361289
A frame made of a metallic wire enclosing a surface area \(A\) is covered with a soap film. If the area of the frame of metallic wire is reduced by \(50 \%\), the energy of the soap film will be changed by
1 \(75 \%\)
2 \(100 \%\)
3 \(25 \%\)
4 \(50 \%\)
Explanation:
As surface energy \(=\) Surface tension x surface area i.e., \(\quad E=S \times 2 A\) New surface energy, \(E_{1}=S \times 2(A / 2)=S \times A\) \(\%\) decreases in surface energy \(\dfrac{E-E_{1}}{E} \times 100\) \(=\dfrac{2 S A-S A}{2 S A} \times 100=50 \%\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361290
The surface energy of a liquid drop is \(U\). It is sprayed into 1000 equal droplets. Then its surface energy becomes
1 10 \(U\)
2 \(U\)
3 1000 \(U\)
4 100 \(U\)
Explanation:
Given, \(U=S \times 4 \pi R^{2}\); when droplet is splitted into 1000 droplets each of radius \(r\), then \(\dfrac{4}{3} \pi R^{3}=1000 \times \dfrac{4}{3} \pi r^{3} \text { or } r=R / 10\) \(\therefore\) Surface energy of all droplets \(\begin{aligned}=S & \times 1000 \times 4 \pi r^{2}=S \times 1000 \times 4 \pi(R / 10)^{2} \\= & 10\left(S \times 4 \pi R^{2}\right)=10 U\end{aligned}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361291
The work done in blowing a soap bubble of surface tension \(0.06\,\,N{m^{ - 1}}\) from \(2\;cm\) radius to \(5\;cm\) radius is
1 \(0.004168\;\,\,J\)
2 \({\rm{0}}{\rm{.003168}}\,\,J\)
3 \({\rm{0}}{\rm{.003158}}\,\,{\rm{ }}J\)
4 \({\rm{0}}{\rm{.004568}}\,\,J\)
Explanation:
Here, \(S = 0.06\,\,N{m^{ - 1}}\) \({r_1} = 2\;cm = 0.02\;m,\,\,{r_2} = 5\;cm = 0.05\;m\) Since, bubble has two surfaces. Initial surface area of the bubble \(=2 \times 4 \pi r_{1}^{2}=2 \times 4 \pi(0.02)^{2}\) \( = 32\pi \times {10^{ - 4}}\;{m^2}\) Final surface area of the bubble \(=2 \times 4 \pi r_{2}^{2}=2 \times 4 \pi(0.05)^{2}\) \( = 200\pi \times {10^{ - 4}}\;{m^2}\) Increase in surface area \(=200 \pi \times 10^{-4}-32 \pi \times 10^{-4}\) \( = 168\,\pi \times {10^{ - 4}}\;{m^2}\) \(\therefore\) Work done \(=\) Surface tension \((S)\) \(\times\) Increase in surface area \(0.06 \times 168\,\,\pi \times {10^{ - 4}} = 0.003168\;J\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361292
If million small drops of water coalesce into one larger drop, then the ratio of total surface energy of the larger drop to that of the smaller drops combined will be
1 \({1: 10^{4}}\)
2 \({1: 10^{3}}\)
3 \({1: 10^{2}}\)
4 \({1: 10}\)
Explanation:
If radius of small drop is \({r}\), then radius of larger drop, \({R}\) is \({\left(10^{6}\right)^{1 / 3}}\). \(\Rightarrow R=100 r\) \(\therefore \) ratio is \(1:{10^2}\)
361289
A frame made of a metallic wire enclosing a surface area \(A\) is covered with a soap film. If the area of the frame of metallic wire is reduced by \(50 \%\), the energy of the soap film will be changed by
1 \(75 \%\)
2 \(100 \%\)
3 \(25 \%\)
4 \(50 \%\)
Explanation:
As surface energy \(=\) Surface tension x surface area i.e., \(\quad E=S \times 2 A\) New surface energy, \(E_{1}=S \times 2(A / 2)=S \times A\) \(\%\) decreases in surface energy \(\dfrac{E-E_{1}}{E} \times 100\) \(=\dfrac{2 S A-S A}{2 S A} \times 100=50 \%\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361290
The surface energy of a liquid drop is \(U\). It is sprayed into 1000 equal droplets. Then its surface energy becomes
1 10 \(U\)
2 \(U\)
3 1000 \(U\)
4 100 \(U\)
Explanation:
