Three identical solid spheres move down three incline A, B and C are all of the same dimensions. A is without friction, the friction between B and a sphere is sufficient to cause rolling without slipping, the friction between C and a sphere causes rolling with slipping. The kinetic energies of A, B, C at the bottom of the inclines are $E_{A}, E_{B}, E_{C}$ :

E_{A} = E_{B} = E_{C}

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E_{A} = E_{B} > E_{C}

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E_{A} > E_{B} > E_{C}

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E_{A} > E_{B} = E_{C}

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Solution

The correct option is A $E_{A} = E_{B} > E_{C}$
Work-energy theorem for all cases: $W_{a l l f o r c e s} = Δ K . E$
Case A: frictionless inclined plane
$m g h = E_{A}$
Case B: Rough inclined plane and pure rolling states that $W_{f} = 0$
$m g h = E_{B}$
Case C: Rough inclined plane and sliding states that $W_{f} < 0$
$m g h + W_{f} = E_{C}$
Hence $E_{A} = E_{B} > E_{C}$

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If $E_{A}, E_{B}, E_{C} a n d E_{D}$ represent the magnitudes of electric fields at A, B, C and D. Which of the options are correct?

2 A^{⊝} + B_{2} \to 2 B^{⊝} + A_{2}; 2 C^{⊝} + B_{2} \to N o r e a c t i o n; 2 D^{⊝} + A_{2} \to 2 A^{⊝} + D_{2}

△ A B C

\frac{e^{A}}{A} + \frac{e^{B}}{B} + \frac{e^{C}}{C}

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