The tyre of a bicycle has a volume of 2 times | vedclass

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(C) Given: Initial volume of air $V_0 = 0.75 	imes 2 	imes 10^{-3} \ m^3 = 1.5 	imes 10^{-3} \ m^3$.The total pressure $P$ inside the tyre when the

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$21$

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(C) Given: Initial volume of air $V_0 = 0.75 	imes 2 	imes 10^{-3} \ m^3 = 1.5 	imes 10^{-3} \ m^3$.The total pressure $P$ inside the tyre when the rider is on it is the sum of atmospheric pressure $P_0$ and the pressure due to the load: $P = P_0 + \frac{mg}{A} = 10^5 + \frac{120 	imes 10}{24 	imes 10^{-4}} = 10^5 + 5 	imes 10^5 = 6 	imes 10^5 \ N \ m^{-2}$.Assuming temperature $T$ remains constant,we use Boyle's Law $(P_1 V_1 = P_2 V_2)$:$P_0 V_{initial} = P V_{final}$$10^5 	imes V_{initial} = 6 	imes 10^5 	imes 2 	imes 10^{-3}$$V_{initial} = 12 	imes 10^{-3} \ m^3$.The volume of air to be added is $\Delta V = V_{initial} - V_0 = 12 	imes 10^{-3} - 1.5 	imes 10^{-3} = 10.5 	imes 10^{-3} \ m^3$.Volume per stroke $v = 500 \ cm^3 = 500 	imes 10^{-6} \ m^3 = 0.5 	imes 10^{-3} \ m^3$.Number of strokes $n = \frac{\Delta V}{v} = \frac{10.5 	imes 10^{-3}}{0.5 	imes 10^{-3}} = 21$.

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