Physics Test-24

Sodium chlorideis not a non-electrolyte.

Sodium chloride is an ionic compound that is made up of sodium cation and the chloride anion. When dissolved in water or some other polar solvent, this compound is known to dissociate into the Na+ and the Cl- ions. Therefore, sodium chloride is not a nonelectrolyte (it is an electrolyte).

Isotopes of an element must have same atomic number (Z) but different mass number (A).

Number of protons is equal to the atomic number.

So, isotopes of an element have same number of protons.

Mass number (A) is equal to the sum of number of protons (p) and number of neutrons (n) i.e.

A = p + n

As isotopes of an element have different mass number but same number of protons, so they must have different number of neutrons.

Electromagnet waves have electric field as well as magnetic field which are perpendicular to each other and the electromagnetic waves propagate in a direction which is perpendicular to both the fields.

Thus, the propagation vector of \(E M\) waves \(\vec{k}=\vec{E} \times \vec{B}\)

Dimension of at \(=\) Dimension of \(F\)

\([\mathrm{at}]=[\mathrm{F}]\)

\([\mathrm{a}]=\left[\frac{\mathrm{F}}{\mathrm{t}}\right]\)

\([\mathrm{a}]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~T}}\right]\)

\([\mathrm{a}]=\left[\mathrm{MLT}^{-3}\right]\)

Dimension of \(\mathrm{bt}^{2}=\) Dimension of \(\mathrm{F}\)

\(\left[\mathrm{bt}^{2}\right]=[\mathrm{F}]\)

\([\mathrm{b}]=\left[\frac{\mathrm{F}}{\mathrm{t}^{2}}\right]\)

\([\mathrm{b}]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~T}^{2}}\right]\)

\([\mathrm{b}]=\left[\mathrm{MLT}^{-4}\right]\)

Let, Emf of one cell is \(\mathrm{E}_1=1.25 \mathrm{~V}\),

Emf of other cell is \(\mathrm{E}_2=0.75 \mathrm{~V}\)

'r' be the internal resistance of both the cells,

\(r=r_1=r_2\) (since, both the cells have equal internal resistance) Both the cells are connected in parallel, hence the effective emf is

\(E_{\text {eff }}=\frac{E_1}{r_1}+\frac{E_2}{r_2}\)

\(E_{\text {eff }}=\frac{E_1 r_2+E_2 r_1}{r_1+r_2}\)

\(E_{\text {eff }}=\frac{E_1 r+E_2 r}{r+r}\)

\(E_{\text {eff }}=r \frac{E_1+E_2}{2 r}\)

\(E_{\text {eff }}=\frac{E_1+E_2}{2}\)

\(E_{\text {eff }}=\frac{1.25+0.75}{2}\)

\(E_{\text {eff }}=\frac{2}{2}=1 \mathrm{~V}\)

The cell will be,

\(\mathrm{Al}(\mathrm{s})\left|\mathrm{Al}^{3+}(\mathrm{aq}) \| \mathrm{Cu}^{2+}(\mathrm{aq})\right| \mathrm{Cu}(\mathrm{s})\)

\(\mathrm{E}_{\text {cell }}^{\mathrm{o}}=\mathrm{E}_{\text {cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^{\circ}\)

\(=\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}-\mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{A}}^{\mathrm{o}}\)

\(=+0.34-(-1.66)\)

\(=+2.00\) V

For the net gravitational force on a particle.

\(F=\frac{G M^{2}}{a^{2}} \).....(1)

\(F_{1}=\frac{G M^{2}}{(a \sqrt{2})^{2}}\)

\(=\frac{G M^{2}}{2 a^{2}}\).....(2)

According to fig.:

For net force, \(F=\sqrt{F^{2}+F^{2}}+F_{1}\)

\(F_{net}=F \sqrt{2}+F_{1}\).....(3)

Put values from (1) and (2) in (3).

\(F_{net}=\frac{G M^{2}}{a^{2}} \sqrt{2}+\frac{G M^{2}}{2 a^{2}}\)

This force will act as centripetal force. Distance of particle from centre of circle is \(\frac{\mathrm{a}}{\sqrt{2}}\).

\(\mathrm{~F}_{\mathrm{C}}=\frac{\mathrm{Mv}^{2}}{\mathrm{r}} \)

\(\mathrm{r}=\frac{\mathrm{a}}{\sqrt{2}}\)

\(F_{C}=F_{net}\)

\(\frac{\mathrm{Mv}^{2}}{\frac{\mathrm{a}}{\sqrt{2}}}=\frac{\mathrm{GM}^{2}}{\mathrm{a}^{2}}\left(\frac{1}{2}+\sqrt{2}\right) \)

\(\Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{a}}\left(\frac{1}{2 \sqrt{2}}+1\right) \)

\(\Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{a}}(1.35) \)

\(\Rightarrow \mathrm{v}=1.16 \sqrt{\frac{\mathrm{GM}}{\mathrm{a}}}\)

Let \(A\) and \(I\) be the area and length of each wire respectively.

Then equivalent resistance of the wires in series, \(\quad R_{e q}=\rho_{e q} \frac{2l}{A}\) ....(1)

Resistance of wire having resistivity \(\rho_1\),\(\mathrm{R}_1=\rho_1 \frac{l}{\mathrm{~A}}\)(2)

Resistance of wire having resistivity \(\rho_2, R_2=\rho_2 \frac{l}{A}\) (3)

Since the wires are in series their equivalent resistance, \(\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2\)

Substituting the values from (1), (2) and (3),

\(\rho_{\text {eq }} \frac{2l}{A}=\rho_1 \frac{l}{A}+\rho_2 \frac{l}{A}\)

\(\Rightarrow \rho_{e q}=\frac{\rho_1+\rho_2}{2}\)

Radiation is a method of heat transfer that does not rely upon any contact between the heat source and the heated object It depends on nature of the body, its temperature and kind & extent of its surface.