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JEE Main Previous Year Questions (2025): Rotational Motion
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Page 1 JEE Main Previous Year Questions (2025): Rotational Motion Q1: The position vectors of two 1 kg particles, (A) and (B), are given by ?? ?? ??? = (?? ?? ?? ?? ??ˆ + ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? and ?? ?? ??? = (?? ?? ??ˆ ??ˆ + ?? ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? , respectively; (?? ?? = ?? ?? /?? ?? ,?? ?? = ?? ???? /?? ,?? ?? = ?? ?? /?? ,?? ?? = ?? ?? /?? ,?? ?? = -?? ?? /?? ?? ,?? ?? = ?????? /?? ) , where ?? is time, ?? and ?? are constants. At ?? = ?? ?? ,?? ?? ????? = ?? ?? ????? and velocities ?? ? ? ?? and ?? ? ? ?? of the particles are orthogonal to each other. At ?? = ?? ?? , the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is v?? ?????? ?? ?? -?? . The value of ?? is JEE Main 2025 (Online) 22nd January Morning Shift Ans: 90 Solution: Given, ?? ?? = 1???? = ?? ?? ?? ?? ??? = (?? 1 ?? 2 ??ˆ + ?? 2 ?? ??ˆ + ?? 3 ?? ?? ^ )?? ?? ?? ??? = (?? 1 ?? ??ˆ + ?? 2 ?? 2 ??ˆ + ?? 3 ?? ?? ^ )?? (?? 1 = 1?? /?? 2 ,?? 2 = 3???? /?? ,?? 3 = 2?? /?? ?? 1 = 2?? /?? ,?? 2 = -1?? /?? 2 ,?? 3 = 4???? /?? ) ?? ?? ???? = ?? ?? ?? ??? ???? = 2?? ??ˆ + 3?? ??ˆ + 2?? ^ ?? ?? ???? = ?? ?? ?? ??? ???? = 2??ˆ - 2?? ??ˆ + 4?? ?? ^ ?? ?? ???? · ?? ?? ???? = 0( As ?? ?? ???? ? ?? ?? ???? given ) ? 4?? - 6???? + 8?? = 0 At ?? = 1,4- 6?? + 8?? = 0 ? 2- 3?? + 4?? = 0 ? 3?? = 2+ 4?? ……(1) given, ?? ?? ???? = ?? ?? ???? ? 4+ 9?? 2 + 4 = 4+ 4+ 16?? 2 ? (2+ 4?? ) 2 = 16?? 2 (from (1) ? 4+ 16?? 2 + 16?? = 16?? 2 ? ?? = - 1 4 ? 3?? = 2+ 4(- 1 4 ) = 1 ? ?? = 1 3 Now, ?? ? = ?? ?? (?? ?? /?? × ?? ?? ) at ?? = 1sec , Page 2 JEE Main Previous Year Questions (2025): Rotational Motion Q1: The position vectors of two 1 kg particles, (A) and (B), are given by ?? ?? ??? = (?? ?? ?? ?? ??ˆ + ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? and ?? ?? ??? = (?? ?? ??ˆ ??ˆ + ?? ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? , respectively; (?? ?? = ?? ?? /?? ?? ,?? ?? = ?? ???? /?? ,?? ?? = ?? ?? /?? ,?? ?? = ?? ?? /?? ,?? ?? = -?? ?? /?? ?? ,?? ?? = ?????? /?? ) , where ?? is time, ?? and ?? are constants. At ?? = ?? ?? ,?? ?? ????? = ?? ?? ????? and velocities ?? ? ? ?? and ?? ? ? ?? of the particles are orthogonal to each other. At ?? = ?? ?? , the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is v?? ?????? ?? ?? -?? . The value of ?? is JEE Main 2025 (Online) 22nd January Morning Shift Ans: 90 Solution: Given, ?? ?? = 1???? = ?? ?? ?? ?? ??? = (?? 