Rotational Motion MCQs with Answers – Part 1 (NEET Physics)

Explanation: For two particles, the centre-of-mass coordinate is \[ x_{\rm CM}=\frac{m_1x_1+m_2x_2}{m_1+m_2}. \] The coordinates are therefore weighted according to the corresponding masses rather than simply averaged. If \(m_1=m_2\), the weights are equal and the expression reduces to the ordinary midpoint \((x_1+x_2)/2\). If the masses are unequal, the heavier particle contributes more strongly to the weighted mean. Hence the centre of mass shifts toward the heavier particle. For two positive masses, the weighted mean must lie between \(x_1\) and \(x_2\). It does not move outside the joining segment merely because the masses are unequal. Thus the decisive ideas are mass weighting, an internal location between the two particles, and greater proximity to the heavier mass.

Explanation: \(\textbf{Given:}\) \[ m_1=2\,\text{kg},\quad x_1=-2\,\text{m},\quad m_2=6\,\text{kg},\quad x_2=4\,\text{m} \] \(\textbf{Centre-of-mass relation:}\) \[ x_{\rm CM}=\frac{m_1x_1+m_2x_2}{m_1+m_2} \] The negative sign of \(x_1\) must be retained because \(x_1\) is a signed coordinate. \(\textbf{Substitution:}\) \[ x_{\rm CM}=\frac{2(-2)+6(4)}{2+6} \] \[ x_{\rm CM}=\frac{-4+24}{8}=\frac{20}{8}=2.5\,\text{m} \] \(\textbf{Physical check:}\) The result lies between \(-2\,\text{m}\) and \(4\,\text{m}\), as it must for two positive masses. It is also much closer to the \(6\,\text{kg}\) particle, which is consistent with the greater weighting of the heavier particle. \(\textbf{Final answer:}\) \[ \boxed{x_{\rm CM}=2.5\,\text{m}} \]

Explanation: \(\textbf{Vector relation:}\) For two particles, \[ \vec R_{\rm CM}=\frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2} \] Here, \[ \vec r_1=4\hat i\,\text{m},\qquad \vec r_2=4\hat j\,\text{m} \] \(\textbf{Substitution:}\) \[ \vec R_{\rm CM} =\frac{m(4\hat i)+3m(4\hat j)}{m+3m} \] \[ \vec R_{\rm CM} =\frac{4m\hat i+12m\hat j}{4m} \] Therefore, \[ \vec R_{\rm CM}=(\hat i+3\hat j)\,\text{m} \] The \(x\)-coordinate is only \(1\,\text{m}\) because only the lighter particle contributes to the positive \(x\)-coordinate. The \(y\)-coordinate is \(3\,\text{m}\) because the \(3m\) particle has the larger weighting in that direction. \(\textbf{Final answer:}\) \[ \boxed{\vec R_{\rm CM}=(\hat i+3\hat j)\,\text{m}} \]

Explanation: \(\textbf{Known positions:}\) \[ x_1=0,\qquad x_2=12\,\text{m},\qquad x_{\rm CM}=4\,\text{m} \] \(\textbf{Governing relation:}\) \[ x_{\rm CM}=\frac{m_1x_1+m_2x_2}{m_1+m_2} \] \(\textbf{Substitution:}\) \[ 4=\frac{m_1(0)+12m_2}{m_1+m_2} \] Multiplying by \(m_1+m_2\), \[ 4m_1+4m_2=12m_2 \] Hence, \[ 4m_1=8m_2, \] so \[ \frac{m_1}{m_2}=2 \] Thus, \[ m_1:m_2=2:1 \] \(\textbf{Physical check:}\) The centre of mass is \(4\,\text{m}\) from \(m_1\) but \(8\,\text{m}\) from \(m_2\), so it is nearer \(m_1\); therefore \(m_1\) should indeed be the heavier particle.

