The critical minimum speed (\(v_{c}\)) required for a mass \(m\) to complete a vertical circle of radius \(R\) is given by the equation:
\(v_{c} = \sqrt{gR}\).


We are given the initial critical speed \(v_{c1} = \sqrt{gR}\).


The radius is increased by a factor of 4, so the new radius is \(R' = 4R\).


The new critical speed \(v_{c}'\) is:
\(v_{c}' = \sqrt{gR'} = \sqrt{g(4R)}\).

\(v_{c}' = 2\sqrt{gR}\).


Since \(v_{c} = \sqrt{gR}\), we have \(v_{c}' = 2v_{c}\).


The circular speed at the top is increased by a factor of 2.
Quick Tip: The critical speed required to complete a vertical loop depends only on the radius of the loop and the gravitational acceleration (\(v_c = \sqrt{gR}\)), independent of the mass \(m\).

\(O_2\) is a diatomic gas (\(n_1 = 1\) mol): \(C_{v1} = 5R/2\).
\(He\) is a monoatomic gas (\(n_2 = 2\) mol): \(C_{v2} = 3R/2\).


The molar specific heat capacity at constant volume for the mixture (\(C_{v, mix}\)) is:
\(C_{v, mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}\).

\(C_{v, mix} = \frac{(1)(\frac{5R}{2}) + (2)(\frac{3R}{2})}{1 + 2}\).

\(C_{v, mix} = \frac{5R/2 + 3R}{3} = \frac{11R/2}{3} = \frac{11R}{6}\).


The molar specific heat capacity at constant pressure for the mixture (\(C_{p, mix}\)) is derived using Mayer's relation \(C_p = C_v + R\):
\(C_{p, mix} = C_{v, mix} + R = \frac{11R}{6} + R = \frac{17R}{6}\).


The required ratio \(\gamma_{mix} = C_p/C_v\) is:
\(\gamma_{mix} = \frac{17R/6}{11R/6} = \frac{17}{11}\).
Quick Tip: The specific heat ratio \(\gamma = C_p/C_v\) for a mixture depends on the weighted average of \(C_v\) values of the components. Remember \(C_v = fR/2\), where \(f\) is the degrees of freedom (3 for monoatomic, 5 for diatomic).

The circuit is a Wheatstone bridge configuration. Let \(C_1=3 \muF\), \(C_2=6 \muF\), \(C_3=1 \muF\), \(C_4=2 \muF\).


We check the balance condition: \(C_1/C_3\) versus \(C_2/C_4\).
\(C_1/C_3 = 3/1 = 3\).
\(C_2/C_4 = 6/2 = 3\).


Since \(C_1/C_3 = C_2/C_4\), the bridge is balanced.


Capacitors \(C_5\) (2 \(\muF\)) and \(C_6\) (2 \(\muF\)) are connected between points of equal potential, so they carry no charge and can be ignored.


The equivalent capacitance is calculated by treating the upper branch (\(C_1\) and \(C_2\) in series) and the lower branch (\(C_3\) and \(C_4\) in series) as being in parallel.


Upper branch equivalent capacitance (\(C_{upper}\)):
\(C_{upper} = \frac{C_1 C_2}{C_1 + C_2} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \muF\).


Lower branch equivalent capacitance (\(C_{lower}\)):
\(C_{lower} = \frac{C_3 C_4}{C_3 + C_4} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \muF\).


Total effective capacitance (\(C_{eq}\)):
\(C_{eq} = C_{upper} + C_{lower} = 2 + \frac{2}{3}\).

\(C_{eq} = \frac{6}{3} + \frac{2}{3} = \frac{8}{3} \muF\).
Quick Tip: In a balanced capacitive Wheatstone bridge, the central connection (between the intermediate points) does not affect the overall capacitance and can be disregarded for simplification.

The circuit consists of a 2 V voltage source connected across two vertices of a delta (\(\Delta\)) configuration of three identical \(30 \Omega\) resistors.


Let the terminals be connected across resistor \(R_1 = 30 \Omega\).


The remaining two resistors, \(R_2 = 30 \Omega\) and \(R_3 = 30 \Omega\), are connected in series between the same two vertices.
\(R_{series} = R_2 + R_3 = 30 \Omega + 30 \Omega = 60 \Omega\).


