Q.A hollow spherical shell of radius $R$ rests with its lower half embedded in sand and its upper half exposed to air; the interior of the lower hemisphere is packed with sand while the interior of the upper hemisphere contains only air. Four candidate points lie on the vertical diameter of the shell: point A lies slightly above the centre, point B lies exactly at the geometric centre, point C lies slightly below the centre (just inside the sand-filled region), and point D lies well below the centre near the bottom of the shell. Which of these points is the likely position of the centre of mass of the whole system?

Step 1 — Identify the forces at the junction O.

Three strings meet at $O$, so $O$ is a massless point in equilibrium under three tensions:

• $T_1$ — along the horizontal string that runs to the block on the floor (it pulls $O$ towards that block, i.e. horizontally to the left).
• $T_2$ — along the inclined string going up to the wall at angle $\theta$ above the horizontal (it pulls $O$ up and to the right).
• $T_3$ — along the vertical string, pulling $O$ straight down. The hanging block of mass $M$ is in equilibrium, so

Step 2 — Equilibrium of the junction O.

Vertical direction: only the inclined string has an upward component, and it must hold up the hanging weight.

Horizontal direction: the inclined string's horizontal component is balanced by the horizontal string.

Step 3 — Equilibrium of the block on the floor.

The horizontal string pulls this block towards $O$ with force $T_1 = Mg\cot\theta$. The only force that can oppose it is static friction from the floor. The string is horizontal, so it adds no vertical component and the normal reaction is just the block's own weight:

Friction available at most is $f_{\max} = \mu N = \mu Mg$. For the block to just stay in equilibrium (the limiting case the question asks for):

Step 4 — Solve.

Sanity check: as $\theta \to 90^\circ$ the inclined string becomes vertical, it pulls $O$ straight up, the horizontal string goes slack, and indeed $\cot 90^\circ = 0$ — no friction needed. As $\theta$ becomes small the inclined string is nearly horizontal and must be pulled very taut, demanding a huge $\mu$ — and $\cot\theta \to \infty$. The formula behaves correctly at both limits.

ANSWER: A

Step 1 — The concept and formula

The centre of mass of a system of discrete point masses is the mass-weighted average position:

It is a weighted average because each particle "pulls" the balance point toward itself in proportion to its mass. Note the answer depends entirely on our choice of origin — and the question fixes that for us: vertex A is the origin.

Step 2 — Set up the coordinates of the triangle

Take A at the origin and let side AB lie along the $+x$ axis. For an equilateral triangle of side $a = 1$ m:

• $\mathbf{A} = (0,\, 0)$ — carries $m_A = 1$ kg
• $\mathbf{B} = (a,\, 0) = (1,\, 0)$ — carries $m_B = 2$ kg
• $\mathbf{C} = \left(\dfrac{a}{2},\ \dfrac{\sqrt{3}}{2}a\right) = \left(\dfrac{1}{2},\ \dfrac{\sqrt{3}}{2}\right)$ — carries $m_C = 3$ kg

Why C sits there: C is equidistant from A and B, so it lies above the midpoint of AB, at $x = a/2$. Its height is found from Pythagoras on the half-triangle:

Step 3 — Total mass

Step 4 — The $x$-coordinate

Step 5 — The $y$-coordinate

Only C is off the x-axis, so only it contributes:

(Equivalently $\frac{\sqrt{3}}{4} \approx 0.433$ m.)

Step 6 — Sanity check

$x_{cm} = 7/12 \approx 0.583$ m. The heaviest mass (3 kg at C, $x = 0.5$) and the 2 kg at $x=1$ pull the centre right of the origin, but the 1 kg at A holds it back — a value a little past the middle is exactly what we expect. Option (D), the origin, would require all the mass at A, so it is clearly wrong.

ANSWER: A

The key idea is that the center of mass divides the bond in inverse proportion to the masses. The heavier atom (oxygen) is closer to the center of mass, so the lighter carbon atom is farther from it. We can find the exact distance by setting the torques (mass × distance) about the center of mass equal.

1. Define the system and variables

   Let the carbon atom have mass $m_C = 12$ u (atomic mass units) and the oxygen atom have mass $m_O = 16$ u. The bond length (distance between nuclei) is $d = 0.12$ nm. Let $x$ be the distance from the carbon atom to the center of mass. Then the distance from oxygen to the center of mass is $d - x$.

