Step 1: Understanding the Concept:

We need to find the dimensional formula of the universal gravitational constant \(G\) and match it with an expression given in terms of \(h\), \(L\), \(M\), and \(T\).


Step 2: Key Formula or Approach:

From Newton's law of gravitation, \(F = \frac{G M_1 M_2}{r^2} \implies G = \frac{F r^2}{M^2}\).

From the energy of a photon, \(E = \frac{h c}{\lambda} \implies h = E \cdot T\).

Find the base dimensions of \(G\) and \(h\) in terms of standard M, L, T.

Then substitute \(h\) into the given options to see which one equals the dimensions of \(G\).


Step 3: Detailed Explanation:

First, find the dimensions of \(G\):
\[ [G] = \frac{[Force] \times [Distance]^2}{[Mass]^2} = \frac{(M L T^{-2}) (L^2)}{M^2} = [M^{-1} L^3 T^{-2}] \]
Next, find the dimensions of \(h\):
\[ [h] = [Energy] \times [Time] = (M L^2 T^{-2}) (T) = [M L^2 T^{-1}] \]
Now, let's test the options to see which one simplifies to \([M^{-1} L^3 T^{-2}]\).

Let's test Option (B): \([h T^{-1} L M^{-2}]\).

Substitute the dimensional formula for \(h\):
\[ [h T^{-1} L M^{-2}] = (M L^2 T^{-1}) \cdot T^{-1} \cdot L \cdot M^{-2} \]
Group the like terms:
\[ = (M \cdot M^{-2}) \cdot (L^2 \cdot L) \cdot (T^{-1} \cdot T^{-1}) \] \[ = M^{-1} L^3 T^{-2} \]
This perfectly matches the dimensional formula for \(G\).


Step 4: Final Answer:

The dimensions of \(G\) are \([h T^{-1} L M^{-2}]\).
Quick Tip: Instead of setting up a complex system of linear equations (\(G = k \cdot h^a L^b M^c T^d\)), evaluating the given options by substituting the dimensions of \(h\) is significantly faster.

Step 1: Understanding the Concept:

For a mass to remain stuck to the inner wall of a rotating cylinder, the upward frictional force must balance the downward gravitational force.

The frictional force is provided by the normal force, which in turn is generated by the centripetal acceleration of the rotating drum.


Step 2: Key Formula or Approach:

Normal force exerted by the wall on the mass: \(N = m r \omega^2\).

Static friction force: \(f_s \le \mu_s N\).

For the mass to not fall, the friction must counteract gravity: \(f_s \ge mg\).

At the minimum rotational speed, friction is at its limiting value: \(\mu_s N = mg\).


Step 3: Detailed Explanation:

Substitute the expression for Normal force into the limiting friction condition:
\[ \mu_s (m r \omega_{min}^2) = mg \]
The mass \(m\) cancels out from both sides:
\[ \mu_s r \omega_{min}^2 = g \]
Rearrange the equation to solve for the coefficient of friction \(\mu_s\):
\[ \mu_s = \frac{g}{r \omega_{min}^2} \]
Given values are \(g = 10 m/s^2\), \(r = 4 m\), and \(\omega_{min} = 5 rad/s\).

Substitute these values into the formula:
\[ \mu_s = \frac{10}{4 \times (5)^2} \] \[ \mu_s = \frac{10}{4 \times 25} = \frac{10}{100} = 0.1 \]

Step 4: Final Answer:

The coefficient of friction is \(0.1\).
Quick Tip: In problems involving "rotor" rides or centrifuges, note that the mass of the object cancels out. The minimum angular velocity required solely depends on the radius, gravity, and coefficient of friction.

Step 1: Understanding the Concept:

We have an Atwood machine. The two blocks will accelerate with the same magnitude but in opposite directions.

After finding the acceleration of the individual blocks, we can calculate the net acceleration of the center of mass of the two-block system.

Using kinematics, we can then find the displacement of the center of mass over the given time interval.


Step 2: Key Formula or Approach:

Acceleration of blocks in an Atwood machine: \(a = \frac{m_1 - m_2}{m_1 + m_2} g\).

Acceleration of the center of mass: \(\vec{a}_{cm} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2}\).

Displacement under constant acceleration from rest: \(S = \frac{1}{2} a_{cm} t^2\).


Step 3: Detailed Explanation:

Let \(m_1 = 2 kg\) and \(m_2 = 1 kg\).

Calculate the common acceleration \(a\) of the blocks:
\[ a = \frac{2 - 1}{2 + 1} (10) = \frac{1}{3} \times 10 = \frac{10}{3} m/s^2 \]
The heavier block \(m_1\) moves downwards (let's define this as the positive direction), so \(\vec{a}_1 = \frac{10}{3}\).

The lighter block \(m_2\) moves upwards, so \(\vec{a}_2 = -\frac{10}{3}\).

Now, find the acceleration of the center of mass:
\[ a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} \] \[ a_{cm} = \frac{(2)\left(\frac{10}{3}\right) + (1)\left(-\frac{10}{3}\right)}{2 + 1} \] \[ a_{cm} = \frac{\frac{20}{3} - \frac{10}{3}}{3} = \frac{\frac{10}{3}}{3} = \frac{10}{9} m/s^2 \]
Now, calculate the distance traversed by the center of mass in \(t = 2 s\):
\[ S_{cm} = \frac{1}{2} a_{cm} t^2 = \frac{1}{2} \left(\frac{10}{9}\right) (2)^2 \] \[ S_{cm} = \frac{1}{2} \left(\frac{10}{9}\right) (4) = \frac{20}{9} m \]
Converting the fraction to a decimal:
\[ \frac{20}{9} \approx 2.22 m \]

Step 4: Final Answer:

The distance traversed by the center of mass is \(2.22 m\).
Quick Tip: For an Atwood machine, the acceleration of the center of mass is always downwards and its magnitude simplifies directly to \(a_{cm} = g \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2\).

Step 1: Understanding the Concept:

A charged particle moving in both an electric and magnetic field experiences the Lorentz force.

We can express the unknown velocity as a vector in the \(x\)-\(y\) plane (\(\vec{v} = v_x\hat{i} + v_y\hat{j}\)) and substitute it into the Lorentz force equation to solve for its components.


Step 2: Key Formula or Approach:

The Lorentz force equation is \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\).

Given values:
\(q = 10^{-9} C\)
\(\vec{E} = 0.4 \hat{j} N/C\)
\(\vec{B} = 4 \times 10^{-3} \hat{k} T\)
\(\vec{F} = (4\hat{i} + 2\hat{j}) \times 10^{-10} N\)


Step 3: Detailed Explanation:

Let the velocity be \(\vec{v} = v_x \hat{i} + v_y \hat{j}\) since the particle is moving in the \(x\)-\(y\) plane.

