Q.What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of $10\ m\ s^{-1}$?

Step 1 — de Broglie's hypothesis

Louis de Broglie proposed that every moving particle of mass $m$ and velocity $\nu$ has an associated "matter wave" whose wavelength $\lambda$ is inversely proportional to its momentum $p = m\nu$.

Step 2 — The equation

This gives $\lambda = \dfrac{h}{p} = \dfrac{h}{m\nu}$, where $h$ is Planck's constant.

Step 3 — Eliminating the other options

Option (A) has an incorrect square root on mass and velocity; option (C) uses "mp" (mass times momentum), which is dimensionally wrong; option (D) does not represent momentum correctly — none of these are the actual de Broglie relation.

1. The concept — Huygens' principle.

Huygens' geometrical construction says:

• Every point on a primary wavefront behaves as a fresh source of disturbance, emitting secondary wavelets.
• These wavelets travel in the forward direction with the speed of the wave in that medium.
• After a time $t$, the new position of the wavefront is the forward envelope (common tangent) of all the secondary wavelets.

2. Why the two speeds must be identical here.

The new wavefront is constructed as the envelope of wavelets of radius $vt$. If the wavelets moved at some speed different from $v$, the envelope they build would advance at that different speed — the wavefront's speed simply is the wavelets' speed. In a homogeneous medium the speed $v = c/n$ is the same at every point and in every direction, so:

3. Why the other options fail.

Options (A), (B) and (D) all claim an inequality between the two speeds. Any such inequality would break Huygens' construction itself — the envelope of the wavelets is the wavefront, so they cannot outrun or lag one another in a uniform medium. (Only in a non-homogeneous medium do wavelets from different points have different radii, and even there each wavelet moves at the local wave speed.)

ANSWER: C

Step 1 — Recognise the statement.

The sentence "heat cannot of itself flow from a body at lower temperature to a body at higher temperature" is the classic Clausius statement of the Second Law of Thermodynamics. The two crucial words are "of itself" — meaning spontaneously, with no external agency.

Step 2 — Why the First Law cannot be the answer.

The First Law is just energy conservation:

It is completely indifferent to direction. If $100\ \mathrm{J}$ flowed spontaneously from a cold body to a hot one, energy would still be perfectly conserved — the First Law would be entirely satisfied. Yet such a process is never observed. So the First Law is not sufficient to explain the one-way nature of heat flow; something more is needed, and that something is the Second Law.

Step 3 — What the Second Law adds: direction.

The Second Law supplies the arrow of time for thermal processes, quantified by entropy: for any spontaneous process in an isolated system,

If heat $Q$ passed spontaneously from a cold reservoir at $T_C$ to a hot one at $T_H$ (with $T_H > T_C$):

since $T_H > T_C$. A negative total entropy change violates the Second Law — so the process is forbidden. Heat flowing the natural way (hot → cold) gives $\Delta S_{\text{total}} > 0$ and is allowed.

Step 4 — The everyday consequence.

This is precisely why a refrigerator needs a compressor and an electricity supply: heat can be moved from cold to hot, but never "of itself" — external work must be done to pay the entropy bill.

Options (A) and (D) — conservation of mass and of momentum — are unrelated conservation principles and say nothing about the direction of heat flow.

ANSWER: C

1. Why $p = h/\lambda$

A photon has zero rest mass, so its momentum cannot be found from $p = mv$. From special relativity, a massless particle satisfies $E = pc$; and from Planck's hypothesis its energy is $E = h\nu$. Combining, and using $c = \nu\lambda$:

(This is also exactly de Broglie's relation, read backwards — it is the same equation that assigns a wavelength to matter.)

Notice that $c$ has cancelled: we do not need the speed of light, which is why the question only supplies $h$ and $\lambda$.

2. Substitute

3. Arithmetic

Mantissas: $\dfrac{6.6}{2.2} = 3$

Exponents: $10^{-34-(-11)} = 10^{-23}$

4. Unit check

The distractors come from multiplying $h$ by $\lambda$ (option A, $1.452\times10^{-44}$) or from inverting the ratio (option B), so it is worth re-reading which quantity sits in the denominator.

ANSWER: C

The key idea is that simple harmonic motion (SHM) arises only when the restoring force (or torque) is proportional to the displacement from equilibrium. For a pendulum, the restoring torque is $mgL\sin\theta$, where $\theta$ is the angular displacement. This is proportional to $\sin\theta$, not $\theta$ itself. The approximation $\sin\theta \approx \theta$ (in radians) holds only for small angles — typically less than about $10^\circ$. That’s why condition (II) is necessary.

