142830
One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is
1 $1: 1000$
2 $1: 10$
3 $1: 100$
4 $1: 1$
Explanation:
B Given, Let, Radius of small water drop $=\mathrm{r}$ Radius of big water drop $=\mathrm{R}$ According to question, Volume of big drop $=1000 \times$ volume of small drops. $\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{r}^{3} \times 1000$ $\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}=1000$ $\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=(10)^{3}$ $\frac{\mathrm{R}}{\mathrm{r}}=10$ We know that, $\text { Surface energy of drop }(\mathrm{U})=\mathrm{A} \times \mathrm{T}$ For big drop - $\mathrm{U}_{1}=4 \pi \mathrm{R}^{2} \mathrm{~T}$ Surface energy of small drops - $\mathrm{U}_{2}=1000 \times 4 \pi \mathrm{r}^{2} \mathrm{~T}$ Ratio of final surface energy to the total initial surface energy, $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{4 \pi \mathrm{T} \times \mathrm{R}^{2}}{1000 \times 4 \pi \mathrm{T} \times \mathrm{r}^{2}}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{2} \times \frac{1}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=(10)^{2} \times \frac{1}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{100}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{1}{10}$ $\text { or } \mathrm{U}_{1}: \mathrm{U}_{2}=1: 10$
MHT-CET 2020
Mechanical Properties of Fluids
142831
The excess pressure inside a spherical drop of water is three time that of another drop of water. The ratio of their surface area is
1 $3: 1$
2 $6: 1$
3 $1: 3$
4 $1: 9$
Explanation:
D Given that, Pressure inside a spherical drop $\left(\mathrm{P}_{1}\right)=3 \mathrm{P}_{2}$ Pressure inside drop water $=\mathrm{P}_{2}$ We know that, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{r}}$ According to question $\mathrm{P}_{1}=3 \mathrm{P}_{2}$ $\frac{2 T}{r_{1}}=3 \times \frac{2 T}{r_{2}}$ $r_{2}=3 r_{1}$ $A=4 \pi r^{2}$ $\frac{A_{1}}{A_{2}}= \frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2} \quad\left(r_{1}=\frac{r_{2}}{3}\right)$ $\frac{A_{1}}{A_{2}}=\left(\frac{r_{2}}{3 r_{2}}\right)^{2}$ $\frac{A_{1}}{A_{2}}=\frac{1}{9}$
MHT-CET 2020
Mechanical Properties of Fluids
142832
Two small drops of mercury each of radius ' $R$ ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is
1 $2^{2 / 3}: 1$
2 $2^{1 / 3}: 1$
3 $2: 1$
4 $\sqrt{2}: 1$
Explanation:
B Given that, radius of large single drops $=\mathrm{R}^{\prime}$ According to question, Volume of two small drops $=$ volume of single largedrop $2 \mathrm{~V}=\mathrm{V}^{\prime}$ $2 \times \frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{R}^{\prime 3}$ $2 \mathrm{R}^{3}=\mathrm{R}^{\prime 3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)^{3}=\frac{1}{2} \Rightarrow\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)=\left(\frac{1}{2}\right)^{1 / 3}$ $\therefore$ We know that surface energy for water drops- $\mathrm{U}=4 \pi \mathrm{R}^{2} \times \mathrm{T}$ According to question- $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}}{4 \pi \mathrm{R}^{\prime 2} \times \mathrm{T}}=2\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)^{2}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2 \times\left[\left(\frac{1}{2}\right)^{1 / 3}\right]^{2}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{1} \times \frac{1}{2^{2 / 3}}=2^{1} \times 2^{-2 / 3}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{\left(1-\frac{2}{3}\right)}=2^{\left(\frac{3-2}{3}\right)}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{1 / 3}$ But officially option (a) in correct.