First, compute the cross product \(\vec{v} \times \vec{B}\):
\[ \vec{v} \times \vec{B} = (v_x \hat{i} + v_y \hat{j}) \times (4 \times 10^{-3} \hat{k}) \]
Using the cross product rules \(\hat{i} \times \hat{k} = -\hat{j}\) and \(\hat{j} \times \hat{k} = \hat{i}\):
\[ \vec{v} \times \vec{B} = -v_x(4 \times 10^{-3})\hat{j} + v_y(4 \times 10^{-3})\hat{i} = (4 \times 10^{-3} v_y)\hat{i} - (4 \times 10^{-3} v_x)\hat{j} \]
Now substitute everything into the Lorentz force equation divided by \(q\):
\[ \frac{\vec{F}}{q} = \vec{E} + \vec{v} \times \vec{B} \] \[ \frac{10^{-10} (4\hat{i} + 2\hat{j})}{10^{-9}} = 0.4 \hat{j} + (4 \times 10^{-3} v_y)\hat{i} - (4 \times 10^{-3} v_x)\hat{j} \] \[ 0.4 \hat{i} + 0.2 \hat{j} = (4 \times 10^{-3} v_y)\hat{i} + (0.4 - 4 \times 10^{-3} v_x)\hat{j} \]
Equate the \(\hat{i}\) components:
\[ 0.4 = 4 \times 10^{-3} v_y \implies v_y = \frac{0.4}{4 \times 10^{-3}} = \frac{400}{4} = 100 m/s \]
Equate the \(\hat{j}\) components:
\[ 0.2 = 0.4 - 4 \times 10^{-3} v_x \] \[ 4 \times 10^{-3} v_x = 0.4 - 0.2 = 0.2 \] \[ v_x = \frac{0.2}{4 \times 10^{-3}} = \frac{200}{4} = 50 m/s \]
Thus, the velocity vector is \(\vec{v} = 50\hat{i} + 100\hat{j}\).


Step 4: Final Answer:

The velocity of the particle is \(50\hat{i} + 100\hat{j}\).
Quick Tip: Always divide the force by the charge immediately on the left side of the Lorentz equation to work with cleaner, smaller numbers like \(0.4\) instead of \(10^{-10}\).

Step 1: Understanding the Concept:

The circuit diagram consists exclusively of NAND gates (D-shaped symbols with a small inversion bubble at the output).

We need to trace the boolean logic through each stage of the circuit from inputs \(X\) and \(Y\) to the final output.


Step 2: Key Formula or Approach:

The boolean expression for a NAND gate with inputs \(A\) and \(B\) is \(\overline{A \cdot B}\).

If both inputs of a NAND gate are tied together (i.e., \(A = B\)), it acts as a NOT gate: \(\overline{A \cdot A} = \overline{A}\).

De Morgan's Theorem: \(\overline{A \cdot B} = \overline{A} + \overline{B}\) and \(\overline{A + B} = \overline{A} \cdot \overline{B}\).


Step 3: Detailed Explanation:

Stage 1: The input \(X\) is fed into both terminals of the first NAND gate.

Output of the top gate = \(\overline{X \cdot X} = \overline{X}\).

Similarly, the input \(Y\) is fed into both terminals of the second NAND gate.

Output of the bottom gate = \(\overline{Y \cdot Y} = \overline{Y}\).

Stage 2: These two signals, \(\overline{X}\) and \(\overline{Y}\), are used as inputs for the third NAND gate.

Output of the third gate = \(\overline{\overline{X} \cdot \overline{Y}}\).

Using De Morgan's theorem, we simplify this:
\(\overline{\overline{X} \cdot \overline{Y}} = \overline{\overline{X}} + \overline{\overline{Y}} = X + Y\).

(Notice that this intermediate stage creates an OR gate).

Stage 3: The output \(X + Y\) is fed into both terminals of the final fourth NAND gate.

Output of the final gate = \(\overline{(X + Y) \cdot (X + Y)} = \overline{X + Y}\).

The boolean expression \(\overline{X + Y}\) represents the logic of a NOR gate.


Step 4: Final Answer:

The given circuit works as a NOR gate.
Quick Tip: Recognize standard building blocks: A NAND gate with shorted inputs is a NOT gate. Two inverted inputs into a NAND gate give an OR gate. An OR gate followed by a NOT gate gives a NOR gate.

Step 1: Understanding the Concept:

When an object falls from infinity to the surface of the Earth, it loses gravitational potential energy and gains kinetic energy.

By the principle of conservation of mechanical energy, the velocity attained is exactly equal to the escape velocity of the Earth.


Step 2: Key Formula or Approach:

The escape velocity from the Earth's surface is given by \(v = \sqrt{2gR}\), where \(g\) is the acceleration due to gravity and \(R\) is the Earth's radius.

The kinetic energy is given by \(k = \frac{1}{2} m v^2\).


Step 3: Detailed Explanation:

Given values:
\(m = 1 kg\)
\(g = 9.8 m/s^2\)
\(R = 6400 km = 6.4 \times 10^6 m\)

First, calculate the velocity \(v\):
\[ v = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \] \[ v = \sqrt{19.6 \times 6.4 \times 10^6} \] \[ v = \sqrt{125.44 \times 10^6} \] \[ v = 11.2 \times 10^3 m/s = 11.2 km/s \]
Next, calculate the kinetic energy \(k\):
\[ k = \frac{1}{2} m v^2 = \frac{1}{2} (1) (11.2 \times 10^3)^2 \] \[ k = \frac{1}{2} (125.44 \times 10^6) \] \[ k = 62.72 \times 10^6 J \]
Convert this into scientific notation matching the options:
\[ k = 6.272 \times 10^7 J \approx 6.27 \times 10^7 J \]

Step 4: Final Answer:

The values of \(v\) and \(k\) respectively are \(11.2 km/s\) and \(6.27 \times 10^7 J\).
Quick Tip: Escape velocity from Earth (\(\approx 11.2 km/s\)) is a standard constant you should memorize. Knowing this instantly eliminates options C and D without any calculation.

Step 1: Understanding the Concept:

The diameter measured by a screw gauge is the sum of the Main Scale Reading (MSR) and the Circular Scale Reading (CSR) multiplied by the Least Count (LC).

A zero error must also be accounted for. If the zero of the circular scale is below the reference line (the 5th division coincides), the zero error is positive.

True Reading = Observed Reading - Zero Error.


