Step 1: Understanding the Question:

The question asks for the value of a constant \(\alpha\) given a relationship between the root mean square (rms) speed and the average speed of oxygen molecules at a certain temperature.


Step 2: Key Formula or Approach:

The formulas for the rms speed (\(v_{rms}\)) and average speed (\(v_{avg}\)) of gas molecules are:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \] \[ v_{avg} = \sqrt{\frac{8RT}{\pi M}} \]
where R is the universal gas constant, T is the absolute temperature, and M is the molar mass of the gas.

We need to find the ratio \(\frac{v_{rms}}{v_{avg}}\) and equate it to the given expression.


Step 3: Detailed Explanation:

First, let's find the theoretical ratio of the rms speed to the average speed:
\[ \frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}} = \sqrt{\frac{3RT}{M} \times \frac{\pi M}{8RT}} = \sqrt{\frac{3\pi}{8}} \]
The question states that this ratio is equal to \(\sqrt{\frac{\alpha+5}{\alpha}}\).

So, we can set up the equation:
\[ \sqrt{\frac{3\pi}{8}} = \sqrt{\frac{\alpha+5}{\alpha}} \]
Squaring both sides of the equation gives:
\[ \frac{3\pi}{8} = \frac{\alpha+5}{\alpha} \]
Now, we substitute the given value of \(\pi = \frac{22}{7}\):
\[ \frac{3 \times \frac{22}{7}}{8} = \frac{\alpha+5}{\alpha} \] \[ \frac{66}{56} = \frac{\alpha+5}{\alpha} \]
Simplifying the fraction on the left side:
\[ \frac{33}{28} = \frac{\alpha+5}{\alpha} \]
Now, we can cross-multiply to solve for \(\alpha\):
\[ 33\alpha = 28(\alpha+5) \] \[ 33\alpha = 28\alpha + 140 \] \[ 33\alpha - 28\alpha = 140 \] \[ 5\alpha = 140 \] \[ \alpha = \frac{140}{5} = 28 \]

Step 4: Final Answer:

The value of \(\alpha\) is 28. This corresponds to option (D).
Quick Tip: Remember the standard formulas for different molecular speeds: \(v_{rms} = \sqrt{\frac{3RT}{M}}\), \(v_{avg} = \sqrt{\frac{8RT}{\pi M}}\), and most probable speed \(v_{mp} = \sqrt{\frac{2RT}{M}}\). The ratio between them is a common topic in kinetic theory of gases. The temperature of 300 K is extra information not needed for the calculation.

Step 1: Understanding the Question:

We are comparing the time of descent for an object on a smooth and a rough inclined plane of the same angle (45\(^{\circ}\)). We need to find the coefficient of kinetic friction \(\mu\) in terms of the ratio of the times, n.


Step 2: Key Formula or Approach:

We will use the equations of motion. Let the length of the inclined plane be L. The distance is given by \(L = \frac{1}{2}at^2\), where 'a' is the acceleration and 't' is the time. We need to find the acceleration for both the smooth and rough cases.

For an inclined plane with angle \(\theta\):

- Acceleration on a smooth surface: \(a_{smooth} = g \sin\theta\)

- Acceleration on a rough surface: \(a_{rough} = g \sin\theta - \mu g \cos\theta\)


Step 3: Detailed Explanation:

Let \(t_s\) be the time taken on the smooth plane and \(t_r\) be the time taken on the rough plane.

Given: \(t_r = n t_s\) and \(\theta = 45^{\circ}\).


For the smooth incline:
\(a_s = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}\).

The distance L is covered in time \(t_s\): \(L = \frac{1}{2} a_s t_s^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t_s^2\). (Equation 1)


For the rough incline:
\(a_r = g \sin 45^{\circ} - \mu g \cos 45^{\circ} = \frac{g}{\sqrt{2}} - \mu \frac{g}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1-\mu)\).

The distance L is covered in time \(t_r\): \(L = \frac{1}{2} a_r t_r^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}}(1-\mu)\right) t_r^2\). (Equation 2)


Since the distance L is the same, we can equate Equation 1 and Equation 2:
\[ \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t_s^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}}(1-\mu)\right) t_r^2 \]
Canceling out the common terms \(\frac{1}{2} \frac{g}{\sqrt{2}}\):
\[ t_s^2 = (1-\mu) t_r^2 \]
Now, substitute \(t_r = n t_s\):
\[ t_s^2 = (1-\mu) (n t_s)^2 \] \[ t_s^2 = (1-\mu) n^2 t_s^2 \]
Cancel \(t_s^2\) from both sides (since \(t_s \neq 0\)):
\[ 1 = (1-\mu) n^2 \]
Rearrange to solve for \(\mu\):
\[ \frac{1}{n^2} = 1 - \mu \] \[ \mu = 1 - \frac{1}{n^2} \]

Step 4: Final Answer:

The coefficient of kinetic friction is \(1 - \frac{1}{n^2}\). This corresponds to option (A).
Quick Tip: For problems comparing motion on smooth and rough surfaces, setting up a ratio is often the quickest method. Note that \(t = \sqrt{2L/a}\), so \(t \propto 1/\sqrt{a}\). This means \(t_r/t_s = n = \sqrt{a_s/a_r}\). Squaring this gives \(n^2 = a_s/a_r\), which leads to the same result faster.

Step 1: Understanding the Question:

We need to find the ratio of the de-Broglie wavelengths of an alpha particle and a proton. Both particles are accelerated from rest by the same electric potential V. The ratio is given in a specific format, and we need to find the value of 'm'.


Step 2: Key Formula or Approach:

The de-Broglie wavelength (\(\lambda\)) is given by \(\lambda = h/p\), where h is Planck's constant and p is the momentum.

When a charged particle 'q' with mass 'm' is accelerated by a potential 'V', its kinetic energy K is given by \(K = qV\).

The momentum can be expressed in terms of kinetic energy as \(p = \sqrt{2mK}\).

Combining these, the de-Broglie wavelength is:
\[ \lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}} \]

Step 3: Detailed Explanation:

Let's denote the properties of the proton with subscript 'p' and the alpha particle with subscript '\(\alpha\)'.

For a proton:

- Mass: \(m_p\)

- Charge: \(q_p = e\)

The de-Broglie wavelength of the proton is:
\[ \lambda_p = \frac{h}{\sqrt{2m_p q_p V}} = \frac{h}{\sqrt{2m_p e V}} \]

For an alpha particle (which is a helium nucleus, \(^4_2He\)):

- Mass: \(m_{\alpha} \approx 4m_p\)

- Charge: \(q_{\alpha} = 2e\)

The de-Broglie wavelength of the alpha particle is:
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V}} = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{h}{\sqrt{16m_p e V}} \]

Now, we find the ratio \(\frac{\lambda_{\alpha}}{\lambda_p}\):
\[ \frac{\lambda_{\alpha}}{\lambda_p} = \frac{\frac{h}{\sqrt{16m_p e V}}}{\frac{h}{\sqrt{2m_p e V}}} = \frac{\sqrt{2m_p e V}}{\sqrt{16m_p e V}} = \sqrt{\frac{2m_p e V}{16m_p e V}} = \sqrt{\frac{2}{16}} = \sqrt{\frac{1}{8}} = \frac{1}{\sqrt{8}} \]
The question states that this ratio is \(\frac{1}{\sqrt{m}}\).

Comparing our result with the given expression:
\[ \frac{1}{\sqrt{8}} = \frac{1}{\sqrt{m}} \]
Therefore, \(m = 8\).


Step 4: Final Answer:

The value of m is 8. This corresponds to option (A).
Quick Tip: When dealing with ratios of de-Broglie wavelengths for particles accelerated by the same potential, remember that \(\lambda \propto \frac{1}{\sqrt{mq}}\). This allows for a quick calculation: \(\frac{\lambda_{\alpha}}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_{\alpha} q_{\alpha}}} = \sqrt{\frac{m_p e}{(4m_p)(2e)}} = \sqrt{\frac{1}{8}}\).

Step 1: Understanding the Question:

A charge is moved from an initial point P to a final point S in a uniform electric field. We need to calculate the work done by the electric field during this displacement.


Step 2: Key Formula or Approach:

The work done (W) by a constant force (\(\vec{F}\)) over a displacement (\(\vec{d}\)) is given by the dot product \(W = \vec{F} \cdot \vec{d}\).

In an electric field \(\vec{E}\), the force on a charge q is \(\vec{F} = q\vec{E}\).

Therefore, the work done by the electric field is \(W = q\vec{E} \cdot \vec{d}\).

The displacement vector \(\vec{d}\) is the final position vector minus the initial position vector, \(\vec{d} = \vec{r}_S - \vec{r}_P\).


