Q.Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Showing the 12 most recent of 56 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A torque 'T' produces angular acceleration $\alpha$ in a circular disc. If radius of disc is doubled keeping the mass constant, angular acceleration becomes (A) $\alpha$ (B) $\alpha/2$ (C) $\alpha/4$ (D) $4\alpha$

   ›Reveal solutionSolution

   Doubling a disc's radius (mass fixed) quadruples its moment of inertia, so for the same applied torque the angular acceleration drops to a quarter. The answer is $\alpha/4$.

   Concept and Intuition

   Torque and angular acceleration are related by $T=I\alpha$, the rotational analogue of $F=ma$. For a solid disc, $I=\tfrac12MR^2$ depends on $R^2$, so any change in radius has an amplified (squared) effect on the moment of inertia — and hence an inverse-squared effect on $\alpha$ for a fixed torque.

   Step-by-Step Solution

   1. Original disc: $I=\tfrac12MR^2$, and $T=I\alpha \Rightarrow \alpha = \dfrac{T}{I} = \dfrac{2T}{MR^2}$.
   2. New disc: radius doubled ($R'=2R$), mass unchanged ($M$). New moment of inertia: $I'=\tfrac12M(2R)^2 = \tfrac12M(4R^2)=4\left(\tfrac12MR^2\right)=4I$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The acceleration of a uniform disc rolling down an inclined plane of length 1.75 m is $1/3$ times the acceleration due to gravity. If a solid sphere is rolling down from the top of the same inclined plane, then the velocity with which it reaches the bottom of the plane is (A) $2.5\ ms^{-1}$ (B) $5\ ms^{-1}$ (C) $3.5\ ms^{-1}$ (D) $7\ ms^{-1}$

   ›Reveal solutionSolution

   Tests rolling-body dynamics on an incline — first find the incline angle from the disc's given acceleration, then apply the same incline to a sphere. Answer: $3.5\ ms^{-1}$.

   Concept and Intuition

   A body rolling without slipping down an incline shares gravitational PE between translational and rotational KE. Its acceleration is reduced from $g\sin\theta$ by the factor $1+k^2/R^2$, where $k^2/R^2$ depends only on the shape (mass distribution), not on mass or radius. A disc ($k^2/R^2=1/2$) and a sphere ($k^2/R^2=2/5$) rolling down the same incline therefore have different accelerations, but both depend on the same $\sin\theta$ — so the disc's data lets us extract $\theta$, which we then reuse for the sphere.

   Step-by-Step Solution

   1. Uniform disc: $I=\frac12MR^2 \Rightarrow k^2/R^2=\frac12$. Acceleration down the incline: $a=\dfrac{g\sin\theta}{1+\frac12}=\dfrac{2}{3}g\sin\theta$.
   2. Given $a_{disc}=\dfrac{1}{3}g$: $\dfrac{2}{3}g\sin\theta=\dfrac13 g \Rightarrow \sin\theta=\dfrac12$ (i.e. $\theta=30^\circ$). …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A thin wire of length $L$ and uniform linear mass density '$\rho$' is bent into a circular loop with centre at 'O' as shown (diagram: a circle with an axis $XX^1$ drawn as a straight horizontal line tangent to the top of the circle; a dashed vertical radius runs from the centre O up to the point of tangency, meeting axis $XX^1$ at a right angle (90°) marked at that point). The moment of inertia of the loop about the axis $XX^1$ is (A) $\dfrac{\rho L^3}{8\pi^2}$ (B) $\dfrac{\rho L^3}{16\pi^2}$ (C) $\dfrac{5\rho L^3}{16\pi^2}$ (D) $\dfrac{3\rho L^3}{8\pi^2}$

   ›Reveal solutionSolution

   This tests the moment of inertia of a ring about an in-plane tangent axis, combining the standard diameter-axis result with the parallel axis theorem. The answer is $\dfrac{3\rho L^3}{8\pi^2}$.

   Concept and Intuition

   The figure shows axis $XX^1$ as a straight line tangent to the circle at its topmost point, lying in the same plane as the ring (the dashed radius meeting it at 90° confirms it's a tangent, not an axis perpendicular to the plane). For a ring, the moment of inertia about any diameter (an axis through the centre, in the plane of the ring) is $I_{d} = \frac{1}{2}mr^2$ — half of the perpendicular-axis value $mr^2$, by the perpendicular axis theorem applied to the two equal in-plane diameters. The tangent axis is parallel to a diameter but shifted by a distance $r$ (the radius) from the centre, so the parallel axis theorem applies directly.

