Q.Three vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ add up to zero. Find which is false.

Concept and Intuition

This problem is a classic application of the principle of conservation of linear momentum.

1. Initial State: The bomb is initially at rest. This means its initial velocity is zero, and therefore, its total initial linear momentum is also zero.
2. Explosion: An explosion is an internal process. No external forces act on the bomb system during the explosion (neglecting gravity for the brief moment of explosion, which is standard for such problems).
3. Conservation: According to the principle of conservation of linear momentum, if no external forces act on a system, the total linear momentum of the system remains constant. Since the initial momentum was zero, the total final momentum of the three parts must also be zero.

Let the three parts have masses $m_1, m_2, m_3$ and velocities $\vec{v_1}, \vec{v_2}, \vec{v_3}$ respectively. The problem states they move with equal speeds, so $|\vec{v_1}| = |\vec{v_2}| = |\vec{v_3}| = v$.

The conservation of momentum equation is:

This means the vector sum of the final momenta of the three parts must be zero. For three vectors to sum to zero, they must be able to form a closed triangle when placed head-to-tail.

Let $\vec{P_1} = m_1 \vec{v_1}$, $\vec{P_2} = m_2 \vec{v_2}$, and $\vec{P_3} = m_3 \vec{v_3}$.

Then $\vec{P_1} + \vec{P_2} + \vec{P_3} = \vec{0}$.

This implies that the magnitudes of these momentum vectors must satisfy the triangle inequality:

Since the speeds are equal ($v$), we have $|\vec{P_1}| = m_1 v$, $|\vec{P_2}| = m_2 v$, and $|\vec{P_3}| = m_3 v$. Substituting these into the triangle inequality and dividing by $v$ (since $v \ne 0$), we get:

Additionally, the sum of the masses of the three parts must equal the original mass of the bomb:

The phrase "in different directions" usually implies that the three momentum vectors are not collinear, meaning they form a non-degenerate triangle. For a non-degenerate triangle, the strict inequality ($<$) must hold for all three conditions. If equality holds for any condition (e.g., $m_3 = m_1 + m_2$), it means the vectors are collinear, forming a degenerate triangle.

Step-by-step Derivation

1. Total Mass Condition: The sum of the masses of the three parts must be equal to the original mass of the bomb, $m$. We will check this for each option first.

   • (A) $\frac{3m}{11} + \frac{m}{3} + \frac{13m}{33} = \frac{9m}{33} + \frac{11m}{33} + \frac{13m}{33} = \frac{33m}{33} = m$. (Satisfied)
   • (B) $\frac{m}{6} + \frac{m}{3} + \frac{m}{2} = \frac{m}{6} + \frac{2m}{6} + \frac{3m}{6} = \frac{6m}{6} = m$. (Satisfied)
   • (C) $\frac{4m}{19} + \frac{5m}{19} + \frac{10m}{19} = \frac{19m}{19} = m$. (Satisfied)
   • (D) $\frac{6m}{29} + \frac{8m}{29} + \frac{15m}{29} = \frac{29m}{29} = m$. (Satisfied)

   All options satisfy the total mass condition. We must now use the triangle inequality.

2. Triangle Inequality Condition: The magnitudes of the masses ($m_1, m_2, m_3$) must satisfy the triangle inequality. For the parts to move in "different directions" (implying non-collinear motion), the strict inequality must hold:

   $m_1 < m_2 + m_3$

   $m_2 < m_1 + m_3$

   $m_3 < m_1 + m_2$

   Let's check each option:

   • (A) Masses: $\frac{3m}{11}, \frac{m}{3}, \frac{13m}{33}$ To compare easily, let's use a common denominator, $33$: $m_1 = \frac{9m}{33}$, $m_2 = \frac{11m}{33}$, $m_3 = \frac{13m}{33}$.
     • Is $m_1 < m_2 + m_3$? $\frac{9m}{33} < \frac{11m}{33} + \frac{13m}{33} \implies \frac{9m}{33} < \frac{24m}{33}$. (True) …

The question tests your understanding of how ammeters and voltmeters are meant to behave in circuits — and what "ideal" means in each case. The key is to remember what each device does: an ammeter measures current through it, and a voltmeter measures voltage across it. To avoid affecting the circuit, each must have a specific resistance.