The equivalent resistance (\(R_{eq}\)) is found by combining \(R_1\) and \(R_{series}\) in parallel:
\(R_{eq} = \frac{R_1 \times R_{series}}{R_1 + R_{series}} = \frac{30 \times 60}{30 + 60}\).

\(R_{eq} = \frac{1800}{90} = 20 \Omega\).


The total current \(I\) drawn from the source is calculated using Ohm's Law:
\(I = \frac{V}{R_{eq}} = \frac{2 V}{20 \Omega}\).

\(I = \frac{1}{10} A\).
Quick Tip: When calculating the equivalent resistance of a symmetrical \(\Delta\) network connected across two terminals, the resistance across the terminals is parallel to the series combination of the other two resistances.

The magnetic field \(B\) due to an infinitely long straight conductor carrying current \(I\) at a perpendicular distance \(r\) is given by:
\(B = \frac{\mu_0 I}{2 \pi r}\).


Given values: \(I = 25 A\).

Distance \(r = 10 cm = 0.10 m\).

Permeability of free space \(\mu_0 = 4\pi \times 10^{-7} T\cdotm/A\).


Substitute the values:
\(B = \frac{(4\pi \times 10^{-7}) \times 25}{2 \pi \times 0.10}\).


Simplify the constants: \(\frac{4\pi}{2\pi} = 2\).
\(B = \frac{2 \times 10^{-7} \times 25}{0.10}\).

\(B = \frac{50 \times 10^{-7}}{0.1}\).

\(B = 500 \times 10^{-7} T\).

\(B = 5.0 \times 10^{-5} T\).
Quick Tip: The direction of the magnetic field can be found using the Right-Hand Thumb Rule. If the current is upward, the field lines circulate counterclockwise. At a point to the right, the magnetic field points into the page.

The phase difference \(\phi\) in an AC circuit is given by \(\tan \phi = \frac{X_L - X_C}{R}\).


Case 1: \(L\) is removed (RC circuit). \(X_L = 0\). Phase difference \(\phi_1 = \pi/3\).
\(\tan(\pi/3) = \left| \frac{0 - X_C}{R} \right| = \frac{X_C}{R}\).
\(X_C / R = \sqrt{3}\). (1)


Case 2: \(C\) is removed (RL circuit). \(X_C = 0\). Phase difference \(\phi_2 = \pi/3\).
\(\tan(\pi/3) = \left| \frac{X_L - 0}{R} \right| = \frac{X_L}{R}\).
\(X_L / R = \sqrt{3}\). (2)


From equations (1) and (2), we deduce \(X_C = R\sqrt{3}\) and \(X_L = R\sqrt{3}\).

Thus, \(X_L = X_C\).


For the original LCR circuit, the net reactance is zero, \(X_{net} = X_L - X_C = 0\).

The phase difference \(\phi\) of the LCR circuit is:
\(\tan \phi = \frac{X_L - X_C}{R} = \frac{0}{R} = 0\).

This implies \(\phi = 0\). The circuit is in resonance.


The power factor is given by \(\cos \phi\):

Power factor \(= \cos(0) = 1\).
Quick Tip: If removing \(L\) results in the same phase shift as removing \(C\), the capacitive reactance (\(X_C\)) must equal the inductive reactance (\(X_L\)). This means the original LCR circuit is operating at resonance, where the impedance \(Z=R\) and the power factor is \(\cos(0) = 1\).

Since the object is short and placed along the principal axis, the image length is determined using longitudinal magnification.

Step 1: Longitudinal magnification relation

For mirrors, \[ m_L = \frac{l'}{l} = m_T^2 \]
where \(m_T\) is the transverse magnification.

Step 2: Transverse magnification for mirror
\[ m_T = -\frac{v}{u} \]

Using the mirror formula: \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]

Solving for \(v\): \[ v = \frac{uf}{u - f} \]

Step 3: Substitute in magnification
\[ m_T = -\frac{v}{u} = -\frac{f}{u - f} \]

Given object distance \(u = -x\) (sign convention), \[ |m_T| = \frac{f}{x - f} \]

Step 4: Image length
\[ l' = l \, m_T^2 = l \left( \frac{f}{x - f} \right)^2 \]
\[ \boxed{l' = \frac{f^2 l}{(x - f)^2}} \] Quick Tip: For small objects placed axially along the principal axis, the longitudinal magnification \(m_L\) is approximated by the square of the transverse magnification \(m_T\): \(m_L = m_T^2\). Also remember the useful forms of transverse magnification: \(m_T = \frac{f}{f-u} = \frac{v-f}{f}\).

