D The volume of $\mathrm{Hg}$ drops before breaks and after breaks will be same. $\therefore \quad \frac{4}{3} \pi R^{3}=n \times \frac{4}{3} \pi r^{3}$ $\mathrm{R}^{3}=\mathrm{nr}^{3}$ $\Rightarrow \quad \left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=\mathrm{n}$ $\Rightarrow \quad \frac{\mathrm{R}}{\mathrm{r}}=\mathrm{n}^{1 / 3}$ $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{1 / 3}}$
MHT-CET 2020
Mechanical Properties of Fluids
142836
The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{\mathbf{2}}$ ', Then
1 $\mathrm{W}_{2}=0$
2 $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
3 $\mathrm{W}_{2}=4 \mathrm{~W}_{1}$
4 $\mathrm{W}_{2}=\mathrm{W}_{1}$
Explanation:
B The work done to blowing a soap bubble of radius ' $\mathrm{R}$ ' and surface tension ' $\mathrm{T}$ ' is given as- $\mathrm{W}_{1}=2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}=8 \pi \mathrm{R}^{2} \mathrm{~T}$ Now, if radius is doubled then work done to blowing the soap bubble is- $\mathrm{W}_{2}=2 \times 4 \pi(2 \mathrm{R})^{2} \times \mathrm{T}^{\prime}$ $\mathrm{W}_{2}=4 \times 8 \pi \mathrm{R}^{2} \times 4 \mathrm{~T}^{\prime}$ Dividing equation (ii) by (i), we get- $\frac{\mathrm{W}_{2}}{\mathrm{~W}_{1}}=\frac{4 \mathrm{~T}^{\prime}}{\mathrm{T}}$ But, in the question soap bubbles is heated then it's surface tension is decrease so, $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
MHT-CET 2020
Mechanical Properties of Fluids
142837
A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy
1 $4 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
2 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
3 is neither released nor absorbed
4 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is absorbed
Explanation:
B Let the number of small spherical drops be ' $n$ ' and radius 'r'. After coalesce it forms a big drop having radius ' $\mathrm{R}$ ' $\because$ Volume of drops before coalesce will be same after coalesce. $\mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{nr}^{3}=\mathrm{R}^{3}$ $\mathrm{n}=\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}$ When the number of drops coalesce the energy is released. $\therefore \quad$ Initial energy of drops $\left(E_{1}\right)=n \times 4 \pi r^{2} T$ Final energy of drop $\left(E_{2}\right)=4 \pi R^{2} T$ $\therefore$ Energy released $(\mathrm{U})=\mathrm{E}_{2}-\mathrm{E}_{1}$ $U=\left(4 \pi R^{2} \cdot T-n \cdot 4 \pi r^{2} \cdot T\right)$ $U=4 \pi T\left(R^{2}-n r^{2}\right)$ Putting the value of ' $n$ ', we get - $\mathrm{U}=4 \pi \mathrm{T}\left(\mathrm{R}^{2}-\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}} \cdot \mathrm{r}^{2}\right)$ $\mathrm{U}=4 \pi \mathrm{TR}^{3}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right]$ $\mathrm{U}=3 \times \frac{4}{3} \pi \mathrm{R}^{3} \mathrm{~T}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$ $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right] \quad\left(\because \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right)$ $\mathrm{U}=-3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ Hence, $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ is released.
