Step 1: Understanding the Concept:

We use dimensional analysis to match units to their expressions in terms of fundamental constants \(h, G, c\).


Step 2: Detailed Explanation:

Let \([L] = h^a G^b c^d\).

Dimensions: \([h] = ML^2T^{-1}, [G] = M^{-1}L^3T^{-2}, [c] = LT^{-1}\).

For Length: \(M^0 L^1 T^0 = (ML^2T^{-1})^a (M^{-1}L^3T^{-2})^b (LT^{-1})^d\)
\(a - b = 0 \implies a = b\).
\(2a + 3b + d = 1 \implies 5a + d = 1\).
\(-a - 2b - d = 0 \implies -3a - d = 0 \implies d = -3a\).
\(5a - 3a = 1 \implies 2a = 1 \implies a = 1/2\).

So \(b = 1/2, d = -3/2\).

Expression: \(\sqrt{\frac{hG}{c^3}}\). (Matches A-I).

Similarly for Time: \(a=1/2, b=1/2, d=-5/2 \implies \sqrt{\frac{hG}{c^5}}\). (Matches B-II).

For Mass: \(a=1/2, b=-1/2, d=1/2 \implies \sqrt{\frac{hc}{G}}\). (Matches C-III).


Step 3: Final Answer:

The correct matching is A-I, B-II, C-III.
Quick Tip: The expressions for Planck length, Planck time, and Planck mass are fundamental in physics. Memorizing that \( L_p \sim 1/c^{3/2} \) and \( T_p \sim 1/c^{5/2} \) helps save time in dimensional analysis.

Step 1: Understanding the Concept:

The resistance is \(R = V/I\). The absolute error in \(R\) is calculated using the relative errors of \(V\) and \(I\) based on the formula \(\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}\).


Step 2: Detailed Explanation:

Given \(V = 10 V, I = 5 A\).

Least counts (errors): \(\Delta V = 500 mV = 0.5 V\), \(\Delta I = 200 mA = 0.2 A\).

Value of Resistance \(R = \frac{V}{I} = \frac{10}{5} = 2 \Omega\).

Relative error: \(\frac{\Delta R}{R} = \frac{0.5}{10} + \frac{0.2}{5}\)
\(\frac{\Delta R}{2} = 0.05 + 0.04 = 0.09\)

Absolute error \(\Delta R = 2 \times 0.09 = 0.18 \Omega\).


Step 3: Final Answer:

The estimated error is 0.18 \(\Omega\).
Quick Tip: When combining variables in a product or quotient, relative errors are added. Always ensure all units (like mV and V) are consistent before calculation.

Step 1: Understanding the Concept:

Work done is \(W = \vec{f} \cdot \vec{d}\). Since the assembly moves with constant velocity, the net force on the block is zero. Friction must balance the component of gravity.


Step 2: Detailed Explanation:

Mass \(m = 1 kg\), angle \(\theta = 30^\circ\).

The block is at rest relative to the incline.

Static friction \(f = mg \sin \theta\) (acting up the incline).
\(f = 1 \cdot 10 \cdot \sin 30^\circ = 5 N\).

The assembly moves vertically up with \(v = 4 m/s\).

Displacement in 2 s is \(d = v \cdot t = 4 \cdot 2 = 8 m\) (upwards).

Angle between friction force (along incline at \(30^\circ\) to horizontal) and vertical displacement is \(60^\circ\).

Work done \(W = f \cdot d \cdot \cos 60^\circ\)
\(W = 5 \cdot 8 \cdot \frac{1}{2} = 20 J\).


Step 3: Final Answer:

The work done by friction is 20 J.
Quick Tip: Work done depends on the dot product of the force and the displacement. Don't forget to account for the angle between the friction vector (along the plane) and the actual motion vector.

Step 1: Understanding the Concept:

Total distance is the total area under the \(|v|-t\) graph (all areas positive). Average velocity is total displacement divided by total time. Displacement is the algebraic sum of areas (above t-axis positive, below negative).


Step 2: Detailed Explanation:

From the graph:

Area 1 (0 to 20 s): Triangle with base 20 and height 5. Area \(= \frac{1}{2} \cdot 20 \cdot 5 = 50 m\).

Area 2 (20 to 40 s): Triangle with base 20 and height \(-5\). Area \(= \frac{1}{2} \cdot 20 \cdot (-5) = -50 m\).

Total Distance \(= |50| + |-50| = 100 m\).

Total Displacement \(= 50 + (-50) = 0\).

