Physics Test - 28

The direction of the electric field at every point must be radial (outward if \(\lambda>0\), inward if \(\lambda<0\)).

We know that the electric field intensity due to an infinitely long straight uniformly charged wire is given as:

\(E=\frac{\lambda}{2 \pi \epsilon_{o} r}\)

\(\Rightarrow E \propto \lambda \quad \ldots\)(1)

Where, \(\lambda=\) linear charge density, \(\epsilon_{0}=\) permittivity, and \(r=\) distance of the point from the wire

Given:

Self-inductance of first coil \(\left(L_{1}\right)=108 \mathrm{mH}\),

Radius of both coils are same i.e., \(r_{1}=r_{2}=r_{1}\),

Number of turns of the first coil \(\left(N_{1}\right)=200\), and

Number of turns of the second coil \(\left(N_{2}\right)=500\)

The Self-induction for the first coil is

\(\Rightarrow L_{1}=\frac{\mu_{0} N_{1}^{2} \pi r^{2}}{2}\)

The Self-induction for the second coil is

\(\Rightarrow L_{2}=\frac{\mu_{0} N_{2}^{2} \pi r^{2}}{2} \quad \cdots \text { (2) }\)

On dividing equation 1 and 2 , we get

\(\Rightarrow \frac{L_{2}}{L_{1}}=\left(\frac{N_{2}}{N_{1}}\right)^{2} \)

\(\Rightarrow L_{2}=L_{1}\left(\frac{N_{2}}{N_{1}}\right)^{2}\)

\(\Rightarrow L_{2}=108 \times\left(\frac{500}{200}\right)^{2}\)

\(=108 \times 6.25=675 \mathrm{mH}\)

This is a case of AND gate. Input and output are shown below: \(\therefore \mathrm{y}=\overline{\overline{\mathrm{A}}}+\overline{\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}=\mathrm{AB}(\operatorname{since} \overline{\mathrm{A}}+\overline{\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}})\)

The best conductor of heat among liquids is Mercury.

Mercury

• Symbol- Hg
• The Greek name of mercury is hydrargyrum from where its symbol is taken.
• Atomic number: 80
• Atomic mass: 200 u
• Melting point: -38 °C
• Boiling point: 356.7 °C
• It is the only metal element to exist in the liquid state at room temperature i.e. in the range 20 to 25 °C
• It is also called liquid silver.
• It is used in thermometers because it can record the slightest change in temperature.

For mutual inductance, we take two coils. One with \(N_{1}\) turns and the other with \(N_{2}\) turns. When we pass current through coil 1 , the magnetic field created by the coil increases in magnitude and the magnetic field lines will also go through coil 2 and create an increasing magnetic flux. The increase in magnetic flux in one direction causes the induced current to create a magnetic field going in the opposite direction to reduce the increasing magnetic flux.

Therefore, as we increase the current in coil 1, the magnitude of the magnetic field goes up and the flux in one direction increases creating an induced EMF across the second coil.

The induced EMF in coil \(2, \varepsilon_{2}\) is given as:

\(\varepsilon_{2}=M_{12} \frac{d i_{1}}{d t}\)

Where,

\(M_{12}=\) mutual inductance coefficient.

\(\frac{d i}{d t}=\) change in current per unit time in coil \(1 .\)

The unit of mutual inductance is Henry.

This mutual inductance is expressed as:

\(M=\frac{\mu_{0} N_{1} N_{2} A}{l}\)

Where,

\(N_{1}=\) number of turns in coil \(1\)

\(N_{2}=\) number of turns in coil \(2\).

\(A=\) area of coils

\(l=\) length of coil

Here, we have

\(N_{1}=300, N_{2}=400, A=10 \mathrm{~cm}^{2}=10 \times 10^{-4} \mathrm{~m}^{2}\) and

\(l=20 \mathrm{~cm}=0.2 \mathrm{~m}\)

Also given that \(\mu_{0}=4 \pi \times 10^{-7} \mathrm{TmA}{ }^{-1}\).

Substituting these values in formula of mutual inductance, we get

\(\Rightarrow M=\frac{\left(4 \pi \times 10^{-7}\right) 300.400\left(10^{-3}\right)}{0.2}\)

On simplifying, we get

\(\Rightarrow M=4 \pi(6) 10^{-5}\)

\(\Rightarrow M=24 \pi \times 10^{-5}\)

\(\therefore M=2.4 \pi \times 10^{-4} H\)

Thus, the mutual inductance is equal to \(2.4 \pi \times 10^{-4} \mathrm{H}\).

