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301. Which of the following is an example of torque in everyday life?
ⓐ. A car moving on a straight road
ⓑ. Opening a door by pushing at its edge
ⓒ. A ball falling freely under gravity
ⓓ. A stone kept on the floor
Correct Answer: Opening a door by pushing at its edge
Explanation: When you push at the edge of the door, you apply a force at a distance from the hinge (axis of rotation). This force produces torque, causing the door to rotate. The farther from the hinge you push, the larger the torque.
302. Why is it easier to loosen a tight nut using a longer spanner rather than a shorter one?
ⓐ. Longer spanner increases applied force
ⓑ. Longer spanner decreases friction
ⓒ. Longer spanner increases torque due to larger lever arm
ⓓ. Longer spanner reduces mass of the nut
Correct Answer: Longer spanner increases torque due to larger lever arm
Explanation: Torque is the product of force and perpendicular distance from the axis. A longer spanner increases the distance, thus increasing torque for the same force, making it easier to loosen the nut.
303. Why does a spinning wheel on a bicycle remain upright more easily than a stationary one?
ⓐ. It has less friction with the ground
ⓑ. Angular momentum provides stability
ⓒ. Its weight is reduced when spinning
ⓓ. Air resistance balances it
Correct Answer: Angular momentum provides stability
Explanation: The spinning wheel has angular momentum. According to the principle of conservation of angular momentum, it resists changes in its orientation, providing gyroscopic stability to the bicycle.
304. Which example illustrates the conservation of angular momentum in space?
ⓐ. A rocket launching from Earth
ⓑ. An astronaut spinning faster when pulling in their arms
ⓒ. A ball rolling down a hill
ⓓ. A satellite re-entering Earth’s atmosphere
Correct Answer: An astronaut spinning faster when pulling in their arms
Explanation: In the absence of external torque in space, angular momentum remains conserved. By reducing moment of inertia, the astronaut’s angular velocity increases, demonstrating conservation of angular momentum.
305. Why do figure skaters extend their arms to slow down a spin?
ⓐ. It reduces their weight
ⓑ. It increases their moment of inertia
ⓒ. It increases torque applied
ⓓ. It decreases their angular momentum
Correct Answer: It increases their moment of inertia
Explanation: Extending arms increases the distance of mass from the axis, increasing moment of inertia. Since angular momentum is conserved, angular velocity decreases, slowing the spin.
306. Why is a tightrope walker often seen carrying a long pole?
ⓐ. To increase weight for balance
ⓑ. To reduce torque from gravity
ⓒ. To increase moment of inertia and improve stability
ⓓ. To counteract wind resistance
Correct Answer: To increase moment of inertia and improve stability
Explanation: The long pole spreads mass farther from the body, increasing moment of inertia. This reduces angular acceleration due to any torque, helping the walker maintain balance.
307. Which of the following devices uses torque for its operation?
ⓐ. Thermometer
ⓑ. Lever
ⓒ. Stopwatch
ⓓ. Microscope
Correct Answer: Lever
Explanation: A lever works on the principle of torque, where applying force at one end produces rotational effect about a pivot. By adjusting lever arms, smaller forces can lift heavier loads efficiently.
308. Why does a spinning gyroscope resist changes in its orientation?
ⓐ. It has low torque
ⓑ. Its angular momentum resists external torque
ⓒ. It has more mass than usual
ⓓ. It uses magnetic forces
Correct Answer: Its angular momentum resists external torque
Explanation: A gyroscope maintains its orientation due to the conservation of angular momentum. External torque is required to change its rotational axis, making it stable and useful in navigation systems.
309. Why does a person experience difficulty in running on a rotating merry-go-round platform?
ⓐ. Because torque due to friction acts on their body
ⓑ. Because angular momentum of the platform interferes with their motion
ⓒ. Because linear velocity is reduced
ⓓ. Because mass of the platform changes
Correct Answer: Because angular momentum of the platform interferes with their motion
Explanation: The rotating merry-go-round has angular momentum, and when the person moves, they must deal with the conservation laws. This causes unusual forces to act on them, making running difficult.
310. Why does a ballet dancer spin gracefully faster when bringing arms and legs close to the body?
ⓐ. Because their mass decreases
ⓑ. Because their angular momentum decreases
ⓒ. Because their moment of inertia decreases, increasing angular velocity
ⓓ. Because torque applied by the floor increases
Correct Answer: Because their moment of inertia decreases, increasing angular velocity
Explanation: Bringing arms and legs closer reduces moment of inertia. Since angular momentum is conserved in the absence of external torque, angular velocity increases, allowing the dancer to spin faster and with more control.
311. Which of the following is a necessary condition for translational equilibrium of a rigid body?
ⓐ. Net torque must be zero
ⓑ. Net external force must be zero
ⓒ. Net momentum must be zero
ⓓ. Angular velocity must be constant
Correct Answer: Net external force must be zero
Explanation: For a rigid body to be in translational equilibrium, the vector sum of all external forces acting on it must be zero. This ensures there is no linear acceleration and the body remains at rest or continues in uniform motion.
312. Which of the following is a necessary condition for rotational equilibrium of a rigid body?
ⓐ. Net force must be zero
ⓑ. Net torque must be zero
ⓒ. Moment of inertia must be zero
ⓓ. Linear velocity must be constant
Correct Answer: Net torque must be zero
Explanation: For rotational equilibrium, the algebraic sum of all torques acting on a body about any axis must be zero. This ensures that the body has no angular acceleration and either remains at rest or continues to rotate uniformly.
313. For complete equilibrium of a rigid body, which conditions must be satisfied simultaneously?
ⓐ. Net force = 0 and net torque = 0
ⓑ. Net velocity = 0 and net acceleration = 0
ⓒ. Net torque = 0 and net energy = constant
ⓓ. Net force = 0 and net momentum = 0
Correct Answer: Net force = 0 and net torque = 0
Explanation: A rigid body is in complete equilibrium when both translational equilibrium (net force = 0) and rotational equilibrium (net torque = 0) are satisfied. This ensures no linear or angular acceleration occurs.
