Moment of Inertia — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Moment of Inertia MCQs with step-by-step solutions (15 questions). Part of System of Particles and Rotational Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Moment of Inertia · medium · numerical
From a circular ring of mass $M$ and radius $R$, an arc corresponding to a $90^\circ$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is $K$ times $MR^2$. Then, the value of $K$ is
A. $\dfrac{3}{4}$ ✓ Correct
B. $\dfrac{7}{8}$
C. $\dfrac{1}{4}$
D. $\dfrac{1}{8}$
Solution: Moment of inertia of the full ring, $I=MR^2$. A $90^\circ$ sector (one-fourth of the ring) is removed, so the removed mass is $M'=\dfrac{M}{4}$ with moment of inertia $I'=M'R^2=\dfrac{MR^2}{4}$. Hence the remaining part has $I''=I-I'=MR^2-\dfrac{MR^2}{4}=\dfrac{3MR^2}{4}$. Comparing with $I''=KMR^2$ gives $K=\dfrac{3}{4}$.
Q2 — Moment of Inertia · medium · numerical
From a disc of radius $R$ and mass $M$, a circular hole of diameter $R$, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?
A. $\dfrac{13MR^2}{32}$ ✓ Correct
B. $\dfrac{11MR^2}{32}$
C. $\dfrac{9MR^2}{32}$
D. $\dfrac{15MR^2}{32}$
Solution: $I_{\text{remain}}=I-I_{(R/2)}$. For the full disc $I=\dfrac{MR^2}{2}$. The removed disc of radius $\dfrac{R}{2}$ has mass $\dfrac{M}{4}$, and by the parallel axes theorem about the centre $I_{(R/2)}=\dfrac{\frac{M}{4}\left(\frac{R}{2}\right)^2}{2}+\dfrac{M}{4}\left(\dfrac{R}{2}\right)^2=\dfrac{MR^2}{32}+\dfrac{MR^2}{16}=\dfrac{3MR^2}{32}$. Hence $I_{\text{remain}}=\dfrac{MR^2}{2}-\dfrac{3MR^2}{32}=\dfrac{16MR^2-3MR^2}{32}=\dfrac{13MR^2}{32}$.
Q3 — Moment of Inertia · medium · numerical
A light rod of length $l$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
A. $\dfrac{m_1 m_2}{m_1+m_2}l^2$ ✓ Correct
B. $\dfrac{m_1+m_2}{m_1 m_2}l^2$
C. $(m_1+m_2)l^2$
D. $\sqrt{m_1 m_2}\,l^2$
Solution: The centre of mass divides the rod so that the distances of the masses from it are $r_1=\dfrac{m_2 l}{m_1+m_2}$ and $r_2=\dfrac{m_1 l}{m_1+m_2}$. Then $I=m_1 r_1^2+m_2 r_2^2=m_1\left(\dfrac{m_2 l}{m_1+m_2}\right)^2+m_2\left(\dfrac{m_1 l}{m_1+m_2}\right)^2=\dfrac{m_1 m_2 l^2}{(m_1+m_2)^2}(m_2+m_1)=\dfrac{m_1 m_2}{m_1+m_2}l^2$.
Q4 — Moment of Inertia · medium · numerical
Three identical spherical shells, each of mass $m$ and radius $r$ are placed as shown in figure. Consider an axis $XX'$, which is touching two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about $XX'$ axis is
A. $\dfrac{11}{5}mr^2$
B. $3mr^2$
C. $\dfrac{16}{5}mr^2$
D. $4mr^2$ ✓ Correct
Solution: $I=I_1+I_2+I_3$. For the shell whose diameter lies on the axis, $I_1=\dfrac{2}{3}mr^2$. For each of the two shells only touched by the axis, by the parallel axes theorem $I_2=I_3=\dfrac{2}{3}mr^2+mr^2=\dfrac{5}{3}mr^2$. Hence $I=\dfrac{2}{3}mr^2+2\times\dfrac{5}{3}mr^2=mr^2\left(\dfrac{2}{3}+\dfrac{10}{3}\right)=4mr^2$.