Step 3: Detailed Explanation:

Given values:

- Charge, \(q = 2 \times 10^{-2}\) C

- Electric field, \(\vec{E} = 30 \hat{i}\) NC\(^{-1}\) (directed along positive x-axis)

- Initial position (P), \(\vec{r}_P = (1\hat{i} + 2\hat{j} + 0\hat{k})\) m

- Final position (S), \(\vec{r}_S = (0\hat{i} + 0\hat{j} + 0\hat{k})\) m


First, calculate the displacement vector \(\vec{d}\):
\[ \vec{d} = \vec{r}_S - \vec{r}_P = (0-1)\hat{i} + (0-2)\hat{j} + (0-0)\hat{k} = -1\hat{i} - 2\hat{j} \]
Now, calculate the work done using the formula \(W = q(\vec{E} \cdot \vec{d})\):
\[ W = (2 \times 10^{-2}) \left( (30\hat{i}) \cdot (-1\hat{i} - 2\hat{j}) \right) \]
The dot product is calculated as:
\[ \vec{E} \cdot \vec{d} = (30 \times -1) + (0 \times -2) + (0 \times 0) = -30 \]
Now, substitute this back into the work equation:
\[ W = (2 \times 10^{-2}) \times (-30) = -60 \times 10^{-2} J = -0.6 J \]
The options are given in millijoules (mJ). We convert Joules to millijoules (1 J = 1000 mJ):
\[ W = -0.6 \times 1000 mJ = -600 mJ \]

Step 4: Final Answer:

The work done by the electric field is -600 mJ. This corresponds to option (A).
Quick Tip: In a uniform electric field, the work done depends only on the displacement parallel to the field. Here, the field is along the x-axis. The displacement along the x-axis is \(x_{final} - x_{initial} = 0 - 1 = -1\) m. So, work done is \(W = F_x \Delta x = (qE_x) \Delta x = (2 \times 10^{-2} \times 30) \times (-1) = -0.6\) J. This method is faster as it ignores the displacement in y and z directions, which are perpendicular to the field.

Step 1: Understanding the Question:

We need to find the work done to pull a conductive loop out of a uniform magnetic field. The work done by the external agent is converted into heat energy dissipated in the loop's resistance due to the induced current.


Step 2: Key Formula or Approach:

1. Calculate the initial magnetic flux (\(\Phi_i\)) through the loop.

2. The final flux (\(\Phi_f\)) is zero as the loop is outside the field.

3. Use Faraday's law of induction to find the induced electromotive force (EMF), \(\mathcal{E} = -\frac{\Delta \Phi}{\Delta t}\).

4. Calculate the induced current using Ohm's law, \(I = \frac{\mathcal{E}}{R}\).

5. The work done is equal to the energy dissipated as heat, \(W = P \times \Delta t = I^2 R \Delta t\).


Step 3: Detailed Explanation:

Given values:

- Area, \(A = 25 cm^2 = 25 \times 10^{-4} m^2\)

- Resistance, \(R = 10 \, \Omega\)

- Magnetic field, \(B = 40.0\) T

- Time interval, \(\Delta t = 1.0\) s


Step 1: Initial magnetic flux. Since the plane of the loop is perpendicular to the field, the angle between the area vector and the magnetic field is 0\(^{\circ}\).
\[ \Phi_i = B \cdot A \cos(0^{\circ}) = 40 \times (25 \times 10^{-4}) = 1000 \times 10^{-4} = 0.1 Wb \]

Step 2: Final magnetic flux. The loop is pulled out of the field.
\[ \Phi_f = 0 Wb \]

Step 3: Change in flux and induced EMF.
\[ \Delta \Phi = \Phi_f - \Phi_i = 0 - 0.1 = -0.1 Wb \]
The magnitude of the induced EMF is:
\[ |\mathcal{E}| = \left|-\frac{\Delta \Phi}{\Delta t}\right| = \left|-\frac{-0.1}{1.0}\right| = 0.1 V \]

Step 4: Induced current.
\[ I = \frac{|\mathcal{E}|}{R} = \frac{0.1}{10} = 0.01 A = 10^{-2} A \]

Step 5: Work done (Energy dissipated).
\[ W = I^2 R \Delta t = (10^{-2})^2 \times 10 \times 1.0 \] \[ W = 10^{-4} \times 10 = 10^{-3} J \] \[ W = 1.0 \times 10^{-3} J \]

Step 4: Final Answer:

The work done will be \(1.0 \times 10^{-3}\) J. This corresponds to option (A).
Quick Tip: An alternative formula for energy dissipated when flux changes is \(W = \frac{(\Delta \Phi)^2}{R \Delta t}\). This can be derived from \(W = (\frac{\mathcal{E}}{R})^2 R \Delta t = \frac{\mathcal{E}^2}{R} \Delta t = \frac{(-\Delta \Phi / \Delta t)^2}{R} \Delta t = \frac{(\Delta \Phi)^2}{R (\Delta t)^2} \Delta t\). Wait, there's a mistake in this derivation. The correct one is \(W = I^2 R \Delta t = (\frac{\Delta \Phi}{R \Delta t})^2 R \Delta t = \frac{(\Delta \Phi)^2}{R(\Delta t)}\). Using this: \(W = \frac{(0.1)^2}{10 \times 1} = \frac{0.01}{10} = 0.001 = 10^{-3}\) J. This can be a useful shortcut.

Step 1: Understanding the Question:

The question asks for the percentage increase in air speed over the upper surface of a wing compared to the lower surface, which is required to generate enough lift to keep the aircraft in level flight.


Step 2: Key Formula or Approach:

1. For level flight, the lift force must equal the weight of the aircraft: \(F_{lift} = mg\).

2. The lift force is generated by the pressure difference (\(\Delta P\)) between the lower and upper surfaces of the wing: \(F_{lift} = \Delta P \times A\), where A is the wing area.

3. The pressure difference is related to the difference in air speeds (\(v_1\) on lower, \(v_2\) on upper surface) by Bernoulli's principle: \(\Delta P = P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)\).

4. We need to find the fractional increase \(\frac{v_2 - v_1}{v_1}\) and express it as a percentage.


Step 3: Detailed Explanation:

Given values:

- Mass, \(m = 5.4 \times 10^5\) kg

- Wing area, \(A = 500\) m\(^2\)

- Speed of aircraft (speed of air on lower surface), \(v_1 = 1080 km/h\)

- Density of air, \(\rho = 1.2\) kg m\(^{-3}\)

- Acceleration due to gravity, \(g = 10\) m s\(^{-2}\)


First, convert the speed to m/s:
\[ v_1 = 1080 \times \frac{1000 m}{3600 s} = 1080 \times \frac{5}{18} = 60 \times 5 = 300 m/s \]

Next, calculate the required lift force (weight of the aircraft):
\[ F_{lift} = mg = (5.4 \times 10^5 kg) \times (10 m/s^2) = 5.4 \times 10^6 N \]

Now, find the pressure difference needed:
\[ \Delta P = \frac{F_{lift}}{A} = \frac{5.4 \times 10^6 N}{500 m^2} = \frac{540 \times 10^4}{500} = 1.08 \times 10^4 Pa \]

Using Bernoulli's principle:
\[ \Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2) \] \[ 1.08 \times 10^4 = \frac{1}{2} \times 1.2 \times (v_2^2 - 300^2) \] \[ 1.08 \times 10^4 = 0.6 (v_2^2 - 90000) \] \[ v_2^2 - 90000 = \frac{1.08 \times 10^4}{0.6} = 1.8 \times 10^4 = 18000 \] \[ v_2^2 = 90000 + 18000 = 108000 \] \[ v_2 = \sqrt{108000} \approx 328.6 m/s \]

Now, calculate the fractional increase:
\[ Fractional increase = \frac{v_2 - v_1}{v_1} = \frac{328.6 - 300}{300} = \frac{28.6}{300} \approx 0.0953 \]
To express this as a percentage, we multiply by 100:
\[ Percentage increase = 0.0953 \times 100 \approx 9.53% \]
This is approximately 10%.


Approximation method:

We can approximate \(v_2^2 - v_1^2 = (v_2 - v_1)(v_2 + v_1)\). Since \(v_2\) is close to \(v_1\), we can approximate \(v_2 + v_1 \approx 2v_1\).
\[ \Delta P \approx \frac{1{2}\rho(v_2 - v_1)(2v_1) = \rho (v_2 - v_1) v_1 \] \[ \frac{v_2 - v_1}{v_1} = \frac{\Delta P}{\rho v_1^2} = \frac{mg/A}{\rho v_1^2} = \frac{mg}{A \rho v_1^2} \] \[ \frac{v_2 - v_1}{v_1} = \frac{5.4 \times 10^6}{500 \times 1.2 \times (300)^2} = \frac{5.4 \times 10^6}{600 \times 90000} = \frac{5.4 \times 10^6}{54 \times 10^6} = 0.1 \]
Percentage increase = \(0.1 \times 100 = 10%\).


Step 4: Final Answer:

The percentage increase in speed is 10%. This corresponds to option (D).
Quick Tip: For aerodynamic lift problems involving small differences in speed, the approximation \(v_2+v_1 \approx 2v_1\) is usually very effective and simplifies the calculation significantly. It saves you from calculating square roots and leads directly to the fractional change.

Step 1: Understanding the Question:

We are given a certain amount of heat energy and we need to determine the final state (temperature and composition) of a given mass of ice initially at a sub-zero temperature.


Step 2: Key Formula or Approach:

We need to calculate the heat required for each phase of the process and compare it with the supplied heat.

1. Heat required to raise the temperature of ice from \(-12^\circ\)C to \(0^\circ\)C: \(Q_1 = m c_{ice} \Delta T\).

2. Heat required to melt all the ice at \(0^\circ\)C into water at \(0^\circ\)C: \(Q_2 = m L_f\).

By comparing the supplied heat \(Q_{sup}\) with \(Q_1\) and \(Q_1 + Q_2\), we can determine the final state.