   Step-by-Step Solution

   1. Circumference relation: $L = 2\pi r \Rightarrow r = \dfrac{L}{2\pi}$.
   2. Mass of the wire (loop): $m = \rho L$ (linear density times total length).
   3. Moment of inertia about a diameter (in-plane axis through centre): $I_d = \dfrac{1}{2}mr^2$. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The moment of Inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through which of the following points? [FIGURE: a circle with center A; point B on the circumference at the left end of the horizontal diameter; point D between A and the right edge; point C near the top on the vertical diameter] (A) B (B) C (C) D (D) A

   ›Reveal solutionSolution

   Moment of inertia about a perpendicular axis grows as $Md^2$ with the distance $d$ of the axis from the centre of mass, so it is maximum about the rim point B — option (A).

   Concept and Intuition

   Moment of inertia measures how the mass is spread out from the axis: $I = \sum m_i r_i^2$. Move the axis away from the centre of mass and every mass element (on average) sits farther from it, so $I$ must increase. The parallel-axis theorem makes this exact:

   $I = I_{cm} + M d^2,$

   where $d$ is the distance between the new axis and the parallel axis through the centre of mass. Note two consequences worth remembering:

   • $I$ is minimum about the axis through the centre of mass ($d=0$) — that is point A here.
   • $I$ increases monotonically with $d$, so among candidate points you simply pick the one farthest from the centre.

   Step-by-Step Solution

   1. For a uniform disc of mass $M$, radius $R$, about the perpendicular axis through its centre: $I_{cm} = \tfrac12 MR^2$.
   2. About a parallel perpendicular axis through a point at distance $d$: $I(d) = \tfrac12 MR^2 + M d^2$.
   3. Read the distances from the figure:
      • A: the centre, $d = 0$ → $I = \tfrac12 MR^2$ (the minimum).
      • D: a point inside the disc, a short distance right of the centre, $0 < d < R$.
      • C: on the vertical line above the centre, $d < R$ (or at most $R$). …

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5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$I_1$ represents moment of inertia of a thin, uniform rod about an axis perpendicular to its length and passing through its centre of mass. The same rod is bent into the shape of a ring. If $I_2$ is moment of inertia of ring about an axis that is tangent to the ring and perpendicular to its plane, then $\dfrac{I_1}{I_2}=$ (A) $\dfrac{\pi^2}{6}$ (B) $\dfrac{\pi}{6}$ (C) $6\pi^2$ (D) $6\pi$

   ›Reveal solutionSolution

   Convert the rod's length into the ring's radius via $L = 2\pi R$, then compare the rod's central MI to the ring's tangential MI (found via the parallel axis theorem). The ratio simplifies to $\pi^2/6$.

   Concept and Intuition

   Bending a rod into a ring conserves its mass and its length (the length becomes the ring's circumference), but completely changes how that mass is distributed relative to various axes. To compare the two moments of inertia fairly, we must first express both in terms of a common variable — here, the rod's original length $L$, related to the ring's radius by $L = 2\pi R$. We also need the parallel axis theorem to shift the ring's MI from its centre to a tangent line.

   Step-by-Step Solution

   1. Rod's MI about the centre, perpendicular to its length: $I_1 = \dfrac{ML^2}{12}$.
   2. When bent into a ring, the rod's length becomes the ring's circumference: $L = 2\pi R \Rightarrow R = \dfrac{L}{2\pi}$.
   3. Ring's MI about its own centre (perpendicular to plane): $I_{cm} = MR^2$.
   4. By the parallel axis theorem, MI about a tangent axis (perpendicular to the plane, distance $R$ from the centre): $I_2 = I_{cm} + MR^2 = MR^2 + MR^2 = 2MR^2$. …

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6. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A solid cylinder rolls down an incline without slipping. Its acceleration depends on: (A) Only the mass of the cylinder (B) Only gravitational acceleration (C) Both the mass of the cylinder and the angle of inclination (D) Both the angle of incline and gravitational acceleration

   ›Reveal solutionSolution

   For rolling without slipping, the acceleration down an incline is $a=\dfrac{g\sin\theta}{1+k^2/R^2}$ — mass cancels out entirely, leaving dependence only on $g$ and the incline angle $\theta$.

   Concept and Intuition

   For any rigid body rolling without slipping down an incline, applying Newton's second law (translational) and the torque equation (rotational) about the contact point, the mass $m$ appears in every term and cancels. What survives is the geometric factor $k^2/R^2$ (radius of gyration ratio, which depends only on the body's shape, not its mass or size), $g$, and $\sin\theta$.