MHT-CET 2020
Mechanical Properties of Fluids
142838
A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is
1 $r=\frac{R}{3}$
2 $r=\frac{R}{9}$
3 $r=\frac{R}{27}$
4 $r=3 R$
Explanation:
A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops. $\therefore \quad \frac{4}{3} \pi \mathrm{R}^{3}=27 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3} =27 \mathrm{r}^{3}$ $\mathrm{R}^{3} =(3)^{3} \mathrm{r}^{3}$ $\mathrm{R} =3 \mathrm{r}$ $\mathrm{r} =\frac{\mathrm{R}}{3}$
MHT-CET 2020
Mechanical Properties of Fluids
142839
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(\mathrm{P}=$ outside pressure $)$
A Pressure inside the first bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Pressure inside the second bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ We know that, $\quad P V=n R \theta$ Where, $\theta=$ absolute temperature As mass of the air inside the soap bubble is conserved $\mathrm{n}_{1}=\mathrm{n}_{2}=\mathrm{n}$ and under isothermal condition, temperature $\theta_{1}=\theta_{2}=\theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}=\mathrm{nR} \theta$ $\text { And }\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}=\mathrm{nR} \theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4 \pi}{3} \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}$ $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \mathrm{r}_{2}^{3}$ On solving the equation, we get- $\mathrm{T}=\frac{\mathrm{P}\left(\mathrm{R}^{3}-\mathrm{r}_{1}^{3}-\mathrm{r}_{2}^{3}\right)}{4\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}-\mathrm{R}^{2}\right)}$
D The volume of $\mathrm{Hg}$ drops before breaks and after breaks will be same. $\therefore \quad \frac{4}{3} \pi R^{3}=n \times \frac{4}{3} \pi r^{3}$ $\mathrm{R}^{3}=\mathrm{nr}^{3}$ $\Rightarrow \quad \left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=\mathrm{n}$ $\Rightarrow \quad \frac{\mathrm{R}}{\mathrm{r}}=\mathrm{n}^{1 / 3}$ $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{1 / 3}}$
MHT-CET 2020
Mechanical Properties of Fluids
142836
The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{\mathbf{2}}$ ', Then
1 $\mathrm{W}_{2}=0$
2 $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
3 $\mathrm{W}_{2}=4 \mathrm{~W}_{1}$
4 $\mathrm{W}_{2}=\mathrm{W}_{1}$
Explanation:
B The work done to blowing a soap bubble of radius ' $\mathrm{R}$ ' and surface tension ' $\mathrm{T}$ ' is given as- $\mathrm{W}_{1}=2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}=8 \pi \mathrm{R}^{2} \mathrm{~T}$ Now, if radius is doubled then work done to blowing the soap bubble is- $\mathrm{W}_{2}=2 \times 4 \pi(2 \mathrm{R})^{2} \times \mathrm{T}^{\prime}$ $\mathrm{W}_{2}=4 \times 8 \pi \mathrm{R}^{2} \times 4 \mathrm{~T}^{\prime}$ Dividing equation (ii) by (i), we get- $\frac{\mathrm{W}_{2}}{\mathrm{~W}_{1}}=\frac{4 \mathrm{~T}^{\prime}}{\mathrm{T}}$ But, in the question soap bubbles is heated then it's surface tension is decrease so, $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
MHT-CET 2020
Mechanical Properties of Fluids
142837
A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy
1 $4 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
2 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
3 is neither released nor absorbed
4 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is absorbed
Explanation:
B Let the number of small spherical drops be ' $n$ ' and radius 'r'. After coalesce it forms a big drop having radius ' $\mathrm{R}$ ' $\because$ Volume of drops before coalesce will be same after coalesce. $\mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{nr}^{3}=\mathrm{R}^{3}$ $\mathrm{n}=\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}$ When the number of drops coalesce the energy is released. $\therefore \quad$ Initial energy of drops $\left(E_{1}\right)=n \times 4 \pi r^{2} T$ Final energy of drop $\left(E_{2}\right)=4 \pi R^{2} T$ $\therefore$ Energy released $(\mathrm{U})=\mathrm{E}_{2}-\mathrm{E}_{1}$ $U=\left(4 \pi R^{2} \cdot T-n \cdot 4 \pi r^{2} \cdot T\right)$ $U=4 \pi T\left(R^{2}-n r^{2}\right)$ Putting the value of ' $n$ ', we get - $\mathrm{U}=4 \pi \mathrm{T}\left(\mathrm{R}^{2}-\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}} \cdot \mathrm{r}^{2}\right)$ $\mathrm{U}=4 \pi \mathrm{TR}^{3}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right]$ $\mathrm{U}=3 \times \frac{4}{3} \pi \mathrm{R}^{3} \mathrm{~T}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$ $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right] \quad\left(\because \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right)$ $\mathrm{U}=-3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ Hence, $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ is released.