1 ?? 2 ??ˆ + ?? 2 ?? ??ˆ + ?? 3 ?? ?? ^ )?? ?? ?? ??? = (?? 1 ?? ??ˆ + ?? 2 ?? 2 ??ˆ + ?? 3 ?? ?? ^ )?? (?? 1 = 1?? /?? 2 ,?? 2 = 3???? /?? ,?? 3 = 2?? /?? ?? 1 = 2?? /?? ,?? 2 = -1?? /?? 2 ,?? 3 = 4???? /?? ) ?? ?? ???? = ?? ?? ?? ??? ???? = 2?? ??ˆ + 3?? ??ˆ + 2?? ^ ?? ?? ???? = ?? ?? ?? ??? ???? = 2??ˆ - 2?? ??ˆ + 4?? ?? ^ ?? ?? ???? · ?? ?? ???? = 0( As ?? ?? ???? ? ?? ?? ???? given ) ? 4?? - 6???? + 8?? = 0 At ?? = 1,4- 6?? + 8?? = 0 ? 2- 3?? + 4?? = 0 ? 3?? = 2+ 4?? ……(1) given, ?? ?? ???? = ?? ?? ???? ? 4+ 9?? 2 + 4 = 4+ 4+ 16?? 2 ? (2+ 4?? ) 2 = 16?? 2 (from (1) ? 4+ 16?? 2 + 16?? = 16?? 2 ? ?? = - 1 4 ? 3?? = 2+ 4(- 1 4 ) = 1 ? ?? = 1 3 Now, ?? ? = ?? ?? (?? ?? /?? × ?? ?? ) at ?? = 1sec , ?? ?? ?? ???? = (?? 1 - ?? 1 )??ˆ + (?? 2 - ?? 2 )??ˆ + (?? 3 - ?? 3 )?? ^ ? ?? ?? ?? ???? = (1- 2)??ˆ + (3?? + 1)??ˆ + (2- 4?? )?? ^ = -??ˆ + 2??ˆ + 3?? ^ ?? ? = ??ˆ ??ˆ ?? ˆ -1 2 3 2 1 2 = ??ˆ(4- 3)- ??ˆ(-2- 6)+ ?? ^ (-1- 4) ?? ? = ??ˆ + 8??ˆ - 5?? ˆ ? ?? ? = v1 2 + 8 2 + (-5) 2 = v1 + 64+ 25 ? ?? ? = v90 kg m 2 s -1 = v?? (given) So, L = 90 Q2: The moment of inertia of a solid disc rotating along its diameter is ?? .?? times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is ?? times higher than the moment of inertia of the given ring. Here, ?? = ____ Consider all the bodies have equal masses. JEE Main 2025 (Online) 28th January Morning Shift Ans: 4 Solution: Given, ?? 1 = 2.5?? 2 and ?? 3 = ?? ?? 2 we know, ?? 1 = ?? ?? 1 2 4 ,?? 2 = ?? ?? 2 2 2 ,?? 3 = 2 5 ?? ?? 3 2 ? ?? ?? 1 2 4 = 2.5 ?? ?? 2 2 2 ? ?? 1 2 = 5?? 2 2 Now, ?? 3 = ?? ?? 2 ? 2 5 ?? ?? 1 2 = ?? ?? ?? 2 2 2 ( As ?? 3 = ?? 1 ) ? 2 5 × 5?? 2 2 = ?? ?? 2 2 2 ? ?? = 4 Page 3 JEE Main Previous Year Questions (2025): Rotational Motion Q1: The position vectors of two 1 kg particles, (A) and (B), are given by ?? ?? ??? = (?? ?? ?? ?? ??ˆ + ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? and ?? ?? ??? = (?? ?? ??ˆ ??ˆ + ?? ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? , respectively; (?? ?? = ?? ?? /?? ?? ,?? ?? = ?? ???? /?? ,?? ?? = ?? ?? /?? ,?? ?? = ?? ?? /?? ,?? ?? = -?? ?? /?? ?? ,?? ?? = ?????? /?? ) , where ?? is time, ?? and ?? are constants. At ?? = ?? ?? ,?? ?? ????? = ?? ?? ????? and velocities ?? ? ? ?? and ?? ? ? ?? of the particles are orthogonal to each other. At ?? = ?? ?? , the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is v?? ?????? ?? ?? -?? . The value of ?? is JEE Main 2025 (Online) 22nd January Morning Shift Ans: 90 Solution: Given, ?? ?? = 1???? = ?? ?? ?? ?? ??? = (?? 1 ?? 2 ??ˆ + ?? 2 ?? ??ˆ + ?? 3 ?? ?? ^ )?? ?? ?? ??? = (?? 1 ?? ??ˆ + ?? 2 ?? 2 ??