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Concept

• Due to surface tension, pressure on the concave side of the soap bubble, liquid drop, etc is greater than the pressure on the convex side of them.
• This is why pressure inside the liquid drops, the air bubbles are greater than the pressure outside.

The excess pressure inside a liquid drop is given by:

$P= {2T \over R}$

where T is the surface tension and R is the radius of drop.

Calculation:

r = $\frac{r_{1} \cdot r_{2}}{r_{1}-r_{2}}=\frac{2 \cdot 4}{4-2}$ = 4cm

Calculation: 

v = $\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}}=\frac{-24 \cdot 1}{-24-1}=\frac{24}{25}$

m = $\frac{-\mathrm{v}}{\mathrm{u}}=-\frac{24}{25(-24)}=\frac{1}{25}$

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$

\(\mathrm{v}{\mathrm{I}}=-\mathrm{m}^{2} \cdot \mathrm{v}{0}=\frac{-1}{(25)^{2}} \cdot 25=\frac{-1}{25}\)

Diff. 

\(\frac{-1}{\mathrm{v}^{2}}\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)+\frac{1}{\mathrm{u}^{2}}\left(\frac{\mathrm{du}}{\mathrm{dt}}\right)=0 \quad\left[\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v}{\mathrm{I}} ; \frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{v}{0}\right]\)

\(\frac{+2}{\mathrm{v}^{3}} \cdot\left(\mathrm{v}{\mathrm{I}}\right)^{2}-\frac{1}{\mathrm{v}^{2}} \cdot \mathrm{a}{\mathrm{I}}-\frac{2}{\mathrm{u}^{3}} \cdot\left(\mathrm{v}{0}\right)^{2}+\frac{1}{\mathrm{u}^{2}} \cdot \mathrm{a}{0}=0\)

\(\frac{\mathrm{a}{\mathrm{I}}}{\mathrm{v}^{2}}=\frac{2}{\mathrm{v}^{3}} \cdot \mathrm{v}{\mathrm{I}}^{2}-\frac{2}{\mathrm{u}^{3}} \cdot \mathrm{v}_{0}^{2}\)

\(\mathrm{a}{\mathrm{I}}=\frac{2}{\mathrm{v}} \cdot \mathrm{v}{\mathrm{I}}^{2}-\frac{2 \mathrm{v}^{2}}{\mathrm{u}^{3}} \cdot \mathrm{v}_{0}^{2}\)

= $\frac{2 \cdot 25}{24} \cdot \frac{1}{25} \cdot \frac{1}{25}-\frac{2}{(24)^{3}} \cdot \frac{24}{25} \cdot \frac{24}{25} \cdot 25 \cdot 25$

$\mathrm{a}_{\mathrm{I}}=\frac{2}{24.25}-\frac{2}{24}$

$\mathrm{a}_{\mathrm{I}}=\frac{2}{24} \cdot \frac{-24}{25}=\frac{-2}{25}$

$100 \mathrm{a}_{\mathrm{I}}=\frac{2}{25} \times 100=8$

Calculation:

\(\mathrm{R}{1}=\frac{2 \mathrm{~L}}{\mathrm{~K}{1} \mathrm{~A}}\)

\(\mathrm{R}{2}=\frac{2 \mathrm{~L}}{\mathrm{~K}{2} \mathrm{~A}}\)

\(\mathrm{R}{3}=\frac{\mathrm{L}}{\mathrm{~K}{3} \mathrm{~A}}\)

\(\frac{θ-100}{\frac{\mathrm{R}{1} \mathrm{R}{2}}{\mathrm{R}{1}+\mathrm{R}{2}}}+\frac{θ-0}{\mathrm{R}_{3}}=0\)

∴ θ = 40

Calculation:

$V_{A}=(2 t \hat{i}+3 n \hat{j}+2 \hat{k})$

$\vec{V}_{B}=(2 \hat{i}-2 \hat{t} \hat{j}+4 \hat{p} \hat{k})$

\(\overrightarrow{\mathrm{V}}{\mathrm{A}} \cdot \overrightarrow{\mathrm{~V}}{\mathrm{B}}=0\)

4 – 6n + 8p = 0

2 – 3n + 4p = 0

3n = 2 + 4p

 \(\left|\overrightarrow{\mathrm{V}}{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{V}}{\mathrm{B}}\right|\)

4 + 9n² + 4 = 4 + 4 + 16p²

p = $\frac{-1}{4}$ ⇒ n = $\frac{1}{3}$

\(\overrightarrow{\mathrm{L}}=\mathrm{m}{\mathrm{A}}\left(\overrightarrow{\mathrm{r}}{\mathrm{A} / \mathrm{B}} \times \overrightarrow{\mathrm{V}}_{\mathrm{A}}\right)\)

\(\overrightarrow{\mathrm{r}}{\mathrm{A} / \mathrm{B}}=\left(\alpha{1}-\beta_{1}\right) \hat{\mathrm{i}}+\left(\alpha_{2}-\beta_{2}\right) \hat{\mathrm{j}}+\left(\alpha_{3}-\beta_{3}\right)\)

= $(1-2) \hat{i}+(1+1) \hat{j}+3 \hat{k}$

= $\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\bf{k}} \ -1 & 2 & 3 \ 2 & 1 & 2 \end{array}\right|=\hat{\bf{i}}+8 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$

= $\sqrt{1+64+25}=\sqrt{90}$

Calculation:

S₁ = 30 × 1 - $\frac{1}{2}$ × 10 × 1 = 25

S3 = 30 + $\left(\frac{-10}{2}\right)$ × (2 × 3 - 1) = 5

\(\frac{\mathrm{S}{1}}{\mathrm{~S}{3}}=\frac{25}{5}\) = 5