Explanation: \(\textbf{Change produced by moving one particle:}\) If only one particle is displaced, the corresponding centre-of-mass displacement is its mass fraction times its displacement. The total mass is \[ M=3+2=5\,\text{kg} \] For the \(2\,\text{kg}\) particle, \[ \Delta x_{\rm CM}=\frac{2}{5}(10\,\text{cm})=4\,\text{cm} \] Hence, in the original coordinate system, the centre of mass moves from \(5\,\text{cm}\) to \[ x_{\rm CM,new}=5+4=9\,\text{cm} \] \(\textbf{Effect of changing the origin:}\) The new origin is \(3\,\text{cm}\) to the right of the old one, so the coordinate of every fixed physical point decreases by \(3\,\text{cm}\). Thus, \[ x'_{\rm CM}=9-3=6\,\text{cm} \] The first operation physically moves the centre of mass, whereas the second operation only changes its numerical coordinate. \(\textbf{Final answer:}\) \[ \boxed{x'_{\rm CM}=6\,\text{cm}} \]

Explanation: \(\textbf{Initial configuration:}\) The initial centre of mass is \[ x_{\rm CM}=\frac{m(-6)+2m(3)}{m+2m} \] \[ x_{\rm CM}=\frac{-6m+6m}{3m}=0 \] Thus the original mass moments about the origin cancel. \(\textbf{Changed mass distribution:}\) The mass at \(x=-6\,\text{m}\) becomes \(2m\), while the other mass remains \(2m\). Therefore, \[ x'_{\rm CM} =\frac{2m(-6)+2m(3)}{2m+2m} \] \[ x'_{\rm CM} =\frac{-12m+6m}{4m} =-\frac{6}{4}\,\text{m} \] Hence, \[ x'_{\rm CM}=-1.5\,\text{m} \] The centre of mass moves toward the negative side because additional mass has been introduced at the negative coordinate. \(\textbf{Final answer:}\) \[ \boxed{x'_{\rm CM}=-1.5\,\text{m}} \]

Explanation: Let the centre of mass divide the line joining the particles into distances \(d_1\) from \(m_1\) and \(d_2\) from \(m_2\). Here, \[ d_1=\frac{L}{5} \] Since the total separation is \(L\), \[ d_2=L-\frac{L}{5}=\frac{4L}{5} \] For two particles, the centre of mass satisfies the inverse mass-distance relation \[ m_1d_1=m_2d_2 \] Substituting the two distances, \[ m_1\left(\frac{L}{5}\right)=m_2\left(\frac{4L}{5}\right) \] Cancelling \(L/5\), \[ m_1=4m_2 \] Therefore, \[ m_1:m_2=4:1 \] This is also physically consistent because the centre of mass is much closer to \(m_1\), so \(m_1\) must be heavier.

Explanation: For the first change, let the original particle position vectors be \(\vec r_1\) and \(\vec r_2\). After translating both particles by \(\vec a\), their new position vectors are \(\vec r_1+\vec a\) and \(\vec r_2+\vec a\). Therefore, \[ \vec R'_{\rm CM} =\frac{m_1(\vec r_1+\vec a)+m_2(\vec r_2+\vec a)}{m_1+m_2}. \] Separating the terms gives \[ \vec R'_{\rm CM}=\vec R_{\rm CM}+\vec a. \] Thus Statement I is correct because the physical system itself has been translated. For Statement II, the particles do not move; only the coordinate origin is shifted through \(\vec a\). With parallel axes, a fixed position vector changes according to \[ \vec r'=\vec r-\vec a. \] Consequently, \[ \vec R'_{\rm CM}=\vec R_{\rm CM}-\vec a, \] not \(\vec R_{\rm CM}+\vec a\). The physical centre of mass is unchanged in Statement II; only its coordinate description changes. Hence Statement II is incorrect.

Explanation: \(\textbf{Represent the unknown mass ratio:}\) Let \[ r=\frac{m_2}{m_1} \] \(\textbf{Initial configuration:}\) Since \(m_1\) is at the origin, \[ 6=\frac{m_2a}{m_1+m_2} \] Dividing numerator and denominator by \(m_1\), \[ 6=\frac{ra}{1+r} \] Hence, \[ ar=6(1+r) \] \(\textbf{After doubling the second mass:}\) The mass at \(x=a\) becomes \(2m_2\), so \[ 8=\frac{2m_2a}{m_1+2m_2} =\frac{2ra}{1+2r} \] Therefore, \[ ar=4(1+2r) \] Equating the two expressions for \(ar\), \[ 6(1+r)=4(1+2r) \] \[ 6+6r=4+8r \] Thus, \[ r=1 \] So the original masses were equal. Using \(r=1\) in the initial relation, \[ 6=\frac{a}{2}, \] which gives \[ a=12\,\text{m} \] \(\textbf{Final answer:}\) \[ \boxed{m_2:m_1=1:1,\qquad a=12\,\text{m}} \]