Step 2: Key Formula or Approach:

Least Count (LC) = \(\frac{Pitch}{Number of circular divisions}\).

Observed Reading = MSR + (CSR \(\times\) LC).

Zero Error = (Coinciding division) \(\times\) LC.


Step 3: Detailed Explanation:

First, calculate the Least Count:
\[ LC = \frac{0.1 mm}{100} = 0.001 mm \]
Determine the Zero Error:

Since the zero of the reference line coincides with the 5th division, the 0 mark of the circular scale has crossed the reference line, meaning there is a positive zero error.
\[ Zero Error = +5 \times 0.001 mm = +0.005 mm \]
Now calculate the Observed Reading:

Main Scale Reading (MSR) = \(5 mm\).

Circular Scale Reading (CSR) = \(50\).
\[ Observed Reading = 5 + 50 \times 0.001 = 5.050 mm \]
Finally, calculate the True Reading (Diameter of the sphere):
\[ True Reading = Observed Reading - Zero Error \] \[ True Reading = 5.050 mm - 0.005 mm = 5.045 mm \]

Step 4: Final Answer:

The diameter of the sphere is \(5.045 mm\).
Quick Tip: Always double-check the pitch value provided. Standard screw gauges often have a pitch of \(1 mm\) or \(0.5 mm\), but this specific problem dictates \(0.1 mm\), altering the typical least count from \(0.01 mm\) to \(0.001 mm\).

Step 1: Understanding the Concept:

The work done in expanding a bubble is equal to the surface tension multiplied by the change in the total surface area.

Because a soap bubble has a thin film with two surfaces (an inner surface and an outer surface) in contact with air, the effective surface area is twice the geometric surface area of a sphere.


Step 2: Key Formula or Approach:

Work done \(W = T \cdot \Delta A\).

For a soap bubble, the change in area is \(\Delta A = 2 \times (4\pi R_2^2 - 4\pi R_1^2) = 8\pi(R_2^2 - R_1^2)\), where \(R_1\) and \(R_2\) are the initial and final radii.


Step 3: Detailed Explanation:

Given values:

Surface tension \(T = 0.03 N/m\).

Initial diameter \(D_1 = 2 cm \implies R_1 = 1 cm = 0.01 m\).

Final diameter \(D_2 = 6 cm \implies R_2 = 3 cm = 0.03 m\).

Calculate the change in total surface area:
\[ \Delta A = 8\pi ((0.03)^2 - (0.01)^2) \] \[ \Delta A = 8\pi (0.0009 - 0.0001) \] \[ \Delta A = 8\pi (0.0008) = 0.0064\pi m^2 \]
Now calculate the work done:
\[ W = T \cdot \Delta A = 0.03 \times 0.0064\pi \] \[ W = 0.000192\pi J \]
Write this in the format \(\alpha \pi \times 10^{-4} J\):
\[ W = 1.92 \pi \times 10^{-4} J \]
Comparing this to the given expression, we find \(\alpha = 1.92\).


Step 4: Final Answer:

The value of \(\alpha\) is \(1.92\).
Quick Tip: A common pitfall is forgetting the factor of \(2\) for soap bubbles. Always remember: water drops have 1 surface area (\(4\pi R^2\)), while soap bubbles have 2 surfaces (\(8\pi R^2\)).

Step 1: Understanding the Concept:

We can treat the gas mixture as a single ideal gas to find the total number of moles in the container.

Once the total moles are found, we can use the total mass and the molar masses of the constituent gases to set up a system of linear equations and solve for the individual moles.


Step 2: Key Formula or Approach:

Ideal Gas Law: \(PV = n_{total} RT\).

Total mass equation: \(m_{total} = n_1 M_1 + n_2 M_2\).

Total moles equation: \(n_{total} = n_1 + n_2\).

Molar mass of CO\(_2\) (\(M_1\)) = \(44 g/mol\).

Molar mass of O\(_2\) (\(M_2\)) = \(32 g/mol\).


Step 3: Detailed Explanation:

Convert given values to standard SI units:
\(P = 100 kPa = 10^5 Pa\).
\(V = 8310 cm^3 = 8310 \times 10^{-6} m^3 = 8.31 \times 10^{-3} m^3\).
\(T = 300 K\).

Calculate the total moles using the Ideal Gas Law:
\[ n_{total} = \frac{PV}{RT} = \frac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300} \] \[ n_{total} = \frac{100 \times 8.31}{8.31 \times 300} = \frac{100}{300} = \frac{1}{3} moles \]
Let \(n_1\) be the moles of CO\(_2\) and \(n_2\) be the moles of O\(_2\).

We have two equations:

1) \(n_1 + n_2 = \frac{1}{3} \implies n_2 = \frac{1}{3} - n_1\)

2) \(44n_1 + 32n_2 = 13.2\)

Substitute the first equation into the second:
\[ 44n_1 + 32\left(\frac{1}{3} - n_1\right) = 13.2 \] \[ 44n_1 + \frac{32}{3} - 32n_1 = 13.2 \] \[ 12n_1 = 13.2 - \frac{32}{3} \]
Multiply by 3 to clear the fraction:
\[ 36n_1 = 39.6 - 32 = 7.6 \] \[ n_1 = \frac{7.6}{36} = \frac{76}{360} = \frac{19}{90} \approx 0.211 moles \]
Now find \(n_2\):
\[ n_2 = \frac{1}{3} - \frac{19}{90} = \frac{30}{90} - \frac{19}{90} = \frac{11}{90} \approx 0.122 moles \]
The moles are approximately \(0.21\) and \(0.12\).


Step 4: Final Answer:

The number of moles of carbon dioxide and oxygen are \(0.21 and 0.12\).
Quick Tip: When the volume is given numerically similar to the gas constant \(R\) (like \(8310\) and \(8.31\)), convert volume directly to \(m^3\) and pressure to Pascals. The math will cancel out beautifully without calculators.

Step 1: Understanding the Concept:

When a bubble rises steadily, it has reached its terminal velocity.

At this point, the upward buoyant force is perfectly balanced by the downward weight of the bubble and the downward viscous drag force (Stokes' Law).

Since the density of air is negligible compared to the liquid, we can safely ignore the bubble's weight.


Step 2: Key Formula or Approach:

Force balance equation: \(F_{buoyancy} = F_{weight} + F_{viscous}\).
\(\frac{4}{3} \pi r^3 \rho_{liquid} g = \frac{4}{3} \pi r^3 \rho_{air} g + 6 \pi \eta r v\).

Ignoring \(\rho_{air}\), the equation simplifies to:
\(\eta = \frac{2}{9} \frac{r^2 \rho_{liquid} g}{v}\).

