361259
The work done in increasing the size of a soap film from \(11\;cm \times 7\;cm\) to \(20\;cm \times 8\;cm\) is \(8 \times {10^{ - 2}}\;J\). The surface tension of the film is
361260
If 1000 droplets of water of surface tension \(0.07\;N/m\), having same radius \(1\;mm\) each, combine to form a single drop. In the process the released surface energy is \(\left( {{\rm{Take }}\,\pi = \frac{{22}}{7}} \right)\)
1 \(7.92 \times {10^{ - 6}}\;J\)
2 \(7.92 \times {10^{ - 4}}\;J\)
3 \(9.68 \times {10^{ - 4}}\;J\)
4 \(8.8 \times {10^{ - 5}}\;J\)
Explanation:
Volume of a bigger single drop \(=1000 \times\) volume of a smaller drop \(\Rightarrow \dfrac{4}{3} \pi R^{3}=1000 \times \dfrac{4}{3} \pi r^{3}\) \(\Rightarrow R^{3}=1000 r^{3} \Rightarrow R=10 r\) Decrease in the surface area, \(\Delta A=1000 \times 4 \pi r^{2}-4 \pi R^{2}\) \(=1000 \times 4 \pi r^{2}-4 \pi(10 r)^{2}=900 \times 4 \pi r^{2}\) Here \(r = 1\;mm = {10^{ - 3}}\;m\), \(S = 0.07\;N/m\) \(\Delta E=\)? Required energy loss in the process, \(\Delta E=\) Surface tension \(\times\) change in surface area \(\Delta E=S . \Delta A\) \(\Delta E=0.07 \times 900 \times 4 \times(22 / 7) \times\left(10^{-3}\right)^{2}\) \(\Delta E = 7.92 \times {10^{ - 4}}\;J\)
JEE - 2023
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361261
A liquid drop having surface energy \(E\) is spread into 512 droplets of same size. The final surface energy of the droplets is
1 \(2\,E\)
2 \(4\,E\)
3 \(8\,E\)
4 \(12\,E\)
Explanation:
According to question, the surface area of the liquid drop is given by \(A=4 \pi R^{2}\) and surface energy is \(E\) When the drop of splits into 512 droplets, the surface area becomes \({A_2} = 512 \times 4\pi {r^2}\quad [r = {\text{ radius}}\,{\text{of}}\,{\text{smaller}}\,{\text{drop }}]\) \(r\) can be calculated by comparing the total volume of bigger and all smaller droplets \(\text { i.e., } \dfrac{4}{3} \pi R^{3}=512 \times 4 \pi r^{3} \Rightarrow r=\dfrac{R}{8}\) Hence, total area of smaller droplets is given by \(A_{1}=512 \times 4 \times \pi \times\left(\dfrac{R}{8}\right)^{2}=8 A\) Change in surface area is given by \(A_{2}-A_{1}\) \(\begin{aligned}& =4 \pi \times\left(\dfrac{512 \times R^{2}}{64}-R^{2}\right) \\& =4 \pi\left(8 R^{2}-R^{2}\right)=7 R^{2}\end{aligned}\) Surface energy,\(E = AT[A = \) area, \(T = \) tension \(]\) So, \(\dfrac{E_{n}}{E_{o}}=\dfrac{A_{1} \times T}{A \times T}=\dfrac{8 A}{A}=8\) \(\therefore E_{n}=8 E\)
MHTCET - 2016
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361262
Which one of the following statement is correct?
1 surface energy is potential energy per unit length
2 surface tension is work done per unit area
3 surface tension is work done per unit length
4 surface energy is work done per unit force
Explanation:
Surface tension is the force applied per unit length or work done (or energy) per unit area of a liquid surface. While surface energy is the amount of work done per unit area by the force.