1. Ideal ammeter: An ammeter is placed in series with the component whose current you want to measure. If it had any resistance, it would add to the circuit's total resistance and reduce the current — giving a wrong reading. So an ideal ammeter should offer zero resistance — it should act like a perfect wire. That eliminates option (A) and confirms (B). …

The concept first

For an incompressible fluid in steady flow, the equation of continuity says the volume passing any cross-section per second is the same everywhere:

where $Q$ is the volume flow rate ($\mathrm{m^3\,s^{-1}}$), $A$ the cross-sectional area and $v$ the flow speed. Notice what this equation does not contain: the shape of the cross-section. Only the area matters. That is the whole trick of this question — it offers you two shapes with the correct area and two with the wrong one.

Step-by-step

1. Convert the time to SI.

2. Compute the required volume flow rate.

3. Apply continuity to get the area.

So any duct whose cross-section measures $0.16\ \mathrm{m^2}$ will do.

4. Test each proposed cross-section.

• (A) Square, side $0.4\ \mathrm{m}$: $A = (0.4)^2 = 0.16\ \mathrm{m^2}$ ✓ Correct.
• (B) Rectangle, $25\ \mathrm{cm} \times 64\ \mathrm{cm}$: convert first — $0.25\ \mathrm{m} \times 0.64\ \mathrm{m}$, so

• (C) Square, side $1.26\ \mathrm{m}$: $A = (1.26)^2 = 1.5876 \approx 1.59\ \mathrm{m^2}$ ✗ — nearly ten times too big. …

Concept and Intuition

We need to evaluate three independent physics statements. Statement I involves the Lorentz force on a moving electron in parallel electric and magnetic fields. Statement II is about the superposition of magnetic fields from two parallel currents. Statement III concerns the net force on a current-carrying loop in a uniform field. Each must be judged true or false based on fundamental principles.

Step-by-step reasoning

1. Statement I: Electron velocity in parallel E and B fields

   • An electron has charge $q = -e$. The Lorentz force is $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
   • Here $\vec{E}$ and $\vec{B}$ are in the same direction, say along $+x$. The electron is projected in that same direction, so $\vec{v}$ is also along $+x$.
   • The cross product $\vec{v} \times \vec{B} = 0$ because the vectors are parallel. So the magnetic force is zero.
   • The electric force is $\vec{F}_E = -e \vec{E}$, which is opposite to $\vec{E}$ (since electron charge is negative). Thus the electric force opposes the motion, slowing the electron.
   • Conclusion: The velocity magnitude does decrease. Statement I is true.

2. Statement II: Magnetic field midway between two parallel wires

   • Two infinitely long parallel wires carry current in the same direction.
   • By the right-hand rule, the magnetic field due to each wire at a point midway between them is perpendicular to the line joining the wires.
   • For currents in the same direction, the fields from each wire at the midpoint are equal in magnitude but opposite in direction (one points into the page, the other out).
   • They cancel exactly, giving zero net field.
   • Conclusion: Statement II is true.

3. Statement III: Net force on a rectangular coil in a uniform magnetic field

   • A current-carrying loop in a uniform magnetic field experiences forces on each side. …

Let’s unpack each statement one by one, starting with the concept that matters most: vectors add according to direction, not just size.

1. Statement I: “An object subjected to velocities $\vec{v}_1$ and $\vec{v}_2$ has a resultant velocity with magnitude $|\vec{v}| = |\vec{v}_1| + |\vec{v}_2|$.” This is only true when $\vec{v}_1$ and $\vec{v}_2$ are in the exact same direction. In general, the magnitude of the resultant vector is given by the law of cosines:

where $\theta$ is the angle between them. The sum $|\vec{v}_1| + |\vec{v}_2|$ is the maximum possible resultant magnitude (when $\theta = 0^\circ$). For any other angle, the resultant is smaller. Since the statement presents this as a universal truth, it is false.

