NEET Test Series from KOTA - 10 Papers In MS WORD
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Mechanical Properties of Fluids
142840
If ' $T$ ' is the surface tension of a soap solution, then the work done in blowing a soap bubble from diameter ' $D$ ' to diameter ' $2 D$ ' is
142842
A metal coin of thickness ' $d$ ' and density ' $\rho$ ' is floating on water of surface tension ' $T$ '. The radius of the coin is $g=$ acceleration due to gravity
1 $\frac{4 \mathrm{~T}}{3 \rho g \mathrm{~d}}$
2 $\frac{3 \mathrm{~T}}{4 \rho g \mathrm{~d}}$
3 $\frac{2 \mathrm{~T}}{\rho g \mathrm{~d}}$
4 $\frac{\mathrm{T}}{\rho \mathrm{gd}}$
Explanation:
C Given, thickness of metal coin $=\mathrm{d}$, density $=$ $\rho$, surface tension $=T$ Let, the radius of coin be $\mathrm{r}$ Downward weight $(\mathrm{mg})=\pi \mathrm{r}^{2} \mathrm{~d} \rho \mathrm{g}$ Upward force due to tension $=2 \pi \mathrm{rT}$ Downward weight $=$ upward force $\pi \mathrm{r}^{2} \mathrm{~d} \rho \mathrm{g}=2 \pi \mathrm{rT}$ $r=\frac{2 T}{\rho g d}$
MHT-CET 2020
Mechanical Properties of Fluids
142843
A water drop of radius ' $R$ ' splits into ' $n$ ' smaller drops, each of radius ' $r$ '. The work done in the process is, $(T=$ surface tension of water)
142840
If ' $T$ ' is the surface tension of a soap solution, then the work done in blowing a soap bubble from diameter ' $D$ ' to diameter ' $2 D$ ' is
142842
A metal coin of thickness ' $d$ ' and density ' $\rho$ ' is floating on water of surface tension ' $T$ '. The radius of the coin is $g=$ acceleration due to gravity
1 $\frac{4 \mathrm{~T}}{3 \rho g \mathrm{~d}}$
2 $\frac{3 \mathrm{~T}}{4 \rho g \mathrm{~d}}$
3 $\frac{2 \mathrm{~T}}{\rho g \mathrm{~d}}$
4 $\frac{\mathrm{T}}{\rho \mathrm{gd}}$
Explanation:
C Given, thickness of metal coin $=\mathrm{d}$, density $=$ $\rho$, surface tension $=T$ Let, the radius of coin be $\mathrm{r}$ Downward weight $(\mathrm{mg})=\pi \mathrm{r}^{2} \mathrm{~d} \rho \mathrm{g}$ Upward force due to tension $=2 \pi \mathrm{rT}$ Downward weight $=$ upward force $\pi \mathrm{r}^{2} \mathrm{~d} \rho \mathrm{g}=2 \pi \mathrm{rT}$ $r=\frac{2 T}{\rho g d}$
MHT-CET 2020
Mechanical Properties of Fluids
142843
A water drop of radius ' $R$ ' splits into ' $n$ ' smaller drops, each of radius ' $r$ '. The work done in the process is, $(T=$ surface tension of water)
142840
If ' $T$ ' is the surface tension of a soap solution, then the work done in blowing a soap bubble from diameter ' $D$ ' to diameter ' $2 D$ ' is
142842
A metal coin of thickness ' $d$ ' and density ' $\rho$ ' is floating on water of surface tension ' $T$ '. The radius of the coin is $g=$ acceleration due to gravity
1 $\frac{4 \mathrm{~T}}{3 \rho g \mathrm{~d}}$
2 $\frac{3 \mathrm{~T}}{4 \rho g \mathrm{~d}}$
3 $\frac{2 \mathrm{~T}}{\rho g \mathrm{~d}}$
4 $\frac{\mathrm{T}}{\rho \mathrm{gd}}$
Explanation:
C Given, thickness of metal coin $=\mathrm{d}$, density $=$ $\rho$, surface tension $=T$ Let, the radius of coin be $\mathrm{r}$ Downward weight $(\mathrm{mg})=\pi \mathrm{r}^{2} \mathrm{~d} \rho \mathrm{g}$ Upward force due to tension $=2 \pi \mathrm{rT}$ Downward weight $=$ upward force $\pi \mathrm{r}^{2} \mathrm{~d} \rho \mathrm{g}=2 \pi \mathrm{rT}$ $r=\frac{2 T}{\rho g d}$
MHT-CET 2020
Mechanical Properties of Fluids
142843
A water drop of radius ' $R$ ' splits into ' $n$ ' smaller drops, each of radius ' $r$ '. The work done in the process is, $(T=$ surface tension of water)
142840
If ' $T$ ' is the surface tension of a soap solution, then the work done in blowing a soap bubble from diameter ' $D$ ' to diameter ' $2 D$ ' is
142842
A metal coin of thickness ' $d$ ' and density ' $\rho$ ' is floating on water of surface tension ' $T$ '. The radius of the coin is $g=$ acceleration due to gravity
1 $\frac{4 \mathrm{~T}}{3 \rho g \mathrm{~d}}$
2 $\frac{3 \mathrm{~T}}{4 \rho g \mathrm{~d}}$
3 $\frac{2 \mathrm{~T}}{\rho g \mathrm{~d}}$
4 $\frac{\mathrm{T}}{\rho \mathrm{gd}}$
Explanation:
C Given, thickness of metal coin $=\mathrm{d}$, density $=$ $\rho$, surface tension $=T$ Let, the radius of coin be $\mathrm{r}$ Downward weight $(\mathrm{mg})=\pi \mathrm{r}^{2} \mathrm{~d} \rho \mathrm{g}$ Upward force due to tension $=2 \pi \mathrm{rT}$ Downward weight $=$ upward force $\pi \mathrm{r}^{2} \mathrm{~d} \rho \mathrm{g}=2 \pi \mathrm{rT}$ $r=\frac{2 T}{\rho g d}$
MHT-CET 2020
Mechanical Properties of Fluids
142843
A water drop of radius ' $R$ ' splits into ' $n$ ' smaller drops, each of radius ' $r$ '. The work done in the process is, $(T=$ surface tension of water)