Average Velocity \(= \frac{Displacement}{Time} = \frac{0}{40} = 0\).


Step 3: Final Answer:

Distance is 100 m and average velocity is zero.
Quick Tip: Average velocity is zero whenever the particle returns to its starting point, which corresponds to the area above the time axis being equal to the area below it.

Step 1: Understanding the Concept:

For uniform angular acceleration, we use the equation \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\). Since it starts from rest, \(\omega_0 = 0\).


Step 2: Detailed Explanation:

Let angular acceleration be \(\alpha\).

In first 2 s: \(\theta_1 = \frac{1}{2} \alpha (2)^2 = 2\alpha\).

Total angle in first 4 s: \(\theta_{total} = \frac{1}{2} \alpha (4)^2 = 8\alpha\).

Angle in the next 2 s (from \(t=2\) to \(t=4\)): \(\theta_2 = \theta_{total} - \theta_1 = 8\alpha - 2\alpha = 6\alpha\).

Ratio \(\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3\).


Step 3: Final Answer:

The ratio is 3.
Quick Tip: For constant acceleration starting from rest, the displacements in consecutive equal time intervals follow the ratio of odd numbers: 1 : 3 : 5 : 7 \dots

Step 1: Understanding the Concept:

This problem involves the conservation of mechanical energy for a rolling object. As the object rolls up, its initial total kinetic energy (translational + rotational) is converted into gravitational potential energy at the maximum height.


: Key Formula or Approach:

Total Kinetic Energy \(K_{total} = \frac{1}{2}mv_o^2 + \frac{1}{2}I\omega^2\).

For rolling without slipping, \(\omega = v_o/R\) and \(I = kmR^2\) (where \(k\) is the shape factor).

By conservation of energy: \(K_{total} = mgh_{max}\).


Step 2: Detailed Explanation:

Initial Total Kinetic Energy:
\[ K = \frac{1}{2}mv_o^2 + \frac{1}{2}(kmR^2)\left(\frac{v_o}{R}\right)^2 = \frac{1}{2}mv_o^2(1+k) \]

At maximum height \(h\), Potential Energy \(U = mgh\).

Equating \(K = U\):
\[ \frac{1}{2}mv_o^2(1+k) = mg\left(\frac{7v_o^2}{10g}\right) \]
\[ \frac{1}{2}(1+k) = \frac{7}{10} \]
\[ 1+k = \frac{14}{10} = 1.4 \]
\[ k = 0.4 = \frac{2}{5} \]

For a solid sphere, the moment of inertia is \(I = \frac{2}{5}mR^2\), so \(k = \frac{2}{5}\).


Step 3: Final Answer:

The object is a solid sphere.
Quick Tip: Remember the \(k\) values for common shapes: Ring (\(k=1\)), Disc/Solid Cylinder (\(k=1/2\)), Solid Sphere (\(k=2/5\)), Hollow Sphere (\(k=2/3\)). This helps identify the object instantly from energy equations.

Step 1: Understanding the Concept:

Gravitational potential energy of a mass \(m\) at distance \(r\) from the center of Earth is \(U = -\frac{GM_e m}{r}\). The increase in potential energy is \(\Delta U = U_{final} - U_{initial}\).


: Key Formula or Approach:
\(\Delta U = -\frac{GM_e m}{R_e + h} - \left(-\frac{GM_e m}{R_e}\right)\).

Given \(h = 2R_e\). Also, \(g = \frac{GM_e}{R_e^2} \implies GM_e = gR_e^2\).


Step 2: Detailed Explanation:

At the surface (\(r = R_e\)): \(U_i = -\frac{GM_e m}{R_e}\).

At height \(h = 2R_e\) (\(r = R_e + 2R_e = 3R_e\)): \(U_f = -\frac{GM_e m}{3R_e}\).

Increase in P.E.:
\[ \Delta U = U_f - U_i = -\frac{GM_e m}{3R_e} + \frac{GM_e m}{R_e} \]
\[ \Delta U = \frac{GM_e m}{R_e} \left(1 - \frac{1}{3}\right) = \frac{GM_e m}{R_e} \left(\frac{2}{3}\right) \]

Substitute \(GM_e = gR_e^2\):
\[ \Delta U = \frac{(gR_e^2)m}{R_e} \cdot \frac{2}{3} = \frac{2}{3} mgR_e \]


Step 3: Final Answer:

The increase in potential energy is \(\frac{2}{3} mgR_e\).
Quick Tip: For small heights (\(h \ll R\)), \(\Delta U \approx mgh\). For large heights, always use the formula \(\Delta U = \frac{mgh}{1 + h/R}\). Here, \(\frac{mg(2R)}{1 + 2R/R} = \frac{2mgR}{3}\).

