142847
When ' $n$ ' number of liquid droplets each of radius ' $r$ ' are merged to form a single drop of radius ' $R$ ', the energy loss is ' $4 E$ '. Where $E$ is the energy of the bigger drop then, the number of droplets ' $n$ ' is
1 $\frac{5 R}{r}$
2 $\frac{5 \mathrm{R}}{\mathrm{r}^{2}}$
3 $\frac{4 R^{2}}{r^{2}}$
4 $\frac{5 R^{2}}{r^{2}}$
Explanation:
D According to question- $\frac{4}{3} \pi r^{3} n=\frac{4}{3} \pi R^{3}$ $r^{3} n=R^{3}$ Energy loss $(4 \mathrm{E})=4 \pi\left(\mathrm{r}^{2} n-\mathrm{R}^{2}\right) \mathrm{T}$ $4\left(4 \pi R^{2} T\right) =4 \pi\left(r^{2} n-R^{2}\right) T$ $4 R^{2}+R^{2} =r^{2} n$ $n =\frac{5 R^{2}}{r^{2}}$
MHT-CET 2020
Mechanical Properties of Fluids
142849
Two soap bubbles have radii in the ratio $4: 3$. What is the ratio of work done to blow these bubbles ?
1 $3: 2$
2 $4: 3$
3 $8: 3$
4 $16: 9$
Explanation:
D Let $r_{1}$ and $r_{2}$ be the radii of two soap bubbles then $\mathrm{r}_{1}: \mathrm{r}_{2}=4: 3$ Now, work done to blow both bubbles is given by $\mathrm{W}_{1}=2 \mathrm{~T}\left(4 \pi \mathrm{r}_{1}^{2}\right)$ and $\quad \mathrm{W}_{2}=2 \mathrm{~T}\left(4 \pi \mathrm{r}_{2}^{2}\right)$ $\text { Then, } \frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}} =\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2}$ $=\left(\frac{4}{3}\right)^{2}=\frac{16}{9}$ Therefore, the ratio of work done to blow these bubbles is $16: 9$.
MHT-CET 2019
Mechanical Properties of Fluids
142852
The excess of pressure, due to surface tension, on a spherical liquid drop of radius ' $R$ ' is proportional to
1 $\mathrm{R}^{2}$
2 $\mathrm{R}^{-2}$
3 $\mathrm{R}$
4 $\mathrm{R}^{-1}$
Explanation:
D The excess of pressure due to surface tension on a spherical liquid drop is given by - $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ Where, $\mathrm{T}=$ Surface tension $\mathrm{R}=$ Radius of liquid drop From equation (i), we get - $\mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\mathrm{P} \propto \mathrm{R}^{-1}$
MHT-CET 2019
Mechanical Properties of Fluids
142854
A liquid drop having surface energy ' $E$ ' is spread into 512 droplets of same size. The final surface energy of the droplets is
1 $2 \mathrm{E}$
2 $4 \mathrm{E}$
3 $8 \mathrm{E}$
4 $12 \mathrm{E}$
Explanation:
C The surface area of the liquid drop is, $A_{1}=4 \pi R^{2}$ Its surface energy is $\mathrm{E}$. When the drop spread in 512 droplets, the surface area of droplets is $\mathrm{A}_{2}=512 \times 4 \pi \mathrm{r}^{2}$. The volume of bigger drop is $\frac{4}{3} \pi R^{3}$ and volume of small droplets is $512 \times \frac{4}{3} \pi \mathrm{r}^{3}$. Then, $\quad \frac{4}{3} \pi \mathrm{R}^{3}=512 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{r}=\frac{\mathrm{R}}{8}$ $\mathrm{~A}_{2}=512 \times 4 \pi \mathrm{r}^{2}$ $512 \times 4 \pi\left(\frac{\mathrm{R}}{8}\right)^{2}=8 \mathrm{~A}_{1}$ Surface energy $(\mathrm{E})=\mathrm{A}$. $\mathrm{T}$ Where, $A=$ area, $T=$ surface tension $\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{A}_{2} \cdot \mathrm{T}}{\mathrm{A}_{1} \cdot \mathrm{T}}=\frac{8 \mathrm{~A}_{1}}{\mathrm{~A}_{1}}=8$ $\mathrm{E}_{2}=8 \mathrm{E}_{1}$ Then $\mathrm{E}_{\mathrm{n}}=8 \mathrm{E}$
142847
When ' $n$ ' number of liquid droplets each of radius ' $r$ ' are merged to form a single drop of radius ' $R$ ', the energy loss is ' $4 E$ '. Where $E$ is the energy of the bigger drop then, the number of droplets ' $n$ ' is
1 $\frac{5 R}{r}$
2 $\frac{5 \mathrm{R}}{\mathrm{r}^{2}}$
3 $\frac{4 R^{2}}{r^{2}}$
4 $\frac{5 R^{2}}{r^{2}}$
Explanation:
D According to question- $\frac{4}{3} \pi r^{3} n=\frac{4}{3} \pi R^{3}$ $r^{3} n=R^{3}$ Energy loss $(4 \mathrm{E})=4 \pi\left(\mathrm{r}^{2} n-\mathrm{R}^{2}\right) \mathrm{T}$ $4\left(4 \pi R^{2} T\right) =4 \pi\left(r^{2} n-R^{2}\right) T$ $4 R^{2}+R^{2} =r^{2} n$ $n =\frac{5 R^{2}}{r^{2}}$
MHT-CET 2020
Mechanical Properties of Fluids
142849
Two soap bubbles have radii in the ratio $4: 3$. What is the ratio of work done to blow these bubbles ?