MHT-CET 2020
Mechanical Properties of Fluids
142833
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is
C Given that, Radius of first bubbles $=\mathrm{r}_{1}$ Radius of second bubbles $=r_{2}$ $\therefore \quad$ The excess pressure $\left(\mathrm{P}_{1}\right)$ developed inside the first bubble is given by- $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} .$ Again, for the $2^{\text {nd }}$ bubble- $\mathrm{P}_{2}=\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ Let the radius of final bubble is ' $\mathrm{R}$ ' then the pressure inside it- $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \ldots \ldots$ According to question bubbles are coalesce in isothermal condition- $\therefore \quad \mathrm{PV}=\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}$ $\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \times \frac{4}{3} \pi \mathrm{R}^{3}=\left(\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\left(\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}$ $\left(4 \mathrm{~T} \times \frac{4}{3} \pi\right) \times \frac{\mathrm{R}^{3}}{\mathrm{R}}=4 \mathrm{~T} \times \frac{4 \pi}{3}\left[\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{1}}+\frac{\mathrm{r}_{2}^{3}}{\mathrm{r}_{2}}\right]$ $\mathrm{R}^{2}=\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}$ $\mathrm{R}=\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}\right)^{1 / 2}$
142830
One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is
1 $1: 1000$
2 $1: 10$
3 $1: 100$
4 $1: 1$
Explanation:
B Given, Let, Radius of small water drop $=\mathrm{r}$ Radius of big water drop $=\mathrm{R}$ According to question, Volume of big drop $=1000 \times$ volume of small drops. $\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{r}^{3} \times 1000$ $\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}=1000$ $\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=(10)^{3}$ $\frac{\mathrm{R}}{\mathrm{r}}=10$ We know that, $\text { Surface energy of drop }(\mathrm{U})=\mathrm{A} \times \mathrm{T}$ For big drop - $\mathrm{U}_{1}=4 \pi \mathrm{R}^{2} \mathrm{~T}$ Surface energy of small drops - $\mathrm{U}_{2}=1000 \times 4 \pi \mathrm{r}^{2} \mathrm{~T}$ Ratio of final surface energy to the total initial surface energy, $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{4 \pi \mathrm{T} \times \mathrm{R}^{2}}{1000 \times 4 \pi \mathrm{T} \times \mathrm{r}^{2}}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{2} \times \frac{1}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=(10)^{2} \times \frac{1}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{100}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{1}{10}$ $\text { or } \mathrm{U}_{1}: \mathrm{U}_{2}=1: 10$
MHT-CET 2020
Mechanical Properties of Fluids
142831
The excess pressure inside a spherical drop of water is three time that of another drop of water. The ratio of their surface area is
1 $3: 1$
2 $6: 1$
3 $1: 3$
4 $1: 9$
Explanation:
D Given that, Pressure inside a spherical drop $\left(\mathrm{P}_{1}\right)=3 \mathrm{P}_{2}$ Pressure inside drop water $=\mathrm{P}_{2}$ We know that, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{r}}$ According to question $\mathrm{P}_{1}=3 \mathrm{P}_{2}$ $\frac{2 T}{r_{1}}=3 \times \frac{2 T}{r_{2}}$ $r_{2}=3 r_{1}$ $A=4 \pi r^{2}$ $\frac{A_{1}}{A_{2}}= \frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2} \quad\left(r_{1}=\frac{r_{2}}{3}\right)$ $\frac{A_{1}}{A_{2}}=\left(\frac{r_{2}}{3 r_{2}}\right)^{2}$ $\frac{A_{1}}{A_{2}}=\frac{1}{9}$
MHT-CET 2020
Mechanical Properties of Fluids
142832
Two small drops of mercury each of radius ' $R$ ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is
1 $2^{2 / 3}: 1$
2 $2^{1 / 3}: 1$
3 $2: 1$
4 $\sqrt{2}: 1$
Explanation:
B Given that, radius of large single drops $=\mathrm{R}^{\prime}$ According to question, Volume of two small drops $=$ volume of single largedrop $2 \mathrm{~V}=\mathrm{V}^{\prime}$ $2 \times \frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{R}^{\prime 3}$ $2 \mathrm{R}^{3}=\mathrm{R}^{\prime 3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)^{3}=\frac{1}{2} \Rightarrow\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)=\left(\frac{1}{2}\right)^{1 / 3}$ $\therefore$ We know that surface energy for water drops- $\mathrm{U}=4 \pi \mathrm{R}^{2} \times \mathrm{T}$ According to question- $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}}{4 \pi \mathrm{R}^{\prime 2} \times \mathrm{T}}=2\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)^{2}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2 \times\left[\left(\frac{1}{2}\right)^{1 / 3}\right]^{2}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{1} \times \frac{1}{2^{2 / 3}}=2^{1} \times 2^{-2 / 3}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{\left(1-\frac{2}{3}\right)}=2^{\left(\frac{3-2}{3}\right)}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{1 / 3}$ But officially option (a) in correct.