Step 1: Understanding the Concept:

When drops coalesce, the total volume remains constant but the total surface area decreases. This reduction in surface area leads to the release of surface energy (\(E = S \cdot \Delta A\)).


Step 2: Detailed Explanation:

Let \(R\) be the radius of the big drop.

Volume conservation: \(8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R^3 = 8r^3 \implies R = 2r\).

Initial surface area \(A_i = 8 \times 4\pi r^2 = 32\pi r^2\).

Final surface area \(A_f = 4\pi R^2 = 4\pi (2r)^2 = 16\pi r^2\).

Decrease in area \(\Delta A = A_i - A_f = 32\pi r^2 - 16\pi r^2 = 16\pi r^2\).

Energy released \(E = S \cdot \Delta A = 16\pi r^2 S\).


Step 3: Final Answer:

The energy released is \(16\pi r^2 S\).
Quick Tip: When \(n\) identical drops coalesce into one, the change in area is \(\Delta A = 4\pi r^2 (n - n^{2/3})\). For \(n=8\), \(\Delta A = 4\pi r^2 (8 - 4) = 16\pi r^2\).

Step 1: Understanding the Concept:

The coefficient of volume expansion is defined as \(\gamma = \frac{1}{V} \frac{dV}{dT}\). We need to express \(V\) as a function of \(T\) using the given process equation and the ideal gas law (\(PV = nRT\)).


Step 2: Detailed Explanation:

Given \(PT^3 = C\).

From the ideal gas law, \(P = \frac{nRT}{V}\).

Substitute \(P\) into the process equation:
\[ \left(\frac{nRT}{V}\right)T^3 = C \implies \frac{nRT^4}{V} = C \]
\[ V = \frac{nR}{C} T^4 \]

Differentiate \(V\) with respect to \(T\):
\[ \frac{dV}{dT} = \frac{nR}{C} (4T^3) \]

Now, calculate \(\gamma\):
\[ \gamma = \frac{1}{V} \frac{dV}{dT} = \frac{1}{(nR/C)T^4} \cdot \frac{nR}{C} (4T^3) \]
\[ \gamma = \frac{4T^3}{T^4} = \frac{4}{T} \]


Step 3: Final Answer:

The coefficient of volume expansion is \(\frac{4}{T}\).
Quick Tip: For a polytropic process \(TV^{a} = const\), the expansion coefficient is \(\gamma = \frac{1}{aT}\). For this problem, \(V \propto T^4 \implies VT^{-4} = const \implies a = -1/4\) doesn't fit directly. Always derive using \(\frac{1}{V} \frac{dV}{dT}\).

Step 1: Understanding the Concept:

SHM follows the form \(x = A \sin(\omega t + \phi)\). Periodic motion repeats after \(T\), but doesn't necessarily follow the sine/cosine linear form. Non-periodic motions do not repeat.


Step 2: Detailed Explanation:

A. \(\sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}\). It is periodic with frequency \(2\omega\). Period \(T = \frac{2\pi}{2\omega} = \pi/\omega\). Since it has a squared term/constant, it's not SHM. \(\to\) (I)

B. \(\sin^3 (2\omega t) = \frac{3 \sin 2\omega t - \sin 6\omega t}{4}\). It's a combination of two SHMs. Periodic with period determined by lowest frequency \(2\omega\). \(T = 2\pi/2\omega = \pi/\omega\). (Checking options: Image says \(T=2\pi/\omega\) for B? Usually \(2\pi/2\omega = \pi/\omega\). Let's stick to standard logic).

C. \(\sin(\omega t) + \cos(\pi \omega t)\). The ratio of frequencies is \(\omega / (\pi \omega) = 1/\pi\) (irrational). Hence, it is non-periodic. \(\to\) (IV)

D. \(\cos \omega t + \cos 2\omega t\). Sum of two periodic functions with frequencies in ratio \(1:2\). Resulting period is \(T = \frac{2\pi}{GCD(\omega, 2\omega)} = \frac{2\pi}{\omega}\). Not SHM. \(\to\) (II)


Step 3: Final Answer:

Matches are A-I, B-II (based on specific options), C-IV, D-II.
Quick Tip: A function \(f(t) = f_1(t) + f_2(t)\) is periodic only if the ratio of their periods \(T_1/T_2\) is a rational number. If irrational (like \(1/\pi\)), it's non-periodic.