2. Apply the center of mass condition

   The center of mass is the point where the weighted distances balance:

This says the “mass × distance” on one side equals that on the other.

3. Solve for $x$

4. Interpret the result The carbon atom is about 0.07 nm from the center of mass, and oxygen is $0.12 - 0.07 = 0.05$ nm away. This makes sense: oxygen is heavier, so it sits closer to the balance point.

The centre of mass (COM) is a conceptual point where the entire mass of a system can be considered to be concentrated for many mechanical purposes. To decide which statement is correct, we need to recall its definition and properties.

Definition: For a system of particles, the position vector of the centre of mass is

where $m_i$ are the masses and $\mathbf{r}_i$ their positions. This formula makes clear that the COM depends on both the masses and their locations — i.e., the mass distribution.

Now evaluate each option:

1. Option (A): “Remains the same in any translatory motion.”

   • If the whole system undergoes pure translation (all particles move by the same displacement), the COM also moves by that displacement. So its position changes relative to a fixed origin. This statement is false unless “remains the same” means relative to the system itself — but the wording “in any translatory motion” implies motion relative to an external frame, so it’s incorrect.

2. Option (B): “Depends on the choice of coordinates.”

   • The numerical coordinates of the COM do change if you shift the origin or rotate axes, but the physical point in space is independent of coordinate choice. The statement is ambiguous; usually “depends on the choice of coordinates” is considered false because the COM is a physical property, not a coordinate artifact.

3. Option (C): “Does not depend on the shape and size of the system.”

   • Shape and size directly affect where masses are located. For example, a uniform rod’s COM is at its midpoint; bend it into a ring, and the COM shifts to the centre (outside the material). So shape and size matter — this is false.

4. Option (D): “Depends on the distribution of its mass.”

   • This is exactly what the definition says: different mass distributions (e.g., dense at one end vs. uniform) give different COM positions. This is true.

Concept. The centre of mass of a body with a piece removed is found by modelling the removed piece as a body of negative mass located at its own centre, then combining it with the full body.

Step 1 — Geometry and masses.

Original disc radius $2R \Rightarrow$ area $=\pi(2R)^2 = 4\pi R^2$.

The smaller disc has the larger disc's radius $2R$ as its diameter, so its radius is $R$ and area $=\pi R^2$. Its centre lies at distance $R$ from the centre $O$.

Taking surface mass density $\sigma$:

Step 2 — Set up coordinates.

Put $O$ at the origin and the hole's centre at $(R,0)$. The full disc's centre of mass is at the origin, and it equals the mass-weighted average of the remaining part and the removed part:

Step 3 — Solve for the remaining part.

The magnitude of the shift is $\dfrac{R}{3}$, directed away from the hole.

The centre of mass (COM) is a point that represents the average location of the total mass of a system. The key idea is that it is a physical property of the mass distribution itself — not a mathematical artifact of where we place our axes. So any statement that says the COM depends on the coordinate system is false, because shifting the origin just shifts the COM coordinates by the same amount; the point in space remains the same relative to the body.

Let’s examine each option carefully.

1. Option (A): “The centre of mass depends on the size and shape but does not depend on the distribution of mass of the body.” This is false. The COM formula is

for discrete masses, or

for continuous bodies. Both explicitly involve how mass is arranged (the distribution). For example, a uniform rod and a rod with a heavy end have the same size and shape but different COM positions. So distribution matters.

2. Option (B): “The centre of mass depends on the coordinate system.”

   This is a classic trap. If you change the coordinate system, the coordinates of the COM change, but the physical point in space does not. The COM is a property of the system, not of the axes. So this statement is false.

3. Option (C): “The centre of mass of a system depends on the size and shape of the body but independent of the co-ordinate system.”

   This is correct. The COM depends on how mass is spread (size, shape, distribution) but is independent of the coordinate system — because shifting or rotating axes just transforms the coordinates of the same point. The physical location relative to the body is fixed.

4. Option (D): “The centre of mass of a body always lies inside the body.”

   This is false. For example, a hollow ring or a horseshoe has its COM outside the material of the body. The COM can be in empty space.

For two particles of masses $m_1,m_2$ at coordinates $x_1,x_2$ from the origin, the centre of mass is

Option (A) has $x_2$ twice (a misprint) and (B), (D) are dimensionally wrong.

Concept: centre of mass of a body with a portion removed (negative-mass method).