Step 3: Detailed Explanation:

Given values:

- Heat supplied, \(Q_{sup} = 184 kJ = 184000 J\)

- Mass of ice, \(m = 600 g = 0.6 kg\)

- Initial temperature, \(T_i = -12^\circ\)C

- Specific heat of ice, \(c_{ice} = 2222.3 J kg^{-1} K^{-1}\)

- Latent heat of fusion, \(L_f = 336 kJ/kg = 336000 J/kg\)


Calculation 1: Heat to reach 0\(^\circ\)C

Heat required to raise the temperature of ice from \(-12^\circ\)C to \(0^\circ\)C:
\[ Q_1 = m c_{ice} \Delta T = 0.6 \times 2222.3 \times (0 - (-12)) = 0.6 \times 2222.3 \times 12 \] \[ Q_1 = 16000.56 J \]

Calculation 2: Heat to melt all ice at 0\(^\circ\)C

Heat required to melt all 0.6 kg of ice at \(0^\circ\)C:
\[ Q_2 = m L_f = 0.6 \times 336000 = 201600 J \]

Analysis:

The total heat supplied is \(Q_{sup} = 184000\) J.

- Since \(Q_{sup} > Q_1\) (184000 J > 16000.56 J), all the ice will reach \(0^\circ\)C.

- The total heat required to convert all ice at \(-12^\circ\)C to water at \(0^\circ\)C is \(Q_{total} = Q_1 + Q_2 = 16000.56 + 201600 = 217600.56\) J.

- Since \(Q_{sup} < Q_{total}\) (184000 J < 217600.56 J), not all the ice will melt.

- This means the final temperature of the system will be \(0^\circ\)C, and it will be a mixture of ice and water.

- Therefore, statement A is correct and statements B and E are incorrect.


Calculation 3: Amount of ice melted

Heat remaining after raising the ice temperature to \(0^\circ\)C:
\[ Q_{rem} = Q_{sup} - Q_1 = 184000 - 16000.56 = 167999.44 J \]
This remaining heat will melt a certain mass of ice (\(m_{melted}\)):
\[ m_{melted} = \frac{Q_{rem}}{L_f} = \frac{167999.44}{336000} \approx 0.5 kg = 500 g \]

Final Composition:

- Mass of water formed: \(m_{water} = m_{melted} = 500\) g.

- Mass of ice remaining: \(m_{ice} = m_{total} - m_{melted} = 600 g - 500 g = 100 g\).

The ratio of the mass of ice to the mass of water is \(m_{ice} : m_{water} = 100 : 500 = 1:5\).

- Therefore, statement D is correct and statement C is incorrect.


Step 4: Final Answer:

The correct statements are A and D. This corresponds to option (D).
Quick Tip: In calorimetry problems, always calculate the energy required for each step (temperature change, phase change) sequentially. Compare the supplied energy at each stage to determine the final state. Do not assume the final state beforehand.

Step 1: Understanding the Question:

We are given two radioactive substances with their initial masses, atomic masses, and half-lives. We need to find the total number of atoms of both substances remaining after a specific time.


Step 2: Key Formula or Approach:

1. Calculate the number of half-lives (\(n\)) for each substance using \(n = t/T_{1/2}\), where t is the elapsed time and \(T_{1/2}\) is the half-life.

2. Calculate the mass remaining (\(m\)) for each substance using \(m = m_0 (1/2)^n\), where \(m_0\) is the initial mass.

3. Calculate the number of moles remaining for each substance.

4. Calculate the number of atoms (\(N\)) remaining for each substance using \(N = (moles) \times N_A\), where \(N_A\) is Avogadro's number (\(6.022 \times 10^{23}\) mol\(^{-1}\)).

5. Sum the number of atoms of A and B.


Step 3: Detailed Explanation:

Given values:

- For substance A: \(M_A = 16\) g/mol, \(T_{1/2, A} = 1\) day, \(m_{0,A} = 320\) g.

- For substance B: \(M_B = 32\) g/mol, \(T_{1/2, B} = 0.5\) day, \(m_{0,B} = 320\) g.

- Elapsed time, \(t = 2\) days.


For Substance A:

- Number of half-lives: \(n_A = \frac{t}{T_{1/2, A}} = \frac{2 days}{1 day} = 2\).

- Mass remaining: \(m_A = m_{0,A} \left(\frac{1}{2}\right)^{n_A} = 320 \left(\frac{1}{2}\right)^2 = 320 \times \frac{1}{4} = 80\) g.

- Number of atoms remaining: \(N_A = \frac{m_A}{M_A} \times N_A = \frac{80}{16} \times N_A = 5 N_A\).


For Substance B:

- Number of half-lives: \(n_B = \frac{t}{T_{1/2, B}} = \frac{2 days}{0.5 day} = 4\).

- Mass remaining: \(m_B = m_{0,B} \left(\frac{1}{2}\right)^{n_B} = 320 \left(\frac{1}{2}\right)^4 = 320 \times \frac{1}{16} = 20\) g.

- Number of atoms remaining: \(N_B = \frac{m_B}{M_B} \times N_A = \frac{20}{32} \times N_A = \frac{5}{8} N_A = 0.625 N_A\).


Total Atoms Remaining:

- Total atoms \(N_{total} = N_A + N_B = 5 N_A + 0.625 N_A = 5.625 N_A\).

- Substitute \(N_A = 6.022 \times 10^{23}\):
\[ N_{total} = 5.625 \times (6.022 \times 10^{23}) \] \[ N_{total} \approx 33.873 \times 10^{23} = 3.3873 \times 10^{24} \]

Step 4: Final Answer:

The total number of atoms left is approximately \(3.38 \times 10^{24}\). This corresponds to option (C).
Quick Tip: It's often useful to first calculate the initial number of atoms and then apply the decay formula \(N = N_0(1/2)^n\). \(N_{0,A} = (320/16)N_A = 20 N_A\). \(N_A = 20 N_A (1/2)^2 = 5 N_A\). \(N_{0,B} = (320/32)N_A = 10 N_A\). \(N_B = 10 N_A (1/2)^4 = 10/16 N_A = 0.625 N_A\). The result is the same, but this approach can sometimes prevent confusion between mass and number of atoms.

Step 1: Understanding the Question:

We need to evaluate the correctness of two statements regarding the properties of electromagnetic (EM) waves.


Step 3: Detailed Explanation:

Analyzing Statement I:

"Electromagnetic waves are not deflected by electric and magnetic field."

Electromagnetic waves are streams of photons. Photons are fundamental particles that are electrically neutral (they have no charge). Forces from electric and magnetic fields act on charged particles (Lorentz force, \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\)). Since photons have zero charge (\(q=0\)), they do not experience any force from external electric or magnetic fields and thus are not deflected. Therefore, Statement I is true.


Analyzing Statement II:

"The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as \(E_0 = cB_0\)."

In an electromagnetic wave propagating in a vacuum, the magnitudes of the electric field (\(E\)) and magnetic field (\(B\)) at any point and any time are related by \(E = cB\), where c is the speed of light in vacuum. This relationship also holds for their amplitudes (maximum values), \(E_0\) and \(B_0\). Thus, the relation \(E_0 = cB_0\) is correct. Therefore, Statement II is true.

*(Note: The question paper OCR might have shown the relation differently, but the standard and correct relation is \(E_0 = cB_0\).)*


Step 4: Final Answer:

Both Statement I and Statement II are true statements about electromagnetic waves. This corresponds to option (A).
Quick Tip: Remember the key properties of EM waves: they are transverse, travel at the speed of light in vacuum, carry energy and momentum, are produced by accelerating charges, and are not deflected by E or B fields because they are chargeless. The relation \(E=cB\) is fundamental.

Step 1: Understanding the Question:

We have a wire of a certain length that is first wound into a 4-turn coil and then into a 1-turn coil. We are given the magnetic field at the center for the first case and asked to find it for the second case, assuming the current is the same.


Step 2: Key Formula or Approach:

The magnetic field (magnetic induction) at the center of a circular coil with N turns and radius r, carrying current I, is given by:
\[ B = \frac{\mu_0 N I}{2r} \]
The total length of the wire, L, is constant. The length is related to the number of turns and radius by \(L = N \times (2\pi r)\), which means \(r = \frac{L}{2\pi N}\).

We can substitute this expression for r into the formula for B to see how B depends on N.


Step 3: Detailed Explanation:

Let's express B in terms of N and the constant length L.
\[ B = \frac{\mu_0 N I}{2 \left(\frac{L}{2\pi N}\right)} = \frac{\mu_0 \pi I N^2}{L} \]
Since \(\mu_0\), \(\pi\), I, and L are all constant for this problem, we can see that the magnetic field B is directly proportional to the square of the number of turns, N.
\[ B \propto N^2 \]
We can write this as a ratio:
\[ \frac{B_2}{B_1} = \left(\frac{N_2}{N_1}\right)^2 \]
Given values:

- Initial number of turns, \(N_1 = 4\)

- Initial magnetic field, \(B_1 = 32\) T

- Final number of turns, \(N_2 = 1\)

We need to find the final magnetic field, \(B_2\).