The lens is as shown in the figure.

To determine the radius of curvature \(R\) of the curved surface, we can write,

\(R^{2}=3^{2}+\left(R-\frac{3}{10}\right)^{2}\)

Solving this, we get,

\(R=15.15 \mathrm{~cm}\)

Speed of light in any medium is \(\frac{c}{\mu}\) where \(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\) and \(\mu\) is the refractive index.

For the lens, \(\frac{\mathbf{c}}{\mu}=2 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

\(\mu=\frac{3 \times 10^{8}}{2 \times 10^{8}}=1.5\)

Putting these values in the lens makers formula for plano convex lens, we have,

\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R}\right)\)

\(=(1.5-1)\left(\frac{1}{15.15}\right) c m^{-1}\)

Solving this, we get \(f \approx 30 \mathrm{~cm}\)

We know, \(\beta_{\mathrm{ac}}=\frac{\Delta \mathrm{I}_{\mathrm{c}}}{\Delta \mathrm{I}_{\mathrm{b}}}\).....(i)

Now, at \(V_{C E}=10 \mathrm{~V}\), we read two values of \(\mathrm{I}_{\mathrm{C}}\) from the graph.

Then, \(\Delta I_{b}=10 \mu \mathrm{A}, \Delta I_{c}=1.5 \mathrm{~mA}\)

Put all the values in (i),

Therefore, \(\beta_{\mathrm{ac}}=\frac{1.5 \mathrm{~mA}}{10 \mu \mathrm{A}}\)

\(=150\)

Electromagnetic waves are generated by accelerated charge.

• An accelerating charge produces a changing electric field which in turn produces a magnetic field. These alternatively changing magnetic and electric field give rise to electromagnetic waves.
• Whereas, the static force generate only electric field and moving particle generates only magnetic field.

As we know that from mirror formula,

\(\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \).....(1)

\(u= \) distance of object

\(v=\) distance of object

\(f=\) focal length

Differentiating (1), we get

\(-v^{-2} \mathrm{dv}-\mathrm{u}^{-2} \mathrm{du}=0\)

\(|\mathrm{d} v|=\left|\frac{v^{2}}{\mathrm{u}^{2}}\right| \mathrm{du}\).....(2)

Here \(|\mathrm{dv}|=\) size of image,

\(|\mathrm{du}|=\) size of object \((\mathrm{b})\)

From the equation (1), we write

\(\frac{\mathrm{u}}{\mathrm{v}}+1=\frac{\mathrm{u}}{\mathrm{f}}\)

Squaring both sides, we get

\(\frac{v^{2}}{\mathrm{v}^{2}}=\left(\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}\right)^{2}\).....(3)

Substituting equation (3) in equation (2) we get,

Size of the image \(\mathrm{dv}=\left(\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}\right)^{2}~\mid du \mid\)

Let \(|\mathrm{du}|=\) size of object \((\mathrm{b})\)

Size of the image \(\mathrm{dv}=\mathrm{b}\left(\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}\right)^{2}\)

As we know that, \(F=\frac{1}{4 \pi \epsilon 0} \frac{q_1 q_2}{r^2}\)
Where \(\epsilon_0\) is the permittivity of free space \(\left(8.854 \times 10^{-12} C ^2 N ^{-2} m ^{-2}\right)\).
The value of \(k=\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 Nm ^2 C^{-2}\)
So, initial the force between two charges \(q_1\) and \(q_2\) is \(200 N\).
\(F=k \frac{q_1 q_2}{r^2}=200 N-\text { ......(1) }\)
If new distance \(r^{\prime}=2 r\)
New Force is \(F^{\prime}=k \frac{q_1 q_2}{r^{\prime 2}}=k \frac{q_1 q_2}{(2 r)^2} \) \( \Rightarrow F^{\prime}=k \frac{q_1 q_2}{4 r^2}=\frac{1}{4} k \frac{q_1 q_2}{(r)^2} \quad \text {.....(2) }\)
From (1) and (2) \(\Rightarrow F^{\prime}=\frac{F}{4}\) or \(\Rightarrow F^{\prime}=\frac{200 N }{4}=50 N\)