1. Initial Potential Energy: Initial potential energy (PE) = mgh, where m is the mass, g is acceleration due to gravity, and h is the initial height (10 m).
2. Kinetic Energy Just Before Impact: By conservation of energy, the potential energy is converted to kinetic energy (KE) just before the object strikes the ground: KE = mgh.
3. Kinetic Energy After Impact: The object loses 50% of its KE after striking the ground, so the remaining KE is 0.5(mgh).
4. Rebound Height: The remaining KE is converted back to potential energy as the object rebounds. Let h’ be the rebound height. 0.5(mgh) = mgh’ h’ = 0.5h = 0.5(10 m) = 5 m
5. Conclusion: The rebound height is 5 m.
The terminal voltage of the battery, whose emf is and internal resistance, when connected through an external resistance of as shown in the figure is:
List-I | List-II | ||
(A) | 1 mol of H2O to O2 | (I) | 3F |
(B) | 1 mol of MnO-4 to Mn2+ | (II) | 2F |
(C) | 1.5 mol of Ca from molten CaCl2 | (III) | 1F |
(D) | 1 mol of FeO to Fe2O3 | (IV) | 5F |
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The potential (V) at any axial point, at 2 m distance(r) from the centre of the dipole of dipole moment vector
of magnitude, 4 × 10-6 C m, is ± 9 × 103 V.
(Take SI units)
Reason R : , where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below :