314. Which example best illustrates both translational and rotational equilibrium?
ⓐ. A book sliding on a frictionless table
ⓑ. A ladder resting against a wall without slipping
ⓒ. A ball freely rolling down an incline
ⓓ. A rocket moving upward with acceleration
Correct Answer: A ladder resting against a wall without slipping
Explanation: The ladder is in equilibrium when the net force on it is zero (translational equilibrium) and the net torque about any point is zero (rotational equilibrium). The wall and floor provide balancing forces and torques.
315. If a uniform rod is balanced horizontally on a knife edge at its midpoint, it is in:
ⓐ. Translational equilibrium only
ⓑ. Rotational equilibrium only
ⓒ. Both translational and rotational equilibrium
ⓓ. Neither translational nor rotational equilibrium
Correct Answer: Both translational and rotational equilibrium
Explanation: The uniform rod has equal weight on both sides of the fulcrum, so net force is zero. Also, moments about the knife edge are equal and opposite, so net torque is zero.
316. A beam is supported at two ends and loaded in the middle. For equilibrium, which condition must hold true?
ⓐ. Upward forces equal downward force of load
ⓑ. Net torque about any point must be zero
ⓒ. Both net force and net torque must be zero
ⓓ. The beam must have zero weight
Correct Answer: Both net force and net torque must be zero
Explanation: The upward reactions at the supports balance the downward load (net force = 0). Also, moments of the reactions about any point balance the load torque (net torque = 0). Both conditions ensure equilibrium.
317. Which of the following best explains why a person carrying two equal weights in both hands remains balanced?
ⓐ. Net force is maximum
ⓑ. Net torque is zero as weights balance each other
ⓒ. Net torque is maximum
ⓓ. Net angular velocity is constant
Correct Answer: Net torque is zero as weights balance each other
Explanation: Equal weights in both hands produce equal and opposite torques about the body’s center. This balances rotational effects. Additionally, vertical forces balance the person’s weight, keeping net force = 0.
318. A seesaw with two children of equal mass sitting at equal distances from the fulcrum is in equilibrium because:
ⓐ. Net force = 0 only
ⓑ. Net torque = 0 only
ⓒ. Both net force and net torque = 0
ⓓ. Moment of inertia is zero
Correct Answer: Both net force and net torque = 0
Explanation: The seesaw experiences equal downward forces balanced by the upward force of the support (net force = 0). Torques due to the children’s weights are equal and opposite, so net torque = 0.
319. If a rigid body is acted upon by three coplanar forces and is in equilibrium, then:
ⓐ. They must be parallel
ⓑ. Their resultant must be zero and they must be concurrent
ⓒ. Their torques must be maximum
ⓓ. Their magnitudes must be equal
Correct Answer: Their resultant must be zero and they must be concurrent
Explanation: For three coplanar forces in equilibrium, the forces must be concurrent (meet at a point) and vector sum of forces must be zero. This ensures both translational and rotational equilibrium.
320. Which mathematical condition represents equilibrium of a rigid body?
ⓐ. \(\sum \vec{F} = 0\) and \(\sum \tau = 0\)
ⓑ. \(\sum \vec{v} = 0\) and \(\sum E = 0\)
ⓒ. \(\sum \vec{a} = 0\) and \(\sum \vec{p} = 0\)
ⓓ. \(\sum I = 0\) and \(\sum \omega = 0\)
Correct Answer: \(\sum \vec{F} = 0\) and \(\sum \tau = 0\)
Explanation: For equilibrium of a rigid body, the vector sum of all external forces must be zero (ensuring no translation) and the sum of all torques about any point must be zero (ensuring no rotation). These are the two fundamental equilibrium conditions.
321. Which type of equilibrium does a ball at the bottom of a bowl represent?
ⓐ. Stable equilibrium
ⓑ. Unstable equilibrium
ⓒ. Neutral equilibrium
ⓓ. Dynamic equilibrium
Correct Answer: Stable equilibrium
Explanation: When the ball is slightly displaced from the bottom of the bowl, it experiences a restoring force and returns to its original position. This tendency to return shows stable equilibrium.
322. Which type of equilibrium does a ball balanced on the top of a hill represent?
ⓐ. Stable equilibrium
ⓑ. Unstable equilibrium
ⓒ. Neutral equilibrium
ⓓ. Translational equilibrium
Correct Answer: Unstable equilibrium
Explanation: When the ball is slightly displaced from the top of the hill, it experiences a force that moves it further away from the original position. This indicates unstable equilibrium.
323. Which type of equilibrium is shown by a ball resting on a flat horizontal surface?
ⓐ. Stable equilibrium
ⓑ. Unstable equilibrium
ⓒ. Neutral equilibrium
ⓓ. Rotational equilibrium
Correct Answer: Neutral equilibrium
Explanation: A ball on a flat surface, when displaced, neither returns to its original position nor moves further away. It simply settles at the new position, indicating neutral equilibrium.
324. What determines the stability of an object in equilibrium?
ⓐ. Mass of the object only
ⓑ. Height of the center of mass relative to base
ⓒ. Surface area of the object
ⓓ. Direction of torque applied
Correct Answer: Height of the center of mass relative to base
Explanation: The lower the center of mass, the more stable an object is. A higher center of mass makes the object easier to topple, reducing stability.
325. Why does a wide base make an object more stable?
ⓐ. It reduces torque
ⓑ. It lowers the center of gravity
ⓒ. It increases the area of support
ⓓ. It increases the mass of the object
Correct Answer: It increases the area of support
Explanation: A wider base increases the support area, meaning the vertical line through the center of mass is less likely to fall outside the base. This prevents toppling and increases stability.
326. Which condition must be satisfied for a body to remain in stable equilibrium?
ⓐ. Center of mass must rise when displaced
ⓑ. Center of mass must lower when displaced
ⓒ. Center of mass must remain at same height when displaced
ⓓ. Mass must be maximum
Correct Answer: Center of mass must rise when displaced
Explanation: In stable equilibrium, displacement causes the center of mass to rise. A restoring torque then acts, pulling the body back to its original position.
327. Why are loaded ships designed with ballast at the bottom?
ⓐ. To increase the ship’s mass
ⓑ. To lower the center of gravity for stability
ⓒ. To reduce friction with water
ⓓ. To make it float higher
Correct Answer: To lower the center of gravity for stability
Explanation: Ballast lowers the ship’s center of gravity, making it more stable and resistant to toppling when waves or winds act on it.
