Physics Test-36

At resonance, the frequency of supply power equals to naturan frequency of LCR circuit.

Impedence, \(Z=\sqrt{R^2+(\omega L-\frac{1 }{ \omega C})}\)

\(\omega_{ L }=\frac{1}{\omega C }\)

so, \( Z=R \)

\( I = \frac{V }{ R} =\frac{200 }{20}=10 A\)

Power, \(W = V \times I \)

\(=200 \times 10\)

\(=2000 W\)

Given: Radius of Coil \(A=a\)

Current in coil \(A=1\)

Field due to Coil A at center,

\(B_{A}=\frac{\mu_{0} I}{2a} \quad \ldots\) (i)

Radius of Coil \(B=2 a\)

Current in coil \(B=1\)

Field due to Coil B at center,

\(B_{B}=\frac{\mu_{0} I }{2(2 a )}=\frac{1}{2}\left(\frac{\mu_{0} I }{2 a }\right)\quad \ldots\) (ii)

Comparing (i) and (ii),

\(B_{B}=\frac{B_{A}}{2}\)

\(\Rightarrow B _{ A }: B _{ B }=2: 1\)

So, the required ratio is \(2: 1\).

Given that,

Mass of Nitrogen, \(m=2.0 \times 10^{-2} kg =20 g\).

Rise in temperature, \(\Delta T=45^{\circ} C\).

Molecular mass of \(N_{2}, M=28\)

Universal gas constant, \(R=8.3 J \operatorname{mol}^{-1} K^{-1}\)

Number of moles, \(n=\frac{m}{M}\)

\(n=\frac{2 \times 10^{-2} \times 10^{3}}{28}\)

\( n=0.714\)

Now, molar specific heat at constant pressure for nitrogen,

\(C_{p}=\frac{7}{2} R \)

\(C_{p}=\frac{7}{2} \times 8.3 \)

\( C_{p}=29.05 J mol ^{-1} K ^{-1}\)

The total amount of heat to be supplied is given by the relation:

\(\Delta Q=n C_{p} \Delta T \)

\(\Delta Q=0.714 \times 29.05 \times 45 \)

\( \Delta Q=933.38 J\)

Clearly, the amount of heat to be supplied is \(933.38 J\).

Hydrogen bomb is based on nuclear fusion. A large amount of nuclear energy is released by fusion of two light elements (elements with low atomic numbers). In a hydrogen bomb, two isotopes of hydrogen, deuterium (1 proton, 1 neutron) and tritium (1 proton, 2 neutron) are fused to form a nucleus of helium and a neutron. This fusion releases \(17.6 \mathrm{MeV}\) of energy. Also, there is no limit on the amount of the fusion that can occur.

Nano - (symbol n) is a unit prefix meaning "one billionth".

• It is used primarily with the metric system.
• This prefix denotes a factor of 10⁻⁹ or 0.000000001.
• The prefix "nano" comes from a Greek word meaning dwarf.

We know that:

When the time period of a simple pendulum is \(T\).

\(T^{2}=4 \pi^{2}\left(\frac{l}{g}\right) \ldots(i)\)

When the length is increased by \(10 \mathrm{~cm}\)

\(T_{1}^{2}=4 \pi^{2}\left(\frac{l+10}{g}\right) \ldots(ii)\)

When the length is decreased by \(10 \mathrm{~cm}\)

\(T_{2}^{2}=4 \pi^{2}\left(\frac{l-10}{g}\right) \ldots(iii)\)

Adding Eqs. (ii) and (iii), we get

\(T_{1}^{2}+T_{2}^{2}=4 \pi^{2}\left[\frac{2 l}{g}\right] \)

\( 2\left(4 \pi^{2}\right)\left(\frac{l}{g}\right)=2 T^{2}\)

Given:

Escape velocity: The minimum velocity required to escape the gravitational field of earth is called escape velocity. It is denoted by \(V_{ e }\).

The escape velocity on earth is given by:

\(V_{e}=\sqrt{\frac{2 G M}{R}}=11.2\) km/s.....(1)

Where, \(G\) is universal gravitational constant, \(M\) is mass of the earth and \(R\) is radius of the earth.

Mass of the planet \(\left(M^{\prime}\right)=2 M\)

Radius of the planet \(\left(R^{\prime}\right)=\frac{R }{ 2}\)

So, The escape velocity on the planet is given by:

\(V_{e}^{\prime}=\sqrt{\frac{2 G M^{\prime}}{R^{\prime}}}\)

\(=\sqrt{\frac{2 G \times 2 M}{\frac{R}{2}}}\)

\(=2 \times \sqrt{\frac{2 G M}{R}}\)

From (1).

\(=2 \times 11.2\)

\(=22.4\) km/s

X-rays are different from UV light because it has different wavelength from UV light.

• All EM radiations travel at the same speed.
• So, X-rays and UV light travels at the same speed.
• Since every EM radiations have different wavelengths according to their energy level.
• So, the X-rays are different from UV light because it has a different wavelength from UV light.

The SI Unit of Charge is Coulomb. It is denoted as C.

The coulomb is defined as the quantity of electricity transported in one second by a current of one ampere.

The internal resistance of the cell is given by:

\(\Rightarrow r=\left(\frac{L_1}{L_2}-1\right) \times R\)

\(\Rightarrow r=\left(\frac{2 L_1}{L_1}-1\right) \times R\)

\(\Rightarrow r=(2-1) \times R\)

\(\Rightarrow r=R=2 \Omega\)