As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
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**Q1.**The position of a particle is given by r="i" ̂ + 2"j" ̂ - k ̂ and its linear momentum is given by p = 3 "i" ̂ + 4 "j" ̂ - 2 +k ̂ then its angular momentum, about the origin is perpendicular to?

**Q2.**If the earth is treated as a sphere of radius R and mass M; its angular momentum about axis of rotation with period T is?

Solution

Angular momentum is given by J=IÏ‰=((2MR

Angular momentum is given by J=IÏ‰=((2MR

^{2})/5)Ï‰ =(2MR^{2})/5×2Ï€/T=(4Ï€MR^{2})/5T**Q3.**A pulley of radius 2m is rotated about its axis by a force F=(20t-5t

^{2}) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10kg m

^{2}, the number of rotations made by the pulley before its direction of motion if reversed, is?

**Q4.**Three point masses, each of mass m are placed at the corners of an equilateral triangle of side l. Moment of inertia of this system about an axis along one side of triangle is?

Solution

Distance of corner mass from opposite side r=√(l

Distance of corner mass from opposite side r=√(l

^{2}-(l/2)^{2})=√3/2 l I=mr^{2}=3/4 ml^{2}**Q5.**A wheel is rolling along the ground with a speed of 2 ms

^{-1}. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to?

**Q6.**If the angular momentum of a rotating body about a fixed axis is increased by 10%. Its kinetic energy will be increased by?

Solution

Kinetic energy K=J

Kinetic energy K=J

^{2}/2I where J is angular momentum and I the moment of inertia. ∴K_{1}=J^{2}/2I , K_{2}=(J+10/100 J)^{2}/2I ∴K_{1}/K_{2}=(100)^{2}/(110)^{2}=100/121 % change =(K_{2}-K_{1})/K_{1}=K_{2}/K_{1}-1 =121/100-1=21%**Q7.**The diameter of a flywheel is increased by 1%. Increase in its moment of inertia about the central axis is?

Solution

I=MR

I=MR

^{2}∴ logI=logM+2logR Differentiating, we get dI/I=0+2dR/R ∴ dI/I×100=2(dR/R)×100 =2×1/100×100=2%**Q8.**A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now the platform is given an angular velocity Ï‰

_{0}. When the tortoise moves along a chord of the platform with a constant velocity (w.r.t. the platform), the angular velocity of the platform will vary with the time t as?

Solution

As there is no external torque, angular momentum will remain constant. When the tortoise moves from A to C, figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from C to B, moment of inertia increases. Therefore, angular velocity decreases

If, M=mass of platform R=radius of platform m=mass of tortoise moving along the chord AB a=perpendicular distance of O from AB Initial angular momentum, I

As there is no external torque, angular momentum will remain constant. When the tortoise moves from A to C, figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from C to B, moment of inertia increases. Therefore, angular velocity decreases

If, M=mass of platform R=radius of platform m=mass of tortoise moving along the chord AB a=perpendicular distance of O from AB Initial angular momentum, I

_{1}=mR^{2}+(MR^{2})/2 At any time t, let the tortoise reach D moving with velocity v ∴ AD=vt AC=√(R^{2}-a^{2}) As DC=AC-AD=(√(R^{2}-a^{2})-vt) ∴ OD=r=a^{2}+[√(R^{2}-a^{2})-vt]^{2}Angular momentum at time t I_{2}=mr^{2}+(MR^{2})/2 As angular momentum is conserved ∴ I_{1}Ï‰_{0}=I_{2}Ï‰(t) This shows that variation of Ï‰(t) with time is nonlinear. Choice (c) is correct**Q9.**A solid sphere is rotating about a diameter at an angular velocity Ï‰. If it cools so that its radius reduces to 1/n of its original value, its angular velocity becomes?

Solution

On applying law of conservation of angular momentum I

On applying law of conservation of angular momentum I

_{1}Ï‰_{1}=I_{2}Ï‰_{2}For solid sphere, I=2/5 mr^{2}⇒2/5 mr_{1}^{2}Ï‰_{1}=2/5 mr_{2}^{2}Ï‰_{2}r^{2}Ï‰=(r/n)^{2}Ï‰_{2}⇒Ï‰_{2}=n^{2}Ï‰**Q10.**Three identical spheres of mass M each are placed at the corners of an equilateral triangle of side 2 m. Taking one of the corners as the origin, the position vector of the centre of mass is?