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(B) The total pressure required to blow an air bubble at a depth $h$ is given by the sum of the hydrostatic pressure and the excess pressure due to su

Vedclass · Accepted Answer

$0.0066 	imes 10^5$

Vedclass · Answer

(B) The total pressure required to blow an air bubble at a depth $h$ is given by the sum of the hydrostatic pressure and the excess pressure due to surface tension.$P = P_{atm} + h ho g + \frac{2T}{r}$.The excess pressure required over the atmospheric pressure is $P_{excess} = h ho g + \frac{2T}{r}$.Given:$r = 0.025 \ cm = 2.5 	imes 10^{-4} \ m$$h = 1 \ cm = 0.01 \ m$$T = 7 	imes 10^{-2} \ N/m$$ho = 10^3 \ kg/m^3$$g = 10 \ m/s^2$Calculating hydrostatic pressure: $h ho g = 0.01 	imes 10^3 	imes 10 = 100 \ N/m^2$.Calculating surface tension pressure: $\frac{2T}{r} = \frac{2 	imes 7 	imes 10^{-2}}{2.5 	imes 10^{-4}} = \frac{14 	imes 10^{-2}}{2.5 	imes 10^{-4}} = 5.6 	imes 10^2 = 560 \ N/m^2$.Total excess pressure $P_{excess} = 100 + 560 = 660 \ N/m^2$.$660 \ N/m^2 = 0.0066 	imes 10^5 \ N/m^2$.

$A$. $Pa \cdot s$	$(i)$ $[L^2 \ T^{-2} \ K^{-1}]$
$B$. $N \cdot m \cdot K^{-1}$	$(ii)$ $[M \ L \ T^{-3} \ K^{-1}]$
$C$. $J \cdot kg^{-1} \cdot K^{-1}$	$(iii)$ $[M \ L^{-1} \ T^{-1}]$
$D$. $W \cdot m^{-1} \cdot K^{-1}$	$(iv)$ $[M \ L^2 \ T^{-2} \ K^{-1}]$

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