Given, \(U=S \times 4 \pi R^{2}\); when droplet is splitted into 1000 droplets each of radius \(r\), then \(\dfrac{4}{3} \pi R^{3}=1000 \times \dfrac{4}{3} \pi r^{3} \text { or } r=R / 10\) \(\therefore\) Surface energy of all droplets \(\begin{aligned}=S & \times 1000 \times 4 \pi r^{2}=S \times 1000 \times 4 \pi(R / 10)^{2} \\= & 10\left(S \times 4 \pi R^{2}\right)=10 U\end{aligned}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361291
The work done in blowing a soap bubble of surface tension \(0.06\,\,N{m^{ - 1}}\) from \(2\;cm\) radius to \(5\;cm\) radius is
1 \(0.004168\;\,\,J\)
2 \({\rm{0}}{\rm{.003168}}\,\,J\)
3 \({\rm{0}}{\rm{.003158}}\,\,{\rm{ }}J\)
4 \({\rm{0}}{\rm{.004568}}\,\,J\)
Explanation:
Here, \(S = 0.06\,\,N{m^{ - 1}}\) \({r_1} = 2\;cm = 0.02\;m,\,\,{r_2} = 5\;cm = 0.05\;m\) Since, bubble has two surfaces. Initial surface area of the bubble \(=2 \times 4 \pi r_{1}^{2}=2 \times 4 \pi(0.02)^{2}\) \( = 32\pi \times {10^{ - 4}}\;{m^2}\) Final surface area of the bubble \(=2 \times 4 \pi r_{2}^{2}=2 \times 4 \pi(0.05)^{2}\) \( = 200\pi \times {10^{ - 4}}\;{m^2}\) Increase in surface area \(=200 \pi \times 10^{-4}-32 \pi \times 10^{-4}\) \( = 168\,\pi \times {10^{ - 4}}\;{m^2}\) \(\therefore\) Work done \(=\) Surface tension \((S)\) \(\times\) Increase in surface area \(0.06 \times 168\,\,\pi \times {10^{ - 4}} = 0.003168\;J\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361292
If million small drops of water coalesce into one larger drop, then the ratio of total surface energy of the larger drop to that of the smaller drops combined will be
1 \({1: 10^{4}}\)
2 \({1: 10^{3}}\)
3 \({1: 10^{2}}\)
4 \({1: 10}\)
Explanation:
If radius of small drop is \({r}\), then radius of larger drop, \({R}\) is \({\left(10^{6}\right)^{1 / 3}}\). \(\Rightarrow R=100 r\) \(\therefore \) ratio is \(1:{10^2}\)
361289
A frame made of a metallic wire enclosing a surface area \(A\) is covered with a soap film. If the area of the frame of metallic wire is reduced by \(50 \%\), the energy of the soap film will be changed by
1 \(75 \%\)
2 \(100 \%\)
3 \(25 \%\)
4 \(50 \%\)
Explanation:
As surface energy \(=\) Surface tension x surface area i.e., \(\quad E=S \times 2 A\) New surface energy, \(E_{1}=S \times 2(A / 2)=S \times A\) \(\%\) decreases in surface energy \(\dfrac{E-E_{1}}{E} \times 100\) \(=\dfrac{2 S A-S A}{2 S A} \times 100=50 \%\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361290
The surface energy of a liquid drop is \(U\). It is sprayed into 1000 equal droplets. Then its surface energy becomes
1 10 \(U\)
2 \(U\)
3 1000 \(U\)
4 100 \(U\)
Explanation:
Given, \(U=S \times 4 \pi R^{2}\); when droplet is splitted into 1000 droplets each of radius \(r\), then \(\dfrac{4}{3} \pi R^{3}=1000 \times \dfrac{4}{3} \pi r^{3} \text { or } r=R / 10\) \(\therefore\) Surface energy of all droplets \(\begin{aligned}=S & \times 1000 \times 4 \pi r^{2}=S \times 1000 \times 4 \pi(R / 10)^{2} \\= & 10\left(S \times 4 \pi R^{2}\right)=10 U\end{aligned}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361291
The work done in blowing a soap bubble of surface tension \(0.06\,\,N{m^{ - 1}}\) from \(2\;cm\) radius to \(5\;cm\) radius is
1 \(0.004168\;\,\,J\)
2 \({\rm{0}}{\rm{.003168}}\,\,J\)
3 \({\rm{0}}{\rm{.003158}}\,\,{\rm{ }}J\)
4 \({\rm{0}}{\rm{.004568}}\,\,J\)
Explanation:
Here, \(S = 0.06\,\,N{m^{ - 1}}\) \({r_1} = 2\;cm = 0.02\;m,\,\,{r_2} = 5\;cm = 0.05\;m\) Since, bubble has two surfaces. Initial surface area of the bubble \(=2 \times 4 \pi r_{1}^{2}=2 \times 4 \pi(0.02)^{2}\) \( = 32\pi \times {10^{ - 4}}\;{m^2}\) Final surface area of the bubble \(=2 \times 4 \pi r_{2}^{2}=2 \times 4 \pi(0.05)^{2}\) \( = 200\pi \times {10^{ - 4}}\;{m^2}\) Increase in surface area \(=200 \pi \times 10^{-4}-32 \pi \times 10^{-4}\) \( = 168\,\pi \times {10^{ - 4}}\;{m^2}\) \(\therefore\) Work done \(=\) Surface tension \((S)\) \(\times\) Increase in surface area \(0.06 \times 168\,\,\pi \times {10^{ - 4}} = 0.003168\;J\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361292
If million small drops of water coalesce into one larger drop, then the ratio of total surface energy of the larger drop to that of the smaller drops combined will be
1 \({1: 10^{4}}\)
2 \({1: 10^{3}}\)
3 \({1: 10^{2}}\)
4 \({1: 10}\)
Explanation:
If radius of small drop is \({r}\), then radius of larger drop, \({R}\) is \({\left(10^{6}\right)^{1 / 3}}\). \(\Rightarrow R=100 r\) \(\therefore \) ratio is \(1:{10^2}\)
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PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361289
A frame made of a metallic wire enclosing a surface area \(A\) is covered with a soap film. If the area of the frame of metallic wire is reduced by \(50 \%\), the energy of the soap film will be changed by
1 \(75 \%\)
2 \(100 \%\)
3 \(25 \%\)
4 \(50 \%\)
Explanation:
As surface energy \(=\) Surface tension x surface area i.e., \(\quad E=S \times 2 A\) New surface energy, \(E_{1}=S \times 2(A / 2)=S \times A\) \(\%\) decreases in surface energy \(\dfrac{E-E_{1}}{E} \times 100\) \(=\dfrac{2 S A-S A}{2 S A} \times 100=50 \%\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361290
The surface energy of a liquid drop is \(U\). It is sprayed into 1000 equal droplets. Then its surface energy becomes
1 10 \(U\)
2 \(U\)
3 1000 \(U\)
4 100 \(U\)
Explanation:
Given, \(U=S \times 4 \pi R^{2}\); when droplet is splitted into 1000 droplets each of radius \(r\), then \(\dfrac{4}{3} \pi R^{3}=1000 \times \dfrac{4}{3} \pi r^{3} \text { or } r=R / 10\) \(\therefore\) Surface energy of all droplets \(\begin{aligned}=S & \times 1000 \times 4 \pi r^{2}=S \times 1000 \times 4 \pi(R / 10)^{2} \\= & 10\left(S \times 4 \pi R^{2}\right)=10 U\end{aligned}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361291
The work done in blowing a soap bubble of surface tension \(0.06\,\,N{m^{ - 1}}\) from \(2\;cm\) radius to \(5\;cm\) radius is
1 \(0.004168\;\,\,J\)
2 \({\rm{0}}{\rm{.003168}}\,\,J\)
3 \({\rm{0}}{\rm{.003158}}\,\,{\rm{ }}J\)
4 \({\rm{0}}{\rm{.004568}}\,\,J\)
Explanation:
Here, \(S = 0.06\,\,N{m^{ - 1}}\) \({r_1} = 2\;cm = 0.02\;m,\,\,{r_2} = 5\;cm = 0.05\;m\) Since, bubble has two surfaces. Initial surface area of the bubble \(=2 \times 4 \pi r_{1}^{2}=2 \times 4 \pi(0.02)^{2}\) \( = 32\pi \times {10^{ - 4}}\;{m^2}\) Final surface area of the bubble \(=2 \times 4 \pi r_{2}^{2}=2 \times 4 \pi(0.05)^{2}\) \( = 200\pi \times {10^{ - 4}}\;{m^2}\) Increase in surface area \(=200 \pi \times 10^{-4}-32 \pi \times 10^{-4}\) \( = 168\,\pi \times {10^{ - 4}}\;{m^2}\) \(\therefore\) Work done \(=\) Surface tension \((S)\) \(\times\) Increase in surface area \(0.06 \times 168\,\,\pi \times {10^{ - 4}} = 0.003168\;J\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361292
If million small drops of water coalesce into one larger drop, then the ratio of total surface energy of the larger drop to that of the smaller drops combined will be
1 \({1: 10^{4}}\)
2 \({1: 10^{3}}\)
3 \({1: 10^{2}}\)
4 \({1: 10}\)
Explanation:
If radius of small drop is \({r}\), then radius of larger drop, \({R}\) is \({\left(10^{6}\right)^{1 / 3}}\). \(\Rightarrow R=100 r\) \(\therefore \) ratio is \(1:{10^2}\)