Explanation: \(\textbf{Initial centre of mass:}\) \[ \vec R_i=(\hat i-\hat j)\,\text{m} \] The displacement of the centre of mass equals the mass-weighted mean of the individual displacements. Thus, \[ \Delta\vec R_{\rm CM} =\frac{2m\,\Delta\vec r_1+3m\,\Delta\vec r_2}{2m+3m} \] \(\textbf{Substitute the two displacements:}\) \[ \Delta\vec R_{\rm CM} =\frac{2(5\hat i-5\hat j)+3(-5\hat i+5\hat j)}{5} \] The numerator becomes \[ (10\hat i-10\hat j)+(-15\hat i+15\hat j) =-5\hat i+5\hat j \] Hence, \[ \Delta\vec R_{\rm CM}=(-\hat i+\hat j)\,\text{m} \] \(\textbf{Final position:}\) \[ \vec R_f=\vec R_i+\Delta\vec R_{\rm CM} \] \[ \vec R_f=(\hat i-\hat j)+(-\hat i+\hat j)=\vec 0 \] Option B represents the displacement of the centre of mass, not its final position. \(\textbf{Final answer:}\) The centre of mass finally lies at the origin.

Explanation: The centre of mass is \[ x_{\rm CM}=\frac{m_1x_1+m_2x_2}{m_1+m_2}. \] For positive masses, this is a weighted mean of \(x_1\) and \(x_2\), so it lies between the two coordinates; therefore Statement I is correct. If one mass is increased while the positions remain fixed, that coordinate receives a larger statistical weight in the numerator, shifting the centre of mass toward that particle; therefore Statement II is correct. If both masses are multiplied by the same positive factor \(k\), then \[ x'_{\rm CM}=\frac{km_1x_1+km_2x_2}{{k(m_1+m_2)}}=x_{\rm CM}, \] so Statement III is correct. If both coordinates are replaced by their negatives, \[ x'_{\rm CM} =\frac{-m_1x_1-m_2x_2}{m_1+m_2} =-x_{\rm CM}. \] Thus the coordinate generally changes sign rather than remaining unchanged. Statement IV is therefore incorrect. Hence only Statements I, II and III are correct.

Explanation: \(\textbf{Infer the original mass relation:}\) Let the original masses of \(P\) and \(Q\) be \(m_P\) and \(m_Q\). The initial centre of mass is at the origin, so \[ m_P(-a)+m_Q(2a)=0 \] Cancelling \(a\), \[ -m_P+2m_Q=0, \] hence \[ m_P=2m_Q \] Let \(m_Q=m\). Then initially \(m_P=2m\). \(\textbf{Changed configuration:}\) A further mass \(m\) is attached at \(P\), so the total mass at \(x=-a\) becomes \[ 2m+m=3m \] The mass at \(x=+2a\) remains \(m\). The new centre of mass is therefore \[ x'_{\rm CM} =\frac{3m(-a)+m(2a)}{3m+m} \] \[ x'_{\rm CM} =\frac{-3ma+2ma}{4m} =-\frac{a}{4} \] The negative sign is physically sensible because additional mass was placed on the negative-\(x\) side. \(\textbf{Final answer:}\) \[ \boxed{x'_{\rm CM}=-\frac{a}{4}} \]

Explanation: The centre of mass is determined by the complete mass distribution and need not coincide with a point occupied by matter. A homogeneous circular ring possesses symmetry about its geometric centre. Consider any small element of mass \(dm\) at position vector \(\vec r\) measured from the centre. Homogeneity and circular symmetry guarantee an equal element \(dm\) at the diametrically opposite position \(-\vec r\). The pair contributes \[ dm\,\vec r+dm(-\vec r)=\vec 0 \] to the mass-weighted position sum. The same cancellation occurs for every such pair around the ring. Consequently the total first moment of mass about the geometric centre is zero, locating the centre of mass there. The fact that the geometric centre is empty space does not invalidate the result because centre of mass is a point characterising the distribution, not necessarily a material particle. Thus both statements are true, and the Reason directly explains the Assertion.