ˆ + ?? 3 ?? ?? ^ )?? (?? 1 = 1?? /?? 2 ,?? 2 = 3???? /?? ,?? 3 = 2?? /?? ?? 1 = 2?? /?? ,?? 2 = -1?? /?? 2 ,?? 3 = 4???? /?? ) ?? ?? ???? = ?? ?? ?? ??? ???? = 2?? ??ˆ + 3?? ??ˆ + 2?? ^ ?? ?? ???? = ?? ?? ?? ??? ???? = 2??ˆ - 2?? ??ˆ + 4?? ?? ^ ?? ?? ???? · ?? ?? ???? = 0( As ?? ?? ???? ? ?? ?? ???? given ) ? 4?? - 6???? + 8?? = 0 At ?? = 1,4- 6?? + 8?? = 0 ? 2- 3?? + 4?? = 0 ? 3?? = 2+ 4?? ……(1) given, ?? ?? ???? = ?? ?? ???? ? 4+ 9?? 2 + 4 = 4+ 4+ 16?? 2 ? (2+ 4?? ) 2 = 16?? 2 (from (1) ? 4+ 16?? 2 + 16?? = 16?? 2 ? ?? = - 1 4 ? 3?? = 2+ 4(- 1 4 ) = 1 ? ?? = 1 3 Now, ?? ? = ?? ?? (?? ?? /?? × ?? ?? ) at ?? = 1sec , ?? ?? ?? ???? = (?? 1 - ?? 1 )??ˆ + (?? 2 - ?? 2 )??ˆ + (?? 3 - ?? 3 )?? ^ ? ?? ?? ?? ???? = (1- 2)??ˆ + (3?? + 1)??ˆ + (2- 4?? )?? ^ = -??ˆ + 2??ˆ + 3?? ^ ?? ? = ??ˆ ??ˆ ?? ˆ -1 2 3 2 1 2 = ??ˆ(4- 3)- ??ˆ(-2- 6)+ ?? ^ (-1- 4) ?? ? = ??ˆ + 8??ˆ - 5?? ˆ ? ?? ? = v1 2 + 8 2 + (-5) 2 = v1 + 64+ 25 ? ?? ? = v90 kg m 2 s -1 = v?? (given) So, L = 90 Q2: The moment of inertia of a solid disc rotating along its diameter is ?? .?? times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is ?? times higher than the moment of inertia of the given ring. Here, ?? = ____ Consider all the bodies have equal masses. JEE Main 2025 (Online) 28th January Morning Shift Ans: 4 Solution: Given, ?? 1 = 2.5?? 2 and ?? 3 = ?? ?? 2 we know, ?? 1 = ?? ?? 1 2 4 ,?? 2 = ?? ?? 2 2 2 ,?? 3 = 2 5 ?? ?? 3 2 ? ?? ?? 1 2 4 = 2.5 ?? ?? 2 2 2 ? ?? 1 2 = 5?? 2 2 Now, ?? 3 = ?? ?? 2 ? 2 5 ?? ?? 1 2 = ?? ?? ?? 2 2 2 ( As ?? 3 = ?? 1 ) ? 2 5 × 5?? 2 2 = ?? ?? 2 2 2 ? ?? = 4 Q3: Two iron solid discs of negligible thickness have radii ?? ?? and ?? ?? and moment of intertia ?? ?? and ?? ?? , respectively. For ?? ?? = ?? ?? ?? , the ratio of ?? ?? and ?? ?? would be ?? /?? , where ?? = ____ ? JEE Main 2025 (Online) 28th January Morning Shift Ans: 16 Solution: Let surface mass density = ?? So, ?? 1 = ?? × ?? ?? 1 2 ?? 2 = ?? × ?? ?? 2 2 = ???? (2?? 1 ) 2 ( As ?? 2 = 2?? 1 = 4???? ?? 1 2 ? ?? 2 = 4?? 1 ?? 1 ?? 2 = ?? 1 ?? 1 2 2 ?? 2 ?? 2 2 2 = ( ?? 1 ?? 2 )( ?? 1 ?? 2 ) 2 ? ?? 1 ?? 2 = ( ?? 1 4?? 1 )( ?? 1 2?? 1 ) 2 = 1 4 × 1 4 = 1 16 = 1 ?? Hence, ?? = 16 Q4: The coordinates of a particle with respect to origin in a given reference frame is (?? ,?? ,?? ) meters. If a force of ?? ? ? = ??ˆ - ??ˆ + ?? ˆ acts on the particle, then the magnitude of torque (with respect to origin) in ?? -direction is ____ . JEE Main 2025 (Online) 29th January Morning Shift Ans: 2 Solution: The torque ?? acting on the particle with respect to the origin can be calculated using the cross product of the position vector ?? and the force vector ?? : Page 4 JEE Main Previous Year Questions (2025): Rotational Motion Q1: The position vectors of two 1 kg particles, (A) and (B), are given by ?? ?? ??? = (?? ?? ?? ?? ??