1 $3: 2$
2 $4: 3$
3 $8: 3$
4 $16: 9$
Explanation:
D Let $r_{1}$ and $r_{2}$ be the radii of two soap bubbles then $\mathrm{r}_{1}: \mathrm{r}_{2}=4: 3$ Now, work done to blow both bubbles is given by $\mathrm{W}_{1}=2 \mathrm{~T}\left(4 \pi \mathrm{r}_{1}^{2}\right)$ and $\quad \mathrm{W}_{2}=2 \mathrm{~T}\left(4 \pi \mathrm{r}_{2}^{2}\right)$ $\text { Then, } \frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}} =\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2}$ $=\left(\frac{4}{3}\right)^{2}=\frac{16}{9}$ Therefore, the ratio of work done to blow these bubbles is $16: 9$.
MHT-CET 2019
Mechanical Properties of Fluids
142852
The excess of pressure, due to surface tension, on a spherical liquid drop of radius ' $R$ ' is proportional to
1 $\mathrm{R}^{2}$
2 $\mathrm{R}^{-2}$
3 $\mathrm{R}$
4 $\mathrm{R}^{-1}$
Explanation:
D The excess of pressure due to surface tension on a spherical liquid drop is given by - $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ Where, $\mathrm{T}=$ Surface tension $\mathrm{R}=$ Radius of liquid drop From equation (i), we get - $\mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\mathrm{P} \propto \mathrm{R}^{-1}$
MHT-CET 2019
Mechanical Properties of Fluids
142854
A liquid drop having surface energy ' $E$ ' is spread into 512 droplets of same size. The final surface energy of the droplets is
1 $2 \mathrm{E}$
2 $4 \mathrm{E}$
3 $8 \mathrm{E}$
4 $12 \mathrm{E}$
Explanation:
C The surface area of the liquid drop is, $A_{1}=4 \pi R^{2}$ Its surface energy is $\mathrm{E}$. When the drop spread in 512 droplets, the surface area of droplets is $\mathrm{A}_{2}=512 \times 4 \pi \mathrm{r}^{2}$. The volume of bigger drop is $\frac{4}{3} \pi R^{3}$ and volume of small droplets is $512 \times \frac{4}{3} \pi \mathrm{r}^{3}$. Then, $\quad \frac{4}{3} \pi \mathrm{R}^{3}=512 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{r}=\frac{\mathrm{R}}{8}$ $\mathrm{~A}_{2}=512 \times 4 \pi \mathrm{r}^{2}$ $512 \times 4 \pi\left(\frac{\mathrm{R}}{8}\right)^{2}=8 \mathrm{~A}_{1}$ Surface energy $(\mathrm{E})=\mathrm{A}$. $\mathrm{T}$ Where, $A=$ area, $T=$ surface tension $\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{A}_{2} \cdot \mathrm{T}}{\mathrm{A}_{1} \cdot \mathrm{T}}=\frac{8 \mathrm{~A}_{1}}{\mathrm{~A}_{1}}=8$ $\mathrm{E}_{2}=8 \mathrm{E}_{1}$ Then $\mathrm{E}_{\mathrm{n}}=8 \mathrm{E}$
142847
When ' $n$ ' number of liquid droplets each of radius ' $r$ ' are merged to form a single drop of radius ' $R$ ', the energy loss is ' $4 E$ '. Where $E$ is the energy of the bigger drop then, the number of droplets ' $n$ ' is
1 $\frac{5 R}{r}$
2 $\frac{5 \mathrm{R}}{\mathrm{r}^{2}}$
3 $\frac{4 R^{2}}{r^{2}}$
4 $\frac{5 R^{2}}{r^{2}}$
Explanation:
D According to question- $\frac{4}{3} \pi r^{3} n=\frac{4}{3} \pi R^{3}$ $r^{3} n=R^{3}$ Energy loss $(4 \mathrm{E})=4 \pi\left(\mathrm{r}^{2} n-\mathrm{R}^{2}\right) \mathrm{T}$ $4\left(4 \pi R^{2} T\right) =4 \pi\left(r^{2} n-R^{2}\right) T$ $4 R^{2}+R^{2} =r^{2} n$ $n =\frac{5 R^{2}}{r^{2}}$
MHT-CET 2020
Mechanical Properties of Fluids
142849
Two soap bubbles have radii in the ratio $4: 3$. What is the ratio of work done to blow these bubbles ?
