Physics Test-28

Given,

\(\mathrm{L}=50 \mathrm{mH}\)

\(=50 \times 10^{-3} \mathrm{H}\)

\(\mathrm{I}=4 \mathrm{~A}\)

Now,

We know that,

Then,

\(U=\frac{1}{2}LI^{2}\)

The magnetic potential energy of inductor \((U)\) is:

\(U=\frac{1}{2} \times 50 \times 10^{-3} \times(4)^{2}\)

\(\Rightarrow U=\frac{1}{2} \times 50 \times 10^{-3} \times 16\)

\(\Rightarrow U=400 \times 10^{-3} \mathrm{H}\)

\(\Rightarrow U=0.4 \mathrm{~J}\)

Let \(B_{0}\) be the initial magnetic field and \(B=\) New magnetic field after changing the radius.

The magnetic field at the center of the circular loop before changing the radius is given by

\(B_{0}=\frac{\mu_{0} I}{2 \pi R}\)

After changing the radius into half the new magnetic field can be written as,

\( B=\frac{\mu_{0} I}{2 \pi \frac{R}{2}}\)

\(\Rightarrow B=2 \frac{\mu_{0} I}{2 \pi R}\)

\(\Rightarrow B=2 B_{0}\)

According to Charles' law if the pressure of the gas is constant,

\(V \propto T\)

Charles' law: Charles' law states that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature if the pressure remains constant.

Given, kinetic energy \(( KE )\) of \(e ^{\prime \prime}\) in first orbit of \(H\)-atom is \(13.6 eV\).

we know that -

\(K E= \frac{13.6 z^2}{ n^2} \quad n \rightarrow n^{\text {th }}\) orbit

\(z \Rightarrow\) atomic number.

we also know that -

(Total Energy) \(=-(\) kinetic Energy \()\)

\(\Rightarrow \quad TE =- KE\)

\(\Rightarrow TE =-13.6 \frac{ z ^2}{ n ^2} eV\)

Now, for \(He ^{+}\)atom \((z=2)\) in second orbit \(( n =2)\) \(TE =-13.6 eV \frac{4}{4}=-13.6 eV\)

[Capacitance A]\(=\left[M^{-1} L 6-2 T^{2} Q^{2}\right]\)

[Magnetic induction C] \(=\left[M T^{-1} Q^{-1}\right]\)

\([C]^{2}=\left[M^{2} T^{-2} Q^{-2}\right]\)

Given, \(A=3 B C^{2}\) or \(B=\frac{A}{3 C^{2}} \Rightarrow [B]=\frac{[A]}{[C]^{2}}\)

\(\therefore [B]=\frac{\left[M^{-1} L^{-2} T^{2} Q^{2}\right]}{\left[M^{2} T^{-1} Q^{-2}\right]}=\left[M^{-3} L^{-2} T^{4} Q^{4}\right]\)

For the first reaction,

\(\mathrm{K}_{1}=\frac{\left[\mathrm{HgCl}_{2} \mid\right.}{\left[\mathrm{HgCl}^{+}\right]\left[\mathrm{Cl}^{-}\right]}\)

For the second reaction,

\(\mathrm{K}_{2}=\frac{\left[\mathrm{HgCl}_{3}\right]}{\left[\mathrm{HgCl}_{2}\right]\left[\mathrm{Cl}^{-}\right]}\)

Finally,

\(\mathrm{K}_{3}=\frac{\left[\mathrm{HgCl}^{+}\right]\left[\mathrm{HgCl}_{3}^{-}\right]}{\left[\left.\mathrm{HgCl}_{2}\right|^{2}\right.}\)

\(\Rightarrow \mathrm{K}_{3}=\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}\)

\(\Rightarrow \mathrm{K}_{3}=\frac{9}{3 \times 10^{6}}\)

\(\Rightarrow \mathrm{K}_{3}=3 \times 10^{-6}\)

Escape velocity is also defined as the speed at which the magnitude kinetic energy of the object is equal to the gravitational potential energy of the body. The body which achieves this escape velocity in nowhere close to the orbit of the body. This velocity of the object keeps decreasing as it moves away from the body, it goes close to zero but doesn’t become zero.

It is given by \(v_{e}=\sqrt{\frac{2 G M}{r}}\) where \(G\) is the gravitational constant, \(M\) is the mass of the body and \(r\) is the distance of the object from the center of the body.

Thus escape velocity is a function of the mass of the body and the distance of the object from the center of the body. Here, since the question says, the escape velocity of a body depends upon its mass as, i.e., the mass of the body which is projected. But we know that the escape velocity is independent of the mass of the body.

Given,

The frequency of Electromagnetic wave along \(y\) - direction

\(\mathrm{\nu}=30 \mathrm{~MHz}\)

The electric field component of the wave along \(y\) - direction.

\(E=6 \mathrm{~vm}^{-1}\)

The speed of light in vacuum \(c = 3 \times 10^8\) m/sec

In Electromagnetic, the ratio of the amplitudes of electric and magnetic field is always constant and it is equal to velocity of the Electromagnetic Waves.

\(\frac{E}{B}=c\)

\(B=\frac{E}{c}=\frac{6}{3 \times 10^{8}}\)

\(B=2 \times 10^{-8} \mathrm{~T}\)

From the given relation, \(D=\frac{n\left(x_{2}-x_{1}\right)}{n_{2}-n_{1}}\)

Here, \([n]=\left[\frac{1}{\text { area } \times \text { time }}\right]=\frac{1}{\left[L^{2} T\right]}-\left[L^{-2} T^{-1}\right]\)

\(x_{2}-x_{1}=[L]\) and \(n_{2}-n_{1}=\left[\frac{1}{\text { volume }}\right]=\left[\frac{1}{L^{3}}\right]\)

\(=\left[L^{-3}\right]\)

So, \([D]=\frac{\left[L^{-2} T^{-1} L\right]}{\left[L^{-3}\right]}=\left[L^{2} T^{-1}\right]\)

Current density\((J=\frac{i}A)\),drift velocity\(v_d=\frac{I}{n e A}\),and Electricfield\(E=\frac{v_d m}{({eT})}\)are depending on the area of the wire.

The current should remain the same along any cross-section as there is no accumulation of charge in the conductor. The current is a flow of charge per unit time hence current will remain constant.