MHT-CET 2020
Mechanical Properties of Fluids
142838
A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is
1 $r=\frac{R}{3}$
2 $r=\frac{R}{9}$
3 $r=\frac{R}{27}$
4 $r=3 R$
Explanation:
A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops. $\therefore \quad \frac{4}{3} \pi \mathrm{R}^{3}=27 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3} =27 \mathrm{r}^{3}$ $\mathrm{R}^{3} =(3)^{3} \mathrm{r}^{3}$ $\mathrm{R} =3 \mathrm{r}$ $\mathrm{r} =\frac{\mathrm{R}}{3}$
MHT-CET 2020
Mechanical Properties of Fluids
142839
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(\mathrm{P}=$ outside pressure $)$
A Pressure inside the first bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Pressure inside the second bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ We know that, $\quad P V=n R \theta$ Where, $\theta=$ absolute temperature As mass of the air inside the soap bubble is conserved $\mathrm{n}_{1}=\mathrm{n}_{2}=\mathrm{n}$ and under isothermal condition, temperature $\theta_{1}=\theta_{2}=\theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}=\mathrm{nR} \theta$ $\text { And }\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}=\mathrm{nR} \theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4 \pi}{3} \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}$ $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \mathrm{r}_{2}^{3}$ On solving the equation, we get- $\mathrm{T}=\frac{\mathrm{P}\left(\mathrm{R}^{3}-\mathrm{r}_{1}^{3}-\mathrm{r}_{2}^{3}\right)}{4\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}-\mathrm{R}^{2}\right)}$
D The volume of $\mathrm{Hg}$ drops before breaks and after breaks will be same. $\therefore \quad \frac{4}{3} \pi R^{3}=n \times \frac{4}{3} \pi r^{3}$ $\mathrm{R}^{3}=\mathrm{nr}^{3}$ $\Rightarrow \quad \left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=\mathrm{n}$ $\Rightarrow \quad \frac{\mathrm{R}}{\mathrm{r}}=\mathrm{n}^{1 / 3}$ $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{1 / 3}}$
MHT-CET 2020
Mechanical Properties of Fluids
142836
The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{\mathbf{2}}$ ', Then
1 $\mathrm{W}_{2}=0$
2 $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
3 $\mathrm{W}_{2}=4 \mathrm{~W}_{1}$
4 $\mathrm{W}_{2}=\mathrm{W}_{1}$
Explanation:
B The work done to blowing a soap bubble of radius ' $\mathrm{R}$ ' and surface tension ' $\mathrm{T}$ ' is given as- $\mathrm{W}_{1}=2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}=8 \pi \mathrm{R}^{2} \mathrm{~T}$ Now, if radius is doubled then work done to blowing the soap bubble is- $\mathrm{W}_{2}=2 \times 4 \pi(2 \mathrm{R})^{2} \times \mathrm{T}^{\prime}$ $\mathrm{W}_{2}=4 \times 8 \pi \mathrm{R}^{2} \times 4 \mathrm{~T}^{\prime}$ Dividing equation (ii) by (i), we get- $\frac{\mathrm{W}_{2}}{\mathrm{~W}_{1}}=\frac{4 \mathrm{~T}^{\prime}}{\mathrm{T}}$ But, in the question soap bubbles is heated then it's surface tension is decrease so, $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
MHT-CET 2020
Mechanical Properties of Fluids
142837
A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy
1 $4 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
2 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
3 is neither released nor absorbed
4 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is absorbed
Explanation:
B Let the number of small spherical drops be ' $n$ ' and radius 'r'. After coalesce it forms a big drop having radius ' $\mathrm{R}$ ' $\because$ Volume of drops before coalesce will be same after coalesce. $\mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{nr}^{3}=\mathrm{R}^{3}$ $\mathrm{n}=\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}$ When the number of drops coalesce the energy is released. $\therefore \quad$ Initial energy of drops $\left(E_{1}\right)=n \times 4 \pi r^{2} T$ Final energy of drop $\left(E_{2}\right)=4 \pi R^{2} T$ $\therefore$ Energy released $(\mathrm{U})=\mathrm{E}_{2}-\mathrm{E}_{1}$ $U=\left(4 \pi R^{2} \cdot T-n \cdot 4 \pi r^{2} \cdot T\right)$ $U=4 \pi T\left(R^{2}-n r^{2}\right)$ Putting the value of ' $n$ ', we get - $\mathrm{U}=4 \pi \mathrm{T}\left(\mathrm{R}^{2}-\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}} \cdot \mathrm{r}^{2}\right)$ $\mathrm{U}=4 \pi \mathrm{TR}^{3}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right]$ $\mathrm{U}=3 \times \frac{4}{3} \pi \mathrm{R}^{3} \mathrm{~T}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$ $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right] \quad\left(\because \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right)$ $\mathrm{U}=-3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ Hence, $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ is released.
