Q.In which of the following examples of motion, can the body be considered approximately a point object:

The question asks which statement about the nature of physical laws is NOT correct. Let’s examine each option carefully, using core principles from physics.

Concept and Intuition:

Physical laws are built on symmetries—they are universal in space and time (homogeneity and isotropy) and are the same everywhere. However, conservation laws come in different mathematical types: some quantities are scalars (like energy), but others are vectors (like momentum) or even tensors. A common mistake is to assume all conserved quantities are scalars, which is false.

1. Option (A): “All conserved quantities are necessarily scalars”

   • In physics, conserved quantities arise from symmetries via Noether’s theorem. For example:
     • Time symmetry → conservation of energy (a scalar).
     • Translational symmetry → conservation of linear momentum (a vector).
     • Rotational symmetry → conservation of angular momentum (a vector).
   • Since momentum and angular momentum are vectors, this statement is false.
   • Pitfall: Many students think only of energy, forgetting momentum and angular momentum.

2. Option (B): “The laws of nature do not change with time”

   • This is the principle of time invariance—a cornerstone of physics. Experiments today give the same results as centuries ago (e.g., Newton’s laws still hold).
   • This is correct.

3. Option (C): “The laws of nature are same everywhere in the universe”

   • This is the cosmological principle—the laws are universal. Gravity works the same on Earth, in distant galaxies, and at the edge of the observable universe.
   • This is correct.

4. Option (D): “The law of gravitation is the same both on the moon and the earth”

   • Newton’s law of gravitation applies universally. The gravitational acceleration differs (due to different masses and radii), but the law itself (inverse-square relation) is identical.
   • This is correct.

Step 1 — The definition.

For a rigid body,

where $r_i$ is the perpendicular distance of the mass element from the axis of rotation. Every quantity in that definition — the masses, and their distances from the chosen axis — matters.

Step 2 — Test each option.

• (A) "does not depend on its shape" — false. A ring and a disc of the same mass and radius have $MR^2$ and $\frac12 MR^2$ respectively: different shape, different $I$.
• (C) "does not depend on its size" — false. $I\propto R^2$ for a given shape; a bigger disc has a larger $I$.
• (D) "does not depend on its mass" — false. $I$ is directly proportional to $M$.
• (B) "depends on the position of the axis of rotation" — true. This is exactly the content of the parallel-axis theorem:

where $d$ is the distance of the new axis from the parallel axis through the centre of mass. Move the axis ($d$ changes) and $I$ changes, even though the body is unchanged. E.g. for a rod of length $L$: $I=\frac{ML^2}{12}$ about the centre but $\frac{ML^2}{3}$ about one end.

ANSWER: B

The concept: what "elastic" actually conserves

For two bodies colliding with no external force:

Momentum conservation follows from Newton's third law: the internal forces are equal and opposite and cancel. It holds for every collision, elastic or not.

An elastic collision additionally satisfies

The balls do deform momentarily, storing energy elastically, but they give all of it back as KE; none is degraded into heat, sound or permanent deformation.

Step 1 — Judge the Assertion

"Both kinetic energy and linear momentum remain conserved" is exactly the pair of equations above. True.

Step 2 — Judge the Reason

"...as the collision is elastic there is no exchange of energy." This is false, and instructively so. Take the standard head-on elastic collision of two equal masses, one at rest:

The moving ball stops dead and the struck ball moves off with the full speed — a complete transfer of kinetic energy from one ball to the other, in a perfectly elastic collision. Elasticity constrains the total, not the individual shares.

Step 3 — Conclude

A true, R false $\Rightarrow$ the printed option "(A) is true but (R) is false", which is (C).

ANSWER: C

Step 1 — Why the graph is a straight line.

In pure rolling there is no slipping, so the contact point is instantaneously at rest and $v_{cm}=\omega R$. The total energy of the object is entirely kinetic (translation of the centre of mass + rotation about it):

Step 2 — Express $I$ through the radius of gyration.

Write $I=mk^2$, where $k$ is the radius of gyration. Using $\omega=v_{cm}/R$,

So $E$ is directly proportional to $v_{cm}^{2}$ — which is exactly why the printed graph, whose $x$-axis is $v_{cm}^{2}$ (not $v_{cm}$), is a straight line through the origin. Its slope is

Step 3 — Read the slope off the graph.

