Question-155032

Question Number 155032 by peter frank last updated on 24/Sep/21

Answered by peter frank last updated on 25/Sep/21

Commented by peter frank last updated on 27/Sep/21

surface tension of soap(γ_1 )= 3×10^(−2) Nm^(−1) surface tension of bubble(γ_2 )= 7×10^(−2) Nm^(−1) diameter of soap bubble(d_1 )= 10mm r=(d_1 /2) diameter of tube (d_2 )=1mm r=(d_2 /2) H=Atmospheric pressure P=pressure inside the bubble and the tube P_1 =pressure bellow the meniscuss r_1 =radius of soap bubble Excess pressure for the soap bubble P−H=((4γ_1 )/r_1 ) .....(i) r_1 =radius of soap bubble Pressure difference on the meniscuss of water P−P_2 =((2γ_2 )/r_2 ) ....(ii) But P_2 =H−ρgh...(iii) h=1mm P−(H−ρgh) =((2γ_2 )/r_2 ) P−H=((2γ_2 )/r_2 )−ρgh ....(iv) (i)=(iv) P−H=((4γ_1 )/r_1 ) =P−H=((2γ_2 )/r_2 )−ρgh ((2γ_2 )/r_2 )−ρgh =((4γ_1 )/r_1 ) ρgh=((2γ_2 )/r_2 )−((4γ_1 )/r_1 ) h=(1/(gh))(((2γ_2 )/r_2 )−((4γ_1 )/r_1 ))

$$\mathrm{surface}\:\mathrm{tension}\:\mathrm{of}\:\mathrm{soap}\left(\gamma_{\mathrm{1}} \right)= \\ $$$$\mathrm{3}×\mathrm{10}^{−\mathrm{2}} \mathrm{Nm}^{−\mathrm{1}} \\ $$$$\mathrm{surface}\:\mathrm{tension}\:\mathrm{of}\:\:\mathrm{bubble}\left(\gamma_{\mathrm{2}} \right)= \\ $$$$\mathrm{7}×\mathrm{10}^{−\mathrm{2}} \mathrm{Nm}^{−\mathrm{1}} \\ $$$$\mathrm{diameter}\:\mathrm{of}\:\mathrm{soap}\:\mathrm{bubble}\left(\mathrm{d}_{\mathrm{1}} \right)= \\ $$$$\mathrm{10mm}\:\:\:\:\mathrm{r}=\frac{\mathrm{d}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\mathrm{diameter}\:\mathrm{of}\:\mathrm{tube}\:\left(\mathrm{d}_{\mathrm{2}} \right)=\mathrm{1mm} \\ $$$$\mathrm{r}=\frac{\mathrm{d}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{H}=\mathrm{Atmospheric}\:\mathrm{pressure} \\ $$$$\mathrm{P}=\mathrm{pressure}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{bubble} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{tube} \\ $$$$\mathrm{P}_{\mathrm{1}} =\mathrm{pressure}\:\mathrm{bellow}\:\mathrm{the}\:\mathrm{meniscuss} \\ $$$$\mathrm{r}_{\mathrm{1}} =\mathrm{radius}\:\mathrm{of}\:\mathrm{soap}\:\mathrm{bubble} \\ $$$$\mathrm{Excess}\:\mathrm{pressure}\:\mathrm{for}\:\mathrm{the}\:\mathrm{soap} \\ $$$$\mathrm{bubble} \\ $$$$\mathrm{P}−\mathrm{H}=\frac{\mathrm{4}\gamma_{\mathrm{1}} }{\mathrm{r}_{\mathrm{1}} }\:\:…..\left(\mathrm{i}\right) \\ $$$$\mathrm{r}_{\mathrm{1}} =\mathrm{radius}\:\mathrm{of}\:\mathrm{soap}\:\mathrm{bubble} \\ $$$$\mathrm{Pressure}\:\mathrm{difference}\:\mathrm{on}\:\mathrm{the} \\ $$$$\:\mathrm{meniscuss}\:\mathrm{of}\:\mathrm{water} \\ $$$$\mathrm{P}−\mathrm{P}_{\mathrm{2}} =\frac{\mathrm{2}\gamma_{\mathrm{2}} }{\mathrm{r}_{\mathrm{2}} }\:\:\:\:\:….\left(\mathrm{ii}\right) \\ $$$$\mathrm{But}\:\:\mathrm{P}_{\mathrm{2}} =\mathrm{H}−\rho\mathrm{gh}…\left(\mathrm{iii}\right) \\ $$$$\mathrm{h}=\mathrm{1mm} \\ $$$$\mathrm{P}−\left(\mathrm{H}−\rho\mathrm{gh}\right)\:=\frac{\mathrm{2}\gamma_{\mathrm{2}} }{\mathrm{r}_{\mathrm{2}} }\:\:\: \\ $$$$\mathrm{P}−\mathrm{H}=\frac{\mathrm{2}\gamma_{\mathrm{2}} }{\mathrm{r}_{\mathrm{2}} }−\rho\mathrm{gh}\:….\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{i}\right)=\left(\mathrm{iv}\right) \\ $$$$\mathrm{P}−\mathrm{H}=\frac{\mathrm{4}\gamma_{\mathrm{1}} }{\mathrm{r}_{\mathrm{1}} }\:=\mathrm{P}−\mathrm{H}=\frac{\mathrm{2}\gamma_{\mathrm{2}} }{\mathrm{r}_{\mathrm{2}} }−\rho\mathrm{gh}\: \\ $$$$\frac{\mathrm{2}\gamma_{\mathrm{2}} }{\mathrm{r}_{\mathrm{2}} }−\rho\mathrm{gh}\:=\frac{\mathrm{4}\gamma_{\mathrm{1}} }{\mathrm{r}_{\mathrm{1}} } \\ $$$$\rho\mathrm{gh}=\frac{\mathrm{2}\gamma_{\mathrm{2}} }{\mathrm{r}_{\mathrm{2}} }−\frac{\mathrm{4}\gamma_{\mathrm{1}} }{\mathrm{r}_{\mathrm{1}} } \\ $$$$\mathrm{h}=\frac{\mathrm{1}}{\mathrm{gh}}\left(\frac{\mathrm{2}\gamma_{\mathrm{2}} }{\mathrm{r}_{\mathrm{2}} }−\frac{\mathrm{4}\gamma_{\mathrm{1}} }{\mathrm{r}_{\mathrm{1}} }\right) \\ $$$$ \\ $$

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