Q5 — Moment of Inertia · medium · numerical
The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its mid-point and perpendicular to its length is $I_0$. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is
A. $I_0+\dfrac{ML^2}{4}$ ✓ Correct
B. $I_0+2ML^2$
C. $I_0+ML^2$
D. $I_0+\dfrac{ML^2}{2}$
Solution: By the parallel axes theorem, $I=I_{\text{CM}}+Mh^2$. Shifting the axis from the mid-point to an end gives $h=\dfrac{L}{2}$, so $I=I_0+M\left(\dfrac{L}{2}\right)^2=I_0+\dfrac{ML^2}{4}$.
Q6 — Moment of Inertia · medium · numerical
Four identical thin rods each of mass $M$ and length $l$, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is
A. $\dfrac{4}{3}Ml^2$ ✓ Correct
B. $\dfrac{2}{3}Ml^2$
C. $\dfrac{13}{3}Ml^2$
D. $\dfrac{1}{3}Ml^2$
Solution: For one rod about the axis through the centre of the square, using the parallel axes theorem, $I_{\text{rod}}=\dfrac{Ml^2}{12}+M\left(\dfrac{l}{2}\right)^2=\dfrac{Ml^2}{3}$. Summing over the four rods of the frame, $I=\dfrac{Ml^2}{3}\times4=\dfrac{4}{3}Ml^2$.
Q7 — Moment of Inertia · medium · numerical
A thin rod of length $L$ and mass $M$ is bent at its mid-point into two halves so that the angle between them is $90^\circ$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is
A. $\dfrac{ML^2}{24}$
B. $\dfrac{ML^2}{12}$ ✓ Correct
C. $\dfrac{ML^2}{6}$
D. $\dfrac{\sqrt{2}ML^2}{24}$
Solution: Each half has mass $\dfrac{M}{2}$ and length $\dfrac{L}{2}$. The moment of inertia of each half about an axis through its end (the bending point) and perpendicular to it is $\dfrac{\text{mass}\times(\text{length})^2}{3}$. Hence $I=2\times\dfrac{1}{3}\times\dfrac{M}{2}\times\left(\dfrac{L}{2}\right)^2=\dfrac{ML^2}{12}$.
Q8 — Moment of Inertia · medium · numerical
The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is
A. $\sqrt{3}:\sqrt{2}$
B. $1:\sqrt{2}$ ✓ Correct
C. $\sqrt{2}:1$
D. $\sqrt{2}:\sqrt{3}$
Solution: Radius of gyration $k=\sqrt{\dfrac{I}{m}}$. So $\dfrac{k_{\text{ring}}}{k_{\text{disc}}}=\sqrt{\dfrac{I_{\text{ring}}}{I_{\text{disc}}}}=\sqrt{\dfrac{MR^2}{\frac{1}{2}MR^2}}=\sqrt{2}$, hence $\dfrac{k_{\text{disc}}}{k_{\text{ring}}}=\dfrac{1}{\sqrt{2}}$, i.e. the disc-to-ring ratio is $1:\sqrt{2}$.
Q9 — Moment of Inertia · medium · numerical
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is
A. $\dfrac{1}{2}MR^2$
B. $MR^2$
C. $\dfrac{7}{2}MR^2$
D. $\dfrac{3}{2}MR^2$ ✓ Correct
Solution: The moment of inertia of the disc about its central axis (normal to its plane, through the centre) is $\dfrac{1}{2}MR^2$. By the parallel axes theorem, about a parallel normal axis through the edge (distance $R$): $I=\dfrac{1}{2}MR^2+MR^2=\dfrac{3}{2}MR^2$.
Q10 — Moment of Inertia · medium · numerical
Three particles, each of mass $m$ grams situated at the vertices of an equilateral $\triangle ABC$ of side $l$ cm (as shown in the figure). The moment of inertia of the system about a line $AX$ perpendicular to $AB$ and in the plane of $ABC$ in g-cm$^2$ units will be
A. $\left(\dfrac{3}{4}\right)ml^2$
B. $2ml^2$
C. $\left(\dfrac{5}{4}\right)ml^2$ ✓ Correct
D. $\left(\dfrac{3}{2}\right)ml^2$
Solution: The perpendicular distances of the vertices from the axis $AX$ are $r_A=0$, $r_B=l$ and $r_C=l\sin30^\circ$. Then $I=m_A r_A^2+m_B r_B^2+m_C r_C^2=m(0)^2+m(l)^2+m(l\sin30^\circ)^2=ml^2+\dfrac{ml^2}{4}=\dfrac{5}{4}ml^2$.