Remember to convert the final answer from SI units (\(Pa\cdots\)) to CGS units (Poise), where \(1 Pa\cdots = 10 Poise\).


Step 3: Detailed Explanation:

Given values:

Radius \(r = \frac{2 mm}{2} = 1 mm = 10^{-3} m\).

Density of liquid \(\rho = 2000 kg/m^3\).

Terminal velocity \(v = 0.5 cm/s = 0.005 m/s\).

Gravity \(g = 10 m/s^2\).

Substitute these values into the derived formula:
\[ \eta = \frac{2}{9} \frac{(10^{-3})^2 \times 2000 \times 10}{0.005} \] \[ \eta = \frac{2}{9} \frac{10^{-6} \times 20000}{0.005} \] \[ \eta = \frac{2}{9} \frac{0.02}{0.005} = \frac{2}{9} (4) = \frac{8}{9} Pa\cdots \]
Calculate the value in SI units:
\[ \eta \approx 0.888 Pa\cdots \]
Convert \(Pa\cdots\) to Poise by multiplying by 10:
\[ \eta_{Poise} = 0.888 \times 10 = 8.88 Poise \]
This matches the \(8.8\) option closely depending on the \(\pi\) or \(g\) strict rounding, but strictly \(8.88 \approx 8.8\).


Step 4: Final Answer:

The coefficient of viscosity is \(8.8 Poise\).
Quick Tip: Pay strict attention to the requested units. Viscosity is frequently asked in "Poise" (CGS unit). Always solve in SI units first (\(N\cdots/m^2\) or \(Pa\cdots\)) and then multiply by 10 to get Poise.

Step 1: Understanding the Concept:

The ball starts from rest, falls, penetrates the sand, and comes to rest again. Thus, its total change in kinetic energy is zero.

By the Work-Energy Theorem, the net work done by all forces on the ball must equal zero.

The two forces acting on the ball are gravity (acting over the entire fall and penetration distance) and the resisting force of the sand (acting only during penetration).


Step 2: Key Formula or Approach:

Work-Energy Theorem: \(W_{net} = \Delta K = 0\).
\(W_{gravity} + W_{sand} = 0\).
\(W_{gravity} = mg(h + d)\), where \(h\) is the fall height and \(d\) is the penetration depth.
\(W_{sand} = -F_{avg} \cdot d\), where \(F_{avg}\) is the upward average force exerted by the sand.


Step 3: Detailed Explanation:

Given values:

Mass \(m = 2 kg\)

Height \(h = 10 m\)

Penetration depth \(d = 10 cm = 0.1 m\)

Gravity \(g = 10 m/s^2\)

Calculate the total work done by gravity:
\[ W_{gravity} = mg(h + d) = 2 \times 10 \times (10 + 0.1) \] \[ W_{gravity} = 20 \times 10.1 = 202 J \]
Calculate the work done by the sand:
\[ W_{sand} = -F_{avg} \times 0.1 \]
Apply the Work-Energy Theorem:
\[ 202 - 0.1 F_{avg} = 0 \] \[ 0.1 F_{avg} = 202 \implies F_{avg} = \frac{202}{0.1} = 2020 N \]

Step 4: Final Answer:

The average force exerted by the sand is \(2020 N\).
Quick Tip: A common mistake is neglecting the work done by gravity *during* the penetration phase (\(mgd\)). Using the Work-Energy theorem over the entire journey prevents this error entirely compared to using separate kinematic equations.

Step 1: Understanding the Concept:

In an electromagnetic wave, the electric field \(\vec{E}\), magnetic field \(\vec{B}\), and the direction of propagation \(\vec{v}\) are all mutually perpendicular.

Their relationship is given by the cross product \(\hat{E} \times \hat{B} = \hat{v}\).

The magnitude of the electric field is tied to the magnetic field by the speed of light \(c\).


Step 2: Key Formula or Approach:

Magnitude relationship: \(E = cB\), where \(c = 3 \times 10^8 m/s\).

Directional relationship: The direction of wave propagation is given by \(\vec{E} \times \vec{B}\).


Step 3: Detailed Explanation:

First, calculate the magnitude of the electric field:
\[ E = (3 \times 10^8 m/s) \times (2 \times 10^{-7} T) = 60 V/m \]
Next, determine the direction of the electric field.

The wave travels along the \(x\)-direction, so the propagation vector is \(+\hat{i}\).

The magnetic field is along the \(y\)-direction, so \(\hat{B} = \hat{j}\).

We need to find the unit vector \(\hat{E}\) such that \(\hat{E} \times \hat{j} = \hat{i}\).

Recall the standard cross product rules for Cartesian unit vectors: \(\hat{k} \times \hat{j} = -\hat{i}\).

Therefore, \((-\hat{k}) \times \hat{j} = \hat{i}\).

This means the electric field must be directed along the negative \(z\)-axis, so \(\hat{E} = -\hat{k}\).

Combining magnitude and direction:
\[ \vec{E} = -60\hat{k} V/m \]

Step 4: Final Answer:

The corresponding electric field is \(-60\hat{k}\).
Quick Tip: The right-hand rule makes determining direction trivial: Point your fingers in the direction of \(\vec{E}\), curl them towards \(\vec{B}\), and your thumb will point in the direction of wave propagation.

Step 1: Understanding the Concept:

The bulb acts as a resistor in the circuit. We first need to find its resistance using its power rating.

Connecting the bulb across the \(400\ \Omega\) resistor places them in parallel.

We then find the equivalent resistance of the entire circuit to find the main current, which will allow us to calculate the voltage drop across the parallel combination.


Step 2: Key Formula or Approach:

Resistance from power rating: \(R = \frac{V_{rated}^2}{P_{rated}}\).

Equivalent resistance for two parallel resistors: \(R_p = \frac{R_1 R_2}{R_1 + R_2}\).

Ohm's Law: \(V = I R_{eq}\).


Step 3: Detailed Explanation:

Calculate the resistance of the bulb:
\[ R_{bulb} = \frac{(200)^2}{100} = \frac{40000}{100} = 400\ \Omega \]
The bulb is connected in parallel with the \(400\ \Omega\) resistor. Calculate the equivalent resistance of this parallel pair:
\[ R_p = \frac{400 \times 400}{400 + 400} = \frac{160000}{800} = 200\ \Omega \]
This parallel combination is in series with the \(200\ \Omega\) resistor. Calculate the total resistance of the circuit:
\[ R_{total} = 200\ \Omega + R_p = 200 + 200 = 400\ \Omega \]
Now, calculate the total current drawn from the 100 V battery:
\[ I_{total} = \frac{V_{battery}}{R_{total}} = \frac{100}{400} = 0.25 A \]
The potential drop across the bulb is the same as the potential drop across the entire parallel combination.
\[ V_{bulb} = I_{total} \times R_p = 0.25 \times 200 = 50 V \]

Step 4: Final Answer:

The potential drop across the bulb is \(50 V\).
Quick Tip: Because the parallel combination (\(200\ \Omega\)) matches the series resistor (\(200\ \Omega\)), the 100 V supply voltage is split exactly equally between them. You can bypass the current calculation entirely: \(100/2 = 50 V\).

