Physics Test - 27

Henry’s Law states that “the partial pressure applied by any gas on a liquid surface is directly proportional to its mole fraction present in a liquid solvent.”

Henry’s law is given by:

P ∝ C (or) P = k_H C

Where ‘P’ is the partial pressure of the gas, ‘C’ is the concentration of the dissolved gas and ‘k_H’ is Henry’s law constant of the gas.

Limitations of Henry’s law

It is applicable when the system is in equilibrium.

It is applicable for those gases only which do not react with water (solvent).

Gas shouldn’t cause any chemical change in the solution.

Henry’s law is applicable only when

• Pressure is not high
• Temperature is not too low
• Gas is extremely soluble.

So, when the pressure of gas is high, Henry's law is not applicable.

The electric field produced by a point charge at a point in the electric field is given as,

\(E=\frac{k Q}{r^{2}}\)

\(\Rightarrow E \propto \frac{1}{r^{2}} \quad \ldots\)(1)

Electric potential due to a point charge at a point in the electric field is given as,

\(V=\frac{k Q}{r}\)

\(\Rightarrow V \propto \frac{1}{r}\quad \ldots\)(2)

Electric field due to an electric dipole on the equatorial line is given as,

\(E=\frac{k P}{r^{3}}\)

\(\Rightarrow E \propto \frac{1}{r^{3}}\quad \ldots\)(3)

Electric field due to an electric dipole on the axial line is given as,

\(E=\frac{2 k P}{r^{3}}\)

\(\Rightarrow E \propto \frac{1}{r^{3}}\quad \ldots\)(4)

From equation (1), equation (2), equation (3), and equation (4) it is clear that only the electric field produced by a point charge is inversely proportional to the square of the distance. So it follows the inverse-square law.

Given:

\(R =\) radius of the sphere

\(E _{1}= E\)

\(q _{1}= Q\)

\(q _{2}=2\)

\(q _{1}=2 Q\)

\(r _{1}= r\)

\(r _{2}=2 r\)

Since, \(r>R\)

So the point is present outside the sphere.

We know that the electric field intensity at a point outside the metallic sphere is given as,

\(E=\frac{k q}{r^{2}} \quad \ldots\)(1)

Where, \(E=\) electric field intensity, \(k =9 \times 10^{9} N - m ^{2} / C ^{2}, q =\) charge on the sphere, and \(r =\) distance of the point from the center of the sphere

When, \(q _{1}= Q\) and \(r _{1}= r\)

\(E_{1}=\frac{k q_{1}}{r_{1}^{2}}\)

\(\Rightarrow E_{1}=\frac{k Q}{r^{2}}\quad \ldots\)(2)

When, \(q _{2}=2 Q\) and \(r _{2}=2 r\)

\(E_{2}=\frac{k q_{2}}{r_{2}^{2}}\)

\(\Rightarrow E_{2}=\frac{2 \times k Q}{(2 r)^{2}}\)

\(\Rightarrow E_{2}=\frac{k Q}{2 r^{2}}\quad \ldots\)(3)

By equation (2) and equation (3),

\(E_{2}=\frac{E_{1}}{2}\)

Since \(E_{1}=E\)

\(E_{2}=\frac{E}{2}\)

Given:

\(\mathrm{y}=\frac{10 \mathrm{~L}}{\pi} \sin \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \).....(i)

According to the question co-ordinates of the point where a horizontal ray becomes vertical after reflection.

Differentiating equation (i) w.r.t. to \(x\). we get:

\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=\frac{\ 10 \mathrm{~L}}{\pi} \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \times \frac{ \pi}{ \mathrm{5L}} \times 2\)

\(=2 \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \)

Since the horizontal ray becomes vertical \(\frac{\mathrm{dy}}{\mathrm{dx}}=1\)

So,

\(2 \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=1 \)

\(\Rightarrow \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\frac{1}{2} \)

\(\Rightarrow \cos\left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\cos \frac{\pi}{3} \)

\(\Rightarrow \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\frac{\pi}{3} \)

\(\mathrm{x}=\frac{5 \mathrm{~L}}{3} \)

Put the value of \(x\) in equation (i),

\(\mathrm{y}=\frac{10 \mathrm{~L}}{\pi} \operatorname{sin}\left(\frac{\pi \times 5 \mathrm{~L}}{3 \times 5 \mathrm{~L}}\right) \)

\(y=\frac{5 \sqrt{3}L}{\pi}\)

Auto reduction is a process in which sulphide ores of less electropositive metals like \(\mathrm{Cu}, \mathrm{Hg}, \mathrm{Pb}\) etc. are heated in presence of air. It is done for the conversion of the ore into sulphate or oxide, which then reacts with the remaining ore of the sulphate in the absence of air to give the less electropositive metal and a gas - sulphur dioxide.

