- 1(current)
Q. No.
(take \(\theta(x)=\sin \theta(x)=\tan \theta(x)=\dfrac{d y}{d x}, g\) is the acceleration due to gravity)
| 1. | \(\dfrac{d^2 y}{d x^2}=\sqrt{\dfrac{\rho g}{S}}\) | 2. | \(\dfrac{d y}{d x}=\sqrt{\dfrac{\rho g}{S}} x\) |
| 3. | \(\dfrac{d^2 y}{d x^2}=\dfrac{\rho g}{S} x\) | 4. | \(\dfrac{d^2 y}{d x^2}=\dfrac{\rho g}{S} y\) |
1. \(198~\text{N}\)
2. \(1.98~\text{mN}\)
3. \(99~\text{N}\)
4. \(19.8~\text{mN}\)
(the surface tension of liquid is \(T\))
| 1. | \(\sqrt { \dfrac{T}{\rho g}}\) | 2. | \(\sqrt { \dfrac{2T}{\rho g}}\) |
| 3. | \(\sqrt{\dfrac{2T\rho}{\pi g}}\) | 4. | \(\sqrt{\dfrac{2T} {\pi \rho g}}\) |
\((\)the surface tension of soap solution \(0.03~\text{Nm}^{-1})\)
1. \(50.1 \times 10^{-4}~\text{J}\)
2. \(30.16 \times 10^{-4}~ \text{J}\)
3. \(5.06 \times 10^{-4} ~\text{J}\)
4. \(3.01\times 10^{-4} ~\text{J}\)
1. is equal to the atmospheric pressure
2. decreases
3. increases
4. remains the same
Air is pushed carefully into a soap bubble of radius \(r\) to double its radius. If the surface tension of the soap solution is \(T,\) then the work done in the process is:
| 1. | \(12\pi r^2T\) | 2. | \(24\pi r^2T \) |
| 3. | \(4\pi r^2T\) | 4. | \(8\pi r^2T\) |
A soap bubble, having a radius of \(1~\text{mm}\), is blown from a detergent solution having a surface tension of \(2.5\times 10^{-2}~\text{N/m}\). The pressure inside the bubble equals at a point \(Z_0\) below the free surface of the water in a container. Taking \(g = 10~\text{m/s}^{2}\), the density of water \(= 10^{3}~\text{kg/m}^3\), the value of \(Z_0\) is:
1. \(0.5~\text{cm}\)
2. \(100~\text{cm}\)
3. \(10~\text{cm}\)
4. \(1~\text{cm}\)
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 \(3\times 10^{-4}~\text J,\) then the value of the surface tension of the liquid is:
| 1. | \(0.250~\text{Nm}^{-1}\) | 2. | \(0.125~\text{Nm}^{-1}\) |
| 3. | \(0.2~\text{Nm}^{-1}\) | 4. | \(8.0~\text{Nm}^{-1}\) |
A certain number of spherical drops of a liquid of radius \({r}\) coalesce to form a single drop of radius \({R}\) and volume \({V}.\) If \({T}\) is the surface tension of the liquid, then:
| 1. | the energy \(= 4{VT}\left( \frac{1}{{r}} - \frac{1}{{R}}\right)\) is released. |
| 2. | the energy \(={ 3{VT}\left( \frac{1}{{r}} + \frac{1}{{R}}\right)}\) is released. |
| 3. | the energy \(={ 3{VT}\left( \frac{1}{{r}} - \frac{1}{{R}}\right)}\) is released. |
| 4. | the energy is neither released nor absorbed. |
- 1(current)
Q. No.
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