ˆ + ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? and ?? ?? ??? = (?? ?? ??ˆ ??ˆ + ?? ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? , respectively; (?? ?? = ?? ?? /?? ?? ,?? ?? = ?? ???? /?? ,?? ?? = ?? ?? /?? ,?? ?? = ?? ?? /?? ,?? ?? = -?? ?? /?? ?? ,?? ?? = ?????? /?? ) , where ?? is time, ?? and ?? are constants. At ?? = ?? ?? ,?? ?? ????? = ?? ?? ????? and velocities ?? ? ? ?? and ?? ? ? ?? of the particles are orthogonal to each other. At ?? = ?? ?? , the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is v?? ?????? ?? ?? -?? . The value of ?? is JEE Main 2025 (Online) 22nd January Morning Shift Ans: 90 Solution: Given, ?? ?? = 1???? = ?? ?? ?? ?? ??? = (?? 1 ?? 2 ??ˆ + ?? 2 ?? ??ˆ + ?? 3 ?? ?? ^ )?? ?? ?? ??? = (?? 1 ?? ??ˆ + ?? 2 ?? 2 ??ˆ + ?? 3 ?? ?? ^ )?? (?? 1 = 1?? /?? 2 ,?? 2 = 3???? /?? ,?? 3 = 2?? /?? ?? 1 = 2?? /?? ,?? 2 = -1?? /?? 2 ,?? 3 = 4???? /?? ) ?? ?? ???? = ?? ?? ?? ??? ???? = 2?? ??ˆ + 3?? ??ˆ + 2?? ^ ?? ?? ???? = ?? ?? ?? ??? ???? = 2??ˆ - 2?? ??ˆ + 4?? ?? ^ ?? ?? ???? · ?? ?? ???? = 0( As ?? ?? ???? ? ?? ?? ???? given ) ? 4?? - 6???? + 8?? = 0 At ?? = 1,4- 6?? + 8?? = 0 ? 2- 3?? + 4?? = 0 ? 3?? = 2+ 4?? ……(1) given, ?? ?? ???? = ?? ?? ???? ? 4+ 9?? 2 + 4 = 4+ 4+ 16?? 2 ? (2+ 4?? ) 2 = 16?? 2 (from (1) ? 4+ 16?? 2 + 16?? = 16?? 2 ? ?? = - 1 4 ? 3?? = 2+ 4(- 1 4 ) = 1 ? ?? = 1 3 Now, ?? ? = ?? ?? (?? ?? /?? × ?? ?? ) at ?? = 1sec , ?? ?? ?? ???? = (?? 1 - ?? 1 )??ˆ + (?? 2 - ?? 2 )??ˆ + (?? 3 - ?? 3 )?? ^ ? ?? ?? ?? ???? = (1- 2)??ˆ + (3?? + 1)??ˆ + (2- 4?? )?? ^ = -??ˆ + 2??ˆ + 3?? ^ ?? ? = ??ˆ ??ˆ ?? ˆ -1 2 3 2 1 2 = ??ˆ(4- 3)- ??ˆ(-2- 6)+ ?? ^ (-1- 4) ?? ? = ??ˆ + 8??ˆ - 5?? ˆ ? ?? ? = v1 2 + 8 2 + (-5) 2 = v1 + 64+ 25 ? ?? ? = v90 kg m 2 s -1 = v?? (given) So, L = 90 Q2: The moment of inertia of a solid disc rotating along its diameter is ?? .?? times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is ?? times higher than the moment of inertia of the given ring. Here, ?? = ____ Consider all the bodies have equal masses. JEE Main 2025 (Online) 28th January Morning Shift Ans: 4 Solution: Given, ?? 1 = 2.5?? 2 and ?? 3 = ?? ?? 2 we know, ?? 1 = ?? ?? 1 2 4 ,?? 2 = ?? ?? 2 2 2 ,?? 3 = 2 5 ?? ?? 3 2 ? ?? ?? 1 2 4 = 2.5 ?? ?? 2 2 2 ? ?? 1 2 = 5?? 2 2 Now, ?? 3 = ?? ?? 2 ? 2 5 ?? ?? 1 2 = ?? ?? ?? 2 2 2 ( As ?? 3 = ?? 1 ) ? 2 5 × 5?? 2 2 = ?? ?? 