max} = 4 \, m/s^2 \): \[ F_{max} = 5 \times 4 = 20 \, N \]

Thus, the maximum force exerted by the spring is \( 20 \, N \). Quick Tip: To calculate the maximum force in simple harmonic motion, use the formula \( F_{max} = m \times a_{max} \), where \( a_{max} = \omega^2 A \).

To determine the velocity \(v\) of the electron, we follow these steps:

1. Electric Field due to an Infinite Cylindrical Wire:

The electric field \(E\) at a distance \(r\) from an infinite wire with linear charge density \(\lambda\) is given by:
\[ E = \frac{\lambda}{2\pi\epsilon_0 r} \]
where \(\epsilon_0\) is the permittivity of free space (\(\epsilon_0 \approx 8.85 \times 10^{-12}\) C\(^2\)/N·m\(^2\)).

2. Force on the Electron:

The electrostatic force \(F\) acting on the electron is:
\[ F = eE = \frac{e\lambda}{2\pi\epsilon_0 r} \]
where \(e\) is the charge of the electron (\(e \approx 1.6 \times 10^{-19}\) C).

3. Centripetal Force:

For the electron to move in a circular path, the electrostatic force must provide the necessary centripetal force:
\[ F = \frac{mv^2}{r} \]
Equating the two expressions for \(F\):
\[ \frac{e\lambda}{2\pi\epsilon_0 r} = \frac{mv^2}{r} \]
Simplifying, we get:
\[ v^2 = \frac{e\lambda}{2\pi\epsilon_0 m} \]
\[ v = \sqrt{\frac{e\lambda}{2\pi\epsilon_0 m}} \]

4. Substitute the Given Values:

Plugging in the values:
\[ v = \sqrt{\frac{(1.6 \times 10^{-19})(2 \times 10^{-8})}{2\pi(8.85 \times 10^{-12})(9 \times 10^{-31})}} \]
\[ v = \sqrt{\frac{3.2 \times 10^{-27}}{5.0 \times 10^{-41}}} \]
\[ v = \sqrt{6.4 \times 10^{13}} \]
\[ v \approx 8 \times 10^6 m/s \]

Final Answer
The velocity of the electron is \(\boxed{8 \times 10^{6}}\) m/s. Quick Tip: In uniform circular motion under electrostatic force, use \( \frac{mv^2}{r} = \frac{e^2}{4 \pi \epsilon_0 r^2} \) to solve for the velocity of the particle.

Step 1: Image formed by \( L_1 \).
For the first lens \( L_1 \), we have:
- Object distance \( u_1 = -6 \, cm \),
- Focal length \( f_1 = 24 \, cm \).

Using the lens formula: \[ \frac{1}{v_1} = \frac{1}{u_1} + \frac{1}{f_1} \]
Substitute the values: \[ \frac{1}{v_1} = \frac{1}{-6} + \frac{1}{24} \] \[ \frac{1}{v_1} = \frac{-4 + 1}{24} = \frac{-3}{24} \quad \Rightarrow \quad v_1 = -8 \, cm \]
So, the image formed by \( L_1 \) is at \( v_1 = -8 \, cm \), which means it is 8 cm in front of the first lens.

Step 2: Image formed by \( L_2 \).
For the second lens \( L_2 \), the object distance is the distance between the two lenses minus the image distance from \( L_1 \): \[ u_2 = 10 \, cm - 8 \, cm = 2 \, cm \]
The focal length of \( L_2 \) is \( f_2 = 9 \, cm \).

Using the lens formula for \( L_2 \): \[ \frac{1}{v_2} = \frac{1}{u_2} + \frac{1}{f_2} \]
Substitute the values: \[ \frac{1}{v_2} = \frac{1}{2} + \frac{1}{9} \] \[ \frac{1}{v_2} = \frac{9 + 2}{18} = \frac{11}{18} \quad \Rightarrow \quad v_2 = \frac{18}{11} \approx 1.636 \, cm \]
So the image formed by \( L_2 \) is located at \( v_2 \approx 18 \, cm \) from the object. Quick Tip: For a system of two lenses, first calculate the image formed by the first lens using the lens formula. Then, use this image as the object for the second lens and apply the lens formula again to find the final image distance.

Given:

- Mass of the elevator system (including passengers) = 1400 kg

- Velocity (\(V\)) = 3 m/s (constant)

- Frictional force (\(f\)) = 2000 N


Since the elevator is moving at a constant speed, the net force on it is zero. This means the upward force (tension \(T\)) must exactly balance the downward forces, which include the gravitational force (\(Mg\)) and the frictional force.

1. Tension in the string: \[ T = Mg + f = 1400 \times 10 + 2000 = 14000 + 2000 = 16000 \, N \]

2. The maximum power used by the motor is calculated as: \[ Maximum Power = F \times V = T \times V = 16000 \times 3 = 48000 \, W = 48 \, kW \]

Thus, the maximum power consumed by the motor is 48 kW. Quick Tip: To calculate maximum power, multiply the tension in the string (which balances all forces) by the velocity of the elevator. Remember, power is the rate of doing work, and it is the product of force and velocity.