Step 1: Understanding the Concept:

This problem is based on the Millikan oil-drop experiment principle.

For the charged droplets to not descend, the upward electrostatic force must exactly balance the downward gravitational force.

Since the charge on the droplet is positive (\(5 nC\)), the electric field must point upwards. Therefore, the lower plate (B) must be at a higher potential than the upper plate (A).


Step 2: Key Formula or Approach:

Force balance: \(qE = mg\).

Mass of a spherical droplet: \(m = \frac{4}{3}\pi r^3 \rho\).

Electric field between parallel plates: \(E = \frac{V_B - V_A}{d}\).


Step 3: Detailed Explanation:

First, calculate the mass of a single oil droplet. Convert units to standard SI:
\(r = 1 mm = 10^{-3} m\)
\(\rho = 1.5 g/cm^3 = 1500 kg/m^3\)
\[ m = \frac{4}{3} \pi (10^{-3})^3 (1500) = \frac{4}{3} \pi (10^{-9}) (1500) = 2000 \pi \times 10^{-9} = 2\pi \times 10^{-6} kg \]
Now, equate the electric force to the gravitational force to find the required electric field \(E\):
\[ qE = mg \] \[ (5 \times 10^{-9}) E = (2\pi \times 10^{-6}) (10) \] \[ (5 \times 10^{-9}) E = 2\pi \times 10^{-5} \] \[ E = \frac{2\pi \times 10^{-5}}{5 \times 10^{-9}} = 0.4\pi \times 10^4 = 4000\pi V/m \]
Next, find the potential difference required across the plates. The distance \(d = \frac{12}{\pi} cm = \frac{0.12}{\pi} m\).
\[ V_B - V_A = E \times d = 4000\pi \times \frac{0.12}{\pi} = 480 V \]
The potential of plate B must be 480 V higher than plate A.

Now, review the given options to find the pair with a difference of +480 V:

Option (A): \(V_B - V_A = 580 - 100 = 480 V\). (Matches)

Option (B): \(V_B - V_A = 100 - 580 = -480 V\).

Option (C): \(V_B - V_A = 400 - 60 = 340 V\).

Option (D): \(V_B - V_A = -200 - 0 = -200 V\).


Step 4: Final Answer:

The values of \(V_A\) and \(V_B\) are \(100 V and 580 V\).
Quick Tip: Always establish the direction of the required electric field first. Positive charges need an upward field to levitate, meaning the bottom plate must be positive (higher potential) relative to the top plate.

Step 1: Understanding the Concept:

According to Gauss's Law, the total electric flux through a closed surface is directly proportional to the net electric charge enclosed by that surface.

Charges outside the surface do not contribute to the net flux.


Step 2: Key Formula or Approach:

Gauss's Law: \(\Phi = \frac{q_{enclosed}}{\epsilon_0}\).

Identify which charges are located inside each of the given spheres.


Step 3: Detailed Explanation:

Let's analyze the first sphere:

Radius \(R_1 = 3 cm\), centered at the origin.

The charge \(q_1 = 8\ \muC\) is at \(x = 2 cm\), which is inside the sphere (\(2 < 3\)).

The charge \(q_2 = -2\ \muC\) is at \(x = 4 cm\), which is outside the sphere (\(4 > 3\)).

So, the net charge enclosed by the first sphere is \(q_{encl,1} = 8\ \muC\).

The flux through the first sphere is \(\Phi_1 = \frac{8\ \muC}{\epsilon_0}\).

Now, let's analyze the second sphere:

Radius \(R_2 = 5 cm\), centered at the origin.

Both the charge \(q_1 = 8\ \muC\) (at \(x = 2 cm\)) and the charge \(q_2 = -2\ \muC\) (at \(x = 4 cm\)) are inside this sphere (\(2 < 5\) and \(4 < 5\)).

So, the net charge enclosed by the second sphere is \(q_{encl,2} = 8\ \muC + (-2\ \muC) = 6\ \muC\).

The flux through the second sphere is \(\Phi_2 = \frac{6\ \muC}{\epsilon_0}\).

Finally, find the ratio of the two fluxes:
\[ Ratio = \frac{\Phi_1}{\Phi_2} = \frac{8\ \muC / \epsilon_0}{6\ \muC / \epsilon_0} = \frac{8}{6} = \frac{4}{3} \]

Step 4: Final Answer:

The ratio of electric flux is \(4:3\).
Quick Tip: Gauss's Law relies strictly on net enclosed charge. The position of the charges inside the boundary does not change the total flux, so you only need to determine if a coordinate is less than the radius.

Step 1: Understanding the Concept:

For a light ray to undergo Total Internal Reflection (TIR) at an interface, it must travel from a denser medium to a rarer medium.

The angle of incidence must be greater than or equal to the critical angle (\(i \ge \theta_c\)).

Based on the geometric setup of an equilateral prism, a horizontally entering ray hits the second face at a specific angle of incidence, allowing us to find the required refractive index.


Step 2: Key Formula or Approach:

The critical angle is given by \(\sin \theta_c = \frac{n_2}{n_1}\), where \(n_1 = 1.6\).

For TIR to occur, the condition is \(\sin i \ge \sin \theta_c \implies \sin i \ge \frac{n_2}{n_1}\).

Rearranging gives the maximum possible value for \(n_2\) to allow TIR: \(n_2 \le n_1 \sin i\).


Step 3: Detailed Explanation:

The prism is equilateral, meaning all its angles are \(60^\circ\).

The light ray is incident normally on the first face, meaning it passes through without any deviation.

It then hits the second face. The normal to the second face makes a \(30^\circ\) angle with the horizontal.

By geometry, the angle of incidence \(i\) at the second face is \(60^\circ\).