2 2 2 ? ?? = 4 Q3: Two iron solid discs of negligible thickness have radii ?? ?? and ?? ?? and moment of intertia ?? ?? and ?? ?? , respectively. For ?? ?? = ?? ?? ?? , the ratio of ?? ?? and ?? ?? would be ?? /?? , where ?? = ____ ? JEE Main 2025 (Online) 28th January Morning Shift Ans: 16 Solution: Let surface mass density = ?? So, ?? 1 = ?? × ?? ?? 1 2 ?? 2 = ?? × ?? ?? 2 2 = ???? (2?? 1 ) 2 ( As ?? 2 = 2?? 1 = 4???? ?? 1 2 ? ?? 2 = 4?? 1 ?? 1 ?? 2 = ?? 1 ?? 1 2 2 ?? 2 ?? 2 2 2 = ( ?? 1 ?? 2 )( ?? 1 ?? 2 ) 2 ? ?? 1 ?? 2 = ( ?? 1 4?? 1 )( ?? 1 2?? 1 ) 2 = 1 4 × 1 4 = 1 16 = 1 ?? Hence, ?? = 16 Q4: The coordinates of a particle with respect to origin in a given reference frame is (?? ,?? ,?? ) meters. If a force of ?? ? ? = ??ˆ - ??ˆ + ?? ˆ acts on the particle, then the magnitude of torque (with respect to origin) in ?? -direction is ____ . JEE Main 2025 (Online) 29th January Morning Shift Ans: 2 Solution: The torque ?? acting on the particle with respect to the origin can be calculated using the cross product of the position vector ?? and the force vector ?? : ?? = ?? × ?? Given the position vector ?? = (1,1,1)m and the force vector ?? = ??ˆ - ??ˆ + ?? ˆ , we need to calculate the cross product: ?? = ??ˆ ??ˆ ?? ˆ 1 1 1 1 -1 1 | Calculating the determinant, we have: ?? = ??ˆ(1· 1- 1· (-1))- ??ˆ(1· 1- 1 · 1)+ ?? ˆ (1· (-1)- 1· 1) This simplifies to: ?? = ??ˆ(1+ 1)- ??ˆ(1- 1)+ ?? ˆ (-1- 1) ?? = 2??ˆ - 0??ˆ - 2?? ˆ The torque vector is ?? = 2??ˆ - 2?? ˆ . To find the magnitude of the torque in the z -direction, we look at the ?? ˆ component: ?? ?? = -2 The magnitude of torque in the ?? -direction is: |?? ?? | = 2Nm Thus, the magnitude of the torque in the ?? -direction is 2 Newton-meters. Q5: A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of ???? ?? as shown in figure. The established torque produces an angular acceleration of ???????? /?? ?? . Moment of intertia of the wheel is ____ ???? ?? ?? . (Acceleration due to gravity = ???? ?? /?? ?? ) JEE Main 2025 (Online) 2nd April Evening Shift Ans: 1 Solution: Page 5 JEE Main Previous Year Questions (2025): Rotational Motion Q1: The position vectors of two 1 kg particles, (A) and (B), are given by ?? ?? ??? = (?? ?? ?? ?? ??ˆ + ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? and ?? ?? ??? = (?? ?? ??ˆ ??ˆ + ?? ?? ?? ?? ??ˆ + ?? ?? ?? ?? ˆ )?? , respectively; (?? ?? = ?? ?? /?? ?? ,?? ?? = ?? ???? /?? ,?? ?? = ?? ?? /?? ,?? ?? = ?? ?? /?? ,?? ?? = -?? ?? /?? ?? ,?? ?? = ?????? /?? ) , where ?? is time, ?? and ?? are constants. At ?? = ?? ?? ,?? ?? ????? = ?? ?? ????? and velocities ?? ? ? ?? and ?? ? ? ?? of the particles are orthogonal to each other. At ?? = ?? ?? , the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is v?? ?????? ?? ?? -?? . The value of ?? is JEE Main 2025 (Online) 22nd January Morning Shift Ans: 90 Solution: Given, ?? ?? = 1???? = ?? ?? ?? ?? ??? = (?? 