Substituting the values into the ratio:
\[ \frac{B_2}{32} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]
Now, solve for \(B_2\):
\[ B_2 = \frac{32}{16} = 2 T \]

Step 4: Final Answer:

The new magnetic induction at the centre will be 2 T. This corresponds to option (B).
Quick Tip: For a fixed length of wire rewound into coils of different numbers of turns, remember the key relations: radius \(r \propto 1/N\) and magnetic field at the center \(B \propto N^2\). This direct proportionality makes solving such ratio problems very quick.

Step 1: Understanding the Question:

We are asked to determine the output (X) for all possible combinations of inputs (A, B) for the given logic circuit and find the corresponding truth table.


Step 2: Analyzing the Circuit Diagram:

The provided diagram shows two NAND gates.

1. The first gate takes inputs A and B. Its output, let's call it Y, is given by \(Y = \overline{A \cdot B}\).

2. This output Y is then fed into the second NAND gate. Both inputs of the second NAND gate are connected to Y. A NAND gate with its inputs tied together acts as a NOT gate.

3. Therefore, the final output X is the NOT of Y. \(X = \overline{Y} = \overline{\overline{A \cdot B}}\).

4. According to the rules of Boolean algebra (double negation), \(\overline{\overline{Z}} = Z\). So, \(X = A \cdot B\).

The entire circuit is equivalent to a single AND gate.


Step 3: Constructing the Truth Table:

We need to construct the truth table for an AND gate, where the output X is 1 only when both inputs A and B are 1.




Step 4: Comparing with Options and Addressing Discrepancy:

Let's examine the truth tables provided in the options.

- Option (A): X = {1, 0, 0, 0 (This is a NOR gate if A and B are swapped, or some other function)

- Option (B): Complex function.

- Option (C): X = {0, 0, 1, 1 (This is X=A)

- Option (D): The truth table is A=0,B=0,X=0; A=0,B=1,X=1; A=1,B=0,X=1; A=1,B=1,X=1. This corresponds to an OR gate (\(X = A + B\)).


There is a clear discrepancy. The circuit diagram evaluates to an AND gate, but none of the options show the truth table for an AND gate. The official answer key for this question indicates that Option (D) is the correct answer. This implies that the question intended to ask for the truth table of an OR gate, or the provided diagram was incorrect in the exam paper. Such errors can occur in competitive exams.


Assuming the official key is to be followed, we choose the option corresponding to an OR gate.

The truth table for an OR gate is:



This matches the data in Option (D) (as per the image).


Step 5: Final Answer:

Based on the analysis of the given circuit, the result should be an AND gate. However, this is not an option. Following the official answer key, which selects option (D), the correct truth table is that of an OR gate. Thus, we select option (D).
Quick Tip: In exams, if you find that your logically derived answer from the given data does not match any of the options, double-check your work. If the mismatch persists, it's likely an error in the question itself. In such cases, you might have to make an educated guess or mark the question for review. Being aware that questions can be flawed is part of exam strategy. Here, the diagram shows an AND gate, but the correct keyed option is for an OR gate.

Step 1: Understanding the Question:

We need to calculate the modulation index of an Amplitude Modulated (AM) wave given its maximum and minimum peak-to-peak voltages.


Step 2: Key Formula or Approach:

The modulation index (\(\mu\)) is defined in terms of the maximum amplitude (\(A_{max}\)) and minimum amplitude (\(A_{min}\)) of the modulated wave as:
\[ \mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} \]
The amplitude of a wave is half of its peak-to-peak voltage.
\(A_{max} = \frac{V_{pp, max}}{2}\)
\(A_{min} = \frac{V_{pp, min}}{2}\)


Step 3: Detailed Explanation:

Given values:

- Maximum peak-to-peak voltage, \(V_{pp, max} = 14\) mV

- Minimum peak-to-peak voltage, \(V_{pp, min} = 6\) mV


First, calculate the maximum and minimum amplitudes:
\[ A_{max} = \frac{14 mV}{2} = 7 mV \] \[ A_{min} = \frac{6 mV}{2} = 3 mV \]

Now, use the formula for the modulation index:
\[ \mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} = \frac{7 - 3}{7 + 3} \] \[ \mu = \frac{4}{10} = 0.4 \]

Step 4: Final Answer:

The modulation index is 0.4. This corresponds to option (B).
Quick Tip: Note that the formula for modulation index can also be written directly in terms of peak-to-peak voltages, as the factor of 1/2 cancels out: \(\mu = \frac{V_{pp,max} - V_{pp,min}}{V_{pp,max} + V_{pp,min}}\). Using this directly: \(\mu = \frac{14 - 6}{14 + 6} = \frac{8}{20} = 0.4\). This can save a calculation step.

Step 1: Understanding the Question:

We are asked to find the new orbital period of a satellite after its orbital radius is changed. The relationship between orbital period and radius is governed by Kepler's Third Law.


Step 2: Key Formula or Approach:

Kepler's Third Law of planetary motion states that the square of the orbital period (T) of a satellite is directly proportional to the cube of the semi-major axis of its orbit (which is the radius R for a circular orbit).
\[ T^2 \propto R^3 \]
For two different orbits of the same central body, we can write this as a ratio:
\[ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3 \]

Step 3: Detailed Explanation:

Given values:

- Initial time period, \(T_1 = 24\) hours

- The separation is decreased to one fourth of the previous value, so the new radius \(R_2 = \frac{1}{4} R_1\).


We need to find the new time period, \(T_2\).

Using the formula from Kepler's Third Law:
\[ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{\frac{1}{4} R_1}{R_1}\right)^3 \] \[ \left(\frac{T_2}{24}\right)^2 = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \]
Now, take the square root of both sides:
\[ \frac{T_2}{24} = \sqrt{\frac{1}{64}} = \frac{1}{8} \]
Solve for \(T_2\):
\[ T_2 = \frac{24}{8} = 3 hours \]

Step 4: Final Answer:

The new time period will be 3 hours. This corresponds to option (D).
Quick Tip: Remember the relationship \(T^2 \propto R^3\). From this, you can deduce that \(T \propto R^{3/2}\). So, if R becomes \(1/4\) of its original value, T will become \((1/4)^{3/2} = ((1/4)^{1/2})^3 = (1/2)^3 = 1/8\) of its original value. \(T_2 = T_1/8 = 24/8 = 3\) hours.

Step 1: Understanding the Question:

The question asks about the factors affecting the sensitivity of a potentiometer. Sensitivity refers to the ability to measure very small potential differences accurately.


Step 3: Detailed Explanation:

The principle of a potentiometer is that the potential drop across any length of a uniform wire is directly proportional to that length. The potential gradient, k, is the potential drop per unit length of the wire:
\[ k = \frac{V}{L} \]
where V is the total potential drop across the wire of total length L.


A potentiometer is considered more sensitive if it can detect a smaller change in potential difference. This means for a small change in potential, there should be a large change in the balancing length. A smaller potential gradient leads to higher sensitivity.

For a given potential difference \(\Delta V\), the change in balancing length is \(\Delta l = \frac{\Delta V}{k}\). To have a large \(\Delta l\) for a small \(\Delta V\), the potential gradient 'k' must be small.

So, Sensitivity \(\propto \frac{1}{k}\).

This means statement (C) is correct and statement (B) is incorrect.


Now let's see how sensitivity depends on the length of the wire, L.

We have \(k = \frac{V}{L}\). Substituting this into the sensitivity relation:
\[ Sensitivity \propto \frac{1}{V/L} = \frac{L}{V} \]
Assuming the potential drop V across the wire is kept constant, the sensitivity is directly proportional to the length of the potentiometer wire, L. A longer wire will have a smaller potential gradient (for the same total voltage), and thus will be more sensitive.

This means statement (A) is correct and statement (D) is incorrect.


Step 4: Final Answer:

The correct statements are (A) and (C). This corresponds to option (A).
Quick Tip: For maximum potentiometer sensitivity, you need the smallest possible potential gradient (k). This is achieved by using a very long wire (large L) and driving it with the smallest possible current/voltage (small V) that is still larger than the EMF to be measured. Think of it as "stretching out" the voltage scale over a longer distance.

Step 1: Understanding the Question:

We need to compare the RMS current in a purely resistive circuit with the RMS current in a series RLC circuit, powered by the same AC source.


Step 2: Key Formula or Approach:

The RMS current in an AC circuit is given by \(I_{rms} = \frac{V_{rms}}{Z}\), where Z is the impedance of the circuit.

- For a purely resistive circuit (a): \(Z_a = R\).

- For a series RLC circuit (b): \(Z_b = \sqrt{R^2 + (X_L - X_C)^2}\), where \(X_L\) is inductive reactance and \(X_C\) is capacitive reactance.


Step 3: Detailed Explanation:

Let's analyze the currents in both circuits.

Circuit (a):

The impedance is purely resistive, \(Z_a = R\).

The RMS current is \(I_a = \frac{V_{rms}}{R}\).


Circuit (b):

The impedance is \(Z_b = \sqrt{R^2 + (X_L - X_C)^2}\).

The RMS current is \(I_b = \frac{V_{rms}}{Z_b} = \frac{V_{rms}}{\sqrt{R^2 + (X_L - X_C)^2}}\).


Comparison:

The term \((X_L - X_C)^2\) is always greater than or equal to zero.