328. A uniform cone placed on its base shows which type of equilibrium?
ⓐ. Stable equilibrium
ⓑ. Unstable equilibrium
ⓒ. Neutral equilibrium
ⓓ. No equilibrium
Correct Answer: Stable equilibrium
Explanation: Displacing the cone slightly causes its center of mass to rise. Restoring torque brings it back to the original position, so it is in stable equilibrium.
329. A cylinder rolling on its curved surface is in:
ⓐ. Stable equilibrium
ⓑ. Unstable equilibrium
ⓒ. Neutral equilibrium
ⓓ. Dynamic equilibrium
Correct Answer: Neutral equilibrium
Explanation: When displaced, the center of mass remains at the same height, and the cylinder simply stays at the new position. This is neutral equilibrium.
330. Why does a person bend forward while climbing a hill with a load on their back?
ⓐ. To increase speed
ⓑ. To balance torque by shifting center of mass
ⓒ. To reduce mass of load
ⓓ. To reduce friction
Correct Answer: To balance torque by shifting center of mass
Explanation: Bending forward shifts the center of mass so that it lies above the base of support (feet). This prevents toppling backward and maintains stable equilibrium while climbing.
331. Which of the following is an example of stable equilibrium in everyday life?
ⓐ. A book lying flat on a table
ⓑ. A pencil balanced on its sharp tip
ⓒ. A ball on top of a hill
ⓓ. A coin standing on its edge
Correct Answer: A book lying flat on a table
Explanation: A book lying flat has its center of mass low and directly above its broad base of support. Any small displacement does not topple it, showing stable equilibrium.
332. Which of the following represents unstable equilibrium in daily life?
ⓐ. A child’s toy top spinning
ⓑ. A person standing with feet apart
ⓒ. A ladder balanced vertically against a smooth wall
ⓓ. A chair placed firmly on the ground
Correct Answer: A ladder balanced vertically against a smooth wall
Explanation: A vertical ladder can topple with a slight disturbance because the center of mass is high and any displacement lowers it, causing it to fall. This is unstable equilibrium.
333. A ball placed on a flat horizontal floor shows:
ⓐ. Stable equilibrium
ⓑ. Neutral equilibrium
ⓒ. Unstable equilibrium
ⓓ. No equilibrium
Correct Answer: Neutral equilibrium
Explanation: When displaced on a flat surface, the ball’s center of mass remains at the same height. It neither returns to its original position nor moves further away, so it is in neutral equilibrium.
334. Why does a wide-base cooking pot not topple easily?
ⓐ. Because its mass is very large
ⓑ. Because its center of gravity is lowered and base is wide
ⓒ. Because it has more friction with the ground
ⓓ. Because it has less volume
Correct Answer: Because its center of gravity is lowered and base is wide
Explanation: A wide base increases stability as the vertical line through the center of mass is less likely to fall outside the base. Lowering the center of gravity further enhances stability.
335. Why does a person stand with legs apart when trying to balance?
ⓐ. To reduce their mass
ⓑ. To increase their base of support for stability
ⓒ. To lower their center of gravity
ⓓ. To increase torque
Correct Answer: To increase their base of support for stability
Explanation: Standing with legs apart increases the area of support, making it harder for the vertical line through the center of mass to go outside the base. This improves stability and prevents falling.
336. Which of the following is an application of equilibrium in furniture design?
ⓐ. Making chairs with narrow legs
ⓑ. Designing tables with wide bases
ⓒ. Designing stools with a very high center of gravity
ⓓ. Making shelves very tall and narrow
Correct Answer: Designing tables with wide bases
Explanation: Wide bases make furniture more stable by lowering the chances of toppling, ensuring both safety and equilibrium in everyday use.
337. Why are double-decker buses made with a heavier lower section?
ⓐ. To reduce speed
ⓑ. To increase friction with road
ⓒ. To lower the center of gravity and improve stability
ⓓ. To reduce fuel consumption
Correct Answer: To lower the center of gravity and improve stability
Explanation: A heavier lower section ensures that the center of mass is close to the ground. This prevents the bus from toppling easily when turning or facing strong winds.
338. Why does a cone placed on its tip topple easily?
ⓐ. Because its mass is less
ⓑ. Because its center of gravity is high and unstable
ⓒ. Because its surface area is large
ⓓ. Because its weight decreases
Correct Answer: Because its center of gravity is high and unstable
Explanation: A cone on its tip has its center of mass high above a narrow base. A small disturbance shifts the vertical line of center of mass outside the base, causing it to topple.
339. Why do athletes lean forward while running fast?
ⓐ. To increase speed of running
ⓑ. To keep their center of mass within the base of support
ⓒ. To reduce their weight
ⓓ. To reduce air resistance
Correct Answer: To keep their center of mass within the base of support
Explanation: Leaning forward ensures that the line of action of their center of mass passes within the base (feet), preventing them from losing balance during fast motion.
340. Why are cars designed with a low and broad structure for racing?
ⓐ. To reduce air resistance only
ⓑ. To increase fuel efficiency
ⓒ. To lower the center of gravity and widen the base for stability
ⓓ. To reduce torque of wheels
Correct Answer: To lower the center of gravity and widen the base for stability
Explanation: A low center of gravity makes cars less likely to topple when turning at high speeds. A wider base of support further enhances stability, which is crucial in racing conditions.
341. What does the moment of inertia of a body represent?
ⓐ. Resistance offered by a body to linear acceleration
ⓑ. Resistance offered by a body to angular acceleration
ⓒ. Resistance offered by a body to translational equilibrium
ⓓ. Resistance offered by a body to gravitational force
Correct Answer: Resistance offered by a body to angular acceleration
Explanation: The moment of inertia measures how difficult it is to change the rotational state of a body about a given axis. It is the rotational analogue of mass in linear motion.
342. Which of the following best defines the moment of inertia?
ⓐ. Product of force and displacement
ⓑ. Product of torque and angular displacement
ⓒ. Sum of mass times square of distance from axis
ⓓ. Product of mass and velocity
Correct Answer: Sum of mass times square of distance from axis
Explanation: The moment of inertia of a system is defined as \(I = \sum m_i r_i^2\), where \(m_i\) is the mass of each particle and \(r_i\) is its perpendicular distance from the axis of rotation.