1 ?? 2 ??ˆ + ?? 2 ?? ??ˆ + ?? 3 ?? ?? ^ )?? ?? ?? ??? = (?? 1 ?? ??ˆ + ?? 2 ?? 2 ??ˆ + ?? 3 ?? ?? ^ )?? (?? 1 = 1?? /?? 2 ,?? 2 = 3???? /?? ,?? 3 = 2?? /?? ?? 1 = 2?? /?? ,?? 2 = -1?? /?? 2 ,?? 3 = 4???? /?? ) ?? ?? ???? = ?? ?? ?? ??? ???? = 2?? ??ˆ + 3?? ??ˆ + 2?? ^ ?? ?? ???? = ?? ?? ?? ??? ???? = 2??ˆ - 2?? ??ˆ + 4?? ?? ^ ?? ?? ???? · ?? ?? ???? = 0( As ?? ?? ???? ? ?? ?? ???? given ) ? 4?? - 6???? + 8?? = 0 At ?? = 1,4- 6?? + 8?? = 0 ? 2- 3?? + 4?? = 0 ? 3?? = 2+ 4?? ……(1) given, ?? ?? ???? = ?? ?? ???? ? 4+ 9?? 2 + 4 = 4+ 4+ 16?? 2 ? (2+ 4?? ) 2 = 16?? 2 (from (1) ? 4+ 16?? 2 + 16?? = 16?? 2 ? ?? = - 1 4 ? 3?? = 2+ 4(- 1 4 ) = 1 ? ?? = 1 3 Now, ?? ? = ?? ?? (?? ?? /?? × ?? ?? ) at ?? = 1sec , ?? ?? ?? ???? = (?? 1 - ?? 1 )??ˆ + (?? 2 - ?? 2 )??ˆ + (?? 3 - ?? 3 )?? ^ ? ?? ?? ?? ???? = (1- 2)??ˆ + (3?? + 1)??ˆ + (2- 4?? )?? ^ = -??ˆ + 2??ˆ + 3?? ^ ?? ? = ??ˆ ??ˆ ?? ˆ -1 2 3 2 1 2 = ??ˆ(4- 3)- ??ˆ(-2- 6)+ ?? ^ (-1- 4) ?? ? = ??ˆ + 8??ˆ - 5?? ˆ ? ?? ? = v1 2 + 8 2 + (-5) 2 = v1 + 64+ 25 ? ?? ? = v90 kg m 2 s -1 = v?? (given) So, L = 90 Q2: The moment of inertia of a solid disc rotating along its diameter is ?? .?? times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is ?? times higher than the moment of inertia of the given ring. Here, ?? = ____ Consider all the bodies have equal masses. JEE Main 2025 (Online) 28th January Morning Shift Ans: 4 Solution: Given, ?? 1 = 2.5?? 2 and ?? 3 = ?? ?? 2 we know, ?? 1 = ?? ?? 1 2 4 ,?? 2 = ?? ?? 2 2 2 ,?? 3 = 2 5 ?? ?? 3 2 ? ?? ?? 1 2 4 = 2.5 ?? ?? 2 2 2 ? ?? 1 2 = 5?? 2 2 Now, ?? 3 = ?? ?? 2 ? 2 5 ?? ?? 1 2 = ?? ?? ?? 2 2 2 ( As ?? 3 = ?? 1 ) ? 2 5 × 5?? 2 2 = ?? ?? 2 2 2 ? ?? = 4 Q3: Two iron solid discs of negligible thickness have radii ?? ?? and ?? ?? and moment of intertia ?? ?? and ?? ?? , respectively. For ?? ?? = ?? ?? ?? , the ratio of ?? ?? and ?? ?? would be ?? /?? , where ?? = ____ ? JEE Main 2025 (Online) 28th January Morning Shift Ans: 16 Solution: Let surface mass density = ?? So, ?? 1 = ?? × ?? ?? 1 2 ?? 2 = ?? × ?? ?? 2 2 = ???? (2?? 1 ) 2 ( As ?? 2 = 2?? 1 = 4???? ?? 1 2 ? ?? 2 = 4?? 1 ?? 1 ?? 2 = ?? 1 ?? 1 2 2 ?? 2 ?? 2 2 2 = ( ?? 1 ?? 2 )( ?? 1 ?? 2 ) 2 ? ?? 1 ?? 2 = ( ?? 1 4?? 1 )( ?? 1 2?? 1 ) 2 = 1 4 × 1 4 = 1 16 = 1 ?? Hence, ?? = 16 Q4: The coordinates of a particle with respect to origin in a given reference frame is (?? ,?? ,?? ) meters. If a force of ?? ? ? = ??