Therefore, the impedance of the RLC circuit \(Z_b = \sqrt{R^2 + (X_L - X_C)^2}\) will always be greater than or equal to R.
\[ Z_b \geq R \]
The minimum value of \(Z_b\) occurs at resonance, when \(X_L = X_C\). At this point, \(Z_b = \sqrt{R^2 + 0} = R\).

Since the current is inversely proportional to the impedance (\(I_{rms} \propto 1/Z\)), and \(Z_b \geq Z_a\), it follows that:
\[ I_b \leq I_a \]
The maximum possible current in circuit (b) is \(\frac{V_{rms}}{R}\), which is equal to the current in circuit (a). This maximum is achieved only at resonance. At all other frequencies, the current in (b) will be less than the current in (a).


Evaluating the Options:

- (A) At resonance, current in (b) is less than that in (a). False. At resonance, \(Z_b = R\), so \(I_b = I_a\).

- (B) The rms current in circuit (b) can be larger than that in (a). False. As shown, \(I_b\) can at most be equal to \(I_a\).

- (C) The rms current in figure(a) is always equal to that in figure (b). False. They are only equal at resonance.

- (D) The rms current in circuit (b) can never be larger than that in (a). True. This is a direct consequence of \(Z_b \geq R\).


Note on official answer key: Some official answer keys have marked option (B) as correct, which contradicts the principles of a standard series RLC circuit. This suggests a possible error in the question or the provided key. Based on physics principles, option (D) is the only correct statement.


Step 4: Final Answer:

Based on a correct physical analysis, the RMS current in circuit (b) can never be larger than that in (a). This corresponds to option (D).
Quick Tip: Remember that in a series RLC circuit, impedance is minimum at resonance and is equal to R. Therefore, the current is maximum at resonance and is equal to V/R. At any other frequency, the impedance is higher, and the current is lower.

Step 1: Principle of Dimensional Homogeneity:

In any valid physical equation, quantities being added or subtracted must have the same dimensions. In the given equation, `x`, `y`, and `a` have dimensions of length [L]. The dimension of `t` is given as [T\(^{-1}\)].


Step 2: Finding the Dimension of A:

From the term \((x - At)\), the dimension of `x` must be equal to the dimension of `At`.
\[ [x] = [A][t] \]
Substituting the known dimensions:
\[ [L] = [A][T^{-1}] \]
Solving for [A]:
\[ [A] = \frac{[L]}{[T^{-1}]} = [LT] \]

Step 3: Finding the Dimension of B:

From the term \((y - B/t)\), the dimension of `y` must be equal to the dimension of `B/t`.
\[ [y] = \frac{[B]}{[t]} \]
Substituting the known dimensions:
\[ [L] = \frac{[B]}{[T^{-1}]} \]
Solving for [B]:
\[ [B] = [L][T^{-1}] = [LT^{-1}] \]

Step 4: Final Answer and Conclusion:

The correctly derived dimensions are \([A] = [LT]\) and \([B] = [LT^{-1}]\).

Let's check the options:

(A) Incorrect.

(B) Incorrect.

(C) Incorrect.

(D) A=\([LT]\), B=\([L^{-1}T^{-1}]\).

Only option (D) has the correct dimension for A. The dimension for B in option (D) is incorrect. This indicates a typographical error in the question's options. Since the dimension for A is correct, option (D) is the most plausible and intended answer.
Quick Tip: Always apply the principle of dimensional homogeneity strictly. If your derived answer does not match any option, check for unconventional definitions \(like [t] = [T^{-1}]\). If a mismatch still exists, find the option that is partially correct, as question papers can contain errors.

Step 1: Understanding the Question:

The question asks for the method to improve the resolving power of a compound microscope. Resolving power is the ability to distinguish between two very closely spaced points.


Step 2: Key Formula or Approach:

The resolving power (RP) of a microscope is given by the formula:
\[ RP = \frac{1}{d_{min}} = \frac{2n \sin\theta}{\lambda} \]
where:

- \(d_{min}\) is the limit of resolution (the minimum distance between two distinguishable points).

- \(n\) is the refractive index of the medium between the objective lens and the object.

- \(\theta\) is the half-angle of the cone of light that can enter the objective lens.

- \(\lambda\) is the wavelength of the light used for illumination.

To improve the resolving power, we need to increase the value of RP. This means we need to decrease \(d_{min}\).


Step 3: Detailed Explanation:

Based on the formula, to increase the resolving power (RP), we should:

1. Decrease the wavelength (\(\lambda\)): Using light of a shorter wavelength (like blue or ultraviolet light) will improve resolution.

2. Increase the refractive index (\(n\)): This is often achieved by using immersion oil (which has a higher refractive index than air) between the object and the objective lens.

3. Increase the angle (\(\theta\)): This means using an objective lens with a larger aperture. The term \(n \sin\theta\) is called the Numerical Aperture (NA) of the objective.


Now let's evaluate the given options:

- (A) Increase the wave length of the light: This would decrease the resolving power, as RP is inversely proportional to \(\lambda\). So, (A) is incorrect.

- (B) Decrease the diameter of the objective lens: This would decrease the light-gathering angle \(\theta\), which in turn decreases the numerical aperture and thus decreases the resolving power. So, (B) is incorrect.

- (C) Decrease the focal length of the eye piece: The eyepiece is primarily responsible for magnification. While magnification is important, resolving power is a property of the objective lens system and the illumination. Changing the eyepiece's focal length does not directly improve the resolving power. So, (C) is not the best option.

- (D) Increase the refractive index of the medium between the object and objective lens: This directly increases the numerical aperture (\(n \sin\theta\)) and therefore increases the resolving power. This is a standard technique (oil immersion) used to achieve high resolution. So, (D) is correct.


Step 4: Final Answer:

The best way to improve the resolving power is to increase the refractive index of the medium. This corresponds to option (D).
Quick Tip: Remember the formula for resolving power of a microscope. Improving resolution means making the denominator \(\lambda\) smaller and the numerator \(2n \sin\theta\) larger. This simple rule helps quickly evaluate the options in such questions.

Step 1: Understanding the Question:

The problem describes a two-stage motion. In the first stage, a constant force accelerates a body from rest. In the second stage, the force is removed, and the body moves at a constant velocity. We need to find the magnitude of the force.


Step 2: Key Formula or Approach:

We can solve this problem by working backward from the second stage of motion.

1. In the second stage, the motion is uniform (constant velocity) since no force is acting. We can find this constant velocity using \(v = distance / time\).

2. This constant velocity is the final velocity achieved at the end of the first stage of motion.

3. Using the equations of motion for the first stage (\(v = u + at\)) and Newton's second law (\(F = ma\)), we can find the force.


Step 3: Detailed Explanation:

Stage 2: Uniform Motion (from t = 20 s to t = 30 s)

- Distance covered, \(d = 50\) m

- Time taken, \(\Delta t = 10\) s

- Since the force has ceased, the body moves with a constant velocity, \(v\).
\[ v = \frac{d}{\Delta t} = \frac{50 m}{10 s} = 5 m/s \]

Stage 1: Accelerated Motion (from t = 0 s to t = 20 s)

- The velocity achieved at the end of this stage is the constant velocity of stage 2. So, final velocity \(v = 5\) m/s.

- Initial velocity, \(u = 0\) (starts from rest).

- Time duration, \(t = 20\) s.

- Mass of the body, \(m = 20\) kg.


First, find the acceleration 'a' using the first equation of motion:
\[ v = u + at \] \[ 5 = 0 + a(20) \] \[ a = \frac{5}{20} = \frac{1}{4} = 0.25 m/s^2 \]

Now, use Newton's second law to find the force F:
\[ F = ma \] \[ F = 20 kg \times 0.25 m/s^2 = 5 N \]

Step 4: Final Answer:

The value of the force is 5 N. This corresponds to option (A).
Quick Tip: Breaking down multi-stage motion problems is key. Often, information from a later stage is needed to solve for an earlier stage. Here, the constant velocity from the second part of the journey was the crucial piece of information to find the acceleration in the first part.

Step 1: Understanding the Question:

We need to evaluate five statements about work done by various forces and identify the correct ones. The sign of work done depends on the angle between the force vector and the displacement vector. Work is positive if the angle is acute (0 to <90), negative if obtuse (>90 to 180), and zero if perpendicular (90).


Step 3: Detailed Explanation:

Statement A: Work done by a man lifting a bucket.

The man applies an upward force on the rope, and the bucket's displacement is upward. The angle between the force applied by the man and the displacement is 0\(^\circ\). Thus, the work done by the man is positive (\(W = Fd\cos(0^\circ) = Fd\)). Statement A is incorrect.


Statement B: Work done by gravitational force in lifting a bucket.

The gravitational force acts downward, while the bucket's displacement is upward. The angle between the gravitational force and the displacement is 180\(^\circ\). Thus, the work done by gravity is negative (\(W = Fd\cos(180^\circ) = -Fd\)). Statement B is correct.


Statement C: Work done by friction on a body sliding down an inclined plane.

The body is sliding down the incline (displacement is downward along the incline). The force of kinetic friction always opposes motion, so it acts upward along the incline. The angle between the frictional force and the displacement is 180\(^\circ\). Thus, the work done by friction is negative. Statement C is incorrect.


Statement D: Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is zero.

To move with uniform velocity, the net force must be zero. This means the applied force must be equal in magnitude and opposite in direction to the frictional force. The applied force is in the direction of motion (displacement). The angle is 0\(^\circ\). The work done by the applied force is positive, not zero. It is the \textit{net work done on the body that is zero. Statement D is incorrect.


