Physics Test - 29

The correct match is:

Given, a cored solenoid. Length of the solenoid, \(l=0.30 \mathrm{~m}\), Area of cross-section, \(A=25 \mathrm{~cm}^{2}=25 \times 10^{-4} \mathrm{~m}^{2}\), Number of turns on the solenoid, \(\mathrm{N}=500\), Current in the solenoid, \(I=2.5 \mathring{A}\), Current flows for time, \(t=10^{-3} \mathrm{~s}\)
Average back emf, \(e=\frac{d \phi}{d t}\quad\quad\)....(1)
Where, \(d \phi=\) Change in flux \(=\) NAB ... (2)
Where, \(B=\) Magnetic field strength \(=\mu_{0} \frac{N I}{l}\quad\quad\)....(3)
Where, \(\mu_{0}=\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}\)
Using equations (2) and (3) in equation \((1),\) we get \(e =\frac{\mu_{0} N^{2} I A}{l t} \)
\(=\frac{4 \pi \times 10^{-7} \times(500)^{2} \times 2.5 \times 25 \times 10^{-4}}{0.3 \times 10^{-3}}=6.5 \mathrm{~V}\)
Thus, the average back emf induced in the solenoid is \(6.5 \mathrm{~V}\).

A point on \(P - V\) diagram shows the state of the system. Each point on a \(P-V\) diagram corresponds to a different state of the gas. The pressure is given on the vertical axis and the volume is given on the horizontal axis, as seen below. 

Every point on a \(P-V\) diagram represents a different state for the gas (one for every possible volume and pressure). 

Given: \(\lambda=0.1 \mathring{A}=0.1 \times 10^{-10} \mathrm{~m}, \mathrm{e}=1.69 \times 10^{-19} \mathrm{C}, \mathrm{h}=6.626 \times 10^{-34}, \mathrm{~V}=?\)

The wavelength of \(x\)-ray produced in the cathode ray tube is given by:

\(\Rightarrow \lambda=\frac{h c}{e V}\)

The above equation can be rewritten for \(V\) as:

\(\Rightarrow V=\frac{h c}{e \lambda}\)

Substituting the given values in the above equation

\(\Rightarrow V=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 0.1 \times 10^{-10}}=117487.5 \mathrm{~V}\)

\(\Rightarrow V=1.17 \times 10^{4}\) Volt \(\approx 1.24 \times 10^{4}\) Volts

1 mole of \(C=x\) atoms \(=N_{A}\) atoms.

We know that \(N _{ A }\) = the number of atoms present in 12 grams of Carbon-12

So, \(x = N _{ A }\)

The atomic mass of \(Mg\) is \(24\) g/mol i.e., 1 mole of Mg \(=24\) g \(= x\) atoms

The number of moles in \(12\) g of Mg \(=\frac{12}{24}=0.5\) moles

1 mole of \(Mg \rightarrow x\) atoms

\(0.5\) mole of \(Mg \rightarrow 0.5 x\) atoms

Therefore, \(12 g\) of Magnesium has \(0.5 x\) atoms.

Given,

Mass, \((m) =1\) tonne \(=1000 kg\)

Initial velocity, \(u=30 m / s\)

Final velocity, \(v=0 m / s\)

Here, friction does the work, and therefore work done is negative.

So, we can consider \(F =- k \sqrt{ x }\)

From the work-energy theorem,

\( \Delta K=\int_{x_{i}}^{x_{f}} F d x \)

\(\Rightarrow \frac{1}{2} m\left(v^{2}-u^{2}\right)=\int_{0}^{d}-k \sqrt{x} d x \)

\(\Rightarrow \frac{1}{2} \times 1000 \times\left(0^{2}-30^{2}\right)=\int_{0}^{d}-1000 \sqrt{x} d x \)

\(\Rightarrow-450=\left[-\frac{2 x^{\frac{3 }{ 2}}}{3}\right]_{0}^{d} \)

\(\Rightarrow 450=\frac{2 d^{\frac{3 }{ 2}}}{3} \)

\(\Rightarrow \frac{450 \times 3}{2}=(d)^{\frac{3}{2}} \)

\(\Rightarrow 675 =(d)^{\frac{3}{2}} \)

We can write this as:

\( (77)^{\frac{3}{2}}=(d)^{\frac{3}{2}} \)

\(\therefore d=77 m\)

Basically, a coaxial cable consists of a hollow (outer) cylindrical conductor surrounding a single (inner) conductor along is the axis. The two conductors are well insulated from each other. The electric field (E) and magnetic field (H) at the cross-sections are shown by solid lines and dotted lines, respectively. The outer conductor acts as the shield and minimizes interference.

Different kinds of dielectric materials, such as Teflon and polythene are covered over copper wire, it acts as a spacer. In the transmission of power through coaxial cable, the dielectric medium separating the inner conductor from outer one plays a vital role. These dielectric materials are good insulators only at low frequencies. As the frequency increase, the energy loss becomes significant That is why a coaxial cable can be used effectively for transmission up to a frequency of 20 MHz.

A steady signal flowing in a wire uniformly distributes itself throughout the cross-section of the wire. A high-frequency signal, on the other hand, distributes itself uniformly, there being a concentration of current on the outer surface of the conductor. If the frequency of the current is very high, the current is almost wholly confined to the surface layers. This is called 'Skin effect".

Given:

Charge on dipole is \(\pm 5 \mu C=\pm 5 \times 10^{-6} C\)

Distance between the charges \(=1\) mm \(=10^{-3} \) m

We know that:

Dipole moment is given by:

\(P=q(2 a)=qd \)

\(=5 \times 10^{-6} \times 10^{-3} \)

\(=5 \times 10^{-9}\) Cm

Given:

Displacement equation is given as \(\mathrm{Y}=0.08 \sin \left(3 \pi \mathrm{t}+\frac{\pi}{4}\right)\)

On comparing with \(\mathrm{Y}=\mathrm{A} \sin (\mathrm{\omega t}+\pi)\)

We get, \(\mathrm{\omega}=3 \pi\) and \(\phi=\frac{\pi}{4}\)

Time period \(\mathrm{T}=\frac{2 \pi}{\mathrm{\omega}}\)

\(=\frac{2 \pi}{3 \pi}\)

\(=\frac{2}{3} \mathrm{~s}\)

Initial phase \(\phi=\frac{\pi}{4}\)

\(\text { At } \mathrm{t}=\frac{7}{36} \mathrm{~s} \)

\( \mathrm{Y}=0.08 \sin \left(3 \pi \times \frac{7}{36}+\frac{\pi}{4}\right)\)

\(=0.08 \sin (\frac{5 \pi }{ 6}) \)

\(\Rightarrow \mathrm{Y}=0.08 \times 0.5\)

\(=0.04 \mathrm{~m}\)

The answer is it will take \(1\) the second time.

Since the particles begin their SHM from extreme it will already have a difference of \(900\).

\(\Rightarrow X=A \sin (\omega t+900)\)

\(X=A \cos (\omega t)\)

Now, Time period \((t)=6=\omega 2 x\)

\(\Rightarrow \omega=3 \pi\)

At, \(X=2 A\)

\(\Rightarrow 2 A=A \cos (3 \pi t)\)

\(\Rightarrow \cos (3 \pi t)=21\)

\(\Rightarrow 3 \pi t=3 \pi\)

\(\Rightarrow t=1.0 s\)