MHT-CET 2020
Mechanical Properties of Fluids
142833
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is
C Given that, Radius of first bubbles $=\mathrm{r}_{1}$ Radius of second bubbles $=r_{2}$ $\therefore \quad$ The excess pressure $\left(\mathrm{P}_{1}\right)$ developed inside the first bubble is given by- $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} .$ Again, for the $2^{\text {nd }}$ bubble- $\mathrm{P}_{2}=\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ Let the radius of final bubble is ' $\mathrm{R}$ ' then the pressure inside it- $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \ldots \ldots$ According to question bubbles are coalesce in isothermal condition- $\therefore \quad \mathrm{PV}=\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}$ $\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \times \frac{4}{3} \pi \mathrm{R}^{3}=\left(\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\left(\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}$ $\left(4 \mathrm{~T} \times \frac{4}{3} \pi\right) \times \frac{\mathrm{R}^{3}}{\mathrm{R}}=4 \mathrm{~T} \times \frac{4 \pi}{3}\left[\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{1}}+\frac{\mathrm{r}_{2}^{3}}{\mathrm{r}_{2}}\right]$ $\mathrm{R}^{2}=\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}$ $\mathrm{R}=\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}\right)^{1 / 2}$
142830
One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is
1 $1: 1000$
2 $1: 10$
3 $1: 100$
4 $1: 1$
Explanation:
B Given, Let, Radius of small water drop $=\mathrm{r}$ Radius of big water drop $=\mathrm{R}$ According to question, Volume of big drop $=1000 \times$ volume of small drops. $\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{r}^{3} \times 1000$ $\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}=1000$ $\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=(10)^{3}$ $\frac{\mathrm{R}}{\mathrm{r}}=10$ We know that, $\text { Surface energy of drop }(\mathrm{U})=\mathrm{A} \times \mathrm{T}$ For big drop - $\mathrm{U}_{1}=4 \pi \mathrm{R}^{2} \mathrm{~T}$ Surface energy of small drops - $\mathrm{U}_{2}=1000 \times 4 \pi \mathrm{r}^{2} \mathrm{~T}$ Ratio of final surface energy to the total initial surface energy, $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{4 \pi \mathrm{T} \times \mathrm{R}^{2}}{1000 \times 4 \pi \mathrm{T} \times \mathrm{r}^{2}}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{2} \times \frac{1}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=(10)^{2} \times \frac{1}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{100}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{1}{10}$ $\text { or } \mathrm{U}_{1}: \mathrm{U}_{2}=1: 10$
MHT-CET 2020
Mechanical Properties of Fluids
142831
The excess pressure inside a spherical drop of water is three time that of another drop of water. The ratio of their surface area is
1 $3: 1$
2 $6: 1$
3 $1: 3$
4 $1: 9$
Explanation:
D Given that, Pressure inside a spherical drop $\left(\mathrm{P}_{1}\right)=3 \mathrm{P}_{2}$ Pressure inside drop water $=\mathrm{P}_{2}$ We know that, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{r}}$ According to question $\mathrm{P}_{1}=3 \mathrm{P}_{2}$ $\frac{2 T}{r_{1}}=3 \times \frac{2 T}{r_{2}}$ $r_{2}=3 r_{1}$ $A=4 \pi r^{2}$ $\frac{A_{1}}{A_{2}}= \frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2} \quad\left(r_{1}=\frac{r_{2}}{3}\right)$ $\frac{A_{1}}{A_{2}}=\left(\frac{r_{2}}{3 r_{2}}\right)^{2}$ $\frac{A_{1}}{A_{2}}=\frac{1}{9}$
MHT-CET 2020
Mechanical Properties of Fluids
142832
Two small drops of mercury each of radius ' $R$ ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is
1 $2^{2 / 3}: 1$
2 $2^{1 / 3}: 1$
3 $2: 1$
4 $\sqrt{2}: 1$
Explanation:
B Given that, radius of large single drops $=\mathrm{R}^{\prime}$ According to question, Volume of two small drops $=$ volume of single largedrop $2 \mathrm{~V}=\mathrm{V}^{\prime}$ $2 \times \frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{R}^{\prime 3}$ $2 \mathrm{R}^{3}=\mathrm{R}^{\prime 3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)^{3}=\frac{1}{2} \Rightarrow\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)=\left(\frac{1}{2}\right)^{1 / 3}$ $\therefore$ We know that surface energy for water drops- $\mathrm{U}=4 \pi \mathrm{R}^{2} \times \mathrm{T}$ According to question- $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}}{4 \pi \mathrm{R}^{\prime 2} \times \mathrm{T}}=2\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)^{2}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2 \times\left[\left(\frac{1}{2}\right)^{1 / 3}\right]^{2}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{1} \times \frac{1}{2^{2 / 3}}=2^{1} \times 2^{-2 / 3}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{\left(1-\frac{2}{3}\right)}=2^{\left(\frac{3-2}{3}\right)}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{1 / 3}$ But officially option (a) in correct.
MHT-CET 2020
Mechanical Properties of Fluids
142833
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is
C Given that, Radius of first bubbles $=\mathrm{r}_{1}$ Radius of second bubbles $=r_{2}$ $\therefore \quad$ The excess pressure $\left(\mathrm{P}_{1}\right)$ developed inside the first bubble is given by- $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} .$ Again, for the $2^{\text {nd }}$ bubble- $\mathrm{P}_{2}=\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ Let the radius of final bubble is ' $\mathrm{R}$ ' then the pressure inside it- $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \ldots \ldots$ According to question bubbles are coalesce in isothermal condition- $\therefore \quad \mathrm{PV}=\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}$ $\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \times \frac{4}{3} \pi \mathrm{R}^{3}=\left(\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\left(\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}$ $\left(4 \mathrm{~T} \times \frac{4}{3} \pi\right) \times \frac{\mathrm{R}^{3}}{\mathrm{R}}=4 \mathrm{~T} \times \frac{4 \pi}{3}\left[\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{1}}+\frac{\mathrm{r}_{2}^{3}}{\mathrm{r}_{2}}\right]$ $\mathrm{R}^{2}=\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}$ $\mathrm{R}=\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}\right)^{1 / 2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Mechanical Properties of Fluids
142830
One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is
1 $1: 1000$
2 $1: 10$
3 $1: 100$
4 $1: 1$
Explanation:
B Given, Let, Radius of small water drop $=\mathrm{r}$ Radius of big water drop $=\mathrm{R}$ According to question, Volume of big drop $=1000 \times$ volume of small drops. $\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{r}^{3} \times 1000$ $\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}=1000$ $\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=(10)^{3}$ $\frac{\mathrm{R}}{\mathrm{r}}=10$ We know that, $\text { Surface energy of drop }(\mathrm{U})=\mathrm{A} \times \mathrm{T}$ For big drop - $\mathrm{U}_{1}=4 \pi \mathrm{R}^{2} \mathrm{~T}$ Surface energy of small drops - $\mathrm{U}_{2}=1000 \times 4 \pi \mathrm{r}^{2} \mathrm{~T}$ Ratio of final surface energy to the total initial surface energy, $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{4 \pi \mathrm{T} \times \mathrm{R}^{2}}{1000 \times 4 \pi \mathrm{T} \times \mathrm{r}^{2}}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{2} \times \frac{1}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=(10)^{2} \times \frac{1}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{100}{1000}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{1}{10}$ $\text { or } \mathrm{U}_{1}: \mathrm{U}_{2}=1: 10$
