Physics Test - 3

In weightlessness, there is no interaction between molecules.

Convection: In convection, gravity plays an important role. When liquid is heating the density of the lower molecules decreases. More dense molecules come down. In this example, there is no gravity so the molecule won't perform this motion. 

Radiation: It is the transfer of thermal energy in form of an electromagnetic wave.

It does not require a medium to transfer heat. Here heat is transferred by conduction. the vessel will conduct heat and transfer it to liquid. 

Conduction: In this process, heat is transferred due to temperature differences in neighbouring molecules.

Force on the moving charged particle due to magnetic field:

\(\overrightarrow{{F}}=q(\vec{V} \times \vec{B})\)

\(\overrightarrow{{F}}=q({a} \hat{i} \times(b \hat{j}+c \hat{k}))\)

\(\vec{F}=q|a b(\hat{i} \times \hat{j})+a c(\hat{i}+\hat{k})|\)

\(\vec{F}=q a|b \hat{k}-c \hat{j}|\)

Magnitude of the force is:

\(|\vec{F}|=q a \sqrt{\left|b^{2}+c^{2}\right|}\)

Given,

Diameter of onjective of telescope,

\(\mathrm{D}=2 \mathrm{~m}=100 \times 2 \mathrm{~cm}=200 \mathrm{~cm}\)

The wavelength of light is:

\(\lambda=600 \mathrm{~nm}=6 \times 10^{-5} \mathrm{~cm}\)

Limit of resolution of telescope,

\(\mathrm{d} \theta=\frac{1.22 \lambda}{\mathrm{D}}\)

\(=\frac{1.22 \times 6 \times 10^{-5}}{200}\)

\(=3.66 \times 10^{-7}\) radian

Given,

Length of 1st solenoid \(={L}\) 

Length of 2 nd solenoid \(=2 {L}\)

Number of turns of 1st solenoid \(={N}\) 

Number of turns of 2nd solenoid \(=4 {L}\)

Magnetic field due to the solenoid is given by:

\(B=\frac{\mu_{0} N I}{L}\)

As \(\mu_{o}\) and current \((I)\) is constant, therefore

\(B \propto \frac{N}{L}\)

\(\frac{B_{1}}{B_{2}}=\frac{N_{1}}{N_{2}} \times \frac{L_{2}}{L_{1}}\)

\(=\frac{N}{4 N} \times \frac{2l}{L}\)

\(=\frac{1}{2}\)

Across the diameter, we can see two semi-circles that is one above the diameter and another below the diameter. The length of the wire between two points at the ends of a diameter of a circle is half of the whole wire.

Therefore, the upper semicircle and lower semicircle have resistance \(2.5\) \(\Omega\) and both the resistance are in parallel.

Equivalent resistance is:

\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)

\(\Rightarrow \frac{1}{R}=\frac{1}{2.5 \Omega}+\frac{1}{2.5 \Omega}\)

\(\Rightarrow \frac{1}{R}=\frac{2.5 \Omega+2.5 \Omega}{2.5 \Omega \times 2.5 \Omega}\)

\(\Rightarrow R=\frac{2.5 \Omega \times 2.5 \Omega}{2.5 \Omega+2.5 \Omega}\)

\(\Rightarrow R=\frac{6.25 \Omega}{5 \Omega}\)

\(R=1.25 \Omega\)

Given,

Momentum is the same for light and a heavy body.

The relationship between the kinetic energy (KE) and Linear momentum (p) is given by,

$KE=\frac{1}{2}\frac{p^2}{m}$

As momentum (p) is constant.

$KE\propto \frac{1}{m}$

∴ Kinetic energy is inversely proportional to mass i.e heavier body will have less kinetic energy and vice-versa.

Therefore, a heavy body will have less kinetic energy and a light body will have more kinetic energy.

Given:

Mass of \(_{3} {Li}^{6}\) nucleus \(=6.015126 {~amu}\)

Mass of \(_{1} {H}^{3}\) nucleus \(=3.016049 {~amu}\)

Mass of \(_{2} {He}^{4}\) nucleus \(=4.002604 {~amu}\)

Mass of \(_{0} {n}^{1}=1.008665 {~amu}\)

Mass of the reactant \(=\) mass of \(_{3} {Li}^{6}+\) mass of neutron

\(=(6.015126+1.008665)=7.023791 {~amu}\)

Mass of the product \(=\) mass of \({ }_{2} {He}^{4}+\) mass of \({ }_{1} {H}^{3}\)

\(=(3.016049+4.002604)=7.018653 {~amu}\)

Mass difference \(\Delta {m}=(7.023791-7.018653)\)

\(=0.005138 {~amu}\)

Energy released \(=(0.005138 \times 931) {MeV}=4.783 {~MeV}\)

Mathematically the refractive index can be written as:

Speed of light in a medium \((v)=\frac{\text { Speed of light in vaccum }(C)}{\text { Refrivctive index }(\mu)}\)

As the speed of light in a vacuum is constant, therefore we can say that speed of light in a medium is inversely proportional to the refractive index of the medium.

Here the light is traveling from air to glass and as we know that the refractive index of glass (1.52) and that of air (1.008), means the glass is a denser medium and air is a rarer medium. Whenever light goes from air to glass, the frequency of light and phase of light does not change.

However, the velocity of light and wavelength of light change (decrease) because the refractive inclex of glass (1.52) is more than that of air (1.008).

As we know,

Power \((P_{0})=F v\) 

By putting \(F=ma\), we get

\(\therefore P_{0}=m a v\) 

By putting \(a=\frac{dv}{dt}\), we get

\(\Rightarrow P_{0}=m v \frac{d v}{d t}\)

Integrating both sides, we get

\(\int P _{0} dt =\int mvdv \)

\(\Rightarrow P _{0} t =\frac{ mv ^{2}}{2} \)

\(\Rightarrow v^{2}=\frac{2 P _{0} t }{ m } \)

\(\Rightarrow v \propto \sqrt{t}\)

\(\therefore v \propto t ^{\frac{1 }{ 2}}\)

Temperature can be measured in three different units Fahrenheit, Celsius, and Kelvin. \({ }^{\circ} F\), \({ }^{\circ} C\), and \(K\) are used to represent Fahrenheit, Celsius, and Kelvin units respectively. There are different formulae for their conversions.

In the given problem temperature is given in \({ }^{\circ} F\). It has to covert in \({ }^{\circ} C\). The formula for this conversion is,

\({ }^{\circ} C=\frac{5}{9}\left({ }^{\circ} F-32\right)\)

\(T^{\circ} C=\frac{5}{9}\left(140-32\right)=\frac{15}{9}×108 =60^{\circ} {C}\)

\(T^{\circ} C=60^{\circ} C\)

Temperature \(140^{\circ} F\) is \(60^{\circ} C\) on a Celsius scale.