361259
The work done in increasing the size of a soap film from \(11\;cm \times 7\;cm\) to \(20\;cm \times 8\;cm\) is \(8 \times {10^{ - 2}}\;J\). The surface tension of the film is
361260
If 1000 droplets of water of surface tension \(0.07\;N/m\), having same radius \(1\;mm\) each, combine to form a single drop. In the process the released surface energy is \(\left( {{\rm{Take }}\,\pi = \frac{{22}}{7}} \right)\)
1 \(7.92 \times {10^{ - 6}}\;J\)
2 \(7.92 \times {10^{ - 4}}\;J\)
3 \(9.68 \times {10^{ - 4}}\;J\)
4 \(8.8 \times {10^{ - 5}}\;J\)
Explanation:
Volume of a bigger single drop \(=1000 \times\) volume of a smaller drop \(\Rightarrow \dfrac{4}{3} \pi R^{3}=1000 \times \dfrac{4}{3} \pi r^{3}\) \(\Rightarrow R^{3}=1000 r^{3} \Rightarrow R=10 r\) Decrease in the surface area, \(\Delta A=1000 \times 4 \pi r^{2}-4 \pi R^{2}\) \(=1000 \times 4 \pi r^{2}-4 \pi(10 r)^{2}=900 \times 4 \pi r^{2}\) Here \(r = 1\;mm = {10^{ - 3}}\;m\), \(S = 0.07\;N/m\) \(\Delta E=\)? Required energy loss in the process, \(\Delta E=\) Surface tension \(\times\) change in surface area \(\Delta E=S . \Delta A\) \(\Delta E=0.07 \times 900 \times 4 \times(22 / 7) \times\left(10^{-3}\right)^{2}\) \(\Delta E = 7.92 \times {10^{ - 4}}\;J\)
JEE - 2023
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361261
A liquid drop having surface energy \(E\) is spread into 512 droplets of same size. The final surface energy of the droplets is
1 \(2\,E\)
2 \(4\,E\)
3 \(8\,E\)
4 \(12\,E\)
Explanation:
According to question, the surface area of the liquid drop is given by \(A=4 \pi R^{2}\) and surface energy is \(E\) When the drop of splits into 512 droplets, the surface area becomes \({A_2} = 512 \times 4\pi {r^2}\quad [r = {\text{ radius}}\,{\text{of}}\,{\text{smaller}}\,{\text{drop }}]\) \(r\) can be calculated by comparing the total volume of bigger and all smaller droplets \(\text { i.e., } \dfrac{4}{3} \pi R^{3}=512 \times 4 \pi r^{3} \Rightarrow r=\dfrac{R}{8}\) Hence, total area of smaller droplets is given by \(A_{1}=512 \times 4 \times \pi \times\left(\dfrac{R}{8}\right)^{2}=8 A\) Change in surface area is given by \(A_{2}-A_{1}\) \(\begin{aligned}& =4 \pi \times\left(\dfrac{512 \times R^{2}}{64}-R^{2}\right) \\& =4 \pi\left(8 R^{2}-R^{2}\right)=7 R^{2}\end{aligned}\) Surface energy,\(E = AT[A = \) area, \(T = \) tension \(]\) So, \(\dfrac{E_{n}}{E_{o}}=\dfrac{A_{1} \times T}{A \times T}=\dfrac{8 A}{A}=8\) \(\therefore E_{n}=8 E\)
MHTCET - 2016
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361262
Which one of the following statement is correct?
1 surface energy is potential energy per unit length
2 surface tension is work done per unit area
3 surface tension is work done per unit length
4 surface energy is work done per unit force
Explanation:
Surface tension is the force applied per unit length or work done (or energy) per unit area of a liquid surface. While surface energy is the amount of work done per unit area by the force.
361259
The work done in increasing the size of a soap film from \(11\;cm \times 7\;cm\) to \(20\;cm \times 8\;cm\) is \(8 \times {10^{ - 2}}\;J\). The surface tension of the film is
361260
If 1000 droplets of water of surface tension \(0.07\;N/m\), having same radius \(1\;mm\) each, combine to form a single drop. In the process the released surface energy is \(\left( {{\rm{Take }}\,\pi = \frac{{22}}{7}} \right)\)
1 \(7.92 \times {10^{ - 6}}\;J\)
2 \(7.92 \times {10^{ - 4}}\;J\)
3 \(9.68 \times {10^{ - 4}}\;J\)
4 \(8.8 \times {10^{ - 5}}\;J\)
Explanation:
Volume of a bigger single drop \(=1000 \times\) volume of a smaller drop \(\Rightarrow \dfrac{4}{3} \pi R^{3}=1000 \times \dfrac{4}{3} \pi r^{3}\) \(\Rightarrow R^{3}=1000 r^{3} \Rightarrow R=10 r\) Decrease in the surface area, \(\Delta A=1000 \times 4 \pi r^{2}-4 \pi R^{2}\) \(=1000 \times 4 \pi r^{2}-4 \pi(10 r)^{2}=900 \times 4 \pi r^{2}\) Here \(r = 1\;mm = {10^{ - 3}}\;m\), \(S = 0.07\;N/m\) \(\Delta E=\)? Required energy loss in the process, \(\Delta E=\) Surface tension \(\times\) change in surface area \(\Delta E=S . \Delta A\) \(\Delta E=0.07 \times 900 \times 4 \times(22 / 7) \times\left(10^{-3}\right)^{2}\) \(\Delta E = 7.92 \times {10^{ - 4}}\;J\)
JEE - 2023
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361261
A liquid drop having surface energy \(E\) is spread into 512 droplets of same size. The final surface energy of the droplets is
1 \(2\,E\)
2 \(4\,E\)
3 \(8\,E\)
4 \(12\,E\)
Explanation:
According to question, the surface area of the liquid drop is given by \(A=4 \pi R^{2}\) and surface energy is \(E\) When the drop of splits into 512 droplets, the surface area becomes \({A_2} = 512 \times 4\pi {r^2}\quad [r = {\text{ radius}}\,{\text{of}}\,{\text{smaller}}\,{\text{drop }}]\) \(r\) can be calculated by comparing the total volume of bigger and all smaller droplets \(\text { i.e., } \dfrac{4}{3} \pi R^{3}=512 \times 4 \pi r^{3} \Rightarrow r=\dfrac{R}{8}\) Hence, total area of smaller droplets is given by \(A_{1}=512 \times 4 \times \pi \times\left(\dfrac{R}{8}\right)^{2}=8 A\) Change in surface area is given by \(A_{2}-A_{1}\) \(\begin{aligned}& =4 \pi \times\left(\dfrac{512 \times R^{2}}{64}-R^{2}\right) \\& =4 \pi\left(8 R^{2}-R^{2}\right)=7 R^{2}\end{aligned}\) Surface energy,\(E = AT[A = \) area, \(T = \) tension \(]\) So, \(\dfrac{E_{n}}{E_{o}}=\dfrac{A_{1} \times T}{A \times T}=\dfrac{8 A}{A}=8\) \(\therefore E_{n}=8 E\)
MHTCET - 2016
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361262
Which one of the following statement is correct?
