Physics Test - 17

Given that radius of earth is \(6400\) km. \(1 \%\) decrease in gravity.

Acceleration due to gravity above the surface of earth at height \(\mathrm{h}\) is given

\(g^{\prime}=g\left(1-\frac{2 h}{R_{e}}\right)\) .....(i)

Here, \(g^{\prime}=0.99 \mathrm{~g}\)

Put the given values in (i).

\( 0.99=1-\frac{2 h}{R_{e}}\)

\(\Rightarrow \frac{2 h}{R_{e}}=0.01\)

\(\Rightarrow h=\frac{0.01}{2} R_{e}\)

Put the value of \(R_{e}\) in above equation.

\( h=\frac{0.01}{2} \times 6400\)

\(=32 \mathrm{~km}\)

\(\mathrm{h}=32 \mathrm{~km}\)

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 \(f \approx 30 \mathrm{~cm}\)

Given,

\(KE _{2}=9 KE _{1}\)

We know that the kinetic energy of a body is given as,

\( K E=\frac{1}{2} m v^{2}\)...(1)

The linear momentum of a body is given as,

\( P = mv\)...(2)

By equation (1) and equation (2) the relation between the kinetic energy and the momentum is given as,

\( K E=\frac{P^{2}}{2 m}\)...(3)

By equation (3) for the initial position,

\( K E_{1}=\frac{P_{1}^{2}}{2 m}\)...(4)

By equation (3) for the final position,

\( K E_{2}=\frac{P_{2}^{2}}{2 m} \)

\(\Rightarrow 9 \times K E_{1}=\frac{P_{2}^{2}}{2 m}\)...(5)

By equation (4) and equation (5),

\(\frac{P_{2}^{2}}{2 m} \times \frac{2 m}{P_{1}^{2}}=\frac{9 \times K E_{1}}{K E_{1}}\)

\(\Rightarrow \frac{P_{2}^{2}}{P_{1}^{2}}=9 \)

\(\Rightarrow \frac{P_{2}}{P_{1}}=3 \)

\(\Rightarrow P_{2}=3 P_{1}\)

So, if the kinetic energy of a body is increased nine times then the momentum of the body will be increased by three times.

According to color code of resistances:

Black Brown Red Orange Yellow Green Blue Violet Grey White

0 1 2 3 4 5 6 7 8 9

For tolerance:

Gold Silver No color

5% 10% 20%

So, for the given color code, resistance is given as \(=\mathrm{AB} \times 10^{\mathrm{C}} \pm\) tolerance

So, \(R=47 \times 10^1 \pm 5 \%\)

\(\mathrm{R}=470 \pm 5 \%\)

The negative temperature coefficient of the resistance is only present in the insulators or the semiconductors. In these, the resistance decreases with increase in temperature.

Insulators have a negative temperature coefficient because their resistance decreases as the temperature rises. As the temperature rises, the semiconductor material’s resistivity decreases, resulting in a negative temperature coefficient which is a form of resistance.

\(\mathrm{B_0=\frac{4 \mu_0 I}{4 \pi \frac{L}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)=\frac{2 \sqrt{2} \mu_0 I}{\pi L} }\)
After folding \(\mathrm{\vec{B}_{A B}=\frac{\mu_0 I}{4 \pi \frac{L}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right) \hat{k}=\frac{2 \sqrt{2}}{4 \pi} \frac{\mu_0 I}{L} \hat{k} }\)
\(\mathrm{\vec{B}_{B E}=\frac{\mu_0 I}{4 \pi \frac{L}{2}}\left(\sin 0^{\circ}+\sin 45^{\circ}\right) \hat{k}=\frac{\sqrt{2} \mu_0 I}{4 \pi L} \hat{k}} \)
Similarly we have: \(\mathrm{\vec{B}_{E C}=\frac{\sqrt{2} \mu_0 I}{4 \pi L} \hat{i}} \)
\(\mathrm{\vec{B}_{C D}=\frac{2 \sqrt{2}}{4 \pi} \frac{\mu_0 I}{L} \hat{i} ; \vec{B}_{D F}=\frac{\sqrt{2} \mu_0 I}{4 \pi L} \hat{i} ; \vec{B}_{F A}=\frac{\sqrt{2} \mu_0 I}{4 \pi L} \hat{k}} \)
\(\therefore\mathrm{\vec{B}=\frac{\sqrt{2} \mu_0 I}{\pi L} \hat{i}+\frac{\sqrt{2} \mu_0 I}{\pi L} \hat{k}} \) \(\therefore \mathrm{|\vec{B}|=\frac{2 \mu_0 I}{\pi L}=\frac{B_0}{\sqrt{2}}}\)

\(\mathrm {B=\frac{\mu_0 I}{2 \pi r} }\)
Force on the particle is \(\mathrm {F=q v B=\frac{\mu_0 I q v}{2 \pi r}(\uparrow)}\)
This force is always perpendicular to the velocity. Since deflection is small, the force is nearly in ( \(\uparrow\) ) direction always.
Impulse is:\(\mathrm {F \Delta t =\frac{\mu_0 I q v \Delta t}{2 \pi r}=\frac{\mu_0 I q L}{2 \pi r}} \)\(\mathrm \therefore{\Delta P =\frac{\mu_0 I q L}{2 \pi r}(\uparrow) }\)
\(\mathrm {\theta =\frac{\Delta P}{P}=\frac{\mu_0 I q L}{2 \pi r m V}}\)

Infrared waves are called waves of heat energy.

• The infrared waves present in the light ray coming from the sun is responsible for the heat energy.
• Thus, infrared waves are called as waves of heat energy.
• The Radio waves, Ultraviolet waves and Microwaves don’t have the heat energy.

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

\(=150\)