The square is cut out with a RADIUS of the disc as its diagonal, so:

diagonal d = R -> side a = R/sqrt(2)

area of square = a^2 = R^2/2

The square's centre lies at the midpoint of that radius, i.e. at distance R/2 from the disc's centre.

Take surface density sigma = 1 (it cancels).

Whole disc: area = pi R^2, centre at x = 0

Removed square: area = R^2/2, centre at x = R/2

Centre of mass of the remaining portion (measured from the disc centre, on the side opposite the cut):

x_cm = [ (pi R^2)(0) - (R^2/2)(R/2) ] / [ pi R^2 - R^2/2 ]

= [ -R^3/4 ] / [ R^2 (pi - 1/2) ]

= -R / [ 4(pi - 1/2) ]

= -R / (4pi - 2)

Distance from the centre = R / (4pi - 2).

ANSWER: A

The key idea is that gravitational forces are vectors. The two 8 kg masses at A and B pull the 2 kg mass at the centroid G along the lines GA and GB. Because the triangle is equilateral and G is the centroid, GA = GB = GC = 1 m, and the angle between GA and GB is 120°. The resultant of these two equal forces lies along the median from C (i.e., along CG), pointing from G toward C. To cancel it, the 4 kg mass must be placed on the same line CG, but on the opposite side of G, so that its pull on the 2 kg mass is exactly opposite and equal in magnitude to that resultant.

1. Force from one 8 kg mass The gravitational force on the 2 kg mass due to a single 8 kg mass at distance 1 m is

So each 8 kg mass exerts a force of magnitude $16G$ along GA and GB respectively.

2. Resultant of the two forces The angle between GA and GB is 120° (since the centroid is equidistant from all vertices and the triangle is equilateral). The magnitude of the resultant of two equal forces $F$ with angle $\theta$ is

With $\cos 120^\circ = -1/2$,

So the resultant force on the 2 kg mass from the two 8 kg masses has magnitude $16G$ and points along the median from C toward C (since the two pulls are symmetric about CG).

3. Cancelling the resultant with the 4 kg mass Let the 4 kg mass be placed at a point P on the line CG, on the side opposite to C (i.e., beyond G away from C). The force it exerts on the 2 kg mass is

where $r = PG$ is the distance from G to P. This force must point from G away from C (i.e., opposite to the resultant $R$) and have the same magnitude $16G$:

4. Solve for $r$ Cancel $G$ (non-zero):

The positive root is taken because distance is positive.

Concept: centre of mass of discrete particles, R_cm = (sum m_i r_i)/(sum m_i).

From the figure, take O (the centre of the square, where the diagonals meet) as the origin, with the square of half-diagonal 1 unit. Corners:

A (top-left) = (-1, +1), mass 2m

B (top-right) = (+1, +1), mass 4m

C (bottom-right) = (+1, -1), mass 2m

D (bottom-left) = (-1, -1), mass 2m

Total mass = 10m.

x_cm = [2(-1) + 4(+1) + 2(+1) + 2(-1)]m / 10m = (-2 + 4 + 2 - 2)/10 = +0.2

y_cm = [2(+1) + 4(+1) + 2(-1) + 2(-1)]m / 10m = (2 + 4 - 2 - 2)/10 = +0.2

So the CM is at (0.2, 0.2), i.e. displaced from O along the direction (1, 1) - straight towards corner B, the heavier 4m corner. It lies on OB.

ANSWER: B

Step 1 — Why the perpendicular-axis theorem applies.

The theorem states that for a planar (lamina) body, the moment of inertia about an axis $z$ perpendicular to the plane equals the sum of the moments of inertia about two mutually perpendicular axes $x$ and $y$ lying in the plane and intersecting the $z$-axis:

A thin uniform rectangular plate is exactly such a lamina, and the figure shows $x$, $y$ in the plane meeting at $O$ with $z$ perpendicular through $O$ — every condition of the theorem is met.

Step 2 — Compute $I_z$.

Step 3 — Definition of the radius of gyration.

The radius of gyration $k$ about an axis is the distance from that axis at which the whole mass could be concentrated to give the same moment of inertia:

Step 4 — Substitute (mass $M = 2$ kg, axis = the $z$-axis through $O$).

Step 5 — Convert units (the step the distractors punish).

(Option (B) 5 cm is the decimal-slip trap; (D) 31.6 cm $\approx\sqrt{0.1}$ m would come from forgetting to add $I_x$ and $I_y$; (C) 38.7 cm comes from dividing by the wrong mass.)

ANSWER: A