Step 1: Understanding the Concept:

Induced emf in a rotating rod is calculated by integrating the motional emf \(d\epsilon = B(r)v dr\) along the length of the rod.


: Key Formula or Approach:
\(\epsilon = \int_0^L B(r) (\omega r) dr\).

Substitute \(B(r) = B_o e^{-\lambda r}\).


Step 2: Detailed Explanation:
\[ \epsilon = \int_0^L (B_o e^{-\lambda r}) (\omega r) dr = B_o \omega \int_0^L r e^{-\lambda r} dr \]

Use integration by parts: \(\int u dv = uv - \int v du\).

Let \(u = r \implies du = dr\).

Let \(dv = e^{-\lambda r} dr \implies v = \frac{e^{-\lambda r}}{-\lambda}\).
\[ \epsilon = B_o \omega \left[ r \left(\frac{e^{-\lambda r}}{-\lambda}\right) \right]_0^L - B_o \omega \int_0^L \frac{e^{-\lambda r}}{-\lambda} dr \]
\[ \epsilon = B_o \omega \left[ -\frac{L e^{-\lambda L}}{\lambda} + 0 \right] + \frac{B_o \omega}{\lambda} \left[ \frac{e^{-\lambda r}}{-\lambda} \right]_0^L \]
\[ \epsilon = B_o \omega \left[ -\frac{L e^{-\lambda L}}{\lambda} - \frac{1}{\lambda^2} (e^{-\lambda L} - 1) \right] \]
\[ \epsilon = B_o \omega \left[ \frac{1}{\lambda^2} - \frac{e^{-\lambda L}}{\lambda^2} - \frac{L e^{-\lambda L}}{\lambda} \right] \]
\[ \epsilon = B_o \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right] \]


Step 3: Final Answer:

The induced emf matches option (A).
Quick Tip: For a uniform field, \(\epsilon = \frac{1}{2}B\omega L^2\). Here, since \(B\) varies, you must integrate. The result must reduce to the uniform case if \(\lambda \to 0\).

Step 1: Understanding the Concept:

In steady state, a capacitor acts as an open circuit (infinite resistance). No current flows through the branch containing the capacitor. We calculate the potential at the nodes around it using Ohm's Law.


Step 2: Detailed Explanation:

In the steady state, the \(2\mu F\) branch has zero current.

The total resistance of the main loop is \(R = 6\Omega + 2\Omega = 8\Omega\).

Current in the circuit \(I = \frac{V}{R} = \frac{2 V}{8 \Omega} = 0.25 A\).

The potential difference across the \(2\Omega\) resistor is \(V_{2\Omega} = I \times 2 = 0.25 \times 2 = 0.5 V\).

The capacitor is in parallel with the \(2\Omega\) resistor's effective nodes? Let's check the loop.

The loop starts at the battery, goes through \(6\Omega\), then splits. Current only goes through the \(2\Omega\) resistor.

The potential drop across the \(6\Omega\) resistor is \(0.25 \times 6 = 1.5 V\).

Potential across the remaining branches is \(2 V - 1.5 V = 0.5 V\).

Since the \(4\Omega\) resistor has no current, the entire \(0.5 V\) potential difference appears across the capacitor.


Step 3: Final Answer:

The potential difference across the capacitor is 0.5 V.
Quick Tip: "Steady state" for a DC circuit means "remove the capacitor branches" to find the current, then find the voltage across the gap where the capacitor was.

Step 1: Understanding the Concept:

The work-energy theorem states that the work done by all forces equals the change in kinetic energy (\(\Delta K = W_{total}\)).

Magnetic force (\(\vec{F}_m = q(\vec{v} \times \vec{B})\)) is always perpendicular to velocity, so it does zero work.

Only the electric field does work.


Step 2: Detailed Explanation:

Work done by electric force \(W_e = \int \vec{F}_e \cdot d\vec{r} = \int (q \vec{E}) \cdot d\vec{r}\).

Since \(\vec{E}\) is along \(\hat{x}\), \(d\vec{r} \cdot \hat{x} = dx\).
\[ \Delta K = W_e = \int_0^L q E_o e^{-\lambda x} dx \]
\[ \Delta K = q E_o \left[ \frac{e^{-\lambda x}}{-\lambda} \right]_0^L \]
\[ \Delta K = -\frac{q E_o}{\lambda} ( e^{-\lambda L} - e^0 ) \]
\[ \Delta K = \frac{q E_o}{\lambda} ( 1 - e^{-\lambda L} ) \]


Step 3: Final Answer:

The change in kinetic energy is \(\frac{q E_o}{\lambda} [ 1 - e^{-\lambda L} ]\).
Quick Tip: Always remember: Magnetic fields do no work on moving charges. In crossed field problems, only the electric field component along the path contributes to the change in kinetic energy.