343. Moment of inertia is the rotational analogue of which linear quantity?
ⓐ. Momentum
ⓑ. Mass
ⓒ. Force
ⓓ. Energy
Correct Answer: Mass
Explanation: Just as mass resists linear acceleration in translational motion, moment of inertia resists angular acceleration in rotational motion. Hence, it is the rotational analogue of mass.
344. On which factors does the moment of inertia of a rigid body depend?
ⓐ. Mass only
ⓑ. Mass distribution and axis of rotation
ⓒ. Velocity of the body
ⓓ. Angular displacement of the body
Correct Answer: Mass distribution and axis of rotation
Explanation: The moment of inertia depends not only on the total mass of the body but also on how the mass is distributed relative to the axis of rotation.
345. What is the SI unit of moment of inertia?
ⓐ. kg·m
ⓑ. kg·m²
ⓒ. N·m
ⓓ. J
Correct Answer: kg·m²
Explanation: Since moment of inertia is given by \(I = mr^2\), where mass is in kilograms and distance in meters, its SI unit is kg·m².
346. Which statement is correct about moment of inertia?
ⓐ. It is a scalar quantity
ⓑ. It is a vector quantity
ⓒ. It is always independent of axis of rotation
ⓓ. It decreases when mass is farther from the axis
Correct Answer: It is a scalar quantity
Explanation: Moment of inertia has magnitude but no direction, hence it is a scalar quantity. However, its value depends on the chosen axis of rotation.
347. Why is it easier to spin a skater with arms folded than with arms stretched?
ⓐ. Because the skater’s mass increases
ⓑ. Because the skater’s moment of inertia decreases
ⓒ. Because torque applied increases
ⓓ. Because angular momentum decreases
Correct Answer: Because the skater’s moment of inertia decreases
Explanation: Folding arms brings mass closer to the axis of rotation, reducing moment of inertia. Since angular momentum is conserved, angular velocity increases, making spinning easier.
348. For a given body, how does moment of inertia change if the axis of rotation is shifted parallel to itself by a distance \(d\)?
ⓐ. Increases by \(md^2\)
ⓑ. Decreases by \(md^2\)
ⓒ. Remains the same
ⓓ. Becomes zero
Correct Answer: Increases by \(md^2\)
Explanation: According to the parallel axis theorem, \(I = I_{\text{cm}} + Md^2\). Hence, shifting the axis parallel to itself increases the moment of inertia by \(Md^2\), where \(M\) is the total mass.
349. Which physical quantity does the distribution of mass affect the most in rotational motion?
ⓐ. Torque
ⓑ. Moment of inertia
ⓒ. Angular displacement
ⓓ. Force
Correct Answer: Moment of inertia
Explanation: Even if two objects have the same mass, their moment of inertia differs depending on how mass is distributed relative to the axis of rotation. Distribution of mass plays a crucial role in rotational dynamics.
350. What is the dimensional formula of moment of inertia?
ⓐ. \([M^1L^1T^0]\)
ⓑ. \([M^1L^2T^0]\)
ⓒ. \([M^0L^2T^{-2}]\)
ⓓ. \([M^1L^1T^{-2}]\)
Correct Answer: \([M^1L^2T^0]\)
Explanation: Since \(I = mr^2\), its dimensional formula is \([M^1L^2T^0]\), where \(M\) is mass and \(L\) is length. It does not involve time.
351. What is the moment of inertia of a thin uniform rod of length \(L\) and mass \(M\), about an axis passing through its center and perpendicular to its length?
ⓐ. \(\frac{1}{12}ML^2\)
ⓑ. \(\frac{1}{3}ML^2\)
ⓒ. \(\frac{1}{2}ML^2\)
ⓓ. \(ML^2\)
Correct Answer: \(\frac{1}{12}ML^2\)
Explanation: For a uniform rod, the moment of inertia about an axis passing through its midpoint and perpendicular to its length is \(\frac{1}{12}ML^2\). This formula comes from integrating \(r^2 dm\) along the rod’s length.
352. What is the moment of inertia of a uniform solid sphere of mass \(M\) and radius \(R\) about its diameter?
ⓐ. \(\frac{2}{3}MR^2\)
ⓑ. \(\frac{2}{5}MR^2\)
ⓒ. \(\frac{3}{5}MR^2\)
ⓓ. \(\frac{1}{2}MR^2\)
Correct Answer: \(\frac{2}{5}MR^2\)
Explanation: A solid sphere’s moment of inertia about its diameter is \(\frac{2}{5}MR^2\). This arises because mass is distributed symmetrically in 3D about the axis.
353. What is the moment of inertia of a thin uniform circular ring of mass \(M\) and radius \(R\) about its central diameter?
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(\frac{3}{2}MR^2\)
ⓓ. \(2MR^2\)
Correct Answer: \(\frac{1}{2}MR^2\)
Explanation: A ring about its central diameter (not axis perpendicular to plane) has \(\frac{1}{2}MR^2\), because half of its mass is farther from the axis compared to the perpendicular case.
354. What is the moment of inertia of a solid cylinder of mass \(M\) and radius \(R\) about its central axis?
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(\frac{2}{5}MR^2\)
ⓓ. \(\frac{3}{4}MR^2\)
Correct Answer: \(\frac{1}{2}MR^2\)
Explanation: For a solid cylinder rotating about its central axis, the moment of inertia is \(\frac{1}{2}MR^2\). This is because all mass elements are distributed around the radius symmetrically.
355. What is the moment of inertia of a thin spherical shell of mass \(M\) and radius \(R\) about its diameter?
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(\frac{2}{3}MR^2\)
ⓒ. \(MR^2\)
ⓓ. \(\frac{2}{5}MR^2\)
Correct Answer: \(\frac{1}{2}MR^2\)
Explanation: A spherical shell has all its mass concentrated at the same radius. For rotation about a diameter, the result is \(\frac{1}{2}MR^2\).
356. Moment of inertia of a thin rectangular plate of length \(L\), breadth \(b\), and mass \(M\) about an axis through its center and perpendicular to its plane is:
ⓐ. \(\frac{1}{12}M(L^2+b^2)\)
ⓑ. \(\frac{1}{2}M(L^2+b^2)\)
ⓒ. \(\frac{1}{3}M(L^2+b^2)\)
ⓓ. \(M(L^2+b^2)\)
Correct Answer: \(\frac{1}{12}M(L^2+b^2)\)
Explanation: A rectangular plate has moment of inertia about its central perpendicular axis as \(\frac{1}{12}M(L^2+b^2)\), which accounts for contributions along both dimensions.