Explanation: \(\textbf{Use the rod's symmetry:}\) The rod is uniform and centred at the origin, so its own centre of mass is at \[ x_{\rm rod}=0 \] Its contribution to the first moment about the origin is therefore \[ (4M)(0)=0 \] \(\textbf{Moment due to the attached mass:}\) The \(2M\) mass at \(x=+2a\) contributes \[ (2M)(2a)=4Ma \] Let the additional mass \(M\) be placed at coordinate \(x\). For the final centre of mass to be at the origin, the total mass-weighted position numerator must vanish: \[ (4M)(0)+(2M)(2a)+Mx=0 \] Thus, \[ 4Ma+Mx=0 \] Cancelling \(M\), \[ x=-4a \] The required mass must be placed on the opposite side of the origin because it must cancel the positive first moment produced by the \(2M\) mass. \(\textbf{Final answer:}\) \[ \boxed{x=-4a} \]

Explanation: \(\textbf{Original lamina:}\) The original centre of mass is at the origin, so its total first moment about the origin is \[ (10M)(0)=0 \] \(\textbf{After removing the circular piece:}\) The removed piece has moment \[ (2M)(2a)=4Ma \] Removing it leaves the remaining lamina with moment \[ 0-4Ma=-4Ma \] Its remaining mass is \[ 10M-2M=8M \] Thus, before refilling, its centre of mass would be at \[ x_{\rm rem}=\frac{-4Ma}{8M}=-\frac{a}{2} \] \(\textbf{Refill with half-density material:}\) Because the replacement material has half the original density but occupies the same volume, the refill mass is \[ M \] The refill is centred at \(x=+2a\), so it contributes the moment \[ M(2a)=2Ma \] The final mass is \[ 8M+M=9M, \] and the final first moment is \[ -4Ma+2Ma=-2Ma \] Therefore, \[ x_{\rm CM,final}=\frac{-2Ma}{9M}=-\frac{2a}{9} \] The negative sign is expected because the refilling restores only half of the mass that was originally removed from the positive side. \(\textbf{Final answer:}\) \[ \boxed{x_{\rm CM,final}=-\frac{2a}{9}} \]

Explanation: \(\textbf{Initial composite centre of mass:}\) By symmetry, the centre of mass of the uniform rod is at its geometric centre \(x=0\), and that of the uniform ring is at its own centre \(x=4a\). Hence, \[ x_{\rm CM,i}=\frac{3M(0)+M(4a)}{3M+M} \] \[ x_{\rm CM,i}=\frac{4Ma}{4M}=a \] The new coordinate origin is therefore the physical point whose original coordinate is \(x=a\). \(\textbf{After coating the ring:}\) The rod remains \(3M\), while the uniformly coated ring now has mass \(2M\) and still has its centre at \(x=4a\). Thus, in the original coordinate system, \[ x_{\rm CM,f} =\frac{3M(0)+2M(4a)}{3M+2M} \] \[ x_{\rm CM,f} =\frac{8Ma}{5M} =\frac{8a}{5} \] \(\textbf{Convert to the shifted coordinate system:}\) Since the shifted origin is at old coordinate \(a\), \[ x'_{\rm CM,f}=x_{\rm CM,f}-a \] \[ x'_{\rm CM,f}=\frac{8a}{5}-\frac{5a}{5}=\frac{3a}{5} \] Option A is the correct final coordinate only in the original frame, not in the shifted frame requested. \(\textbf{Final answer:}\) \[ \boxed{x'_{\rm CM,f}=\frac{3a}{5}} \]

