Physics Test - 25

Given:

\(v \propto g^{p} h^{q}\)

We know that:

\(v=\left[\mathrm{L} \mathrm{T}^{-1}\right]\)

\(g=\left[\mathrm{L} \mathrm{T}^{-2}\right]\)

\(h=[\mathrm{L}]\)

By substituting the dimension of each quantity and comparing the powers in both sides we get:

\({\left[\mathrm{L} \mathrm{T}^{-1}\right]=\left[\mathrm{L} \mathrm{T}^{-2}\right]^{\mathrm{p}}[\mathrm{L}]^{\mathrm{q}}} \)

\(\Rightarrow {\left[\mathrm{L} \mathrm{T}^{-1}\right]=\left[\mathrm{L}^{\mathrm{p}+\mathrm{q}} \mathrm{T}^{-2 \mathrm{p}}\right]}\)

On comparing the powers

\( p+q=1,-2 p=-1 \)

\(\therefore p=\frac{1}{2}, q=\frac{1}{2}\)

\(\overrightarrow{\mathrm{A}}={x}_{1} \hat{\mathrm{\imath}}+10 \hat{\mathrm{j}}\)

\(\overrightarrow{\mathrm{B}}={x}_{2} \hat{\mathrm{i}}+10 \hat{\mathrm{j}}\)

\(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{A}}=\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right){\hat{i}}+20 \hat{\mathrm{j}}\)

\(\Rightarrow \frac{20}{\mathrm{x}_{1}+\mathrm{x}_{2}}=\tan 60^{\circ}=\frac{0.87}{0.5}\)

\(\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}=\frac{10}{0.87}\)

\(\Rightarrow|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{\left(\frac{10}{.87}\right)^{2}+20^{2}}=23 m\)

Infraredelectromagnetic wave is used for thermal imaging.

• The electromagnetic spectrum is the arrangement of electromagnetic waves, in the order of their respective wavelength.
• Infrared rays are a type of electromagnetic waves which are emitted by hot bodies.
• Within the electromagnetic spectrum, infrared waves occur at frequencies above microwaves below the wavelength of visible red light, that this type of electromagnetic waves termed infrared.
• Thermal imaging is the technique in which thermal energy and infrared radiation are used to gather information about objects.
  

Given:

A point particle of mass \(0.1 \mathrm{~kg}\) is executing SHM of amplitude \(0.1 \mathrm{~m}\).

\(\mathrm{KE}\) = \(8 \times 10^{-3} J\)

Initial phase of oscillation = \(45^{\circ}=\frac{\pi}{4}\)

We know that:

\(\mathrm{KE}\) at mean position \(=\frac{1}{2} m \omega^{2} a^{2}\)

According to the question:

\(\frac{1}{2} m \omega^{2} a^{2}=8 \times 10^{-3}\).....(1)

Put all the given values in (1).

\(\omega=\left(\frac{2 \times 8 \times 10^{-3}}{m a^{2}}\right)^{\frac{1}{2}}\)

\(=\left[\frac{2 \times 8 \times 10^{-3}}{0.1 \times(0.1)^{2}}\right]^{\frac{1 }{2}}\)

\(=4\)

Equation of SHM is,\(y=a \sin (\omega t+\theta)\)

\(=0.1 \sin \left(4 t+\frac{\pi}{4}\right)\)

Potential energy of a magnetic moment at an angle \(\theta\) with the magnetic field \(=-\mathrm{MB} \cos \theta\)

Thus work done in rotating from angle \(\theta_{1}\) to \(\theta_{2}=\operatorname{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)\)

\(=\operatorname{MB}\left(\cos 0^{\circ}-\cos 90^{\circ}\right)\)

\(=\mathrm{M B}(\mathrm{1}-\mathrm{0})\)

\(=\mathrm{MB}\)

According to Gauss's law, the total charge inside any closed surface is measured by the flow through it. As a result, the Gauss law holds true for closed surfaces. Only symmetric body charge distributions, such as spherical, cylindrical, and plane symmetry, are valid for Gauss's law.The sum of all charges encompassed by the surface is included in the term \(\mathrm{q}\) on the right side of Gauss's law. The charges could be anywhere within the surface.The electrostatic field of a symmetric system can be calculated using Gauss's law. The inverse square dependence on distance inherent in Coulomb's law is the basis for Gauss's law. Any deviation from the inverse square law will be indicated by a violation of Gauss' law.