357. Moment of inertia of a ring of mass \(M\) and radius \(R\) about its central axis (perpendicular to the plane) is:
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(2MR^2\)
ⓓ. \(\frac{3}{2}MR^2\)
Correct Answer: \(MR^2\)
Explanation: For a ring, all the mass is at distance \(R\) from the axis. Therefore, \(I = MR^2\).
358. Moment of inertia of a hollow cylinder of mass \(M\), inner radius \(R_1\), and outer radius \(R_2\) about its central axis is:
ⓐ. \(\frac{1}{2}M(R_1^2+R_2^2)\)
ⓑ. \(M(R_1^2+R_2^2)\)
ⓒ. \(\frac{1}{4}M(R_1^2+R_2^2)\)
ⓓ. \(\frac{2}{5}M(R_1^2+R_2^2)\)
Correct Answer: \(\frac{1}{2}M(R_1^2+R_2^2)\)
Explanation: For a hollow cylinder, the formula accounts for both inner and outer radii: \(I = \frac{1}{2}M(R_1^2+R_2^2)\).
359. Moment of inertia of a uniform disc of radius \(R\) and mass \(M\) about an axis through its center and perpendicular to its plane is:
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(\frac{1}{4}MR^2\)
ⓓ. \(\frac{2}{5}MR^2\)
Correct Answer: \(\frac{1}{2}MR^2\)
Explanation: For a uniform disc about its perpendicular central axis, the moment of inertia is \(\frac{1}{2}MR^2\). This is widely used in rotational dynamics problems.
360. Moment of inertia of a disc of mass \(M\) and radius \(R\) about an axis along its diameter (in the plane of the disc) is:
ⓐ. \(\frac{1}{4}MR^2\)
ⓑ. \(\frac{1}{2}MR^2\)
ⓒ. \(\frac{3}{4}MR^2\)
ⓓ. \(\frac{2}{5}MR^2\)
Correct Answer: \(\frac{1}{4}MR^2\)
Explanation: Using the perpendicular axis theorem, for a disc \(I_x + I_y = I_z\). Since \(I_z = \frac{1}{2}MR^2\), each diameter axis has \(I = \frac{1}{4}MR^2\).
361. What is the moment of inertia of a thin uniform rod of length \(L\) and mass \(M\), about an axis passing through one end and perpendicular to its length?
ⓐ. \(\frac{1}{12}ML^2\)
ⓑ. \(\frac{1}{3}ML^2\)
ⓒ. \(\frac{1}{2}ML^2\)
ⓓ. \(ML^2\)
Correct Answer: \(\frac{1}{3}ML^2\)
Explanation: For a rod pivoted at one end, the moment of inertia is greater compared to its center because the average distance of the mass from the axis is larger. The correct formula is \(I = \frac{1}{3}ML^2\).
362. What is the moment of inertia of a hollow spherical shell of mass \(M\) and radius \(R\) about a diameter?
ⓐ. \(\frac{2}{3}MR^2\)
ⓑ. \(\frac{1}{2}MR^2\)
ⓒ. \(\frac{2}{5}MR^2\)
ⓓ. \(MR^2\)
Correct Answer: \(\frac{2}{3}MR^2\)
Explanation: A hollow sphere has all its mass distributed at the same radius, leading to a larger moment of inertia than a solid sphere. Its formula is \(I = \frac{2}{3}MR^2\).
363. Moment of inertia of a uniform solid cylinder of mass \(M\), radius \(R\), and length \(L\), about its central diameter is:
ⓐ. \(\frac{1}{4}MR^2 + \frac{1}{12}ML^2\)
ⓑ. \(\frac{1}{2}MR^2 + \frac{1}{12}ML^2\)
ⓒ. \(\frac{1}{3}MR^2 + \frac{1}{12}ML^2\)
ⓓ. \(\frac{2}{5}MR^2 + \frac{1}{12}ML^2\)
Correct Answer: \(\frac{1}{4}MR^2 + \frac{1}{12}ML^2\)
Explanation: Using the perpendicular axis theorem, the moment of inertia of a cylinder about its diameter accounts for contributions from both its radius and its length.
364. What is the moment of inertia of a rectangular plate of mass \(M\), length \(L\), and breadth \(b\) about one of its sides as axis?
ⓐ. \(\frac{1}{3}Mb^2\)
ⓑ. \(\frac{1}{3}ML^2\)
ⓒ. \(\frac{1}{12}M(L^2+b^2)\)
ⓓ. \(\frac{1}{2}ML^2\)
Correct Answer: \(\frac{1}{3}Mb^2\)
Explanation: For rotation about one side, the moment of inertia depends only on the perpendicular distance of mass elements from that side, giving \(I = \frac{1}{3}Mb^2\).
365. The moment of inertia of a ring of mass \(M\) and radius \(R\) about any tangent in the plane of the ring is:
ⓐ. \(MR^2\)
ⓑ. \(\frac{3}{2}MR^2\)
ⓒ. \(2MR^2\)
ⓓ. \(\frac{1}{2}MR^2\)
Correct Answer: \(\frac{3}{2}MR^2\)
Explanation: By the parallel axis theorem, \(I = I_{\text{diameter}} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2\).
366. The moment of inertia of a thin uniform rod of length \(L\) and mass \(M\), about an axis along its length (longitudinal axis), is:
ⓐ. \(\frac{1}{12}ML^2\)
ⓑ. 0
ⓒ. \(ML^2\)
ⓓ. Very small but not zero
Correct Answer: 0
Explanation: Since all mass elements are on the axis itself, their perpendicular distance from the axis is zero. Hence the moment of inertia is zero for the longitudinal axis.
367. Moment of inertia of a uniform disc of mass \(M\) and radius \(R\) about a tangent in its plane is:
ⓐ. \(\frac{3}{4}MR^2\)
ⓑ. \(\frac{5}{4}MR^2\)
ⓒ. \(\frac{1}{2}MR^2\)
ⓓ. \(MR^2\)
Correct Answer: \(\frac{3}{4}MR^2\)
Explanation: Using the parallel axis theorem, \(I = I_{\text{diameter}} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2\). Actually, for tangent in plane of disc: \(\frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2\). Correction → Answer: B. \(\frac{5}{4}MR^2\).