Apply the TIR condition:
\[ \sin(60^\circ) \ge \frac{n_2}{1.6} \]
Substitute the value of \(\sin(60^\circ)\):
\[ \frac{\sqrt{3}}{2} \ge \frac{n_2}{1.6} \]
Solve for \(n_2\):
\[ n_2 \le 1.6 \times \frac{\sqrt{3}}{2} \] \[ n_2 \le 0.8\sqrt{3} \]
Convert the decimal to a fraction to match the options:
\[ 0.8 = \frac{8}{10} = \frac{4}{5} \]
So, \(n_2 \le \frac{4\sqrt{3}}{5}\).

The limiting (or maximum boundary) value for \(n_2\) is \(\frac{4\sqrt{3}}{5}\).


Step 4: Final Answer:

The required limiting value of \(n_2\) is \(\frac{4\sqrt{3}}{5}\).
Quick Tip: Pay close attention to geometry when a ray enters normally. In a prism with angle \(A\), a normally incident ray always hits the opposite face at an angle of incidence exactly equal to \(A\).

Step 1: Understanding the Concept:

The circuit has inductors, a capacitor, and a resistor in series, but switches can short out certain components.

Closing \(S_1\) shorts out \(L_2\), leaving \(L_1, C, R\) in the circuit.

Closing \(S_2\) shorts out \(L_1\), leaving \(L_2, C, R\) in the circuit.

We use the phase angle formula for series LCR circuits to establish equations for both cases.


Step 2: Key Formula or Approach:

The phase difference \(\phi\) in a series LCR circuit is given by \(\tan \phi = \frac{|X_L - X_C|}{R}\).

Assuming the circuit remains capacitive (\(X_C > X_L\)), we use \(\tan \phi = \frac{X_C - X_L}{R}\).

(Note: Assuming it is inductive yields the exact same final result for the required expression).


Step 3: Detailed Explanation:

Case 1: \(S_1\) closed, \(S_2\) open. \(L_2\) is shorted.

The phase difference is \(30^\circ\).
\[ \tan 30^\circ = \frac{X_C - X_{L1}}{R} \implies \frac{1}{\sqrt{3}} = \frac{X_C - X_{L1}}{R} \] \[ X_{L1} = X_C - \frac{R}{\sqrt{3}} \quad --- (Eq 1) \]
Case 2: \(S_2\) closed, \(S_1\) open. \(L_1\) is shorted.

The phase difference is \(60^\circ\).
\[ \tan 60^\circ = \frac{X_C - X_{L2}}{R} \implies \sqrt{3} = \frac{X_C - X_{L2}}{R} \] \[ X_{L2} = X_C - R\sqrt{3} \quad --- (Eq 2) \]
We are asked to find the value of \(3L_1 - L_2\).

Since \(X_L = \omega L\), we can write \(3L_1 - L_2 = \frac{3X_{L1} - X_{L2}}{\omega}\).

Let's compute \(3X_{L1} - X_{L2}\) using Eq 1 and Eq 2:
\[ 3X_{L1} - X_{L2} = 3\left(X_C - \frac{R}{\sqrt{3}}\right) - \left(X_C - R\sqrt{3}\right) \] \[ 3X_{L1} - X_{L2} = 3X_C - \sqrt{3}R - X_C + \sqrt{3}R = 2X_C \]
The unknown resistance \(R\) cancels out perfectly!

Now, calculate \(X_C\):

Given \(\omega = 300 rad/s\) and \(C = 100 \muF = 10^{-4} F\).
\[ X_C = \frac{1}{\omega C} = \frac{1}{300 \times 10^{-4}} = \frac{10^4}{300} = \frac{100}{3}\ \Omega \]
Finally, evaluate the required expression:
\[ 3L_1 - L_2 = \frac{2X_C}{\omega} = \frac{2 \times (100/3)}{300} = \frac{200}{900} = \frac{2}{9} H \]

Step 4: Final Answer:

The value of \((3L_1 - L_2)\) is \(\frac{2}{9}\).
Quick Tip: When a problem asks for a specific linear combination of variables (like \(3L_1 - L_2\)) and fails to provide a seemingly necessary constant (like \(R\)), it is a massive hint that the missing constant will algebraically cancel out during evaluation.

Step 1: Understanding the Concept:

Mutual inductance is the ratio of the magnetic flux linked with one circuit to the current in the other circuit.

Because \(L \gg R\), we can assume the magnetic field produced by the large square loop is practically uniform over the entire small area of the inner circular loop.


Step 2: Key Formula or Approach:

The magnetic field at the center of a square loop of side \(L\) carrying current \(I\) is derived using the Biot-Savart law: \(B = \frac{\mu_0 I}{4\pi d} (\sin \theta_1 + \sin \theta_2)\) for each of the 4 sides.

Magnetic flux \(\Phi = B \times A_{circle}\).

Mutual inductance \(M = \frac{\Phi}{I}\).


Step 3: Detailed Explanation:

For the square loop, the perpendicular distance from the center to any side is \(d = L/2\).

The angles made by the ends of a side at the center are \(\theta_1 = 45^\circ\) and \(\theta_2 = 45^\circ\).

The magnetic field due to one side of the square at the center is:
\[ B_{side} = \frac{\mu_0 I}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{2\pi L} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I}{2\pi L} \left(\frac{2}{\sqrt{2}}\right) = \frac{\sqrt{2} \mu_0 I}{\pi L} \]
Since there are 4 identical sides, the total magnetic field at the center is:
\[ B_{total} = 4 \times B_{side} = 4 \left( \frac{\sqrt{2} \mu_0 I}{\pi L} \right) = \frac{4\sqrt{2} \mu_0 I}{\pi L} \]
Wait, let's re-evaluate this carefully.
\(B_{side} = \frac{\mu_0 I}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{2\mu_0 I}{4\pi L} (\sqrt{2}) = \frac{\sqrt{2}\mu_0 I}{2\pi L}\).

Total field \(B_{total} = 4 \times \frac{\sqrt{2}\mu_0 I}{2\pi L} = \frac{2\sqrt{2}\mu_0 I}{\pi L}\).

(Correction applied: The factor of 2 was initially misplaced).

Now, calculate the magnetic flux through the small circular loop of area \(A = \pi R^2\):
\[ \Phi = B_{total} \times A = \left( \frac{2\sqrt{2} \mu_0 I}{\pi L} \right) (\pi R^2) = \frac{2\sqrt{2} \mu_0 R^2 I}{L} \]
Finally, the mutual inductance \(M\) is the flux per unit current:
\[ M = \frac{\Phi}{I} = \frac{2\sqrt{2} \mu_0 R^2}{L} \]

Step 4: Final Answer:

The mutual inductance is \(2\sqrt{2} \mu_0 R^2 / L\).
Quick Tip: Always use the "large to small" approach for mutual inductance when dealing with vastly different sizes: compute the B-field from the larger loop and multiply by the area of the smaller loop. The reciprocity theorem ensures \(M_{12} = M_{21}\).