Gauss's law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface.

Given \(\mathrm{G}=10 \Omega\)

\(\mathrm{I}_{\mathrm{g}}=3 \mathrm{~mA} \)

\( \mathrm{I}=8 \mathrm{~A}\)

In case of conversion of galvanometer into ammeter.

We have \(\mathrm{Ig}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}\)

\( S=\frac{I_g G}{I-I_g} \)

\(S=\frac{\left(3 \times 10^{-3}\right) 10}{8-0.003}=3.75 \times 10^{-3} \Omega\)

The velocity of revolving electron is: \({v}=\frac{{e}^{2}}{2 {nh} \epsilon_{0}}\)

The Rydberg's constant is: \({R}=\frac{m {e}^{4}}{8 {E}_{0}^{2} {~h}^{3} {c}}\)

So, \(v R=\frac{e^{2}}{2 n h \epsilon_{0}} \times \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3} c}\)

\(v R=\frac{m e^{6}}{16 n h^{4} \epsilon_{0}^{3} c}\)

Since, \({n}, {h}, \epsilon_{0}, {~m}, {c}\) and e are constant.

Hence, the \(v R\) is inversely proportional to principal quantum number (\(n\)).

The energy of an electron is:

\({E}=\frac{-m {e}^{4}}{8 {n}^{2} {~h}^{2} {\epsilon}_{0}^{2}}\)

Thus,\(\frac{{v}}{{E}}=\frac{\frac{{e}^{2}}{2 {nh} \epsilon_{0}}}{\frac{-m {e}^{4}}{8 {n}^{2} {~h}^{2} \epsilon_{0}^{2}}}\)

\(\frac{{V}}{{E}}=\frac{{e}^{2}}{2 n h \epsilon_{0}} \times \frac{8 n^{2} {~h}^{2} {E}_{0}^{2}}{-m {e}^{4}} \)

\(⇒\frac{{v}}{{E}}=-\frac{4 {nh} \epsilon_{0}}{{me}^{2}}\)

Since, \({n}, {h}, \epsilon_{0}, {m}\) and e are constant.

So, the \(\frac{{v}}{{E}}\) is proportional to principal quantum number \(n\).

We slip while walking on a path having pond scum or green algae because the friction between the feet and the path is reduced.

Friction is a force between two surfaces that are sliding, or trying to slide, across each other.For example, when you try to push or pull luggage along the floor, friction makes this difficult.Friction always works in the opposite of the direction in which the object is moving or trying to move.Static friction acts on objects when they are resting on a surface.Without this static friction, feet would slip out and making it difficult to walk.

In projectile thrown at angle \(\theta\) Range \(\mathrm{R}\) and maximum height \(\mathrm{H}\) are given as:

Range, \(\mathrm{R}=\frac{\mathrm{u}^{2}(\sin 2 \theta)}{\mathrm{g}}=\frac{\mathrm{u}^{2} 2 \sin \theta \cos \theta}{\mathrm{g}}\)

Maximum Height, \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)

Given: \(\mathrm{H}=\mathrm{R}\)

\(\frac{u^{2}(2 \sin \theta \cos \theta)}{g}=\frac{u^{2} \sin ^{2} \theta}{2 g}\)

\(4 \cos \theta=\sin \theta\)

\(⇒\tan \theta=4\)

\(⇒\theta=76^{\circ}\)