1 $3: 2$
2 $4: 3$
3 $8: 3$
4 $16: 9$
Explanation:
D Let $r_{1}$ and $r_{2}$ be the radii of two soap bubbles then $\mathrm{r}_{1}: \mathrm{r}_{2}=4: 3$ Now, work done to blow both bubbles is given by $\mathrm{W}_{1}=2 \mathrm{~T}\left(4 \pi \mathrm{r}_{1}^{2}\right)$ and $\quad \mathrm{W}_{2}=2 \mathrm{~T}\left(4 \pi \mathrm{r}_{2}^{2}\right)$ $\text { Then, } \frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}} =\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2}$ $=\left(\frac{4}{3}\right)^{2}=\frac{16}{9}$ Therefore, the ratio of work done to blow these bubbles is $16: 9$.
MHT-CET 2019
Mechanical Properties of Fluids
142852
The excess of pressure, due to surface tension, on a spherical liquid drop of radius ' $R$ ' is proportional to
1 $\mathrm{R}^{2}$
2 $\mathrm{R}^{-2}$
3 $\mathrm{R}$
4 $\mathrm{R}^{-1}$
Explanation:
D The excess of pressure due to surface tension on a spherical liquid drop is given by - $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ Where, $\mathrm{T}=$ Surface tension $\mathrm{R}=$ Radius of liquid drop From equation (i), we get - $\mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\mathrm{P} \propto \mathrm{R}^{-1}$
MHT-CET 2019
Mechanical Properties of Fluids
142854
A liquid drop having surface energy ' $E$ ' is spread into 512 droplets of same size. The final surface energy of the droplets is
1 $2 \mathrm{E}$
2 $4 \mathrm{E}$
3 $8 \mathrm{E}$
4 $12 \mathrm{E}$
Explanation:
C The surface area of the liquid drop is, $A_{1}=4 \pi R^{2}$ Its surface energy is $\mathrm{E}$. When the drop spread in 512 droplets, the surface area of droplets is $\mathrm{A}_{2}=512 \times 4 \pi \mathrm{r}^{2}$. The volume of bigger drop is $\frac{4}{3} \pi R^{3}$ and volume of small droplets is $512 \times \frac{4}{3} \pi \mathrm{r}^{3}$. Then, $\quad \frac{4}{3} \pi \mathrm{R}^{3}=512 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{r}=\frac{\mathrm{R}}{8}$ $\mathrm{~A}_{2}=512 \times 4 \pi \mathrm{r}^{2}$ $512 \times 4 \pi\left(\frac{\mathrm{R}}{8}\right)^{2}=8 \mathrm{~A}_{1}$ Surface energy $(\mathrm{E})=\mathrm{A}$. $\mathrm{T}$ Where, $A=$ area, $T=$ surface tension $\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{A}_{2} \cdot \mathrm{T}}{\mathrm{A}_{1} \cdot \mathrm{T}}=\frac{8 \mathrm{~A}_{1}}{\mathrm{~A}_{1}}=8$ $\mathrm{E}_{2}=8 \mathrm{E}_{1}$ Then $\mathrm{E}_{\mathrm{n}}=8 \mathrm{E}$
142847
When ' $n$ ' number of liquid droplets each of radius ' $r$ ' are merged to form a single drop of radius ' $R$ ', the energy loss is ' $4 E$ '. Where $E$ is the energy of the bigger drop then, the number of droplets ' $n$ ' is
1 $\frac{5 R}{r}$
2 $\frac{5 \mathrm{R}}{\mathrm{r}^{2}}$
3 $\frac{4 R^{2}}{r^{2}}$
4 $\frac{5 R^{2}}{r^{2}}$
Explanation:
D According to question- $\frac{4}{3} \pi r^{3} n=\frac{4}{3} \pi R^{3}$ $r^{3} n=R^{3}$ Energy loss $(4 \mathrm{E})=4 \pi\left(\mathrm{r}^{2} n-\mathrm{R}^{2}\right) \mathrm{T}$ $4\left(4 \pi R^{2} T\right) =4 \pi\left(r^{2} n-R^{2}\right) T$ $4 R^{2}+R^{2} =r^{2} n$ $n =\frac{5 R^{2}}{r^{2}}$
MHT-CET 2020
Mechanical Properties of Fluids
142849
Two soap bubbles have radii in the ratio $4: 3$. What is the ratio of work done to blow these bubbles ?