1 surface energy is potential energy per unit length
2 surface tension is work done per unit area
3 surface tension is work done per unit length
4 surface energy is work done per unit force
Explanation:
Surface tension is the force applied per unit length or work done (or energy) per unit area of a liquid surface. While surface energy is the amount of work done per unit area by the force.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361259
The work done in increasing the size of a soap film from \(11\;cm \times 7\;cm\) to \(20\;cm \times 8\;cm\) is \(8 \times {10^{ - 2}}\;J\). The surface tension of the film is
361260
If 1000 droplets of water of surface tension \(0.07\;N/m\), having same radius \(1\;mm\) each, combine to form a single drop. In the process the released surface energy is \(\left( {{\rm{Take }}\,\pi = \frac{{22}}{7}} \right)\)
1 \(7.92 \times {10^{ - 6}}\;J\)
2 \(7.92 \times {10^{ - 4}}\;J\)
3 \(9.68 \times {10^{ - 4}}\;J\)
4 \(8.8 \times {10^{ - 5}}\;J\)
Explanation:
Volume of a bigger single drop \(=1000 \times\) volume of a smaller drop \(\Rightarrow \dfrac{4}{3} \pi R^{3}=1000 \times \dfrac{4}{3} \pi r^{3}\) \(\Rightarrow R^{3}=1000 r^{3} \Rightarrow R=10 r\) Decrease in the surface area, \(\Delta A=1000 \times 4 \pi r^{2}-4 \pi R^{2}\) \(=1000 \times 4 \pi r^{2}-4 \pi(10 r)^{2}=900 \times 4 \pi r^{2}\) Here \(r = 1\;mm = {10^{ - 3}}\;m\), \(S = 0.07\;N/m\) \(\Delta E=\)? Required energy loss in the process, \(\Delta E=\) Surface tension \(\times\) change in surface area \(\Delta E=S . \Delta A\) \(\Delta E=0.07 \times 900 \times 4 \times(22 / 7) \times\left(10^{-3}\right)^{2}\) \(\Delta E = 7.92 \times {10^{ - 4}}\;J\)
JEE - 2023
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361261
A liquid drop having surface energy \(E\) is spread into 512 droplets of same size. The final surface energy of the droplets is
1 \(2\,E\)
2 \(4\,E\)
3 \(8\,E\)
4 \(12\,E\)
Explanation:
According to question, the surface area of the liquid drop is given by \(A=4 \pi R^{2}\) and surface energy is \(E\) When the drop of splits into 512 droplets, the surface area becomes \({A_2} = 512 \times 4\pi {r^2}\quad [r = {\text{ radius}}\,{\text{of}}\,{\text{smaller}}\,{\text{drop }}]\) \(r\) can be calculated by comparing the total volume of bigger and all smaller droplets \(\text { i.e., } \dfrac{4}{3} \pi R^{3}=512 \times 4 \pi r^{3} \Rightarrow r=\dfrac{R}{8}\) Hence, total area of smaller droplets is given by \(A_{1}=512 \times 4 \times \pi \times\left(\dfrac{R}{8}\right)^{2}=8 A\) Change in surface area is given by \(A_{2}-A_{1}\) \(\begin{aligned}& =4 \pi \times\left(\dfrac{512 \times R^{2}}{64}-R^{2}\right) \\& =4 \pi\left(8 R^{2}-R^{2}\right)=7 R^{2}\end{aligned}\) Surface energy,\(E = AT[A = \) area, \(T = \) tension \(]\) So, \(\dfrac{E_{n}}{E_{o}}=\dfrac{A_{1} \times T}{A \times T}=\dfrac{8 A}{A}=8\) \(\therefore E_{n}=8 E\)
MHTCET - 2016
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361262
Which one of the following statement is correct?
1 surface energy is potential energy per unit length
2 surface tension is work done per unit area
3 surface tension is work done per unit length
4 surface energy is work done per unit force
Explanation:
Surface tension is the force applied per unit length or work done (or energy) per unit area of a liquid surface. While surface energy is the amount of work done per unit area by the force.