The one marked point is $E=3\ \text{J}$ at $v_{cm}^{2}=4\ \mathrm{m^2\,s^{-2}}$, so

Step 4 — Solve for $k^{2}/R^{2}$.

With $m=1\ \text{kg}$,

Step 5 — Identify the body.

Standard values of $k^{2}/R^{2}$ for rolling bodies:

Only the disc reproduces the printed point $E=3$ J at $v_{cm}^2=4\ \mathrm{m^2 s^{-2}}$. (Note the ring and the hollow cylinder share $k^2/R^2=1$, so options (B) and (D) are in any case the same physical case and both give $4$ J.)

ANSWER: C

When a body rolls without slipping down an incline, its gravitational potential energy converts into two forms: translational kinetic energy and rotational kinetic energy. The steeper the rotational "cost" (i.e., the larger the moment of inertia relative to mass and radius), the less energy remains for translation — and therefore the lower the final speed.

The key quantity is the radius of gyration $k$, defined by $I = mk^2$. For pure rolling, the ratio $k^2/R^2$ determines how much of the total kinetic energy goes into rotation. A larger $k^2/R^2$ means more energy is "locked" in rotation, leaving less for translation.

1. Write the energy conservation equation. Starting from rest at height $h$, the loss in gravitational potential energy equals the total kinetic energy at the bottom:

For pure rolling without slipping, $\omega = v/R$. Substituting $I = mk^2$:

2. Solve for the final speed $v$. Cancel $m$ (mass does not matter) and rearrange:

Since $g$ and $h$ are the same for all bodies, the final speed depends only on $k^2/R^2$ — and it decreases as this ratio increases.

3. Compare the values of $k^2/R^2$ for the three bodies.

   • Ring: $I = mR^2 \Rightarrow k^2 = R^2 \Rightarrow \frac{k^2}{R^2} = 1$
   • Solid cylinder: $I = \frac{1}{2}mR^2 \Rightarrow k^2 = \frac{R^2}{2} \Rightarrow \frac{k^2}{R^2} = \frac{1}{2}$
   • Solid sphere: $I = \frac{2}{5}mR^2 \Rightarrow k^2 = \frac{2}{5}R^2 \Rightarrow \frac{k^2}{R^2} = \frac{2}{5} = 0.4$

   The ring has the largest $k^2/R^2$, so it has the smallest $v$.

The core rule in dimensional analysis is simple: you can only add or subtract quantities that have the same dimensions. Multiplication and division, on the other hand, are always allowed — they just produce a new dimension. So when you see a minus or plus sign between two terms, check whether those terms have identical dimensions. If they don’t, the expression is physically meaningless.

Let’s go through each option one by one.

1. Option (A): $\frac{P-Q}{R}$

   The numerator contains $P - Q$. Since $P$ and $Q$ have different dimensions, this subtraction is not allowed. The expression is invalid from the start — no matter what $R$ is.

   So (A) can never be a meaningful quantity.

2. Option (B): $PQ - R$

   Here $PQ$ is a product of $P$ and $Q$, so it has some new dimension. $R$ has its own dimension. For the subtraction $PQ - R$ to be meaningful, $PQ$ and $R$ must have the same dimension. Is that possible? Yes — if the dimensions of $P$, $Q$, and $R$ happen to satisfy $[P][Q] = [R]$, then the subtraction is valid. Since the problem only says they have different dimensions, it doesn’t forbid this relation. So (B) can be meaningful for some choices.

3. Option (C): $\frac{PQ}{R}$

   This is just multiplication and division — no addition or subtraction. It is always dimensionally valid. The result will have dimensions $[P][Q]/[R]$, which is fine. So (C) is always meaningful.

4. Option (D): $\frac{PR - Q^2}{R}$

   Look at the numerator: $PR - Q^2$. For this to be valid, $PR$ and $Q^2$ must have the same dimension. That means $[P][R] = [Q]^2$. This is a condition that could be satisfied by some set of quantities with different dimensions. So (D) can be meaningful for appropriate choices.