368. Moment of inertia of a uniform solid sphere of radius \(R\) and mass \(M\) about a tangent to the sphere is:
ⓐ. \(\frac{2}{3}MR^2\)
ⓑ. \(\frac{5}{3}MR^2\)
ⓒ. \(\frac{7}{5}MR^2\)
ⓓ. \(\frac{2}{5}MR^2\)
Correct Answer: \(\frac{7}{5}MR^2\)
Explanation: By the parallel axis theorem, \(I = I_{\text{diameter}} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2\).
369. Moment of inertia of a hollow sphere of mass \(M\) and radius \(R\) about a tangent is:
ⓐ. \(\frac{2}{3}MR^2\)
ⓑ. \(\frac{5}{3}MR^2\)
ⓒ. \(\frac{7}{3}MR^2\)
ⓓ. \(\frac{1}{2}MR^2\)
Correct Answer: \(\frac{5}{3}MR^2\)
Explanation: For a hollow sphere, \(I_{\text{diameter}} = \frac{2}{3}MR^2\). By the parallel axis theorem, \(I = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2\).
370. Moment of inertia of a thin rod of mass \(M\) and length \(L\), about an axis passing through its midpoint but parallel to its length, is:
ⓐ. 0
ⓑ. \(\frac{1}{12}ML^2\)
ⓒ. \(\frac{1}{3}ML^2\)
ⓓ. \(\frac{1}{2}ML^2\)
Correct Answer: 0
Explanation: Since the axis passes through and is parallel to the rod’s length, every particle’s perpendicular distance from the axis is zero, making the moment of inertia zero.
371. According to the parallel axis theorem, if the moment of inertia of a body about an axis through its center of mass is \(I_{\text{cm}}\), then the moment of inertia about a parallel axis at a distance \(d\) is:
ⓐ. \(I = I_{\text{cm}} + Md\)
ⓑ. \(I = I_{\text{cm}} + Md^2\)
ⓒ. \(I = I_{\text{cm}} – Md^2\)
ⓓ. \(I = I_{\text{cm}} + \frac{Md^2}{2}\)
Correct Answer: \(I = I_{\text{cm}} + Md^2\)
Explanation: The parallel axis theorem states that the moment of inertia about any axis parallel to the one through the center of mass is equal to the sum of the moment of inertia about the center of mass axis plus \(Md^2\), where \(M\) is the mass and \(d\) is the perpendicular distance.
372. A uniform rod of mass \(M\) and length \(L\) has a moment of inertia about its center \(\frac{1}{12}ML^2\). Using the parallel axis theorem, the moment of inertia about an end is:
ⓐ. \(\frac{1}{12}ML^2\)
ⓑ. \(\frac{1}{4}ML^2\)
ⓒ. \(\frac{1}{3}ML^2\)
ⓓ. \(\frac{1}{2}ML^2\)
Correct Answer: \(\frac{1}{3}ML^2\)
Explanation: Here, \(d = \frac{L}{2}\). By the theorem, \(I = I_{\text{cm}} + Md^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2\).
373. Moment of inertia of a uniform disc of radius \(R\) about its central axis is \(\frac{1}{2}MR^2\). Using the parallel axis theorem, its moment of inertia about a tangent to the disc is:
ⓐ. \(MR^2\)
ⓑ. \(\frac{3}{2}MR^2\)
ⓒ. \(\frac{5}{4}MR^2\)
ⓓ. \(2MR^2\)
Correct Answer: \(\frac{5}{4}MR^2\)
Explanation: By the theorem, \(I = I_{\text{cm}} + Md^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2\). But since the tangent is in-plane of the disc, we must take diameter axis first: \(I = \frac{1}{4}MR^2\). Then apply theorem: \(I = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2\).
374. Which of the following best demonstrates the use of the parallel axis theorem?
ⓐ. Calculating the moment of inertia of a solid sphere about its diameter.
ⓑ. Calculating the moment of inertia of a rod about an axis through one end.
ⓒ. Calculating the moment of inertia of a thin ring about its central axis.
ⓓ. Calculating the moment of inertia of a hollow sphere about its diameter.
Correct Answer: Calculating the moment of inertia of a rod about an axis through one end.
Explanation: The parallel axis theorem is applied when we need to find the moment of inertia about an axis not passing through the center of mass but parallel to one that does. A rod about its end is a classic example.
375. The parallel axis theorem cannot be applied if:
ⓐ. The new axis is not parallel to the axis through the center of mass.
ⓑ. The axis passes through the center of mass.
ⓒ. The body is irregular in shape.
ⓓ. The body is non-rigid.
Correct Answer: The new axis is not parallel to the axis through the center of mass.
Explanation: The theorem strictly applies only to axes parallel to the axis passing through the center of mass. For non-parallel axes, the perpendicular axis theorem or integration methods must be used.
376. The moment of inertia of a solid sphere about its diameter is \(\frac{2}{5}MR^2\). Using the parallel axis theorem, its moment of inertia about a tangent is:
ⓐ. \(\frac{7}{5}MR^2\)
ⓑ. \(\frac{5}{3}MR^2\)
ⓒ. \(\frac{3}{2}MR^2\)
ⓓ. \(MR^2\)
Correct Answer: \(\frac{7}{5}MR^2\)
Explanation: By the theorem, \(I = I_{\text{cm}} + Md^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2\).
377. The parallel axis theorem is mathematically expressed as:
ⓐ. \(I = I_{\text{cm}} + Md^2\)
ⓑ. \(I = I_{\text{cm}} – Md^2\)
ⓒ. \(I = I_{\text{cm}} \times Md^2\)
ⓓ. \(I = \frac{I_{\text{cm}}}{Md^2}\)
Correct Answer: \(I = I_{\text{cm}} + Md^2\)
Explanation: The correct formula is \(I = I_{\text{cm}} + Md^2\). Here, \(I_{\text{cm}}\) is the moment of inertia through the center of mass axis, \(M\) is the mass, and \(d\) is the perpendicular distance between the two axes.
378. A ring of radius \(R\) and mass \(M\) has a moment of inertia about a tangent in the plane of the ring as:
ⓐ. \(\frac{3}{2}MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(\frac{1}{2}MR^2\)
ⓓ. \(2MR^2\)
Correct Answer: \(\frac{3}{2}MR^2\)
Explanation: About diameter: \(I = \frac{1}{2}MR^2\). By parallel axis theorem, \(I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2\).