Step 1: Understanding the Concept:

Electromagnetic waves carry momentum (\(p = U/c\)). When they hit a surface, they transfer this momentum, creating radiation pressure. They are composed of photons which have zero rest mass.


Step 2: Detailed Explanation:

Assertion: True. Radiation pressure \(P = I/c\) (for total absorption) or \(2I/c\) (for reflection) exists because EM waves carry momentum.

Reason: True. Photons have zero rest mass. Energy is related to frequency, not mass.

Explanation check: The pressure exists because of momentum, not because of the absence of mass. Therefore, (R) is not the reason why (A) is true.


Step 3: Final Answer:

Both are true but (R) is not the correct explanation of (A).
Quick Tip: Radiation pressure is a consequence of the momentum-energy relation \(p = E/c\). If EM waves had mass, they would still exert pressure, likely even more.

Step 1: Understanding the Concept:

For thin lenses in contact, the equivalent power of the combination is the algebraic sum of the individual powers. The sign of the equivalent power determines the nature of the combined lens (positive for converging/convex, negative for diverging/concave).


: Key Formula or Approach:

Equivalent power \( P = P_1 + P_2 \).

In terms of focal lengths: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).

For a convex lens, \( f_1 = +f_{convex} \); for a concave lens, \( f_2 = -f_{concave} \).


Step 2: Detailed Explanation:

The equivalent focal length \( F \) is given by:
\[ \frac{1}{F} = \frac{1}{f_{convex}} - \frac{1}{f_{concave}} = \frac{f_{concave} - f_{convex}}{f_{convex} \cdot f_{concave}} \]

The combination behaves as a concave lens if the equivalent focal length \( F \) is negative.

For \( F < 0 \), we must have:
\[ f_{concave} - f_{convex} < 0 \]
\[ f_{concave} < f_{convex} \]

Taking magnitudes: \( |f_{convex}| > |f_{concave}| \).

Conversely, if \( |f_{convex}| < |f_{concave}| \), the power of the convex lens is higher, and the system behaves as a convex lens.

Interchanging the positions of thin lenses in contact does not change the equivalent focal length.


Step 3: Final Answer:

The combination behaves as a concave lens if \( |f_{convex}| > |f_{concave}| \).
Quick Tip: The combination of lenses in contact always behaves like the lens which is "stronger," meaning the one with the higher power or the smaller focal length.

Step 1: Understanding the Concept:

We analyze the image formation by lens P first. The image formed by lens P then serves as the object for lens Q. The final position and magnification are determined by the lens formula and magnification formula applied sequentially.


: Key Formula or Approach:

Lens Formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).

Magnification: \( m = \frac{v}{u} \).

Total Magnification: \( M = m_1 \times m_2 \).


Step 2: Detailed Explanation:

For Lens P:
\( u_1 = -15 cm, f_1 = +10 cm \).

\[ \frac{1{v_1 - \frac{1{-15 = \frac{1{10 \implies \frac{1{v_1 = \frac{1{10 - \frac{1{15 = \frac{3 - 2{30 = \frac{1{30 \implies v_1 = +30 cm \).

The image \( I_1 \) is formed 30 cm to the right of lens P.

Magnification \( m_1 = \frac{v_1{u_1} = \frac{30}{-15} = -2 \).

For Lens Q:

Lens Q is placed 15 cm to the right of lens P.

The image \( I_1 \) (which is 30 cm right of P) acts as the object for Q.

Its position relative to Q is \( u_2 = 30 - 15 = 15 cm \) to the right of Q.

Since the rays were converging toward this point, it is a virtual object for Q: \( u_2 = +15 cm \).

Given \( f_2 = +15 cm \):

\[ \frac{1{v_2 - \frac{1{+15 = \frac{1{15 \implies \frac{1{v_2 = \frac{1{15 + \frac{1{15 = \frac{2{15 \implies v_2 = +7.5 cm \).

The final image is real (since \( v_2 > 0 \)) and 7.5 cm to the right of Q.

Magnification \( m_2 = \frac{v_2{u_2} = \frac{7.5}{15} = +0.5 \).

Total Magnification \( M = m_1 \times m_2 = (-2) \times (0.5) = -1 \).