MHT-CET 2020
Mechanical Properties of Fluids
142838
A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is
1 $r=\frac{R}{3}$
2 $r=\frac{R}{9}$
3 $r=\frac{R}{27}$
4 $r=3 R$
Explanation:
A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops. $\therefore \quad \frac{4}{3} \pi \mathrm{R}^{3}=27 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3} =27 \mathrm{r}^{3}$ $\mathrm{R}^{3} =(3)^{3} \mathrm{r}^{3}$ $\mathrm{R} =3 \mathrm{r}$ $\mathrm{r} =\frac{\mathrm{R}}{3}$
MHT-CET 2020
Mechanical Properties of Fluids
142839
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(\mathrm{P}=$ outside pressure $)$
A Pressure inside the first bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Pressure inside the second bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ We know that, $\quad P V=n R \theta$ Where, $\theta=$ absolute temperature As mass of the air inside the soap bubble is conserved $\mathrm{n}_{1}=\mathrm{n}_{2}=\mathrm{n}$ and under isothermal condition, temperature $\theta_{1}=\theta_{2}=\theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}=\mathrm{nR} \theta$ $\text { And }\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}=\mathrm{nR} \theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4 \pi}{3} \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}$ $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \mathrm{r}_{2}^{3}$ On solving the equation, we get- $\mathrm{T}=\frac{\mathrm{P}\left(\mathrm{R}^{3}-\mathrm{r}_{1}^{3}-\mathrm{r}_{2}^{3}\right)}{4\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}-\mathrm{R}^{2}\right)}$
D The volume of $\mathrm{Hg}$ drops before breaks and after breaks will be same. $\therefore \quad \frac{4}{3} \pi R^{3}=n \times \frac{4}{3} \pi r^{3}$ $\mathrm{R}^{3}=\mathrm{nr}^{3}$ $\Rightarrow \quad \left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=\mathrm{n}$ $\Rightarrow \quad \frac{\mathrm{R}}{\mathrm{r}}=\mathrm{n}^{1 / 3}$ $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{1 / 3}}$
MHT-CET 2020
Mechanical Properties of Fluids
142836
The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{\mathbf{2}}$ ', Then
1 $\mathrm{W}_{2}=0$
2 $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
3 $\mathrm{W}_{2}=4 \mathrm{~W}_{1}$
4 $\mathrm{W}_{2}=\mathrm{W}_{1}$
Explanation:
B The work done to blowing a soap bubble of radius ' $\mathrm{R}$ ' and surface tension ' $\mathrm{T}$ ' is given as- $\mathrm{W}_{1}=2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}=8 \pi \mathrm{R}^{2} \mathrm{~T}$ Now, if radius is doubled then work done to blowing the soap bubble is- $\mathrm{W}_{2}=2 \times 4 \pi(2 \mathrm{R})^{2} \times \mathrm{T}^{\prime}$ $\mathrm{W}_{2}=4 \times 8 \pi \mathrm{R}^{2} \times 4 \mathrm{~T}^{\prime}$ Dividing equation (ii) by (i), we get- $\frac{\mathrm{W}_{2}}{\mathrm{~W}_{1}}=\frac{4 \mathrm{~T}^{\prime}}{\mathrm{T}}$ But, in the question soap bubbles is heated then it's surface tension is decrease so, $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
MHT-CET 2020
Mechanical Properties of Fluids
142837
A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy
1 $4 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
2 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
3 is neither released nor absorbed
4 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is absorbed
Explanation:
B Let the number of small spherical drops be ' $n$ ' and radius 'r'. After coalesce it forms a big drop having radius ' $\mathrm{R}$ ' $\because$ Volume of drops before coalesce will be same after coalesce. $\mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{nr}^{3}=\mathrm{R}^{3}$ $\mathrm{n}=\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}$ When the number of drops coalesce the energy is released. $\therefore \quad$ Initial energy of drops $\left(E_{1}\right)=n \times 4 \pi r^{2} T$ Final energy of drop $\left(E_{2}\right)=4 \pi R^{2} T$ $\therefore$ Energy released $(\mathrm{U})=\mathrm{E}_{2}-\mathrm{E}_{1}$ $U=\left(4 \pi R^{2} \cdot T-n \cdot 4 \pi r^{2} \cdot T\right)$ $U=4 \pi T\left(R^{2}-n r^{2}\right)$ Putting the value of ' $n$ ', we get - $\mathrm{U}=4 \pi \mathrm{T}\left(\mathrm{R}^{2}-\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}} \cdot \mathrm{r}^{2}\right)$ $\mathrm{U}=4 \pi \mathrm{TR}^{3}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right]$ $\mathrm{U}=3 \times \frac{4}{3} \pi \mathrm{R}^{3} \mathrm{~T}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$ $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right] \quad\left(\because \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right)$ $\mathrm{U}=-3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ Hence, $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ is released.