MHT-CET 2020
Mechanical Properties of Fluids
142831
The excess pressure inside a spherical drop of water is three time that of another drop of water. The ratio of their surface area is
1 $3: 1$
2 $6: 1$
3 $1: 3$
4 $1: 9$
Explanation:
D Given that, Pressure inside a spherical drop $\left(\mathrm{P}_{1}\right)=3 \mathrm{P}_{2}$ Pressure inside drop water $=\mathrm{P}_{2}$ We know that, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{r}}$ According to question $\mathrm{P}_{1}=3 \mathrm{P}_{2}$ $\frac{2 T}{r_{1}}=3 \times \frac{2 T}{r_{2}}$ $r_{2}=3 r_{1}$ $A=4 \pi r^{2}$ $\frac{A_{1}}{A_{2}}= \frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2} \quad\left(r_{1}=\frac{r_{2}}{3}\right)$ $\frac{A_{1}}{A_{2}}=\left(\frac{r_{2}}{3 r_{2}}\right)^{2}$ $\frac{A_{1}}{A_{2}}=\frac{1}{9}$
MHT-CET 2020
Mechanical Properties of Fluids
142832
Two small drops of mercury each of radius ' $R$ ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is
1 $2^{2 / 3}: 1$
2 $2^{1 / 3}: 1$
3 $2: 1$
4 $\sqrt{2}: 1$
Explanation:
B Given that, radius of large single drops $=\mathrm{R}^{\prime}$ According to question, Volume of two small drops $=$ volume of single largedrop $2 \mathrm{~V}=\mathrm{V}^{\prime}$ $2 \times \frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{R}^{\prime 3}$ $2 \mathrm{R}^{3}=\mathrm{R}^{\prime 3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)^{3}=\frac{1}{2} \Rightarrow\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)=\left(\frac{1}{2}\right)^{1 / 3}$ $\therefore$ We know that surface energy for water drops- $\mathrm{U}=4 \pi \mathrm{R}^{2} \times \mathrm{T}$ According to question- $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=\frac{2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}}{4 \pi \mathrm{R}^{\prime 2} \times \mathrm{T}}=2\left(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\right)^{2}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2 \times\left[\left(\frac{1}{2}\right)^{1 / 3}\right]^{2}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{1} \times \frac{1}{2^{2 / 3}}=2^{1} \times 2^{-2 / 3}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{\left(1-\frac{2}{3}\right)}=2^{\left(\frac{3-2}{3}\right)}$ $\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=2^{1 / 3}$ But officially option (a) in correct.
MHT-CET 2020
Mechanical Properties of Fluids
142833
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is
C Given that, Radius of first bubbles $=\mathrm{r}_{1}$ Radius of second bubbles $=r_{2}$ $\therefore \quad$ The excess pressure $\left(\mathrm{P}_{1}\right)$ developed inside the first bubble is given by- $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} .$ Again, for the $2^{\text {nd }}$ bubble- $\mathrm{P}_{2}=\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ Let the radius of final bubble is ' $\mathrm{R}$ ' then the pressure inside it- $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \ldots \ldots$ According to question bubbles are coalesce in isothermal condition- $\therefore \quad \mathrm{PV}=\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}$ $\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \times \frac{4}{3} \pi \mathrm{R}^{3}=\left(\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\left(\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}$ $\left(4 \mathrm{~T} \times \frac{4}{3} \pi\right) \times \frac{\mathrm{R}^{3}}{\mathrm{R}}=4 \mathrm{~T} \times \frac{4 \pi}{3}\left[\frac{\mathrm{r}_{1}^{3}}{\mathrm{r}_{1}}+\frac{\mathrm{r}_{2}^{3}}{\mathrm{r}_{2}}\right]$ $\mathrm{R}^{2}=\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}$ $\mathrm{R}=\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}\right)^{1 / 2}$