Statement E: Work done by the air resistance on an oscillating pendulum.

Air resistance is a dissipative force that always opposes the velocity of the pendulum bob. At any point in its swing, the direction of air resistance is opposite to the direction of motion (displacement). The angle is 180\(^\circ\). Therefore, the work done by air resistance over any part of the swing (and over a full cycle) is always negative. Statement E is correct.


Step 4: Final Answer:

The correct statements are B and E. This corresponds to option (C).
Quick Tip: A simple rule for the sign of work: If a force helps the motion (has a component in the direction of displacement), work is positive. If a force opposes the motion (has a component opposite to the direction of displacement), work is negative. Resistive forces like friction and air resistance almost always do negative work.

Step 1: Understanding the Question:

We are given the velocity of an object at one point in its uniform circular motion. We need to find its velocity and acceleration at another point on the circle.


Step 2: Key Formula or Approach:

1. Analyze the initial information to determine the parameters of the motion: radius, speed, and direction of rotation (clockwise or counter-clockwise).

2. For uniform circular motion, the speed is constant, but the velocity vector changes direction, always being tangent to the circle.

3. The acceleration is always directed towards the center of the circle (centripetal acceleration) and has a constant magnitude of \(a_c = v^2/r\).


Step 3: Detailed Explanation:

Analyzing the initial state:

- The path is a circle centered at the origin.

- When the object is at x = +2 m, its position vector is \(\vec{r}_1 = 2\hat{i}\) m. This means the radius of the circle is \(r=2\) m.

- At this point, the velocity is \(\vec{v}_1 = -4\hat{j}\) m/s.

- The speed of the object is constant, so \(v = |\vec{v}_1| = 4\) m/s.

- At position (2, 0), the object is moving in the -y direction. This indicates a clockwise rotation.


Finding velocity and acceleration at the new point:

- The new point is at x = -2 m. Assuming it's on the x-axis, the position vector is \(\vec{r}_2 = -2\hat{i}\) m. This is the point (-2, 0).


Velocity at \(\vec{r}_2 = -2\hat{i}\):

- The velocity vector must be tangent to the circular path at (-2, 0).

- For a clockwise motion, at the leftmost point (-2, 0), the object will be moving upward, i.e., in the positive y-direction.

- The speed remains constant at 4 m/s.

- Therefore, the new velocity is \(\vec{v}_2 = 4\hat{j}\) m/s.


Acceleration at \(\vec{r}_2 = -2\hat{i}\):

- The acceleration is centripetal, meaning it always points from the object's position towards the center of the circle (the origin).

- At the position \(\vec{r}_2 = -2\hat{i}\), the direction towards the center is along the positive x-axis (\(+\hat{i}\)).

- The magnitude of the centripetal acceleration is:
\[ a_c = \frac{v^2}{r} = \frac{(4 m/s)^2}{2 m} = \frac{16}{2} = 8 m/s^2 \]
- So, the acceleration vector is \(\vec{a}_2 = 8\hat{i}\) m/s\(^2\).


Step 4: Final Answer:

At x = -2 m, the velocity is \(\vec{v}=4\hat{j} m/s\) and the acceleration is \(\vec{a}=8\hat{i} m/s^2\). This corresponds to option (D).
Quick Tip: Visualizing the circular motion is extremely helpful. Draw a simple x-y plane. Mark the initial point (2,0) and the velocity vector pointing down (\(- \hat{j}\)). This immediately tells you the rotation is clockwise. Then, move to the final point (-2,0) and determine the tangent (velocity) and the direction to the center (acceleration) from there.

Step 1: Understanding the Question:

We need to find the uncertainty in the calculated refractive index based on measurements of real and apparent thickness using a travelling microscope with a Vernier scale.


Step 2: Key Formula or Approach:

1. Calculate the least count (LC) of the travelling microscope. The uncertainty in each measurement is equal to the LC.

2. The formula for refractive index is \(\mu = \frac{Real Thickness (t_r)}{Apparent Thickness (t_a)}\).

3. The uncertainty \(\Delta \mu\) is calculated using the formula for propagation of errors:
\[ \frac{\Delta \mu}{\mu} = \frac{\Delta t_r}{t_r} + \frac{\Delta t_a}{t_a} \]
where \(\Delta t_r = \Delta t_a = LC\).


Step 3: Detailed Explanation:

Calculating the Least Count (LC):

- Value of 1 Main Scale Division (MSD): The main scale has 20 divisions in 1 cm.

So, 1 MSD = \( \frac{1}{20} \) cm = 0.05 cm = 0.5 mm.

- Relation between VSD and MSD: 50 Vernier Scale Divisions (VSD) = 49 MSD.

So, 1 VSD = \( \frac{49}{50} \) MSD.

- Least Count = 1 MSD - 1 VSD = \( 1 - \frac{49}{50} \) MSD = \( \frac{1}{50} \) MSD.

LC = \( \frac{1}{50} \times 0.5 \) mm = 0.01 mm.

- The uncertainty in each thickness measurement is \(\Delta t_r = \Delta t_a = LC = 0.01\) mm.


Calculating the Uncertainty in Refractive Index:

- Given values: \(t_r = 5.25\) mm, \(t_a = 5.00\) mm.

- Refractive index, \(\mu = \frac{5.25}{5.00} = 1.05\).

- Now, calculate \(\Delta \mu\):
\[ \Delta \mu = \mu \left( \frac{\Delta t_r}{t_r} + \frac{\Delta t_a}{t_a} \right) \] \[ \Delta \mu = 1.05 \left( \frac{0.01}{5.25} + \frac{0.01}{5.00} \right) \] \[ \Delta \mu = 1.05 \times 0.01 \left( \frac{1}{5.25} + \frac{1}{5.00} \right) \] \[ \Delta \mu = 0.0105 \left( \frac{5.00 + 5.25}{5.25 \times 5.00} \right) = 0.0105 \left( \frac{10.25}{26.25} \right) \] \[ \Delta \mu \approx 0.0105 \times 0.39047 \approx 0.0041 \]
- The uncertainty is approximately \(4.1 \times 10^{-3}\).

The question asks for the value of x, where the uncertainty is \(x \times 10^{-3}\).

So, \(x = 4.1\). Since the answer must be an integer, we take the nearest integer value.

x = 4.


Step 4: Final Answer:

The value of x is 4.
Quick Tip: Error propagation calculations are common. For a quantity \(Z = A/B\), the fractional error is \(\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\). Always remember to first calculate the least count of the measuring instrument, as it represents the uncertainty in the raw measurements.

Step 1: Understanding the Question:

We are given a car in circular motion where tangential acceleration equals centripetal acceleration. We need to find the time for the first quarter revolution.


Step 2: Key Formula or Approach:

1. Given condition: \(a_t = a_c\).

2. Formulas: \(a_t = v \frac{dv}{ds}\) (where s is arc length) and \(a_c = \frac{v^2}{r}\).

3. Set up a differential equation relating speed \(v\) and distance \(s\), and solve it.

4. Use \(v = \frac{ds}{dt}\) to find the time taken.


Step 3: Detailed Explanation:

From the given condition, \(a_t = a_c\):
\[ v \frac{dv}{ds} = \frac{v^2}{r} \]
Separating variables:
\[ \frac{dv}{v} = \frac{ds}{r} \]
Integrate both sides. Let the initial speed be \(v_0\) at \(s=0\) and speed be \(v\) at distance \(s\).
\[ \int_{v_0}^{v} \frac{dv}{v} = \int_0^s \frac{ds}{r} \] \[ [\ln(v)]_{v_0}^{v} = \frac{s}{r} \implies \ln(v) - \ln(v_0) = \frac{s}{r} \] \[ \ln\left(\frac{v}{v_0}\right) = \frac{s}{r} \implies v = v_0 e^{s/r} \]
Now, substitute \(v = \frac{ds}{dt}\):
\[ \frac{ds}{dt} = v_0 e^{s/r} \]
Separate variables again to find time T for a distance S:
\[ \int_0^S e^{-s/r} ds = \int_0^T v_0 dt \] \[ \left[ -r e^{-s/r} \right]_0^S = v_0 T \] \[ -r e^{-S/r} - (-r e^0) = v_0 T \] \[ r(1 - e^{-S/r}) = v_0 T \implies T = \frac{r}{v_0}(1 - e^{-S/r}) \]
Now substitute the given values:

- Radius, \(r = 600\) m.

- Initial speed, \(v_0 = 54 km/hr = 54 \times \frac{5}{18} = 15 m/s\).

- For a quarter revolution, the distance is \(S = \frac{1}{4}(2\pi r) = \frac{\pi r}{2}\).

- So, the exponent is \(-\frac{S}{r} = -\frac{\pi r / 2}{r} = -\frac{\pi}{2}\).

Substituting these into the equation for T:
\[ T = \frac{600}{15} \left(1 - e^{-\pi/2}\right) = 40 \left(1 - e^{-\pi/2}\right) \]
Comparing this with the given form \(t(1-e^{-\pi/2})\), we find that \(t = 40\).