Thus, in simple words we can say the auto reduction process is used in the preparation of less electropositive metals like \(\mathrm{Cu}, \mathrm{Hg}, \mathrm{Pb}\) etc.

Some of the examples are as following,

Copper (Cu) ore extraction: Here, partial roasting of sulphide ore of Cu to give the oxide. After some time, air supply is stopped and the temperature is raised to complete the leftover sulphide ore to form the metal.

\(2 \mathrm{Cu}_{2} \mathrm{~S}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CuO}+2 \mathrm{SO}_{2}\)

\(\mathrm{Cu}_{2} \mathrm{~S}+2 \mathrm{Cu}_{2} \mathrm{O} \rightarrow 6 \mathrm{Cu}+\mathrm{SO}_{2}\)

Lead (Pb) ore extraction:

\(2 \mathrm{PbS}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{PbO}+2 \mathrm{SO}_{2}\)

\(2 P b O+P b S \rightarrow 3 P b+S O_{2}\)

Given:

\(E=10^{4}\) Newton per Coulomb

\(q=1.6 \times 10^{-19}\) C

We know that:

\(F=q E\)

Where, \(F\) is the force due to the electric field, \(q\) is the charge, and \(E\) is the electric field.

\(F=\left(1.6 \times 10^{-19}\right) \times 10^{4}\)

\(\Rightarrow F=1.6 \times 10^{-15}\) Newton

Interference- When two light waves of the same frequency with zero initial phase difference or constant phase difference superimpose over each other, then the resultant amplitude in the region of superimposition is different from the amplitude of individual waves. This modification in intensity in the region of superposition is known as interference.

Condition for Interference

(1) The source must be coherent

(2) The source should be monochromatic (single wavelength)

(3) The sources having a definite phase relationship.

Given that:

\(\frac{\mathrm{I}_{\alpha}}{\mathrm{I}_{\beta}}=100\)

\( \mathrm{R}_{\mathrm{B}}=1000~ \Omega\)

\( \mathrm{R}_{\mathrm{L}}=1000~ \Omega\)

\( \beta=100=\frac{I_{\alpha}}{I_{\beta}}\)

\( \mathrm{V}_{0}=4 \mathrm{~V}\)

We know that,

\( \frac{\mathrm{V}_{0}}{\mathrm{~V}_{\mathrm{i}}}=\beta \frac{\mathrm{R}_{\mathrm{L}}}{\mathrm{R}_{\mathrm{B}}}\).....(i)

Put all the given values in equation (i), we get:

\( \mathrm{V}_{\mathrm{i}}\)\(=\frac{4 \times 1000}{100 \times 2000}\)

\(=0.02\)

\(=20 \mathrm{mV}\)

\(500 \mathrm{gm}\) of \(25 \% \mathrm{w} / \mathrm{w} \mathrm{NaOH}\)

\(\therefore \frac{25}{100} \times 500=125\)

\(500 \mathrm{gm}\) of \(15 \% \mathrm{w} / \mathrm{w} \mathrm{NaOH}\)

\(\therefore \frac{15}{100} \times 500=75\)

\(\therefore\) Total mass of \(\mathrm{NaOH}=125+75=200 \mathrm{~g}\)

Moles of \(\mathrm{NaOH}=\frac{200}{40}=5\)

Total mass of Solvent \(=500+500-200=800 \mathrm{~g}\)

\(=0.8 \mathrm{~kg}\)

\(\therefore\) Molality \(=\frac{\text { Moles }}{\text { Mass of Solvent }}\)

\(=\frac{5}{0.8}\) \(=6.25\)

The circuit consists of two NOT gates connected to each input of a NOR gate.

The correct truth table of the combination will be: