Surface Energy

Q. The work done in blowing a soap bubble of $10\text{cm}$ radius is -
(surface tension of soap $=0.03\text{N/m}$).

1. $6.54\times {10}^{-3}\text{J}$
2. $7.54\times {10}^{-3}\text{J}$
3. $9.54\times {10}^{-3}\text{J}$
4. $8.54\times {10}^{-3}\text{J}$

Q. The surface tension of soap solution is $0.03\text{N/m}$. The work done in blowing to form a soap bubble of surface area $40{\text{cm}}^2$ is:

1. $1.2\times {10}^{-4}\text{J}$
2. $2.4\times {10}^{-4}\text{J}$
3. $12\times {10}^{-4}\text{J}$
4. $24\times {10}^{-4}\text{J}$

Q. The surface tension of soap solution is $T=0.03{\text{Nm}}^{-1}$. If the size of a soap bubble is increased from a radius of $3\text{cm}$ to $5\text{cm}$, the work done in increasing the size of the soap bubble is

1. $4\pi \text{mJ}$
2. $0.2\pi \text{mJ}$
3. $2\pi \text{mJ}$
4. $0.4\pi \text{mJ}$

Q. A spherical liquid drop of radius $R$ is divided into 8 equal droplets. If the surface tension is $T$, then the work done in the process will be:

1. $2\pi R^{2}T$
2. $3\pi R^{2}T$
3. $4\pi R^{2}T$
4. $2\pi R{T}^2$

Q. If $T$ is the surface tension of a liquid, the energy needed to break a liquid drop of radius $R$ into 64 drops is

1. $6\pi R^{2}T$
2. $2\pi R^{2}T$
3. $12\pi R^{2}T$
4. $8\pi R^{2}T$

Q. A drop of liquid of diameter $2.8\text{mm}$ break up into $216$ identical drops. What is the approximate change in energy of bigger drop?
(Given: Surface tension of liquid, $T=75\text{dyne/cm}$)

1. $25\pi \text{erg}$
2. $29.4\pi \text{erg}$
3. $32\pi \text{erg}$
4. $28\pi \text{erg}$

Q. If $T$ is the surface tension of a liquid, then the energy needed to break the liquid drop of radius $R$ into $64$ identical drops is

1. $6\pi R^{2}T$
2. $2\pi R^{2}T$
3. $12\pi R^{2}T$
4. $8\pi R^{2}T$

Q. A rectangular film of liquid is extended from $(4\text{cm}\times 2\text{cm})$ to $(5\text{cm}\times 4\text{cm})$. if the work done is $3\times {10}^{-4}\text{J},$ the value of surface tension of the liquid is

1. $0.250{\text{Nm}}^{-1}$
2. $0.2{\text{Nm}}^{-1}$
3. $8.0{\text{Nm}}^{-1}$
4. $0.125{\text{Nm}}^{-1}$

Q. When water droplets merge to form a bigger drop, energy is.

1. liberated
2. absorbed
3. neither liberated nor absorbed

Q. A circular film having radius $2\text{cm}$ is extended such that its radius becomes $4\text{cm}$. If the value of surface tension of the liquid is $0.25\text{N/m}$, then the work done in extending the circular film is

1. $3\pi \times {10}^{-4}\text{J}$
2. $4\pi \times {10}^{-4}\text{J}$
3. $6\pi \times {10}^{-4}\text{J}$
4. $8\pi \times {10}^{-4}\text{J}$

Q. A drop of liquid of diameter $2.8\text{mm}$ break up into $216$ identical drops. What is the approximate change in energy of bigger drop?
(Given: Surface tension of liquid, $T=75\text{dyne/cm}$)

1. $25\pi \text{erg}$
2. $29.4\pi \text{erg}$
3. $32\pi \text{erg}$
4. $28\pi \text{erg}$

Q. A film of water is formed between two straight parallel wires each $10\text{cm}$ long and at separation $0.5\text{cm}$. Calculate the work required to increase $1\text{mm}$ distance between the wires. Surface tension of water$=72\times {10}^{-3}\text{N/m}$.

1. $144\times {10}^{-7}\text{J}$
2. $104\times {10}^{-7}\text{J}$
3. $72\times {10}^{-7}\text{J}$
4. $288\times {10}^{-7}\text{J}$

Q. The work done in blowing a soap bubble of $10\text{cm}$ radius is -
(surface tension of soap $=0.03\text{N/m}$).

1. $6.54\times {10}^{-3}\text{J}$
2. $7.54\times {10}^{-3}\text{J}$
3. $9.54\times {10}^{-3}\text{J}$
4. $8.54\times {10}^{-3}\text{J}$

Q. Two mercury drops each of radius $r$ merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension of mercury, is

1. ${\left(\sqrt[3]{32}\right)}^8\pi r^{2}T$
2. $2^4\pi r^{2}T$
3. $2\pi r^{2}T$
4. ${\left(\sqrt[3]{32}\right)}^4\pi r^{2}T$

Q.

A spherical liquid drop of radius R is divided into eight equal droplets. If surface tension is T, then the work done in this process will be

1. $2\pi R^{2}T$

2. $3\pi R^{2}T$

3. $4\pi R^{2}T$

4. $2\pi R{T}^2$

Q. Work done on a liquid drop to increase its radius from $3\text{cm}$ to $5\text{cm}$ is $\pi \times {10}^{-4}\text{J}$. Find the surface tension of liquid.

1. $\frac{1}{8}\text{N/m}$
2. $\frac{1}{4}\text{N/m}$
3. $\frac{1}{64}\text{N/m}$
4. $\frac{1}{32}\text{N/m}$

Q. A mercury drop of radius $1\text{cm}$ is sprayed into ${10}^6$ droplets of equal size. Calculate the energy expended ( in $\text{mJ}$ upto two decimals) if surface tension of mercury is $35\times {10}^{-3}\text{N/m}$.

Q. A film of water is formed between two straight parallel wires each $10\text{cm}$ long and at separation $0.5\text{cm}$. Calculate the work required to increase $1\text{mm}$ distance between the wires. Surface tension of water$=72\times {10}^{-3}\text{N/m}$.

1. $144\times {10}^{-7}\text{J}$
2. $104\times {10}^{-7}\text{J}$
3. $72\times {10}^{-7}\text{J}$
4. $288\times {10}^{-7}\text{J}$

Q. A thin, rectangular film of liquid is extended from $(2\text{cm}\times 3\text{cm})$ to $(4\text{cm}\times 6\text{cm})$. If the work done in the process is $6\times {10}^{-4}\text{J}$, the value of surface tension of the liquid is

1. $0.133\text{N/m}$
2. $0.33\text{N/m}$
3. $0.166\text{N/m}$
4. $0.12\text{N/m}$

Q. If $T$ is surface tension of the soap solution, then the amount of work done in blowing a soap bubble from diameter $D$ to diameter $2D$ is

1. $2\pi D^{2}T$
2. $4\pi D^{2}T$
3. $6\pi D^{2}T$
4. $8\pi D^{2}T$