379. Which step is necessary when applying the parallel axis theorem?
ⓐ. Locating the axis through the center of mass.
ⓑ. Measuring the perpendicular distance between the two axes.
ⓒ. Adding \(Md^2\) to the central axis moment of inertia.
ⓓ. All of the above.
Correct Answer: All of the above.
Explanation: To apply the theorem, we must know the center of mass axis, the perpendicular distance \(d\), and then add \(Md^2\) to the central moment of inertia.
380. The parallel axis theorem is most useful for:
ⓐ. Bodies with symmetry about the axis.
ⓑ. Finding the moment of inertia about axes away from the center of mass.
ⓒ. Spherical objects only.
ⓓ. Cylindrical bodies only.
Correct Answer: Finding the moment of inertia about axes away from the center of mass.
Explanation: The theorem simplifies problems where the axis of rotation does not pass through the center of mass but is parallel to such an axis, making calculations easier without direct integration.
381. The perpendicular axis theorem is applicable to which type of bodies?
ⓐ. Three-dimensional bodies
ⓑ. One-dimensional bodies
ⓒ. Planar lamina (two-dimensional bodies)
ⓓ. Only spherical bodies
Correct Answer: Planar lamina (two-dimensional bodies)
Explanation: The perpendicular axis theorem applies only to plane laminae, i.e., flat two-dimensional bodies lying entirely in a plane. It relates the moment of inertia about a perpendicular axis to those about two perpendicular axes lying in the plane.
382. According to the perpendicular axis theorem, if \(I_x\) and \(I_y\) are the moments of inertia about two perpendicular axes in the plane of the lamina, then the moment of inertia about a perpendicular axis \(I_z\) is:
ⓐ. \(I_z = I_x + I_y\)
ⓑ. \(I_z = I_x – I_y\)
ⓒ. \(I_z = I_x \cdot I_y\)
ⓓ. \(I_z = \frac{I_x + I_y}{2}\)
Correct Answer: \(I_z = I_x + I_y\)
Explanation: The perpendicular axis theorem states \(I_z = I_x + I_y\), where \(z\)-axis is perpendicular to the plane of the lamina and passes through the intersection of \(x\) and \(y\) axes.
383. Which of the following is a practical example of the perpendicular axis theorem?
ⓐ. Moment of inertia of a rod about one end
ⓑ. Moment of inertia of a sphere about a tangent
ⓒ. Moment of inertia of a circular disc about its central axis perpendicular to the plane
ⓓ. Moment of inertia of a cylinder about its diameter
Correct Answer: Moment of inertia of a circular disc about its central axis perpendicular to the plane
Explanation: For a disc, using perpendicular axis theorem, \(I_z = I_x + I_y\). Since \(I_x = I_y\), we can find \(I_z\) easily without integration.
384. The perpendicular axis theorem cannot be applied if:
ⓐ. The body is not flat (not a lamina)
ⓑ. The axes chosen do not intersect
ⓒ. The axes are not mutually perpendicular
ⓓ. All of the above
Correct Answer: All of the above
Explanation: For the theorem to hold, the body must be a lamina, the two axes must be in the plane of the lamina, intersect at a point, and be mutually perpendicular.
385. For a uniform circular disc of mass \(M\) and radius \(R\), the moment of inertia about a diameter is:
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(\frac{1}{4}MR^2\)
ⓒ. \(\frac{1}{3}MR^2\)
ⓓ. \(\frac{2}{5}MR^2\)
Correct Answer: \(\frac{1}{4}MR^2\)
Explanation: Using perpendicular axis theorem, \(I_z = I_x + I_y\). For a disc, \(I_z = \frac{1}{2}MR^2\). Since \(I_x = I_y\), each equals \(\frac{1}{4}MR^2\).
386. For a thin circular ring of radius \(R\) and mass \(M\), the moment of inertia about any diameter in its plane is:
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(\frac{1}{4}MR^2\)
ⓓ. \(\frac{2}{3}MR^2\)
Correct Answer: \(\frac{1}{2}MR^2\)
Explanation: For a ring, \(I_z = MR^2\). By perpendicular axis theorem, \(I_z = I_x + I_y\). Since symmetry gives \(I_x = I_y\), each equals \(\frac{1}{2}MR^2\).
387. The perpendicular axis theorem helps in calculating the moment of inertia of:
ⓐ. Solid spheres
ⓑ. Rods
ⓒ. Planar objects like discs and rings
ⓓ. Cylinders
Correct Answer: Planar objects like discs and rings
Explanation: The theorem is strictly applicable to laminar objects (planar), where the relation between axes in the plane and perpendicular axis holds true.
388. For a rectangular lamina of length \(a\) and breadth \(b\), the perpendicular axis theorem can be applied to relate:
ⓐ. Moments of inertia about its diagonals
ⓑ. Moments of inertia about its length, breadth, and perpendicular axis through the center
ⓒ. Only about the length axis
ⓓ. Only about the breadth axis
Correct Answer: Moments of inertia about its length, breadth, and perpendicular axis through the center
Explanation: If \(I_x\) is about length, \(I_y\) about breadth, then \(I_z = I_x + I_y\), where \(I_z\) is perpendicular to the plane through the center.
389. Which law or principle does the perpendicular axis theorem resemble in mathematical form?
ⓐ. Law of conservation of energy
ⓑ. Pythagoras theorem
ⓒ. Principle of conservation of momentum
ⓓ. Hooke’s law
Correct Answer: Pythagoras theorem
Explanation: Just as Pythagoras theorem states the square of the hypotenuse equals the sum of the squares of the other two sides, the perpendicular axis theorem relates the perpendicular axis moment of inertia as a sum of the in-plane moments.
390. If a disc of radius \(R\) has mass \(M\), then the ratio of its moment of inertia about a central axis perpendicular to the plane to that about a diameter is:
ⓐ. 1 : 1
ⓑ. 2 : 1
ⓒ. 1 : 2
ⓓ. 4 : 1
Correct Answer: 2 : 1
Explanation: For a disc, \(I_z = \frac{1}{2}MR^2\) and \(I_{\text{diameter}} = \frac{1}{4}MR^2\). Hence, the ratio is \( \frac{\frac{1}{2}MR^2}{\frac{1}{4}MR^2} = 2:1\).