Explanation: \(\textbf{Changed mass distribution:}\) After the stated changes, the masses are \(4m\) and \(2m\). The first particle moves from \(-2a\) to \[ x_1=-2a+d \] The second particle remains at \[ x_2=3a \] Hence the new centre-of-mass coordinate is \[ x_{\rm CM}=\frac{4m(-2a+d)+2m(3a)}{4m+2m} \] Simplifying the numerator, \[ x_{\rm CM}=\frac{-8ma+4md+6ma}{6m} \] Therefore, \[ x_{\rm CM}=\frac{4d-2a}{6} \] \(\textbf{Threshold condition:}\) The least displacement corresponds to the centre of mass being exactly at the origin: \[ x_{\rm CM}=0 \] Thus, \[ 4d-2a=0, \] which gives \[ d=\frac{a}{2} \] For any larger \(d\), the numerator becomes positive and the centre of mass lies to the right of the origin. \(\textbf{Final answer:}\) \[ \boxed{d_{\min}=\frac{a}{2}} \]

Explanation: Let the fixed position vectors of particles \(A\) and \(B\) be \(\vec r_A\) and \(\vec r_B\). \(\textbf{Initial mass weighting:}\) Initially the total mass is \(3m\), so \[ \frac{m\vec r_A+2m\vec r_B}{3m}=2\hat i+\hat j \] Therefore, \[ \vec r_A+2\vec r_B=6\hat i+3\hat j. \tag{1} \] \(\textbf{After the mass increase at A:}\) The total mass becomes \(5m\), while both position vectors remain unchanged: \[ \frac{3m\vec r_A+2m\vec r_B}{5m} =\frac65\hat i+\frac95\hat j \] Multiplying by \(5\), \[ 3\vec r_A+2\vec r_B=6\hat i+9\hat j. \tag{2} \] Subtracting Eq. (1) from Eq. (2), \[ 2\vec r_A=6\hat j \] Hence, \[ \vec r_A=3\hat j\,\text{m} \] Substitution back into Eq. (1) would give \(\vec r_B=3\hat i\,\text{m}\), which also reproduces both stated centre-of-mass positions. \(\textbf{Final answer:}\) \[ \boxed{\vec r_A=3\hat j\,\text{m}} \]

Explanation: The masses are unchanged between the two configurations, so the first configuration can be used to determine their mass ratio. \(\textbf{Extract the mass ratio from Configuration I:}\) \[ 2=\frac{m_1(-2)+m_2(4)}{m_1+m_2} \] Multiplying through, \[ 2m_1+2m_2=-2m_1+4m_2 \] Hence, \[ 4m_1=2m_2, \] so \[ m_2=2m_1 \] \(\textbf{Use the same masses in Configuration II:}\) \[ 5=\frac{m_1(1)+2m_1X}{m_1+2m_1} \] Therefore, \[ 5=\frac{1+2X}{3} \] Thus, \[ 15=1+2X, \] and \[ X=7\,\text{m} \] A useful check is that the heavier particle is at \(7\,\text{m}\), so the centre of mass at \(5\,\text{m}\) is correctly closer to that particle than to the \(1\,\text{m}\) particle. \(\textbf{Final answer:}\) \[ \boxed{X=7\,\text{m}} \]

Explanation: \(\textbf{Initial centre of mass:}\) With \(m_1\) at \(0\) and \(m_2\) at \(L\), \[ x_i=\frac{m_2L}{m_1+m_2} \] \(\textbf{After interchanging the positions:}\) Now \(m_2\) is at \(0\) and \(m_1\) is at \(L\), so \[ x_f=\frac{m_1L}{m_1+m_2} \] Since \(m_2\gt m_1\), the displacement magnitude of the centre of mass is \[ x_i-x_f=\frac{(m_2-m_1)L}{m_1+m_2} \] The given displacement is \(L/5\), hence \[ \frac{m_2-m_1}{m_1+m_2}=\frac15 \] Multiplying through, \[ 5m_2-5m_1=m_1+m_2 \] Therefore, \[ 4m_2=6m_1 \] Thus, \[ \frac{m_2}{m_1}=\frac32 \] The larger mass initially pulls the centre of mass toward \(x=L\); interchanging the particles shifts it symmetrically toward the opposite side. \(\textbf{Final answer:}\) \[ \boxed{m_2:m_1=3:2} \]