But why does the bob size matter? The formula for the period $T = 2\pi\sqrt{L/g}$ assumes the pendulum is a simple pendulum: a point mass suspended by a massless, inextensible string. If the bob is large, its size is not negligible compared to the length $L$, and the pendulum behaves as a physical pendulum — the centre of mass shifts, and the moment of inertia changes. The motion is still oscillatory, but the simple harmonic approximation fails unless the bob is small enough to treat as a point mass. So condition (I) is also necessary.

Let’s go through the reasoning step by step.

1. The condition for SHM

   For any oscillating system, SHM requires a restoring force (or torque) that is directly proportional to the displacement, and opposite in direction. For a pendulum, the restoring torque about the pivot is $\tau = -mgL\sin\theta$. For small $\theta$, $\sin\theta \approx \theta$ (in radians), so $\tau \approx -mgL\theta$. This is of the form $\tau = -k\theta$, which gives SHM. The approximation is accurate to within about 1% for $\theta < 10^\circ$ (about 0.175 rad). So condition (II) is essential.

2. Why the bob size matters

   The derivation $T = 2\pi\sqrt{L/g}$ assumes the entire mass is concentrated at a point at distance $L$ from the pivot. If the bob has a significant size, the distance from the pivot to the centre of mass is not exactly $L$, and the moment of inertia is not simply $mL^2$. The period becomes $T = 2\pi\sqrt{I/(mgd)}$, where $I$ is the moment of inertia and $d$ is the distance from pivot to centre of mass. This is still oscillatory, but the motion is not simple harmonic unless the amplitude is small — and even then, the formula changes. More importantly, the simple pendulum model (which is what the question refers to) explicitly requires a point mass. So condition (I) is also necessary.

3. Evaluating the statements

   • Statement (I): “the size of the bob is negligible in comparison with the length of the pendulum” — this is required for the simple pendulum approximation. Correct.
   • Statement (II): “the angular amplitude is less than $10^\circ$” — this ensures $\sin\theta \approx \theta$ holds, giving SHM. Correct.

   Both conditions must be satisfied for the pendulum to oscillate simple harmonically. Therefore, both (I) and (II) are correct.

The photoelectric effect is a beautiful demonstration of the particle nature of light. When light of sufficient frequency (above the threshold frequency) strikes a metal surface, it ejects electrons. The key insight is that light behaves as a stream of photons, each carrying energy $h\nu$. One photon interacts with one electron, transferring its entire energy. If that energy exceeds the work function $\phi$ (the minimum energy needed to free the electron), the electron is ejected with kinetic energy $K_{\text{max}} = h\nu - \phi$.

This one-photon–one-electron rule is the foundation. It tells us that the number of ejected electrons depends on how many photons strike the surface per second — that is, the intensity. The energy of each ejected electron depends on the photon energy $h\nu$, not on how many photons arrive.

Let’s examine each option carefully.

1. Option (A): Number of $e^-$ ejected increases with the increase in the intensity of incident light.

   Intensity is energy per unit area per second. For monochromatic light, intensity $I = n h\nu$, where $n$ is the number of photons per second per unit area. If you increase $I$ while keeping frequency $\nu$ fixed, $n$ increases. More photons mean more electrons ejected (provided $\nu$ is above threshold). This is correct.

2. Option (B): Energy of $e^-$ ejected increases with the increase in the intensity of incident light.

   The maximum kinetic energy of an ejected electron is $K_{\text{max}} = h\nu - \phi$. Intensity does not appear here. Changing intensity changes the number of photons, not the energy per photon. So the energy of each electron remains the same. This is false.

3. Option (C): Number of $e^-$ ejected increases with the increase in the frequency of incident light.

   Increasing frequency (above threshold) increases the energy per photon, but does not change the number of photons unless intensity is also adjusted. If intensity is held constant, higher frequency means fewer photons (since $I = n h\nu$). So the number of ejected electrons actually decreases. Even if intensity is not fixed, frequency alone does not control the count — it only determines whether ejection happens at all. This is false.

4. Option (D): Number of $e^-$ ejected increases with the increase in work function.

   Work function $\phi$ is the minimum energy needed to eject an electron. A higher work function makes it harder to eject electrons — fewer electrons will have enough energy to escape, and the threshold frequency rises. So the number ejected decreases, not increases. This is false.