Characteristic X-ray wavelengths follow Moseley’s law: \[ \frac{1}{\lambda} \propto (Z - b)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Since the element is the same, the factor \((Z-b)^2\) is constant. Hence, \[ \frac{1}{\lambda} \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For \(K_\alpha\) line:

Transition: \(n_2 = 2 \rightarrow n_1 = 1\) \[ \frac{1}{\lambda_\alpha} \propto \left(1 - \frac{1}{4}\right) = \frac{3}{4} \]

For \(K_\beta\) line:

Transition: \(n_2 = 3 \rightarrow n_1 = 1\) \[ \frac{1}{\lambda_\beta} \propto \left(1 - \frac{1}{9}\right) = \frac{8}{9} \]

Taking ratio: \[ \frac{\lambda_\alpha}{\lambda_\beta} = \frac{8/9}{3/4} = \frac{32}{27} \]

Given: \(\lambda_\alpha = 0.32\,\AA\)
\[ \lambda_\beta = 0.32 \times \frac{27}{32} = 0.27\,\AA \] Quick Tip: Remember the transitions for K series X-rays: \(K_\alpha\) is \(n=2 \to n=1\), \(K_\beta\) is \(n=3 \to n=1\). Since \(1/\lambda\) is proportional to the difference of inverse squares, the wavelengths are inversely proportional to this difference ratio.

According to Rutherford's scattering formula, the number of alpha particles scattered (\(N\)) through an angle \(\theta\) is inversely proportional to the fourth power of the sine of half the scattering angle.
\(N \propto \frac{1}{\sin^4(\theta/2)}\).


We compare the counts at two different angles \(\theta_1 = 60^\circ\) and \(\theta_2 = 90^\circ\).

Given \(N_1 = 100\) per minute at \(\theta_1 = 60^\circ\). Find \(N_2\) at \(\theta_2 = 90^\circ\).

\(\frac{N_2}{N_1} = \frac{\sin^4(\theta_1/2)}{\sin^4(\theta_2/2)}\).


Calculate half-angles and sine values:
\(\theta_1/2 = 60^\circ/2 = 30^\circ\). \(\sin(30^\circ) = 1/2\).
\(\theta_2/2 = 90^\circ/2 = 45^\circ\). \(\sin(45^\circ) = 1/\sqrt{2}\).


Substitute these values:
\(\frac{N_2}{100} = \frac{(1/2)^4}{(1/\sqrt{2})^4}\).

\(\frac{N_2}{100} = \frac{1/16}{1/4}\).

\(\frac{N_2}{100} = \frac{4}{16} = \frac{1}{4}\).

\(N_2 = 100 / 4 = 25\).

The number of alpha-particles scattered per minute at \(90^\circ\) is 25.
Quick Tip: The highly sensitive dependence on the scattering angle (\(\propto 1/\sin^4(\theta/2)\)) is the key result of the Rutherford model, demonstrating that large angle scattering requires interaction with a concentrated, massive nucleus.

Initially, the diode is forward biased and a steady current of \[ I_F = 200\,mA = 0.2\,A \]
flows through the resistor \(R = 200\,\Omega\).


The corresponding applied voltage is approximately \[ V \approx I_F R = (0.2)(200) = 40\,V \]
(neglecting the small diode forward drop).


At time \(t=0\), the applied voltage is instantaneously reversed. However, the current through the diode does not drop to zero immediately due to the presence of stored minority charge carriers in the diode (reverse recovery effect).


Just after reversal (\(t=0^+\)):

The diode behaves momentarily like a short circuit, and the reverse current is limited only by the external resistor.

\[ I(0^+) = \frac{V}{R} = \frac{40}{200} = 0.2\,A = 200\,mA \]

Thus, the magnitude of the instantaneous current just after reversal is equal to the initial forward current.
Quick Tip: When a diode transitions instantaneously from forward bias to reverse bias, the reverse current initially matches the forward current magnitude (\(I_{R} \approx I_{FB}\)), limited only by the external circuit resistance, before the diode fully blocks the flow during the reverse recovery time (\(t_{rr}\)).