   Step-by-Step Solution

   1. Newton's second law along the incline: $mg\sin\theta-f=ma$ (where $f$ is friction).
   2. Torque about the center of mass: $fR=I\alpha=mk^2\cdot\dfrac{a}{R}$ (using rolling constraint $a=R\alpha$), so $f=\dfrac{mk^2a}{R^2}$.
   3. Substituting into the first equation: $mg\sin\theta-\dfrac{mk^2a}{R^2}=ma \Rightarrow g\sin\theta=a\left(1+\dfrac{k^2}{R^2}\right)$.
   4. So $a=\dfrac{g\sin\theta}{1+k^2/R^2}$ — note $m$ has cancelled completely from both sides. …

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7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The moment of inertia of a rod about an axis passing through its centre and perpendicular to its length is $\frac{1}{12}ML^2$, where M is the mass and L is the length of the rod. The rod is bent in the middle, so that the two halves make an angle of $60^0$. The moment of inertia of the bent rod about the same axis would be (A) $\dfrac{1}{12}ML^2$ (B) $\dfrac{1}{8\sqrt{3}}ML^2$ (C) $\dfrac{1}{24}ML^2$ (D) $\dfrac{1}{48}ML^2$

   ›Reveal solutionSolution

   Bending a rod at its center does not change its moment of inertia about the perpendicular axis through that center point — it stays $\dfrac{1}{12}ML^2$, regardless of the bend angle.

   Concept and Intuition

   The axis in question is perpendicular to the plane containing the rod, and it passes through the exact point where the rod is bent (the original center). For moment of inertia about such an axis, all that matters for each mass element is its straight-line distance from that fixed pivot point. Bending the rod simply rotates each half about this very pivot — like swinging a needle about a pin — which does NOT change any point's distance from the pin. So the distance distribution of mass from the axis is completely unaffected by the bend, and hence so is the moment of inertia.

   Step-by-Step Solution

   1. Before bending: the whole rod (mass $M$, length $L$) has $I=\dfrac{1}{12}ML^2$ about the perpendicular axis through its center.
   2. After bending at the center, split the rod into two halves, each of mass $M/2$ and length $L/2$, both hinged at the same central point (the axis location), just making an angle of $60°$ between them instead of being collinear.
   3. For an axis perpendicular to the plane and passing through one end of a rod of mass $m$, length $l$: $I_{end}=\dfrac13ml^2$ — this formula only depends on distances from that end along the rod, and is unaffected by any rotation of the rod about that end (in the plane perpendicular to the axis). …

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8. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A solid sphere and a thin circular ring of same mass and same radii are in rotational motion with same angular speeds about their diameters. Find the ratio of works done to stop them, $\dfrac{W_{sphere}}{W_{ring}}$ (A) 5:2 (B) 2:5 (C) 4:5 (D) 5:4

   ›Reveal solutionSolution

   Comparing rotational kinetic energies of a solid sphere and a ring (both

   spinning about a diameter) with equal mass, radius and angular speed gives

   the ratio of "work to stop" as their moment-of-inertia ratio, 4:5. Answer: (C).

   Concept and Intuition

   The work required to bring a rotating object to rest equals the rotational

   kinetic energy it currently has: $W = \tfrac12 I \omega^2$. With $M$, $R$

   and $\omega$ identical for both bodies, the ratio of the required work is

   exactly the ratio of their moments of inertia about the given axis (a

   diameter, in this case).

   Step-by-Step Solution

   1. Moment of inertia of a solid sphere about a diameter: $I_{sphere} = \dfrac{2}{5}MR^2$.
   2. Moment of inertia of a thin circular ring about a diameter (NOT the central/perpendicular axis): using the perpendicular-axis theorem, a ring's moment of inertia about its central axis (perpendicular to its plane) is $MR^2$; about any diameter (a "planar" axis, and by symmetry two perpendicular diameters share the load equally) it is half that: $I_{ring,\ diameter} = \dfrac12 MR^2$. …

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9. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If the moment of inertia of solid sphere of mass 2 kg and radius 10 cm about its tangent is I, then the moment of inertia of a uniform disc of mass 3.5 kg and radius 20 cm about its diameter is (A) $3.75\ I$ (B) $2.25\ I$ (C) $1.25\ I$ (D) $1.75\ I$

   ›Reveal solutionSolution

   This tests the parallel-axis theorem applied to a sphere (about its tangent) and the standard MOI formula for a disc about its diameter — the ratio comes out to a clean $1.25$.

   Concept and Intuition

   The moment of inertia of a solid sphere about an axis through its centre is $\tfrac25 MR^2$. To find its MOI about a tangent line (which is parallel to a diameter, displaced by a distance $R$, the radius), we use the parallel-axis theorem: $I_{tangent} = I_{centre} + MR^2 = \tfrac25 MR^2 + MR^2 = \tfrac75 MR^2$. Separately, a uniform disc's MOI about a diameter (an axis in its own plane through the centre) is the standard result $\tfrac14 mr^2$ (half of the perpendicular-axis value $\tfrac12 mr^2$, by the perpendicular axis theorem applied to two diameters). We just compute both numerically and take the ratio.