Q11 — Moment of Inertia · medium · numerical
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is
A. $2:3$
B. $2:1$
C. $\sqrt{5}:\sqrt{6}$ ✓ Correct
D. $1:\sqrt{2}$
Solution: Moment of inertia of a disc about a tangential axis in its plane, $I_d=\dfrac{5}{4}M_d R^2$, and of a ring about a tangential axis in its plane, $I_r=\dfrac{3}{2}M_r R^2$. Since $I=Mk^2\Rightarrow k=\sqrt{\dfrac{I}{M}}$, so $\dfrac{k_d}{k_r}=\sqrt{\dfrac{I_d}{I_r}\times\dfrac{M_r}{M_d}}=\sqrt{\dfrac{(5/4)M_d R^2}{(3/2)M_r R^2}\times\dfrac{M_r}{M_d}}=\sqrt{\dfrac{5}{6}}$. $\therefore k_d:k_r=\sqrt{5}:\sqrt{6}$.
Q12 — Moment of Inertia · medium · theory
A circular disc is to be made using iron and aluminium. To keep its moment of inertia maximum about a geometrical axis, it should be so prepared that
A. aluminium is at the interior and iron surrounds it ✓ Correct
B. iron is at the interior and aluminium surrounds it
C. aluminium and iron layers are in alternate order
D. sheet of iron is used at both external surfaces and aluminium sheet as inner material
Solution: Moment of inertia depends on the distribution of mass about the axis of rotation. Density of iron is more than that of aluminium, therefore for the moment of inertia to be maximum the iron should be far away from the axis. Thus, aluminium should be at the interior and iron surrounds it.
Q13 — Moment of Inertia · medium · theory
$ABC$ is a right angled triangular plate of uniform thickness. The sides are such that $AB>BC$ as shown in figure. $I_1, I_2, I_3$ are moments of inertia about $AB, BC$ and $AC$ respectively. Then, which of the following relations is correct?
A. $I_1=I_2=I_3$
B. $I_2>I_1>I_3$ ✓ Correct
C. $I_3<I_2<I_1$
D. $I_3>I_1>I_2$
Solution: The moment of inertia of a body about an axis depends not only on the mass but also on the distribution of mass from the axis. For a given body the mass is same, so it depends only on the distribution of mass from the axis. The mass is farthest from axis $BC$, so $I_2$ is maximum; the mass is nearest to axis $AC$, so $I_3$ is minimum. Hence the correct sequence is $I_2>I_1>I_3$.
Q14 — Moment of Inertia · medium · numerical
The moment of inertia of a disc of mass $M$ and radius $R$ about a tangent to its rim in its plane is
A. $\dfrac{2}{3}MR^2$
B. $\dfrac{3}{2}MR^2$
C. $\dfrac{4}{5}MR^2$
D. $\dfrac{5}{4}MR^2$ ✓ Correct
Solution: Moment of inertia of a disc about its diameter is $I_d=\dfrac{1}{4}MR^2$. By the perpendicular axis theorem, the moment of inertia about a tangent passing through the rim and lying in the plane of the disc is $I=I_d+MR^2=\dfrac{1}{4}MR^2+MR^2=\dfrac{5}{4}MR^2$.
Q15 — Moment of Inertia · medium · theory
In a rectangle $ABCD$ ($BC=2AB$). The moment of inertia is minimum along axis through
A. $BC$
B. $BD$
C. $HF$
D. $EG$ ✓ Correct
Solution: The magnitude of the moment of inertia depends on the distribution of mass from the axis. About the axis $EG$ the mass lies at the smallest average distance, whereas about $BD$ it lies at the greatest distance. Hence the moment of inertia is minimum about the axis through $EG$.