Physics Test - 67

If a liquid does not wet glass, its angle of contact is obtuse.

Contact angle: The contact angle is a measure of the ability of a liquid to wet the surface of a solid. 

• Adhesive forces: Attractive forces between molecules of different types are called adhesive forces.
• Cohesive forces: Attractive forces between molecules of the same types are called cohesive forces.
• For contact angle 90° the cohesive forces and adhesive forces are in the equilibrium state.

If a liquid doesn’t wet the glass which means the interaction of liquid molecules is greater than the interaction between the liquid and solid molecule, these lead to an obtuse angle of contact.

Here, \(A B=r=90 \mathrm{~cm}=0.9 \mathrm{~m}\)

\(\mathrm{q_{A}=10 \mu C=10 \times 10^{-6} C} \)

\(\mathrm{q_{B}=40 \mu C=40 \times 10^{-6} C, A C=?}\)

At point \(\mathrm{C, E_{A}=E_{B}}\)

\(\mathrm{\frac{q_{A}}{4 \pi \varepsilon_{0}(A C)^{2}}=\frac{q_{B}}{4 \pi \varepsilon_{0}(B C)^{2}}} \)

\(\mathrm{\frac{q_{A}}{(A C)^{2}}=\frac{q_{B}}{(r-A C)^{2}}}\)

\(\mathrm{\frac{10 \times 10^{-6}}{(A C)^{2}}=\frac{40 \times 10^{-6}}{(0.9-A C)^{2}}} \)

\(\mathrm{\frac{1}{(A C)^{2}}=\frac{4}{(0.9-A C)^{2}}}\)

\(\mathrm{\frac{1}{A C}=\frac{2}{(0.9-A C)}} \)

\(\mathrm{0.9-A C=2 A C} \)

\(\mathrm{3 A C=0.9 }\)

\(\mathrm{A C=0.3 \mathrm{~m}=30 \mathrm{~cm}}\)

So at \(30 \mathrm{~cm}\) distance from \(\mathrm{A}\) the electric intensity will be zero.

Using Dot Product principle: \(A × B=AB \cos (\$)\)

Here, \(\$\), means the angle between vector \(A\) and \(B\)

Angle of arbitrary vectors \(A, B\), is defined in range \(\left(0^{\circ}, 180^{\circ}\right)\)

Then, factor \(AB\) dissolved on both sides of equation. We will get \(\cos (\$)=\frac{1}{2}\)

Finally referring to the above range, you can get only one answer:

\(60^{\circ}\)

\(\begin{aligned} & P=\frac{V^2}{R} \Rightarrow R=\frac{V^2}{P} \\ & \left(V_R\right)\left(V_C\right)\end{aligned}\)

\(\begin{aligned} & \mathrm{R}=\frac{100 \times 10^2}{50}=\mathrm{R}=200 \Omega \\ & \mathrm{V}_{\mathrm{R}}^2+\mathrm{V}_{\mathrm{C}}^2=\mathrm{V}^2 \\ & (100)^2+\mathrm{V}_{\mathrm{C}}^2=(200)^2 \\ & \mathrm{i}=\frac{100}{200}=\frac{1}{2} ; \quad \mathrm{V}^2=40000 \\ & \mathrm{~V}=\mathrm{I} \times \mathrm{X}_{\mathrm{C}} ; \quad \mathrm{V}_{\mathrm{C}}{ }^2=30000 \\ & \mathrm{~V}_{\mathrm{C}}=100 \sqrt{3} \\ & \mathrm{X}_{\mathrm{C}}=200 \sqrt{3} \\ & 200 \sqrt{3}=\frac{1}{\omega \mathrm{C}} \\ & \mathrm{C}=\frac{1}{20 \times 50 \times 20 \sqrt{3}}=\frac{50 \times 10^{-6}}{\sqrt{\mathrm{x}}} \\ & \sqrt{\mathrm{x}}=50 \times 10^{-6} \times 100 \times 200 \sqrt{3} \\ & \mathrm{X}=3\end{aligned}\)

Frequency \(=\frac{1}{\text { time constant }}\)

We know that,

Dimensions of \(R=\left[M L^{2} T^{-3} I^{-2}\right]\)

Dimensions of \(L=\left[M L^{2} T^{-2} I^{-2}\right]\)

Dimensions of \(C=\left[M^{-1} L^{-2} T^{4} I^{2}\right]\)

Hence, dimensions of \(R C L=\left[M L^{2} T^{-1} I^{-2}\right]\)

Which are not equal to dimensions of frequency;

The constant is given by \(C R\) and \(L C\).

Alpha particle is a helium nucleus containing two protons and two neutrons so its charge is twice the proton's charge while the mass is about 4 times greater.

Let the charge of the proton be \(+e\), then the charge of the alpha particle will be \(+2 \mathrm{e} .\) Similarly mass of the proton be \(m\), then the mass of the alpha particle will be \(4 \mathrm{~m}\). Specific charge = charge/mass of the substance.

For proton,

Specific charge \(=\frac{e}{m}\)

For alpha particle,

Specific charge \(=\frac{2 \mathrm{e}}{4 \mathrm{~m}}\)

Therefore the ratio is \(=\frac{e}{m} \times \frac{4 m}{2 e}=2: 1\)

So, the charge to mass ratio of an alpha particle is approximate twice the charge to mass ratio of a proton.

We know that:

The Potential energy of the harmonic oscillator is given by \(\mathrm{PE}=\frac{1}{2} \mathrm{kx}^{2}\)

However, there can be initial potential energy, which is a constant.

Thus, the generalized potential energy of a harmonic oscillator is:

\(\mathrm{PE}=\mathrm{V}_{0}+\) \(\frac{1}{2} \mathrm{kx}^{2}\).....(1)

From the given figure,

At \(\mathrm{x}=0, \mathrm{PE}=0.01 \mathrm{~J}=V_{0}\)

At \(\mathrm{x}=0.02 \mathrm{~m}, \mathrm{PE}=0.04 \mathrm{~J}\)

Substituting in the general form of Potential energy, we have \(\mathrm{V}_{0}=0.01 \mathrm{~J}\)

Put all the given values in (1).

Thus, \(0.01+\frac{1}{2} \mathrm{k}(0.02)^{2}=0.04 \)

\( \mathrm{k}=150 \mathrm{~N} / \mathrm{m}\)

Pressure is the force applied per unit area in a direction perpendicular to the surface of the object. The SI unit of pressure is the Pascal (Pa).

Pressure \((P)=\frac{\text { Force }}{\text { Area }}\)

From the above information we can say that pressure is measured In the context of force and area.

Let radius of circular orbit of the Earth and Venus be \(\mathrm{r_e}\) and \(\mathrm{r_v}\) respectively \(\mathrm{\left(\frac{r_c}{r_v}\right)^3=\left(\frac{365}{220}\right)^2 }\) [Kepler's third law] 
\(\Rightarrow\mathrm{ \frac{r_c}{r_v}=(2.75)^{1 / 3}=1.4}\)
From the drawing given in the problem \(\mathrm{\frac{M^{\prime} N^{\prime}}{M N}=\frac{N^{\prime} V}{N V} }\)
\(\mathrm{M^{\prime} N^{\prime}=M N\left(\frac{N^{\prime} V}{N V}\right)=1000 \times \frac{r_v}{r_c-r_v}=1000 \times \frac{1}{\frac{r_e}{r_v}-1}}\)\(=1000 \times \frac{1}{1.4-1}=2500 \mathrm{~km}\)