Q23.A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is 7 × 10-3 J. The speed of the centre of mass of the sphere is ______ cm s-1.
What This Question Tests
This question requires knowledge of the total kinetic energy formula for a solid sphere undergoing rolling without slipping and calculating the center of mass speed.
Concepts Tested
Formulas Used
KE_rolling = 0.5Mv_cm²(1 + I/(MR²))
I_solid sphere = (2/5)MR²
📚 NCERT Sections This Tests
6.11 — Dynamics Of Rotational
Physics Class 11 · Chapter 6
6.11 Dynamics of rotational the motion of extended bodies. motion about a fixed axis A large class of problems with extended bodies can be
2.4 — A Spherical Conductor Of Radius 12 Cm Has A Charge Of 1.6 × 10–7C
Physics Class 11 · Chapter 2
2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?
5.2 — Lists The Kinetic Energies For Various X I
Physics Class 11 · Chapter 5
5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )∆xAnswer The initial kinetic energy of the bullet ∆ x → 0 ∑ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1×1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where ‘lim’ stands for the limit of the sum when 2 ∆x tends to zero. Thus, for a varying force 2 × 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳
📋 Question Details
- Chapter
- Rotation
- Topic
- Kinetic energy of rolling motion
- Year
- 2023
- Shift
- 31 Jan Shift 1
- Q Number
- Q23
- Type
- Numerical
- NCERT Ref
- Class 11 Physics Ch 7: Rotation
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