Physics Test - 27

The maximum flux will pass through the cube if the charge enclosed is the maximum (from Gauss' Law).

According to Gauss' law,

\(\phi=\frac{Q_{i n}}{\epsilon_{0}}\)

So, in order to have maximum electric flux through the cube, we need to have maximum charge inside the cube, which is achieved by coinciding the wire with the space diagonal of the cube. The maximum length of the string which can fit into the cube is \(\sqrt{3}a\), equal to its body diagonal. The total charge inside the cube is \(\sqrt{3} a \lambda\)

\(\phi=\frac{Q_{i n}}{\epsilon_{0}}=\frac{\sqrt{3} \lambda a}{\epsilon_{0}}\)

From Rydberg's relation:

\(\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)\)

\(\Rightarrow v=\mathrm{R}_{\mathrm{H}} \mathrm{c}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)\)

First line of lyman series is from transition \(2 \rightarrow 1\)

Series limit of lyman is from transition \(\infty \rightarrow 1\)

Series limit of balmer is from transition \(\infty \rightarrow 2\)

\(v_{1}=R_{H} c\left(1-\frac{1}{\infty^{2}}\right)\)

\(v_{2}=R_{H} c\left(1-\frac{1}{2^{2}}\right)\)

\(v_{3}=R_{H} c\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)\)

\(v_{1}-v_{2}=R_{H} c\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=v_{3}\)

So,\({v}_{1}-{v}_{2}={v}_{3}\)

\(E=5.5 e V\), \(\phi=4.5 e V\)

Therefore, the kinetic energy will be equal to

\( K . E=E-\phi\)

On substituting the values in above formula, we get

\( K . E=(5.5-4.5) e V \Rightarrow K . E=1 e V\)

Wavelength \((\lambda)=\frac{h}{\sqrt{2 M_{e} \times K \cdot E}}\)

\(\lambda\) is the wavelength.

\(h\) is the Planck's constant  \(=6.6 \times 10^{-34}\).

\(M_{e}\) is the mass of an electron  \(=9.1 \times 10^{-31}\).

\(K . E\), is the kinetic energy  \(=1.16 \times 10^{-19}\).

So on substituting the values in the formula of wavelength, we get 

\( \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31}  \times 1.16 \times 10^{-19}}}=1.24 \times 10^{-9} m\)

The Magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the image formed at least distance of distinct vision to angle subtended at the eye by the object lying in infinity.

\(M=-\frac{f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)\)

fo \(=\) focal length of the object \(=150 \mathrm{~cm}\) fe= focal length of the eyepiece=5cm

\(\mathrm{D}=\) least distance of the distinct vision \(=25 \mathrm{~cm}\)

\(M=\frac{-150}{5}\left(1+\frac{5}{25}\right)=-36\)

\(M=\frac{\beta}{\alpha}=\frac{\tan \beta}{\tan \alpha}(a s\) we know angles are small \()\)

\(\tan \alpha=\frac{H}{u}=\frac{100}{3000}\)

\(=\frac{1}{30}\)

Where \(\mathrm{H}=\) height of the object and \(\mathrm{u}=\) distance of the object from the objective.

\(M=\frac{\tan \beta}{\left(\frac{1}{30}\right)}=\frac{-36}{30}\)

\(=\frac{H^{\prime}}{D}\)

\(H^{\prime}=\) the height of the image and \(\mathrm{D}=\) distance of the image formation

Thus,

\(H^{\prime}=\frac{-36 \times 25}{30}=-30 \mathrm{~cm}\)

Therefore, the negative sign indicates that we get an inverted image.

The heat \(Q\) is converted into the internal energy and work. According to the first law of thermodynamics,

\(Q = U + W\)

\(nC _{ p } \Delta T = nC _{ v } \Delta T + W\)

The defraction of heat converted to work is given as

\(\frac{ W }{ Q }=\frac{ nC _{ p } \Delta T - nC _{ v } \Delta T }{ nC _{ p } \Delta T }\)

\(=\frac{C_{p}-C_{v}}{C_{p}}\)

\(=\frac{\frac{5}{2} R-\frac{3}{2} R}{\frac{5}{2} R}\)

\(=\frac{2}{5}\)

A satellite of mass M is orbiting in an orbit of radius R. Workdone by gravity per revolution on the satellite is given by 0.

Gravity acts towards the center of the earth. So if the satellite moves in a circular orbit with center of earth as the orbit center the work done at any point is zero, since the force and the displacement are perpendicular.

\(\mathrm{s}=\mathrm{v} . \mathrm{dt}\)

F perpendicular to \(\mathrm{v}\) so \(W=0\).

The magnetic moment of a magnet is a vector quantity that determines the torque it will experience in an external magnetic field.It is considered to be a vector having a magnitude and direction. The direction of the magnetic moment points from the South Pole to the North Pole of the magnet.

As we know, water can take a long time to warm up or cool down compared to the air. When water cools down to \(0^{\circ} C\), ice begins to form and floats on top of relatively warmer water; water at the bottom of a lake or river is typically \(4^{\circ} C\).