ˆ - ??ˆ + ?? ˆ acts on the particle, then the magnitude of torque (with respect to origin) in ?? -direction is ____ . JEE Main 2025 (Online) 29th January Morning Shift Ans: 2 Solution: The torque ?? acting on the particle with respect to the origin can be calculated using the cross product of the position vector ?? and the force vector ?? : ?? = ?? × ?? Given the position vector ?? = (1,1,1)m and the force vector ?? = ??ˆ - ??ˆ + ?? ˆ , we need to calculate the cross product: ?? = ??ˆ ??ˆ ?? ˆ 1 1 1 1 -1 1 | Calculating the determinant, we have: ?? = ??ˆ(1· 1- 1· (-1))- ??ˆ(1· 1- 1 · 1)+ ?? ˆ (1· (-1)- 1· 1) This simplifies to: ?? = ??ˆ(1+ 1)- ??ˆ(1- 1)+ ?? ˆ (-1- 1) ?? = 2??ˆ - 0??ˆ - 2?? ˆ The torque vector is ?? = 2??ˆ - 2?? ˆ . To find the magnitude of the torque in the z -direction, we look at the ?? ˆ component: ?? ?? = -2 The magnitude of torque in the ?? -direction is: |?? ?? | = 2Nm Thus, the magnitude of the torque in the ?? -direction is 2 Newton-meters. Q5: A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of ???? ?? as shown in figure. The established torque produces an angular acceleration of ???????? /?? ?? . Moment of intertia of the wheel is ____ ???? ?? ?? . (Acceleration due to gravity = ???? ?? /?? ?? ) JEE Main 2025 (Online) 2nd April Evening Shift Ans: 1 Solution: FR = I?? ? I = FR ?? = 10× 0.2 2 = 1 kg - m 2 Q6: A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is v ?? ?? where ?? = ____ . JEE Main 2025 (Online) 4th April Morning Shift Ans: 4 Solution: To determine the ratio of velocities for a circular ring and a solid sphere rolling down an inclined plane without slipping, we apply the principle of mechanical energy conservation: Conservation of Mechanical Energy: ?? ?? + ?? ?? = ?? ?? + ?? ?? Initially (at the top), we have: Initial kinetic energy, ?? ?? = 0 (since they start from rest) Initial potential energy, ?? ?? = ???? h Finally (at the bottom), we have: Final potential energy, ?? ?? = 0 Final kinetic energy, ?? ?? = 1 2 ?? ?? 2 (1 + ?? 2 ?? 2 ) This gives: ???? h = 1 2 ?? ?? 2 (1+ ?? 2 ?? 2 ) Solving for the velocity ?? :Read More
FAQs on JEE Main Previous Year Questions (2025): Rotational Motion
| 1. What is rotational motion and how does it differ from linear motion? |  |
Ans.Rotational motion refers to the motion of an object about an axis, where different points on the object move in circular paths. In contrast, linear motion involves movement along a straight line. Key differences include that in rotational motion, quantities such as angular displacement, angular velocity, and angular acceleration are used, whereas linear motion uses displacement, velocity, and acceleration. Additionally, in rotational motion, all points on a rigid object rotate about a common axis, while in linear motion, points may move independently.