Since \( |M| = 1 \), the size of the final image is the same as that of object AB.


Step 3: Final Answer:

The final image is real, formed at 7.5 cm right of lens Q, with a size same as that of AB.
Quick Tip: Always keep track of the object type for the second lens. If the image from the first lens falls behind the second lens, it acts as a virtual object (\( u \) is positive).

Step 1: Understanding the Concept:

Intensity at a point on the screen depends on the phase difference between the two waves. The phase difference is determined by the path difference from the slits to that point.


: Key Formula or Approach:

Path difference: \( \Delta x = \frac{dy}{D} \).

Phase difference: \( \phi = \frac{2\pi}{\lambda} \Delta x \).

Resultant Intensity: \( I = I_{max} \cos^2\left(\frac{\phi}{2}\right) \).


Step 2: Detailed Explanation:

The point "exactly opposite to one of the slits" is at a height \( y = d/2 \) from the central axis.

Given: \( d = 5\lambda \), \( D = 10d = 50\lambda \), and \( y = d/2 = 2.5\lambda \).

Calculate path difference \( \Delta x \):
\[ \Delta x = \frac{d \cdot y}{D} = \frac{d \cdot (d/2)}{10d} = \frac{d}{20} \]

Substitute \( d = 5\lambda \):
\[ \Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4} \]

Calculate phase difference \( \phi \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]

Calculate intensity \( I \):
\[ I = I_o \cos^2\left(\frac{\phi}{2}\right) = I_o \cos^2\left(\frac{\pi/2}{2}\right) = I_o \cos^2\left(\frac{\pi}{4}\right) \]
\[ I = I_o \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_o}{2} \]


Step 3: Final Answer:

The intensity of the fringe is \( I_o/2 \).
Quick Tip: A path difference of \( \lambda/4 \) always corresponds to a phase difference of \( 90^\circ \) and results in an intensity which is exactly half of the maximum intensity.

Step 1: Understanding the Concept:

The de Broglie wavelength of a moving particle is inversely proportional to its momentum. Since the mass of the electron remains constant, the wavelength is inversely proportional to its velocity.


: Key Formula or Approach:
\( \lambda = \frac{h}{p} = \frac{h}{mv} \implies \lambda \propto \frac{1}{v} \).


Step 2: Detailed Explanation:

In free space, wavelength \( \lambda_o = \frac{h}{mv} \).

In the medium, velocity \( v' \) is reduced by 20%:
\[ v' = v - 0.20v = 0.80v = \frac{4}{5}v \].

The de Broglie wavelength in the medium is:
\[ \lambda_m = \frac{h}{mv'} = \frac{h}{m(0.8v)} = \frac{\lambda_o}{0.8} \]
\[ \lambda_m = \frac{10}{8} \lambda_o = 1.25 \lambda_o \].

Comparing with \( \lambda_m = \alpha \lambda_o \), we get \( \alpha = 1.25 \).


Step 3: Final Answer:

The value of \( \alpha \) is 1.25.
Quick Tip: When velocity decreases by a fraction \( f \), the wavelength increases by a factor \( 1/(1-f) \). Here \( 1/(1-0.2) = 1/0.8 = 1.25 \).

Step 1: Understanding the Concept:

Mass defect (\( \Delta m \)) is the difference between the total mass of the constituent nucleons (protons and neutrons) and the actual experimental mass of the atom. Carbon-12 has 6 protons and 6 neutrons.


: Key Formula or Approach:
\( \Delta m = [Z \cdot m_p + (A-Z) \cdot m_n] - M_{experimental} \).

Energy equivalent \( E = \Delta m \cdot 931.5 MeV/u \).


Step 2: Detailed Explanation:

For \( {}^{12}_{6}C \):

Number of protons \( Z = 6 \).

Number of neutrons \( A - Z = 12 - 6 = 6 \).

Total mass of nucleons:
\[ M_{nucleons} = 6 \times 1.00727 + 6 \times 1.00866 \]
\[ M_{nucleons} = 6.04362 + 6.05196 = 12.09558 u \].

Experimental mass \( M_{exp} = 12.00000 u \).

Mass defect:
\[ \Delta m = 12.09558 - 12.00000 = 0.09558 u \].

Convert mass defect to MeV/\( c^2 \):
\[ \Delta E = 0.09558 \times 931.5 MeV/c^2 \]
\[ \Delta E \approx 89.03277 MeV/c^2 \].


Step 3: Final Answer:

The mass defect is 89.03 MeV/\( c^2 \).
Quick Tip: Always maintain as many decimal places as possible in atomic mass calculations, as the difference (mass defect) is small but is multiplied by a large factor (931.5).