2. Statement II: “The magnitude of displacement is either less or equal to the path length of an object between two points.” Displacement is the straight-line distance between the initial and final positions — a vector quantity. Path length is the total distance actually travelled along the trajectory — a scalar.
   • If the object moves in a straight line without turning back, displacement equals path length. …

The zeroth law is the foundation of thermometry. Without it, you could never be sure that a thermometer reading in one system means the same thing for another system. The law creates an equivalence relation: if A is in equilibrium with C, and B is in equilibrium with C, then A and B are in equilibrium with each other. This transitivity is what allows us to assign a single number — temperature — to the state of thermal equilibrium.

1. What the zeroth law actually says.

   The Reason (R) states it precisely: two systems separately in thermal equilibrium with a third are in thermal equilibrium with each other. This is not a trivial statement — it is an empirical fact about the world that cannot be derived from other laws. It establishes that "being in thermal equilibrium" is a transitive relation.

2. How this leads to temperature.

   Because of transitivity, all systems that are in thermal equilibrium with each other share a common property. We call that property temperature. If system A and system B are both in equilibrium with a thermometer (system C), then A and B must have the same temperature — even if they never touch each other. This is why a single thermometer can measure the temperature of any object: the reading on the thermometer tells you the temperature of the object it is in equilibrium with.

3. Why the Assertion is true. …

Concept and intuition:

This question tests the fundamental laws of thermal radiation. A black body is an ideal emitter and absorber — its radiation spectrum depends only on temperature, not on the material. Emissivity (ε) and absorptivity (α) are equal for any surface in thermal equilibrium (Kirchhoff’s law). A perfect reflector absorbs nothing, so its emissivity is zero. And because a black body absorbs all incident radiation, it must also emit the maximum possible radiation at a given temperature — any real body emits less.

1. Statement A: “The energy emitted versus wavelength plot of a black body depends on the material of the black body.”

   This is false. Planck’s law gives the spectral radiance of a black body as a function only of temperature and wavelength — no material property appears. The shape and peak of the curve are universal (Wien’s displacement law). So A is incorrect.

2. Statement B: “Emissivity of a completely reflecting surface is zero.”

   True. A perfectly reflecting surface absorbs no radiation (absorptivity = 0). By Kirchhoff’s law, emissivity equals absorptivity for a body in thermal equilibrium, so emissivity = 0. It emits no thermal radiation.

3. Statement C: “Emissivity is equal to (absorptive power)².” …

Concept and Intuition

The fundamental difference in how electric and magnetic fields interact with charges lies in their dependence on the charge's motion.

An electric field is created by charges and exerts a force on any charge, regardless of whether that charge is moving or stationary. The force is directly proportional to the charge and the electric field strength.

A magnetic field, on the other hand, is created by moving charges (currents) and exerts a force only on other moving charges. A stationary charge will not experience a magnetic force. Furthermore, the direction of the magnetic force depends on the direction of the charge's velocity relative to the magnetic field.

Let's evaluate each statement based on these principles.

Step-by-step Evaluation

1. Evaluate Statement A: "Magnetic field exerts force only on a moving charge"

   The force experienced by a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula for the magnetic component:

   $\vec{F}_m = q(\vec{v} \times \vec{B})$

   From this formula, if the charge is stationary, its velocity $\vec{v} = \vec{0}$. Substituting this into the formula, we get $\vec{F}_m = q(\vec{0} \times \vec{B}) = \vec{0}$. This means a stationary charge experiences no magnetic force. Only when $\vec{v} \neq \vec{0}$ can there be a magnetic force (provided $\vec{v}$ is not parallel to $\vec{B}$).

   Therefore, statement A is true.

2. Evaluate Statement B: "Electric field exerts force on both stationary and moving charge"

   The force experienced by a charge $q$ in an electric field $\vec{E}$ is given by:

   $\vec{F}_e = q\vec{E}$

   This formula shows that the electric force depends only on the magnitude of the charge $q$ and the electric field $\vec{E}$. It does not depend on the velocity $\vec{v}$ of the charge. Thus, whether the charge is stationary ($\vec{v} = \vec{0}$) or moving ($\vec{v} \neq \vec{0}$), it will experience an electric force if it is in an electric field.

   Therefore, statement B is true. …