If the mass of the electron (9×10^-31 kg) is taken as unit of mass, the radius of the first Bohr orbit (0.5×10-¹⁰m) as unit of length and 500 newton as
the unit of force, then the unit of time in the new system would be

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15 × 10^–12s

3 × 10^–22s

15 × 10^–20s

45 × 10^–20s

answer is A.

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Detailed Solution

$F = {MLT}^{- 2} = \frac{ML}{T^{2}}$
$T^{2} = \frac{ML}{F} = \frac{9 \times 10^{- 31} \times 5 \times 10^{- 11}}{500}$
$= 9 \times 10^{- 31} \times 10^{- 11} \times 10^{- 2} = 9 \times 10^{- 44}$
$T = 3 \times 10^{- 22} \sec$

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