Step 1: Understanding the Concept:

In a p-n junction, the depletion region must satisfy the condition of charge neutrality. This means the total ionized charge on the p-side must equal the total ionized charge on the n-side.


: Key Formula or Approach:

Charge neutrality condition: \( q \cdot N_A \cdot x_p = q \cdot N_D \cdot x_n \).

Where \( N_A, N_D \) are acceptor/donor concentrations and \( x_p, x_n \) are widths of depletion layers on each side.


Step 2: Detailed Explanation:

Given:
\( N_A (p-side) = 10^{15} cm^{-3} \).
\( N_D (n-side) = 10^{18} cm^{-3} \).

From neutrality: \( N_A \cdot x_p = N_D \cdot x_n \).
\[ \frac{x_p}{x_n} = \frac{N_D}{N_A} = \frac{10^{18}}{10^{15}} = 1000 \].

Since \( x_p = 1000 \cdot x_n \), the depletion region extends much further into the p-side.

This is a general principle: the depletion region is always wider on the more lightly doped side.


Step 3: Final Answer:

The depletion region width is more on the p-side compared to that in the n-side.
Quick Tip: Remember the inverse relationship: \textbf{Depletion Width \( \propto \) 1 / Doping Concentration}. The "thin" side is the "highly doped" side.

Step 1: Understanding the Concept:

Elastic potential energy is the energy stored in a body when it is deformed within its elastic limit. For a stretched wire, this energy is calculated using Young's modulus, strain, and volume.


: Key Formula or Approach:

Energy density (\( u \)) = \( \frac{1}{2} \cdot Y \cdot (strain)^2 \).

Total Potential Energy (\( U \)) = \( u \cdot Volume \).

Strain = \( \frac{\Delta L}{L} \).


Step 2: Detailed Explanation:

Given:
\( L = 3 m \).
\( \Delta L = 3 mm = 3 \times 10^{-3} m \).
\( V = 600 \times 10^{-6} m^3 \).
\( Y = 1.1 \times 10^{11} N/m^2 \).

Calculate Strain:
\[ Strain = \frac{\Delta L}{L} = \frac{3 \times 10^{-3}}{3} = 10^{-3} \].

Calculate Potential Energy:
\[ U = \frac{1}{2} \cdot Y \cdot (Strain)^2 \cdot V \]
\[ U = \frac{1}{2} \times 1.1 \times 10^{11} \times (10^{-3})^2 \times 600 \times 10^{-6} \]
\[ U = \frac{1}{2} \times 1.1 \times 10^{11} \times 10^{-6} \times 600 \times 10^{-6} \]
\[ U = 0.5 \times 1.1 \times 600 \times 10^{-1} \]
\[ U = 0.5 \times 1.1 \times 60 = 33 J \].


Step 3: Final Answer:

The elastic potential energy stored in the wire is 33 J.
Quick Tip: Always double-check units. Here, converting elongation from mm to m and volume from \( \mum^3 \) (if applicable) is essential. The formula \( U = \frac{1}{2} \cdot Stress \cdot Strain \cdot Volume \) is the same as above.

Step 1: Understanding the Concept:

This problem involves the principle of calorimetry where heat lost by one substance equals heat gained by another. One part of the water loses sensible heat during cooling. The other part gains sensible heat to reach boiling point and then gains latent heat to evaporate.


Step 2: Detailed Explanation:

Let \( s = 4200 J/kg\cdotK \) and \( L = 2256 \times 10^3 J/kg \).

Heat Lost by \( x \) grams of water:

Cooling from \( 50^\circC \) to \( 0^\circC \):
\[ Q_{lost} = m \cdot s \cdot \Delta T = \left(\frac{x}{1000}\right) \cdot 4200 \cdot (50 - 0) = 210x J \].

Heat Gained by \( (1000-x) \) grams of water:

1. Heating from \( 50^\circC \) to \( 100^\circC \):
\[ Q_1 = m \cdot s \cdot \Delta T' = \left(\frac{1000-x}{1000}\right) \cdot 4200 \cdot (100 - 50) = 210(1000-x) J \].

2. Evaporation at \( 100^\circC \):
\[ Q_2 = m \cdot L = \left(\frac{1000-x}{1000}\right) \cdot 2256 \times 10^3 = 2256(1000-x) J \].

Total heat gained:
\[ Q_{gained} = (210 + 2256)(1000-x) = 2466(1000-x) J \].