Step 1: Understanding the Concept:

The binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons.

It is calculated by finding the mass defect (the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus) and converting it to energy.

Finally, dividing by the total number of nucleons gives the binding energy per nucleon.


Step 2: Key Formula or Approach:

Number of protons \(Z = 83\), Number of neutrons \(N = A - Z = 209 - 83 = 126\).

Mass defect: \(\Delta m = Z m_p + N m_n - M_{nucleus}\).

Binding Energy: \(BE = \Delta m \times 931 MeV\).

Binding Energy per nucleon: \(BE/A = BE / 209\).


Step 3: Detailed Explanation:

Calculate the total mass of the individual constituent nucleons:

Mass of 83 protons = \(83 \times 1.007825 u = 83.649475 u\).

Mass of 126 neutrons = \(126 \times 1.008665 u = 127.091790 u\).

Total constituent mass = \(83.649475 + 127.091790 = 210.741265 u\).

Now, subtract the actual mass of the Bismuth nucleus to find the mass defect:
\[ \Delta m = 210.741265 u - 208.980388 u = 1.760877 u \]
Convert the mass defect into energy in MeV:
\[ BE = 1.760877 u \times 931 MeV/u = 1639.376487 MeV \]
Calculate the binding energy per nucleon by dividing by \(A = 209\):
\[ \frac{BE}{A} = \frac{1639.376487}{209} \approx 7.8439 MeV/nucleon \]
This rounds to \(7.84 MeV\).


Step 4: Final Answer:

The binding energy per nucleon is \(7.84 MeV\).
Quick Tip: Always maintain at least 6 decimal places during mass defect calculations to prevent severe rounding errors before multiplying by 931.

Step 1: Understanding the Concept:

The given equation describes Simple Harmonic Motion (SHM).

The particle comes to rest when its velocity is zero (at the extreme positions).

The particle has zero acceleration when it passes through the mean position.

We find velocity and acceleration by taking the first and second derivatives of the position function with respect to time.


Step 2: Key Formula or Approach:

Velocity \(v = \frac{dx}{dt}\).

Acceleration \(A = \frac{dv}{dt}\).

Set \(v = 0\) to find \(t_1\) and \(A = 0\) to find \(t_2\).


Step 3: Detailed Explanation:

Given \(x = a \sin(50t + \pi/3)\).

Differentiate to find velocity:
\[ v = \frac{d}{dt} [a \sin(50t + \pi/3)] = 50a \cos(50t + \pi/3) \]
The particle comes to rest when \(v = 0\):
\[ 50a \cos(50t_1 + \pi/3) = 0 \implies \cos(50t_1 + \pi/3) = 0 \]
The first positive time this occurs is when the phase equals \(\pi/2\):
\[ 50t_1 + \frac{\pi}{3} = \frac{\pi}{2} \] \[ 50t_1 = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} \] \[ t_1 = \frac{\pi}{6 \times 50} = \frac{\pi}{300} s \]
Now, differentiate velocity to find acceleration:
\[ A = \frac{d}{dt} [50a \cos(50t + \pi/3)] = -2500a \sin(50t + \pi/3) \]
The acceleration is zero when \(A = 0\):
\[ -2500a \sin(50t_2 + \pi/3) = 0 \implies \sin(50t_2 + \pi/3) = 0 \]
The first positive time this occurs is when the phase equals \(\pi\):
\[ 50t_2 + \frac{\pi}{3} = \pi \] \[ 50t_2 = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \] \[ t_2 = \frac{2\pi}{3 \times 50} = \frac{2\pi}{150} = \frac{\pi}{75} s \]

Step 4: Final Answer:

The times \(t_1\) and \(t_2\) respectively are \(\frac{\pi}{300} s, \frac{\pi}{75} s\).
Quick Tip: In SHM, velocity is zero at extremes (phase = \(\pi/2, 3\pi/2\)) and acceleration is zero at the mean position (phase = \(\pi, 2\pi\)). Setting the phase equal to these standard angles gives the fastest solution.

Step 1: Understanding the Concept:

In an interference pattern, the resultant intensity at any point on the screen depends on the phase difference between the two interfering waves.

Once the phase difference is found, it can be directly converted into the corresponding path difference.


Step 2: Key Formula or Approach:

The intensity formula is \(I = I_{max} \cos^2\left(\frac{\phi}{2}\right)\), where \(\phi\) is the phase difference.

The relation between phase difference and path difference is \(\phi = \frac{2\pi}{\lambda} \Delta x\).


Step 3: Detailed Explanation:

Given that the intensity is \(I = \frac{3}{4} I_{max}\).

Substitute this into the intensity formula:
\[ \frac{3}{4} I_{max} = I_{max} \cos^2\left(\frac{\phi}{2}\right) \] \[ \cos^2\left(\frac{\phi}{2}\right) = \frac{3}{4} \]
Taking the square root of both sides:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{\sqrt{3}}{2} \]
The principal angle whose cosine is \(\frac{\sqrt{3}}{2}\) is \(30^\circ\) or \(\frac{\pi}{6}\) radians.
\[ \frac{\phi}{2} = \frac{\pi}{6} \implies \phi = \frac{\pi}{3} radians \]
Now, convert the phase difference to a path difference \(\Delta x\):
\[ \phi = \frac{2\pi}{\lambda} \Delta x \] \[ \frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x \] \[ \Delta x = \frac{\lambda}{3 \times 2} = \frac{\lambda}{6} \]
Comparing this to the given format \(\frac{\lambda}{x}\), we find that \(x = 6\).


Step 4: Final Answer:

The value of \(x\) is \(6\).
Quick Tip: Remember the standard intensity fractions: \(I = I_{max}\) at \(\Delta x = 0\), \(I = I_{max}/2\) at \(\Delta x = \lambda/4\), and \(I = 3I_{max}/4\) at \(\Delta x = \lambda/6\). Memorizing these saves calculation time.

Step 1: Understanding the Concept:

An electron orbiting a nucleus behaves like a circular current loop.

This loop generates a magnetic field at the center of the orbit.

By expressing the magnetic field in terms of the principal quantum number \(n\), we can easily find the ratio for different orbits.


Step 2: Key Formula or Approach:

The magnetic field at the center of a circular current loop is \(B = \frac{\mu_0 I}{2r}\).

The equivalent current is \(I = \frac{e}{T} = \frac{ev}{2\pi r}\).

Substituting \(I\) gives \(B = \frac{\mu_0 ev}{4\pi r^2}\).