   Step-by-Step Solution

   1. Sphere: $M=2\ kg$, $R=10\ cm = 0.1\ m$. $I_{centre} = \tfrac25(2)(0.1)^2 = 0.4\times0.02=0.008\ \mathrm{kg\,m^2}$.
   2. By parallel axis theorem, $I = I_{centre} + MR^2 = 0.008 + 2(0.01) = 0.008+0.02=0.028\ \mathrm{kg\,m^2}$ (equivalently $I=\tfrac75 MR^2 = 1.4\times2\times0.01=0.028$). …

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10. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
    Q.The ratio of moment of inertia with respect to their diameters of a circular disc and a solid sphere having same radii and same masses is (A) 4 : 5 (B) 5 : 4 (C) 8 : 5 (D) 5 : 8

    ›Reveal solutionSolution

    Comparing the standard diametral moment-of-inertia formulas for a disc ($MR^2/4$) and a solid sphere ($2MR^2/5$) of equal mass and radius gives the ratio 5:8. Answer: (D).

    Concept and Intuition

    Moment of inertia about an axis depends on how mass is distributed relative to that axis. A thin disc's mass lies in a flat plane, so about a diameter (an axis in that plane) it has a smaller moment of inertia than about its own central perpendicular axis. A solid sphere's mass is distributed in three dimensions symmetrically, giving a fixed known coefficient about any diameter. Knowing these two standard formulas by heart lets you answer such ratio questions instantly.

    Step-by-Step Solution

    1. Standard result: moment of inertia of a uniform disc about a diameter $=\dfrac{1}{4}MR^2$ (half of its axial value $\dfrac12MR^2$, by the perpendicular axis theorem).
    2. Standard result: moment of inertia of a uniform solid sphere about a diameter $=\dfrac{2}{5}MR^2$. …

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11. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If the radius of gyration of a thin circular ring about an axis passing through its centre and perpendicular to its plane is $10\sqrt{2}\ \mathrm{cm}$, then its radius of gyration about its diameter is (A) 10 cm (B) 20 cm (C) $10\sqrt{2}$ cm (D) $20\sqrt{2}$ cm

    ›Reveal solutionSolution

    Uses radius of gyration for a ring about its central perpendicular axis vs. its diameter, linked by the perpendicular axis theorem; the diameter value is $10$ cm.

    Concept and Intuition

    For a thin ring, all the mass lies at distance $R$ from the centre, so the moment of inertia about the axis through the centre perpendicular to its plane is simply $I_z = MR^2$ (every mass element is at the same distance from that axis). The perpendicular axis theorem relates this to the two diameters (which are identical by symmetry): $I_z = I_x + I_y = 2I_{diameter}$, so $I_{diameter} = \tfrac12 MR^2$.

    Step-by-Step Solution

    1. About the central perpendicular axis: $I_z = MR^2 = Mk_1^2 \Rightarrow k_1 = R$.
    2. Given $k_1 = 10\sqrt2\ \mathrm{cm}$, so $R = 10\sqrt2\ \mathrm{cm}$. …

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12. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.Three thin uniform rods each of mass M and length L are placed along the three axes of a cartesian coordinate system with one end of all the rods at origin. The moment of inertia of the system of the rods about z-axis is (A) $\dfrac{ML^2}{3}$ (B) $\dfrac{2ML^2}{3}$ (C) $\dfrac{ML^2}{2}$ (D) $ML^2$

    ›Reveal solutionSolution

    Only the rods perpendicular to the $z$-axis (along $x$ and $y$) contribute; each gives $\frac13ML^2$, so the total is $\frac23ML^2$.

    Concept and Intuition

    Moment of inertia about an axis depends only on the perpendicular distance of each mass element from that axis. A rod lying exactly ON the axis of rotation (here, the rod along $z$, being rotated about the $z$-axis) has zero perpendicular distance everywhere, so it contributes nothing. The other two rods, lying along $x$ and $y$, have points whose distance from the $z$-axis equals their own coordinate value — making their moment of inertia about $z$ identical to the standard "rod about a perpendicular axis through one end" result.

    Step-by-Step Solution

    1. Rod along the $z$-axis: every point is $(0,0,z)$, whose perpendicular distance from the $z$-axis is $0$. So its MI about $z$-axis $=0$.
    2. Rod along the $x$-axis: points are $(x,0,0)$, perpendicular distance from $z$-axis is $x$. Its MI about the $z$-axis is the same as a rod's MI about a perpendicular axis through one end: $I_x=\int_0^L x^2\left(\frac{M}{L}dx\right)=\frac{M}{L}\cdot\frac{L^3}{3}=\frac{ML^2}{3}.$ …

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