Equating heat lost and heat gained:
\[ 210x = 2466(1000-x) \]
\[ 210x = 2466000 - 2466x \implies 2676x = 2466000 \]
\[ x = \frac{2466000}{2676} \approx 921.52 \].


Step 3: Final Answer:

The value of \( x \) to the closest integer is 922.
Quick Tip: To "evaporate" water at \( 50^\circC \), you must first heat it to \( 100^\circC \) before the phase change occurs. Don't forget to add both sensible heat and latent heat.

Step 1: Understanding the Concept:

Resonance in a series LCR circuit occurs when the inductive reactance (\( X_L \)) and capacitive reactance (\( X_C \)) become equal. At this state, the angular frequency is called the resonant frequency.


: Key Formula or Approach:

Resonant frequency \( \omega_o = \frac{1}{\sqrt{LC}} \).

Inductive Reactance \( X_L = \omega_o L \).


Step 2: Detailed Explanation:

Given:
\( L = 1.6 H \).
\( C = 40 \times 10^{-6} F \).

Calculate resonant angular frequency \( \omega_o \):
\[ \omega_o = \frac{1}{\sqrt{1.6 \times 40 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} \]
\[ \omega_o = \frac{1}{8 \times 10^{-3}} = 125 rad/s \].

Calculate inductive reactance \( X_L \):
\[ X_L = \omega_o \cdot L = 125 \times 1.6 = 200\ \Omega \].


Step 3: Final Answer:

The inductive reactance at resonant frequency is 200 \(\Omega\).
Quick Tip: At resonance, \( X_L = X_C \). You can also directly calculate \( X_L = \sqrt{\frac{L}{C}} \), which equals \( \sqrt{\frac{1.6}{40 \times 10^{-6}}} = \sqrt{40000} = 200 \).

Step 1: Understanding the Concept:

The current \( I \) in a closed circuit containing a cell of emf \( E \), internal resistance \( r \), and external resistance \( R \) is given by Ohm's law.


: Key Formula or Approach:
\( I = \frac{E}{R + r} \implies E = I(R + r) \).


Step 2: Detailed Explanation:

From the first case:
\( E = 0.25 \cdot (5 + r) \quad \dots(1) \).

From the second case:
\( E = 0.5 \cdot (2 + r) \quad \dots(2) \).

Since the emf \( E \) is constant for the cell, equate (1) and (2):
\[ 0.25(5 + r) = 0.5(2 + r) \]

Divide both sides by 0.25:
\[ 5 + r = 2(2 + r) \implies 5 + r = 4 + 2r \]
\[ r = 1\ \Omega \].


Step 3: Final Answer:

The internal resistance of the cell is 1 \(\Omega\).
Quick Tip: When current doubles, the total resistance must have halved. So \( (5+r)/2 = 2+r \), which leads to \( r=1 \).

Step 1: Understanding the Concept:

According to Faraday's law of induction, a change in magnetic flux through a circuit induces an electromotive force (emf). The induced current can then be found using Ohm's law.


: Key Formula or Approach:

Flux \( \Phi = B \cdot A \cdot \cos \theta \) (here \( \theta = 0^\circ \)).

Induced emf \( e = \left| \frac{d\Phi}{dt} \right| = A \cdot \left| \frac{dB}{dt} \right| \).

Current \( I = e/R \).


Step 2: Detailed Explanation:

Radius \( r = 20 cm = 0.2 m \).

Area \( A = \pi r^2 = \frac{22}{7} \cdot (0.2)^2 = \frac{22 \cdot 0.04}{7} = \frac{0.88}{7} m^2 \).

Magnetic field \( B = 2t^2 + 2t + 3 \).

Rate of change \( \frac{dB}{dt} = 4t + 2 \).

At \( t = 3 s \), \( \frac{dB}{dt} = 4(3) + 2 = 14 T/s \).

Induced emf:
\[ e = A \cdot \frac{dB}{dt} = \frac{0.88}{7} \times 14 = 0.88 \times 2 = 1.76 V \].

Induced current:
\[ I = \frac{e}{R} = \frac{1.76}{2} = 0.88 A \].

Given \( I = \frac{\alpha}{50} \implies 0.88 = \frac{\alpha}{50} \).
\[ \alpha = 0.88 \times 50 = 44 \].


Step 3: Final Answer:

The value of \( \alpha \) is 44.
Quick Tip: Always ensure radius is in meters. "Plane perpendicular to field" means the area vector and field vector are parallel, so \( \cos 0 = 1 \).