391. The moment of inertia of a system of particles is calculated by:
ⓐ. Adding their masses directly
ⓑ. Adding the distances of each particle from the axis
ⓒ. Summing the product of each particle’s mass and the square of its distance from the axis
ⓓ. Taking the product of their masses and distances
Correct Answer: Summing the product of each particle’s mass and the square of its distance from the axis
Explanation: For a system of particles, the moment of inertia is \(I = \sum m_i r_i^2\), where \(m_i\) is the mass of each particle and \(r_i\) is its perpendicular distance from the axis.
392. If two point masses of \(2\,\text{kg}\) each are placed at distances of \(1\,\text{m}\) and \(2\,\text{m}\) from a given axis, the total moment of inertia is:
ⓐ. 2 kg·m²
ⓑ. 6 kg·m²
ⓒ. 8 kg·m²
ⓓ. 10 kg·m²
Correct Answer: 10 kg·m²
Explanation: \(I = \sum m_i r_i^2 = 2(1^2) + 2(2^2) = 2(1) + 2(4) = 2 + 8 = 10 \, \text{kg·m}^2\).
393. The moment of inertia of two equal masses \(m\) connected by a light rod of length \(2a\), about an axis through the midpoint and perpendicular to the rod, is:
ⓐ. \(2ma^2\)
ⓑ. \(ma^2\)
ⓒ. \(4ma^2\)
ⓓ. \(\frac{1}{2}ma^2\)
Correct Answer: \(2ma^2\)
Explanation: Each mass is at a distance \(a\) from the midpoint. Thus, \(I = m a^2 + m a^2 = 2ma^2\).
394. Four point masses \(M\) are placed at the corners of a square of side \(a\). The moment of inertia about an axis perpendicular to the square and passing through its center is:
ⓐ. \(2Ma^2\)
ⓑ. \(4Ma^2\)
ⓒ. \(Ma^2\)
ⓓ. \(\frac{1}{2}Ma^2\)
Correct Answer: \(2Ma^2\)
Explanation: Distance of each mass from the center = \(\frac{\sqrt{2}}{2}a\). So \(I = 4M\left(\frac{a}{\sqrt{2}}\right)^2 = 4M\left(\frac{a^2}{2}\right) = 2Ma^2\).
395. The moment of inertia of a thin rod of mass \(M\) and length \(L\) about an axis perpendicular to one end is:
ⓐ. \(\frac{1}{3}ML^2\)
ⓑ. \(\frac{1}{12}ML^2\)
ⓒ. \(\frac{1}{2}ML^2\)
ⓓ. \(\frac{2}{3}ML^2\)
Correct Answer: \(\frac{1}{3}ML^2\)
Explanation: By direct integration or parallel axis theorem: \(I = I_{\text{cm}} + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2\).
396. For a system of three particles, each of mass \(m\), placed at the vertices of an equilateral triangle of side \(a\), the moment of inertia about an axis passing through the centroid and perpendicular to the plane is:
ⓐ. \(ma^2\)
ⓑ. \(\frac{3}{2}ma^2\)
ⓒ. \(m a^2/3\)
ⓓ. \(ma^2/2\)
Correct Answer: \(ma^2\)
Explanation: Distance of each vertex from centroid = \(\frac{\sqrt{3}}{3}a\). Hence, \(I = 3m\left(\frac{\sqrt{3}}{3}a\right)^2 = 3m\left(\frac{a^2}{3}\right) = ma^2\).
397. If a solid sphere of mass \(M\) and radius \(R\) is considered as made of thin discs, then its moment of inertia about a diameter is obtained by:
ⓐ. Direct summation of masses
ⓑ. Applying the parallel axis theorem
ⓒ. Integrating \(r^2 dm\) over the sphere’s volume
ⓓ. Adding its translational kinetic energy
Correct Answer: Integrating \(r^2 dm\) over the sphere’s volume
Explanation: Moment of inertia of a solid sphere about its diameter is derived by integration: \(I = \int r^2 dm = \frac{2}{5}MR^2\).
398. The moment of inertia of a composite body is:
ⓐ. The product of the moments of inertia of its parts
ⓑ. The sum of the moments of inertia of its parts about the same axis
ⓒ. The average of the moments of inertia of its parts
ⓓ. The square root of the sum of their moments of inertia
Correct Answer: The sum of the moments of inertia of its parts about the same axis
Explanation: By the principle of superposition, the total moment of inertia of a composite system equals the sum of the moments of inertia of its parts about the same axis.
399. The moment of inertia of a uniform semicircular ring of mass \(M\) and radius \(R\) about a perpendicular axis through its center is:
ⓐ. \(MR^2\)
ⓑ. \(\frac{1}{2}MR^2\)
ⓒ. \(\frac{1}{4}MR^2\)
ⓓ. \(\frac{3}{4}MR^2\)
Correct Answer: \(\frac{3}{4}MR^2\)
Explanation: For a semicircular ring, calculation by integration gives \(I = \frac{3}{4}MR^2\).
400. A rectangular plate of mass \(M\), length \(a\), and breadth \(b\) has its moment of inertia about an axis perpendicular to the plate through its center as:
ⓐ. \(\frac{1}{12}M(a^2 + b^2)\)
ⓑ. \(\frac{1}{4}M(a^2 + b^2)\)
ⓒ. \(\frac{1}{3}M(a^2 + b^2)\)
ⓓ. \(\frac{1}{2}M(a^2 + b^2)\)
Correct Answer: \(\frac{1}{12}M(a^2 + b^2)\)
Explanation: For a rectangular plate, the perpendicular axis theorem and direct integration yield \(I = \frac{1}{12}M(a^2 + b^2)\), combining both dimensions.
The chapter System of Particles and Rotational Motion plays a central role in Class 11 Physics (NCERT/CBSE syllabus), with frequent questions in board exams and competitive exams such as JEE, NEET, and state-level tests. This topic develops problem-solving skills through concepts like moment of inertia, parallel and perpendicular axis theorems, and rotational dynamics. The entire series consists of 570 MCQs with solutions, divided into 6 parts for systematic practice. In this part, you will solve the fourth set of 100 MCQs with answers, carefully crafted for exam-level preparation.
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