Physics Test - 2

Band gap,

\(E_{g}=\frac{hc}{\lambda}\)

\(=\frac{\left(6.63 × 10^{-34}\right)\left(3× 10^{8}\right)}{2480 × 10^{-9} × 1.6 × 10^{-19}}~eV\)

\(=0.5 ~eV\)

The volume of the cubic domain is:

\(V=\left(10^{-6} \mathrm{~m}\right)^{3}\)

\(=10^{-18} \mathrm{~m}^{3}\)

\(=10^{-12} \mathrm{~cm}^{3}\)

Its mass is volume \(\times\) density \(=7.9 \mathrm{~g} \mathrm{~cm}^{-3} \times 10^{-12} \mathrm{~cm}^{3}=7.9 \times 10^{-12} \mathrm{~g}\)

It is given that Avagadro number \(\left(6.023 \times 10^{23}\right)\) of iron atoms have a mass of 55 g. Hence, the number of atoms in the domain is

\(N=\frac{7.9 \times 10^{-12} \times 6.023 \times 10^{23}}{55}\)

\(=8.65 \times 10^{10}\) atoms

The maximum possible dipole moment \(m_{\max }\) is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned.

Thus,

\(m_{\max }=\left(8.65 \times 10^{10}\right) \times\left(9.27 \times 10^{-24}\right)\)

\(=8.0 \times 10^{-13} \mathrm{Am}^{2}\)

The consequent magnetisation is

\(M_{\max }=\frac{m_{\max }}{Domain volume}\)

\(=\frac{8.0 \times 10^{-13} \mathrm{Am}^{2}}{10^{-18} \mathrm{~m}^{3}}\)

\(=8.0 \times 10^{5} \mathrm{Am}^{-1}\)

Given,

\(t_{1} = 2\) s

\(t_{2} = 10\) s

If \(t_{1}\) and \(t_{2}\) are the time, when body is at the same height.

Then, 

\(h=\frac{1}{2} g t_{1} t_{2}\)

\(\Rightarrow h=\frac{1}{2} \times g \times 2 \times 10\)

\(\Rightarrow h=10g\)

Given \(\mu_{1}=1.33, d=1 mm =10^{-3} m , D=1.33 m, \lambda=6300 \mathring{A}\) 

\(=6.3 \times 10^{-7} m\)

When the experiment is performed in liquid, \(\lambda\) change to 

\(\lambda^{\prime}=\frac{\lambda}{\mu_{1}}\)

Fringe width, 

\(\beta=\frac{D \lambda}{\mu_{l}}\)

\(\frac{1.33 \times 6.3 \times 10^{-7}}{1.33 \times 10^{-3}}\)

\(=6.3 \times 10^{-4} m\)

\(=0.63~ mm\)

The Einstein's equation for photoelectric effect is,

\(\mathrm{eV}_{0}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}\)

where, \(\mathrm{V}_{0}=\) stopping potential, \(\lambda=\) wavelength of incident light, \(\mathrm{W}=\) work function of metal.

\(E=4 \mathrm{NC}^{-1}, ~d=1m\)

\(\mathrm{V}_{0}=\frac{\mathrm{E} }{ \mathrm{d}}=\frac{4 }{ 1}=4\) volt

\(\lambda=200 \mathrm{~nm}=200 \times 10^{-9} \mathrm{~m}\)

Thus, \(\mathrm{W}=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}_{0}\)

\(=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{200 \times 10^{-9}}-\left(1.6 \times 10^{-19}\right) 4\)

\(=3.53 \times 10^{-19} \mathrm{~J}\) (as \(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\))

\(=\frac{3.53 \times 10^{-19} }{1.6 \times 10^{-19} }=2.2 \mathrm{eV}\)

For the net gravitational force on a particle.

\(F=\frac{G M^{2}}{a^{2}} \).....(1)

\(F_{1}=\frac{G M^{2}}{(a \sqrt{2})^{2}}\)

\(=\frac{G M^{2}}{2 a^{2}}\).....(2)

According to fig.:

For net force, \(F=\sqrt{F^{2}+F^{2}}+F_{1}\)

\(F_{net}=F \sqrt{2}+F_{1}\).....(3)

Put values from (1) and (2) in (3).

\(F_{net}=\frac{G M^{2}}{a^{2}} \sqrt{2}+\frac{G M^{2}}{2 a^{2}}\)

This force will act as centripetal force. Distance of particle from centre of circle is \(\frac{\mathrm{a}}{\sqrt{2}}\).

\(\mathrm{~F}_{\mathrm{C}}=\frac{\mathrm{Mv}^{2}}{\mathrm{r}} \)

\(\mathrm{r}=\frac{\mathrm{a}}{\sqrt{2}}\)

\(F_{C}=F_{net}\)

\(\frac{\mathrm{Mv}^{2}}{\frac{\mathrm{a}}{\sqrt{2}}}=\frac{\mathrm{GM}^{2}}{\mathrm{a}^{2}}\left(\frac{1}{2}+\sqrt{2}\right) \)

\(\Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{a}}\left(\frac{1}{2 \sqrt{2}}+1\right) \)

\(\Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{a}}(1.35) \)

\(\Rightarrow \mathrm{v}=1.16 \sqrt{\frac{\mathrm{GM}}{\mathrm{a}}}\)

When the lift is stationary spring force balances weight:

\({kx}=\mathrm{mg}=49 \mathrm{~N}...(1)\)

Where,

\(k=\) force constant of spring 

\(x=\) elongation 

True weight \(=49 \mathrm{~N}\)

From equation (1), we get

\(k=\frac{49}{x}\)

\(\mathrm{m}=\frac{49}{9.8}=5 \mathrm{~kg}\)

when lift moves with acceleration \(5 \mathrm{~m} / \mathrm{s}^{2}\) downward we have:

\({kx}_{2}=\mathrm{mg}-5 \times \mathrm{m}\)

Where,

Pseudo force in lift frame \(=5m\) upward

\({x}_{2}=\) new elongation

\(\Rightarrow \mathrm{kx}_{2}=49-5 \times 5=24 \mathrm{~N}\)

So new reading in spring balance \(=24\) N

The relation between angular frequency and displacement is given as

\(\mathrm{v}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}\).....(1)

Suppose

\(\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}\)

On differentiating the above equation w.r.t. time we get

\(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t}\)

The maximum value of velocity will be \(v_{\max}=A \omega\)

The displacement for the time when speed is half the maximum is given as

\(\mathrm{v}=\frac{\mathrm{A} \omega}{2} \)

\(\mathrm{~A}^{2} \omega^{2}=4 \omega\left(\mathrm{A}^{2}-\mathrm{x}^{2}\right)\)

By substituting the value in \((1)\) we get the displacement as:

\(\mathrm{x}=\frac{\mathrm{A} \sqrt{3}}{2}\)

Path difference for the rays coming from the two edges of the slit is

\(\Delta=a \sin \theta, a=\) slit width

For the first minimum, \(\alpha=\pi\)

where \( \alpha=\frac{\pi a}{\lambda} \sin \theta=\pi\)

or \( a \sin \theta=\lambda\)

Phase difference \(=\frac{2 \pi}{\lambda} \Delta=2 \pi\)

\(\begin{aligned} & \text { Gain in binding energy }=B \cdot E_f-B_i \\ & =2(121 \times 8.1)-242 \times 7.6 \\ & =121 \mathrm{MeV}\end{aligned}\)