| 2. What is the moment of inertia and why is it important in rotational motion? | |
Ans.The moment of inertia is a property of a body that quantifies its resistance to rotational motion about a given axis. It depends on the mass distribution of the object relative to the axis of rotation. The moment of inertia is crucial because it affects how much torque is needed to achieve a desired angular acceleration. The formula for moment of inertia varies depending on the shape of the object, for example, I = (1/3) m r² for a solid cylinder rotating about its end.
| 3. How is torque defined, and what role does it play in rotational motion? | |
Ans.Torque is defined as the rotational equivalent of linear force. It measures the tendency of a force to rotate an object about an axis and is calculated as the product of the force and the distance from the axis of rotation to the point where the force is applied, expressed as τ = r × F, where τ is torque, r is the distance (lever arm), and F is the force. Torque is essential in determining how quickly an object will rotate when a force is applied.
| 4. Can you explain the concept of angular momentum and its conservation? | |
Ans.Angular momentum is a measure of the quantity of rotation of an object and is given by the product of an object's moment of inertia and its angular velocity, represented as L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity. The principle of conservation of angular momentum states that if no external torque acts on a system, the total angular momentum of that system remains constant. This principle is crucial in various physical scenarios, such as in figure skating, where a skater pulls in their arms to spin faster.
| 5. What is the relationship between linear and angular motion? | |
Ans.Linear motion and angular motion are related through the concepts of radius and angular displacement. For example, the linear velocity (v) of a point on a rotating object is related to its angular velocity (ω) by the equation v = rω, where r is the radius from the axis of rotation to the point. Similarly, linear displacement (s) can be related to angular displacement (θ) through the equation s = rθ. Understanding this relationship allows for the analysis of rotational systems in terms of their linear counterparts.
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