Step 4: Final Answer:

The value of t is 40.
Quick Tip: For circular motion problems where acceleration components are related, setting up and solving the differential equation is the standard method. Using the form \(a_t = v \frac{dv}{ds}\) is often more direct than \(a_t = \frac{dv}{dt}\) if the problem involves distance.

Step 1: Understanding the Condition (Brewster's Angle):

The condition that the reflected and refracted rays are perpendicular to each other means that the light is incident at Brewster's angle (\(i_B\)).


Step 2: Key Formulas:

1. Brewster's Law: The tangent of Brewster's angle is equal to the ratio of the refractive indices of the two media.

\[ \tan(i_B) = n_{21} = \frac{n_2}{n_1} \]
2. Refractive Index of a Dielectric: For a non-magnetic dielectric medium, the refractive index \(n\) is related to the dielectric constant (relative permittivity \(\epsilon_r\)) by the formula:

\[ n = \sqrt{\epsilon_r \mu_r} \]
Since it is a dielectric medium, the relative permeability \(\mu_r \approx 1\), so \(n = \sqrt{\epsilon_r}\).


Step 3: Detailed Calculation:

Given values:

- Dielectric constant of medium 1, \(\epsilon_{r1} = 2.8\).

- Dielectric constant of medium 2, \(\epsilon_{r2} = 6.8\).


First, find the refractive indices of the two media:
\[ n_1 = \sqrt{\epsilon_{r1}} = \sqrt{2.8} \] \[ n_2 = \sqrt{\epsilon_{r2}} = \sqrt{6.8} \]
Now, apply Brewster's Law to find the tangent of the angle of incidence:
\[ \tan(i_B) = \frac{n_2}{n_1} = \frac{\sqrt{6.8}}{\sqrt{2.8}} = \sqrt{\frac{6.8}{2.8}} = \sqrt{\frac{68}{28}} = \sqrt{\frac{17}{7}} \]
The question provides the angle of incidence in a specific format:
\[ i = \tan^{-1}\left(\sqrt{1 + \frac{10}{\theta}}\right) \]
This means that:
\[ \tan(i) = \sqrt{1 + \frac{10}{\theta}} \]
Since the condition is met, \(i = i_B\). We can equate the two expressions for the tangent of the angle:
\[ \sqrt{1 + \frac{10}{\theta}} = \sqrt{\frac{17}{7}} \]
Squaring both sides of the equation:
\[ 1 + \frac{10}{\theta} = \frac{17}{7} \]
Now, solve for \(\theta\):
\[ \frac{10}{\theta} = \frac{17}{7} - 1 \] \[ \frac{10}{\theta} = \frac{17 - 7}{7} = \frac{10}{7} \] \[ \theta = 7 \]

Step 4: Final Answer:

The value of \(\theta\) is 7.
Quick Tip: Brewster's angle problems are a direct application of the formula \(\tan(i_B) = n_2/n_1\). The key is to correctly relate the given properties (like dielectric constant) to the refractive index. For dielectrics, \(n = \sqrt{\epsilon_r}\) is the crucial link. Always set up the equation carefully based on the given form of the angle.

Step 1: Understanding the Question:

We are using a potentiometer to find the internal resistance of a cell. We have two scenarios with two different shunt resistances and their corresponding balancing lengths.


Step 2: Key Formula or Approach:

Let \(E\) be the EMF of the cell and \(l_0\) be the balancing length without any shunt. Let \(k\) be the potential gradient of the potentiometer wire. Then \(E = k l_0\).

When the cell is shunted with a resistance \(S\), the terminal voltage across the cell is \(V = E \frac{S}{S+r}\). The balancing length for this voltage is \(l\), so \(V = k l\).

Combining these, we get \(k l = (k l_0) \frac{S}{S+r} \implies l = l_0 \frac{S}{S+r}\).


Step 3: Detailed Explanation:

We have two cases:

Case 1: Shunt \(S_1 = 5 \, \Omega\), balancing length \(l_1 = 200\) cm.
\[ 200 = l_0 \frac{5}{5+r} \quad (Equation 1) \]
Case 2: Shunt \(S_2 = 15 \, \Omega\), balancing length \(l_2 = 300\) cm.
\[ 300 = l_0 \frac{15}{15+r} \quad (Equation 2) \]
We can solve these two equations for \(r\). Let's divide Equation 2 by Equation 1:
\[ \frac{300}{200} = \frac{l_0 \frac{15}{15+r}}{l_0 \frac{5}{5+r}} \] \[ \frac{3}{2} = \frac{15}{15+r} \times \frac{5+r}{5} \] \[ \frac{3}{2} = \frac{3(5+r)}{15+r} \]
Cancel the 3 from both sides:
\[ \frac{1}{2} = \frac{5+r}{15+r} \]
Cross-multiply:
\[ 1(15+r) = 2(5+r) \] \[ 15 + r = 10 + 2r \] \[ r = 15 - 10 = 5 \, \Omega \]

Step 4: Final Answer:

The internal resistance of the cell is 5 \(\Omega\).
Quick Tip: When you have two scenarios in a potentiometer problem for finding internal resistance, setting up a ratio of the two conditions is the quickest way to solve. This eliminates the unknown potential gradient \(k\) and the open-circuit balancing length \(l_0\).

Step 1: Understanding the Question:

In a series RLC circuit, the current is maximum when the circuit is in resonance. We are asked to find the value of capacitance that brings the circuit to resonance at the given frequency.


Step 2: Key Formula or Approach:

The condition for resonance in a series RLC circuit is when the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)).
\[ X_L = X_C \] \[ 2\pi f L = \frac{1}{2\pi f C} \]
Solving for the capacitance, C:
\[ C = \frac{1}{(2\pi f)^2 L} = \frac{1}{4\pi^2 f^2 L} \]

Step 3: Detailed Explanation:

Given values:

- Inductance, \(L = 2 \, \muH = 2 \times 10^{-6}\) H.

- Frequency, \(f = 7 \, kHz = 7 \times 10^3\) Hz.

- \(\pi = 22/7\)

Substitute these values into the formula for capacitance at resonance:
\[ C = \frac{1}{4\pi^2 f^2 L} \] \[ C = \frac{1}{4 \left(\frac{22}{7}\right)^2 (7 \times 10^3)^2 (2 \times 10^{-6})} \] \[ C = \frac{1}{4 \times \frac{484}{49} \times (49 \times 10^6) \times (2 \times 10^{-6})} \]
Cancel out the common terms:
\[ C = \frac{1}{4 \times 484 \times (10^6) \times (2 \times 10^{-6})} \] \[ C = \frac{1}{4 \times 484 \times 2} \] \[ C = \frac{1}{8 \times 484} \] \[ C = \frac{1}{3872} \, F \]

Step 4: Final Answer:

The problem states that the capacitance is of the form \(C = \frac{1}{x}\) F.

By comparing our result \(C = \frac{1}{3872}\) F with the given form, we find that:
\[ x = 3872 \] Quick Tip: Resonance is a fundamental concept in AC circuits. For a series RLC circuit, resonance means the impedance is at its minimum (\(Z=R\)), and therefore the current is at its maximum. The resonance condition \(X_L = X_C\) is the key to solving for frequency, inductance, or capacitance.

Step 1: Understanding the Question:

We need to find the magnitude of the angular momentum of a projectile about its point of projection after 2 seconds.


Step 2: Key Formula or Approach:

The angular momentum \(\vec{L}\) of a particle about a point is given by \(\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})\), where \(\vec{r}\) is the position vector from the point and \(\vec{p}\) is the linear momentum.

Alternatively, we can use the relation \(\frac{d\vec{L}}{dt} = \vec{\tau}\), where \(\vec{\tau}\) is the torque about the same point. Integrating this gives \(\vec{L}(t) = \int_0^t \vec{\tau}(t') dt'\) since the initial angular momentum is zero.


Step 3: Detailed Explanation (Using Torque Method):

1. The only force acting on the projectile is gravity, \(\vec{F}_g = -mg\hat{j}\).

2. The position vector of the projectile at time t is \(\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}\), where \(x(t) = (v_0 \cos\theta)t\).

3. The torque \(\vec{\tau}\) about the origin (point of projection) is:

\[ \vec{\tau} = \vec{r} \times \vec{F}_g = (x\hat{i} + y\hat{j}) \times (-mg\hat{j}) = -mgx(\hat{i} \times \hat{j}) = -mgx\hat{k} \]
4. Substitute the expression for x(t):

\[ \vec{\tau}(t) = -mg(v_0 \cos\theta)t \hat{k} \]
5. Integrate the torque from t=0 to t=2s to find the angular momentum:

\[ \vec{L}(t) = \int_0^t \vec{\tau}(t') dt' = \int_0^t -mg(v_0 \cos\theta)t' \hat{k} dt' = -mg(v_0 \cos\theta) \left[\frac{t'^2}{2}\right]_0^t \hat{k} \]
\[ \vec{L}(t) = -\frac{1}{2}mg(v_0 \cos\theta)t^2 \hat{k} \]
6. Substitute the given values: \(m = 100 g = 0.1 kg\), \(g = 10 m/s^2\), \(v_0 = 20 m/s\), \(\theta = 45^\circ\), \(t=2\) s.

\[ v_0 \cos\theta = 20 \cos(45^\circ) = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} m/s \]
\[ |\vec{L}(2)| = \frac{1}{2}(0.1)(10)(10\sqrt{2})(2^2) = \frac{1}{2} \times 1 \times 10\sqrt{2} \times 4 = 20\sqrt{2} \, kg m^2/s \]
7. We are given that the magnitude is \(\sqrt{K}\).

\[ \sqrt{K} = 20\sqrt{2} \]
Squaring both sides:

\[ K = (20\sqrt{2})^2 = 400 \times 2 = 800 \]

Step 4: Final Answer:

The value of K is 800.
Quick Tip: Using the torque method (\(\vec{L} = \int \vec{\tau} dt\)) is often simpler for projectile motion angular momentum problems than calculating \(\vec{r} \times \vec{p}\) directly, as it avoids finding all components of position and velocity at the given time.