1 $3: 2$
2 $4: 3$
3 $8: 3$
4 $16: 9$
Explanation:
D Let $r_{1}$ and $r_{2}$ be the radii of two soap bubbles then $\mathrm{r}_{1}: \mathrm{r}_{2}=4: 3$ Now, work done to blow both bubbles is given by $\mathrm{W}_{1}=2 \mathrm{~T}\left(4 \pi \mathrm{r}_{1}^{2}\right)$ and $\quad \mathrm{W}_{2}=2 \mathrm{~T}\left(4 \pi \mathrm{r}_{2}^{2}\right)$ $\text { Then, } \frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}} =\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2}$ $=\left(\frac{4}{3}\right)^{2}=\frac{16}{9}$ Therefore, the ratio of work done to blow these bubbles is $16: 9$.
MHT-CET 2019
Mechanical Properties of Fluids
142852
The excess of pressure, due to surface tension, on a spherical liquid drop of radius ' $R$ ' is proportional to
1 $\mathrm{R}^{2}$
2 $\mathrm{R}^{-2}$
3 $\mathrm{R}$
4 $\mathrm{R}^{-1}$
Explanation:
D The excess of pressure due to surface tension on a spherical liquid drop is given by - $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ Where, $\mathrm{T}=$ Surface tension $\mathrm{R}=$ Radius of liquid drop From equation (i), we get - $\mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\mathrm{P} \propto \mathrm{R}^{-1}$
MHT-CET 2019
Mechanical Properties of Fluids
142854
A liquid drop having surface energy ' $E$ ' is spread into 512 droplets of same size. The final surface energy of the droplets is
1 $2 \mathrm{E}$
2 $4 \mathrm{E}$
3 $8 \mathrm{E}$
4 $12 \mathrm{E}$
Explanation:
C The surface area of the liquid drop is, $A_{1}=4 \pi R^{2}$ Its surface energy is $\mathrm{E}$. When the drop spread in 512 droplets, the surface area of droplets is $\mathrm{A}_{2}=512 \times 4 \pi \mathrm{r}^{2}$. The volume of bigger drop is $\frac{4}{3} \pi R^{3}$ and volume of small droplets is $512 \times \frac{4}{3} \pi \mathrm{r}^{3}$. Then, $\quad \frac{4}{3} \pi \mathrm{R}^{3}=512 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{r}=\frac{\mathrm{R}}{8}$ $\mathrm{~A}_{2}=512 \times 4 \pi \mathrm{r}^{2}$ $512 \times 4 \pi\left(\frac{\mathrm{R}}{8}\right)^{2}=8 \mathrm{~A}_{1}$ Surface energy $(\mathrm{E})=\mathrm{A}$. $\mathrm{T}$ Where, $A=$ area, $T=$ surface tension $\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{A}_{2} \cdot \mathrm{T}}{\mathrm{A}_{1} \cdot \mathrm{T}}=\frac{8 \mathrm{~A}_{1}}{\mathrm{~A}_{1}}=8$ $\mathrm{E}_{2}=8 \mathrm{E}_{1}$ Then $\mathrm{E}_{\mathrm{n}}=8 \mathrm{E}$