MHT-CET 2020
Mechanical Properties of Fluids
142838
A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is
1 $r=\frac{R}{3}$
2 $r=\frac{R}{9}$
3 $r=\frac{R}{27}$
4 $r=3 R$
Explanation:
A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops. $\therefore \quad \frac{4}{3} \pi \mathrm{R}^{3}=27 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3} =27 \mathrm{r}^{3}$ $\mathrm{R}^{3} =(3)^{3} \mathrm{r}^{3}$ $\mathrm{R} =3 \mathrm{r}$ $\mathrm{r} =\frac{\mathrm{R}}{3}$
MHT-CET 2020
Mechanical Properties of Fluids
142839
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(\mathrm{P}=$ outside pressure $)$
A Pressure inside the first bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Pressure inside the second bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ We know that, $\quad P V=n R \theta$ Where, $\theta=$ absolute temperature As mass of the air inside the soap bubble is conserved $\mathrm{n}_{1}=\mathrm{n}_{2}=\mathrm{n}$ and under isothermal condition, temperature $\theta_{1}=\theta_{2}=\theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}=\mathrm{nR} \theta$ $\text { And }\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}=\mathrm{nR} \theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4 \pi}{3} \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}$ $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \mathrm{r}_{2}^{3}$ On solving the equation, we get- $\mathrm{T}=\frac{\mathrm{P}\left(\mathrm{R}^{3}-\mathrm{r}_{1}^{3}-\mathrm{r}_{2}^{3}\right)}{4\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}-\mathrm{R}^{2}\right)}$
D The volume of $\mathrm{Hg}$ drops before breaks and after breaks will be same. $\therefore \quad \frac{4}{3} \pi R^{3}=n \times \frac{4}{3} \pi r^{3}$ $\mathrm{R}^{3}=\mathrm{nr}^{3}$ $\Rightarrow \quad \left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3}=\mathrm{n}$ $\Rightarrow \quad \frac{\mathrm{R}}{\mathrm{r}}=\mathrm{n}^{1 / 3}$ $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{1 / 3}}$
MHT-CET 2020
Mechanical Properties of Fluids
142836
The work done in blowing a soap bubble of radius ' $R$ ' is ' $W_{1}$ ' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius ' $2 R$ ' is blown and the work done is ' $W_{\mathbf{2}}$ ', Then
1 $\mathrm{W}_{2}=0$
2 $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
3 $\mathrm{W}_{2}=4 \mathrm{~W}_{1}$
4 $\mathrm{W}_{2}=\mathrm{W}_{1}$
Explanation:
B The work done to blowing a soap bubble of radius ' $\mathrm{R}$ ' and surface tension ' $\mathrm{T}$ ' is given as- $\mathrm{W}_{1}=2 \times 4 \pi \mathrm{R}^{2} \times \mathrm{T}=8 \pi \mathrm{R}^{2} \mathrm{~T}$ Now, if radius is doubled then work done to blowing the soap bubble is- $\mathrm{W}_{2}=2 \times 4 \pi(2 \mathrm{R})^{2} \times \mathrm{T}^{\prime}$ $\mathrm{W}_{2}=4 \times 8 \pi \mathrm{R}^{2} \times 4 \mathrm{~T}^{\prime}$ Dividing equation (ii) by (i), we get- $\frac{\mathrm{W}_{2}}{\mathrm{~W}_{1}}=\frac{4 \mathrm{~T}^{\prime}}{\mathrm{T}}$ But, in the question soap bubbles is heated then it's surface tension is decrease so, $\mathrm{W}_{2} \lt 4 \mathrm{~W}_{1}$
MHT-CET 2020
Mechanical Properties of Fluids
142837
A fix number of spherical drops of a liquid of radius ' $r$ ' coalesce to form a large drop of radius ' $R$ ' and volume ' $V$ '. If ' $T$ ' is the surface tension then energy
1 $4 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
2 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released
3 is neither released nor absorbed
4 $3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is absorbed