Step 1: Understanding the Question:

We are given the mass, the restoring force law, and the maximum speed of a particle in Simple Harmonic Motion (SHM). We need to find the amplitude of this motion.


Step 2: Key Formula or Approach:

1. The force law for SHM is given by \(F = -kx\), where k is the force constant.

2. The angular frequency \(\omega\) of the oscillation is given by \(\omega = \sqrt{\frac{k}{m}}\).

3. The maximum speed in SHM is related to the amplitude (A) and angular frequency by \(v_{max} = A\omega\).


Step 3: Detailed Explanation:

Given values:

- Mass, \(m = 250 g = 0.25 kg\).

- Force, \(F = -25x\) N.

- Maximum speed, \(v_{max} = 4\) m/s.


1. Find the force constant (k):

By comparing the given force law \(F = -25x\) with the standard SHM equation \(F = -kx\), we get:

\[ k = 25 N/m \]
2. Find the angular frequency (\(\omega\)):

\[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{25}{0.25}} = \sqrt{100} = 10 rad/s \]
3. Find the amplitude (A):

Using the relation for maximum speed:

\[ v_{max} = A\omega \]
\[ A = \frac{v_{max}}{\omega} = \frac{4 m/s}{10 rad/s} = 0.4 m \]
4. Convert amplitude to centimeters:

The question asks for the amplitude in cm.

\[ A = 0.4 m \times 100 \frac{cm}{m} = 40 cm \]

Step 4: Final Answer:

The amplitude of the motion is 40 cm.
Quick Tip: For SHM problems, always start by identifying the force constant `k` from the given force equation. From there, you can find the angular frequency `\(\omega\)` and then relate it to other quantities like period, frequency, velocity, and acceleration.

Step 1: Understanding the Question:

We are given the electric potential inside a spherical charge distribution and need to find the volume charge density.


Step 2: Key Formula or Approach:

The relationship between electric potential \(V\) and volume charge density \(\rho\) is given by Poisson's equation:
\[ \nabla^2 V = -\frac{\rho}{\epsilon_0} \]
For a spherically symmetric potential that depends only on \(r\), the Laplacian operator \(\nabla^2\) is:
\[ \nabla^2 V = \frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dV}{dr}\right) \]
An alternative two-step method is to first find the electric field \(\vec{E} = -\nabla V\) and then use Gauss's Law in differential form, \(\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}\).


Step 3: Detailed Explanation:

Given potential: \(V(r) = 2ar^2 + b\).

1. Find the electric field E(r):

For spherical symmetry, \(E = -\frac{dV}{dr}\).

\[ E = -\frac{d}{dr}(2ar^2 + b) = -4ar \]
The electric field vector is \(\vec{E} = -4ar \hat{r}\).

2. Find the charge density \(\rho\) using Gauss's Law:

Gauss's Law in differential form is \(\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}\). For spherical symmetry:

\[ \nabla \cdot \vec{E} = \frac{1}{r^2} \frac{d}{dr}(r^2 E_r) \]
Here, the radial component of the electric field is \(E_r = -4ar\).

\[ \frac{\rho}{\epsilon_0} = \frac{1}{r^2} \frac{d}{dr}(r^2 (-4ar)) = \frac{1}{r^2} \frac{d}{dr}(-4ar^3) \]
\[ \frac{\rho}{\epsilon_0} = \frac{1}{r^2}(-12ar^2) = -12a \]
3. Solve for \(\rho\):

\[ \rho = -12a\epsilon_0 \]
4. Compare with the given expression:

The problem states that the volume charge density is \(\rho = -\lambda a \epsilon_0\).

Comparing our result with this expression:

\[ -12a\epsilon_0 = -\lambda a \epsilon_0 \]
This gives \(\lambda = 12\).


Step 4: Final Answer:

The value of \(\lambda\) is 12.
Quick Tip: Remember the differential relationships between potential, field, and charge density: \(\vec{E} = -\nabla V\) and \(\nabla \cdot \vec{E} = \rho/\epsilon_0\). Combining them gives Poisson's equation \(\nabla^2 V = -\rho/\epsilon_0\). For spherical symmetry problems, knowing the spherical forms of the gradient and divergence operators is essential.

Step 1: Understanding the Question:

We have two scenarios using a meter bridge with two unknown resistors, R\(_1\) and R\(_2\). We need to use the balancing conditions from both scenarios to find the product R\(_1\)R\(_2\).


Step 2: Key Formula or Approach:

The principle of a balanced meter bridge (a form of Wheatstone bridge) is:
\[ \frac{Resistance in Left Gap}{Resistance in Right Gap} = \frac{Balancing Length (from left)}{100 - Balancing Length (from left)} \]
Or, \(\frac{R_{left}}{R_{right}} = \frac{l}{100-l}\), where l is in cm.


Step 3: Detailed Explanation:

Case 1: R\(_1\) and R\(_2\) in Series

- Resistance in left gap: \(R_S = R_1 + R_2\).

- Resistance in right gap: 10 \(\Omega\).

- Balancing length: \(l_1 = 60\) cm.

Using the meter bridge formula:
\[ \frac{R_1 + R_2}{10} = \frac{60}{100 - 60} = \frac{60}{40} = \frac{3}{2} \] \[ R_1 + R_2 = 10 \times \frac{3}{2} = 15 \, \Omega \quad (Equation 1) \]

Case 2: R\(_1\) and R\(_2\) in Parallel

- Resistance in left gap: \(R_P = \frac{R_1 R_2}{R_1 + R_2}\).

- Resistance in right gap: 3 \(\Omega\).

- Balancing length: \(l_2 = 40\) cm.

Using the meter bridge formula:
\[ \frac{R_P}{3} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \] \[ R_P = 3 \times \frac{2}{3} = 2 \, \Omega \]
So, \(\frac{R_1 R_2}{R_1 + R_2} = 2 \, \Omega \quad (Equation 2)\).


Solving for the Product R\(_1\)R\(_2\):

We have the sum from Equation 1 (\(R_1 + R_2 = 15\)) and a relation involving the product from Equation 2. We can substitute the sum into Equation 2.
\[ \frac{R_1 R_2}{15} = 2 \] \[ R_1 R_2 = 15 \times 2 = 30 \, \Omega^2 \]

Step 4: Final Answer:

The product of R\(_1\)R\(_2\) is 30 \(\Omega^2\).
Quick Tip: In meter bridge problems with two setups, look for ways to combine the equations. Here, the first setup gave the sum of the resistances, which could be directly substituted into the equation from the second setup (which involved the parallel combination) to find the product.

Step 1: Understanding the Question:

We have a block moving at a constant speed on a thin film of liquid. The motion is sustained by a horizontal force. We need to find the speed of the block.


Step 2: Key Formula or Approach:

1. Since the block moves at a constant speed, the net force on it is zero. This means the applied horizontal force is equal in magnitude to the opposing viscous drag force. \(F_{applied} = F_{viscous}\).

2. The viscous force for a fluid layer with a linear velocity profile is given by Newton's law of viscosity:

\[ F_{viscous} = \eta A \frac{v}{d} \]
where \(\eta\) is the coefficient of viscosity, A is the area of contact, v is the speed, and d is the thickness of the film.


Step 3: Detailed Explanation:

Given values:

- Applied force, \(F = 0.1\) N.

- Area, \(A = 0.20\) m\(^2\).

- Viscosity, \(\eta = 5.0 \times 10^{-3}\) Pl (Poiseuille, which is Pa·s or kg m\(^{-1}\)s\(^{-1}\)).

- Film thickness, \(d = 0.25 mm = 0.25 \times 10^{-3}\) m.


Set the applied force equal to the viscous force:
\[ F = \eta A \frac{v}{d} \]
Rearrange the formula to solve for the speed, v:
\[ v = \frac{F \cdot d}{\eta \cdot A} \]
Substitute the given values:
\[ v = \frac{(0.1 N) \times (0.25 \times 10^{-3} m)}{(5.0 \times 10^{-3} Pa·s) \times (0.20 m^2)} \]
The \(10^{-3}\) terms in the numerator and denominator cancel out.
\[ v = \frac{0.1 \times 0.25}{5.0 \times 0.20} = \frac{0.025}{1.0} = 0.025 m/s \]
The question asks for the answer in the format `_____ \(\times 10^{-3}\) m/s`.
\[ 0.025 m/s = 25 \times 10^{-3} m/s \]

Step 4: Final Answer:

The value to be filled in the blank is 25.
Quick Tip: When a body moves at a constant velocity, it is in dynamic equilibrium. This means the driving force is perfectly balanced by the resistive forces (like friction or viscous drag). This is a very common setup in mechanics and fluid dynamics problems.