From Bohr's model, velocity \(v \propto \frac{1}{n}\) and radius \(r \propto n^2\).


Step 3: Detailed Explanation:

Substitute the proportionalities of \(v\) and \(r\) into the magnetic field equation:
\[ B \propto \frac{v}{r^2} \] \[ B \propto \frac{(1/n)}{(n^2)^2} \] \[ B \propto \frac{1/n}{n^4} = \frac{1}{n^5} \]
So, the magnetic field is inversely proportional to the fifth power of the principal quantum number.

We need the ratio of the magnetic fields in the 2nd (\(n_1 = 2\)) and 4th (\(n_2 = 4\)) orbits.
\[ \frac{B_2}{B_4} = \frac{(n_4)^5}{(n_2)^5} \] \[ \frac{B_2}{B_4} = \left(\frac{4}{2}\right)^5 \] \[ \frac{B_2}{B_4} = (2)^5 = 32 \]

Step 4: Final Answer:

The ratio of the magnetic fields is \(32\).
Quick Tip: For Bohr model proportionalities, memorizing \(r \propto n^2/Z\), \(v \propto Z/n\), and \(E \propto Z^2/n^2\) allows you to quickly derive secondary proportionalities like \(T \propto n^3/Z^2\) and \(B \propto Z^3/n^5\).

Step 1: Understanding the Concept:

The change in internal energy of an ideal gas depends only on its temperature change and molar specific heat at constant volume (\(c_v\)), regardless of the process path.

We must ensure all values are in consistent units (calories) before applying the formula.


Step 2: Key Formula or Approach:

Mayer's relation: \(c_p - c_v = R\).

Change in internal energy: \(\Delta U = n c_v \Delta T\).

Mechanical equivalent of heat: \(1 calorie = 4.18 Joules\).


Step 3: Detailed Explanation:

First, convert the universal gas constant \(R\) from Joules to calories.
\[ R = 8.36 J/mol.^\circC = \frac{8.36}{4.18} cal/mol.^\circC = 2 cal/mol.^\circC \]
Next, calculate the molar specific heat at constant volume \(c_v\) using Mayer's relation.
\[ c_v = c_p - R = 8 cal/mol.^\circC - 2 cal/mol.^\circC = 6 cal/mol.^\circC \]
Now, identify the given parameters for the process:

Number of moles \(n = 5\).

Temperature change \(\Delta T = 20^\circC - 10^\circC = 10^\circC\).

Calculate the change in internal energy:
\[ \Delta U = n c_v \Delta T \] \[ \Delta U = 5 \times 6 \times 10 \] \[ \Delta U = 300 calories \]

Step 4: Final Answer:

The change in the internal energy is \(300 calorie\).
Quick Tip: Always check the units of \(R\) and specific heats. Examiners frequently mix Joules and Calories in the same problem to test your attention to dimensional consistency.

Step 1: Understanding the Concept:

Sunlight consists of parallel rays from infinity. When a convex lens focuses parallel rays, they converge exactly at the principal focus.

Since the burning is fastest when focused perfectly, the distance between the lens and the paper is the focal length \(f\).

We can use the Lens Maker's Formula to relate the focal length, radii of curvature, and the refractive index.


Step 2: Key Formula or Approach:

Lens Maker's Formula: \(\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).

For an equiconvex lens (implied when only one radius is given), \(R_1 = R\) and \(R_2 = -R\).


Step 3: Detailed Explanation:

From the problem description, the focal length is \(f = 30 cm\).

The radius of curvature is given as \(R = 60 cm\). So \(R_1 = 60 cm\) and \(R_2 = -60 cm\).

Apply the Lens Maker's Formula:
\[ \frac{1}{30} = (\mu - 1) \left(\frac{1}{60} - \frac{1}{-60}\right) \] \[ \frac{1}{30} = (\mu - 1) \left(\frac{1}{60} + \frac{1}{60}\right) \] \[ \frac{1}{30} = (\mu - 1) \left(\frac{2}{60}\right) \] \[ \frac{1}{30} = \frac{\mu - 1}{30} \]
Multiply both sides by 30:
\[ 1 = \mu - 1 \implies \mu = 2 \]
The problem states the refractive index is \(\frac{\alpha}{10}\).

Equating the values:
\[ \frac{\alpha}{10} = 2 \implies \alpha = 20 \]

Step 4: Final Answer:

The value of \(\alpha\) is \(20\).
Quick Tip: When a problem simply states "the radius of curvature of the lens", safely assume it refers to a symmetric equiconvex or equiconcave lens where \(|R_1| = |R_2| = R\).

Step 1: Understanding the Concept:

The total moment of inertia of the solid sphere about the axis \(AB\) relies on the parallel axis theorem, as it rotates about an axis separated from its center of mass.

We equate this to the moment of inertia of the rod rotating about one of its ends.


Step 2: Key Formula or Approach:

Moment of inertia of a rod about its end: \(I_{rod} = \frac{M L^2}{3}\).

Moment of inertia of a solid sphere about its center of mass: \(I_{cm} = \frac{2}{5} m R^2\).

Parallel axis theorem: \(I_{sphere} = I_{cm} + m d^2\).

Equate the two moments of inertia: \(I_{rod} = I_{sphere}\).


Step 3: Detailed Explanation:

First, calculate the moment of inertia of the rod.

Mass of rod \(M = 40 kg\), Length \(L = 3 m\).
\[ I_{rod} = \frac{40 \times (3)^2}{3} = 40 \times 3 = 120 kg m^2 \]
Next, set up the expression for the moment of inertia of the solid sphere.

Mass of sphere \(m = 10 kg\), distance from axis \(d = 3 m\).
\[ I_{sphere} = \frac{2}{5} m R^2 + m d^2 \] \[ I_{sphere} = \frac{2}{5}(10) R^2 + 10(3)^2 \] \[ I_{sphere} = 4 R^2 + 90 \]
Now, equate the two moments of inertia:
\[ 4 R^2 + 90 = 120 \] \[ 4 R^2 = 120 - 90 = 30 \] \[ R^2 = \frac{30}{4} = \frac{15}{2} \]
Take the square root to find \(R\):
\[ R = \sqrt{\frac{15}{2}} \]
The problem states \(R = \sqrt{\frac{\alpha}{2}}\). By comparing, we get:
\[ \alpha = 15 \]

Step 4: Final Answer:

The value of \(\alpha\) is \(15\).
Quick Tip: Always double-check which axis the object is rotating around. For a rod, \(ML^2/12\) is for the center, but \(ML^2/3\) is for the edge. Misidentifying the axis is the most common error in rigid body dynamics.