Explanation:
B Let the number of small spherical drops be ' $n$ ' and radius 'r'. After coalesce it forms a big drop having radius ' $\mathrm{R}$ ' $\because$ Volume of drops before coalesce will be same after coalesce. $\mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{nr}^{3}=\mathrm{R}^{3}$ $\mathrm{n}=\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}}$ When the number of drops coalesce the energy is released. $\therefore \quad$ Initial energy of drops $\left(E_{1}\right)=n \times 4 \pi r^{2} T$ Final energy of drop $\left(E_{2}\right)=4 \pi R^{2} T$ $\therefore$ Energy released $(\mathrm{U})=\mathrm{E}_{2}-\mathrm{E}_{1}$ $U=\left(4 \pi R^{2} \cdot T-n \cdot 4 \pi r^{2} \cdot T\right)$ $U=4 \pi T\left(R^{2}-n r^{2}\right)$ Putting the value of ' $n$ ', we get - $\mathrm{U}=4 \pi \mathrm{T}\left(\mathrm{R}^{2}-\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}} \cdot \mathrm{r}^{2}\right)$ $\mathrm{U}=4 \pi \mathrm{TR}^{3}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right]$ $\mathrm{U}=3 \times \frac{4}{3} \pi \mathrm{R}^{3} \mathrm{~T}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)$ $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right] \quad\left(\because \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right)$ $\mathrm{U}=-3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ Hence, $\mathrm{U}=3 \mathrm{VT}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ is released.
MHT-CET 2020
Mechanical Properties of Fluids
142838
A mercury drop of radius ' $R$ ' is divided into 27 droplets of same size. The radius ' $r$ ' of each droplet is
1 $r=\frac{R}{3}$
2 $r=\frac{R}{9}$
3 $r=\frac{R}{27}$
4 $r=3 R$
Explanation:
A The volume of mercury drop before dividing into 27 drops will be same to volume of 27 drops. $\therefore \quad \frac{4}{3} \pi \mathrm{R}^{3}=27 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3} =27 \mathrm{r}^{3}$ $\mathrm{R}^{3} =(3)^{3} \mathrm{r}^{3}$ $\mathrm{R} =3 \mathrm{r}$ $\mathrm{r} =\frac{\mathrm{R}}{3}$
MHT-CET 2020
Mechanical Properties of Fluids
142839
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' combine to form a single soap bubble of radius ' $R$ '. The surface tension of soap solution is $(\mathrm{P}=$ outside pressure $)$
A Pressure inside the first bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Pressure inside the second bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ We know that, $\quad P V=n R \theta$ Where, $\theta=$ absolute temperature As mass of the air inside the soap bubble is conserved $\mathrm{n}_{1}=\mathrm{n}_{2}=\mathrm{n}$ and under isothermal condition, temperature $\theta_{1}=\theta_{2}=\theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}=\mathrm{nR} \theta$ $\text { And }\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}=\mathrm{nR} \theta$ $\because\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4 \pi}{3} \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \frac{4 \pi}{3} \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \frac{4 \pi}{3} \mathrm{r}_{2}^{3}$ $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \mathrm{R}^{3}=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}\right) \mathrm{r}_{1}^{3}+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}\right) \mathrm{r}_{2}^{3}$ On solving the equation, we get- $\mathrm{T}=\frac{\mathrm{P}\left(\mathrm{R}^{3}-\mathrm{r}_{1}^{3}-\mathrm{r}_{2}^{3}\right)}{4\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}-\mathrm{R}^{2}\right)}$
