Physics

Step: Find the correct relation between forbidden energies.
The forbidden gap in the conduction, semiconductor, and insulator is shown in the figure below; 
 
Therefore, the correct relation between them is \({EG}_{3}>{EG}_{2}>{EG}_{1}\).
Hence, option (2) is the correct answer.
 

Step 1: Check the assertion.
It is correct that the angular momentum of an isolated system of particles is conserved.

Step 2: Check the reason.
On an isolated system, there is no force acting. So the net torque on an isolated system of particles is zero.
The torque is given by the rate of change of angular momentum. So the rate of change of angular momentum is also equal to zero in an isolated system.

Hint: \(F=mw^2R\)

Step: Find the time period of the revolution.
Given: The  central fictitious force \(F\) that is inversely proportional to \(R^3\) i.e., 
\(F\propto \frac{1}{R^3}\Rightarrow F= \frac{K}{R^3}\) 
\(\Rightarrow \frac{K}{R^3}= m\omega^2 R\Rightarrow \omega^2 = \frac{K}{mR^4}\)
\(\Rightarrow \left(\frac{2\pi}{T}\right)^2= \frac{K}{mR^4}~~~\left[\omega = \frac{2\pi}{T}\right]\)
\(\Rightarrow T^2 \propto R^4\)
\(\Rightarrow T\propto R^2\)
Therefore, the time period of the revolution is proportional to \(R^2.\)
Hence, option (1) is the correct answer.

Path difference=dsin$\mathrm{\theta }=(2\mathrm{n}–1)\frac{\mathrm{\lambda }}{2}$

sin$\mathrm{\theta }$$=(2\mathrm{n}–1)\frac{\mathrm{\lambda }}{2\mathrm{d}}=(2\mathrm{n}–1)\frac{2000\times {10}^{–10}}{2\times 0.0008\times {10}^{–3}}=\frac{(2\mathrm{n}–1)}{8}$

$\mathrm{Maximum}<mspace\; width="1em"></mspace>\mathrm{value}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{n}=4$
$\mathrm{So}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{total}<mspace\; width="1em"></mspace>\mathrm{no}.\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{minimas}=8<mspace\; width="1em"></mspace>(4<mspace\; width="1em"></mspace>\mathrm{above}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>4<mspace\; width="1em"></mspace>\mathrm{below}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{central}<mspace\; width="1em"></mspace>\mathrm{maxima})$

(3) Mass of a uranium nucleus $=92\times 1.6725\times {10}^{–27}+143\times 1.6747\times {10}^{–27}=393.35\times {10}^{–27}<mspace\; width="1em"></mspace>\mathrm{kg}$

Number of nuclei in the given mass $=\frac{1}{393.35\times {10}^{–27}}=2.542\times {10}^{24}$

Energy released $=200\times 2.542\times {10}^{24}<mspace\; width="1em"></mspace>\mathrm{MeV}$

$=5.08\times {10}^{26}<mspace\; width="1em"></mspace>\mathrm{MeV}=8.135\times {10}^{13}\mathrm{J}=8.2\times {10}^{13}\mathrm{J}$

Here,Δm=0.02866 u

∴ The energy liberated per nucleon

=0.02866 x 931/4

=26.7/4 MeV

=6.675MeV

$From<mspace\; width="1em"></mspace>the<mspace\; width="1em"></mspace>figure,<mspace\; width="1em"></mspace>\tan \theta =\frac{{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2}=\frac{x}{4}$
$For<mspace\; width="1em"></mspace>small<mspace\; width="1em"></mspace>\theta <mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>when<mspace\; width="1em"></mspace>\theta <mspace\; width="1em"></mspace>is<mspace\; width="1em"></mspace>counted<mspace\; width="1em"></mspace>in<mspace\; width="1em"></mspace>rad,<mspace\; width="1em"></mspace>\tan \theta \approx \theta \Rightarrow \theta =\frac{x}{4}$
$\Rightarrow \frac{\lambda }{a}=\frac{x}{4}$
$\Rightarrow x=\frac{4\lambda }{a}=2.4<mspace\; width="1em"></mspace>mm$

Step 1: Find the change in the volume of the sphere. 
The bulk modulus of the sphere is given by;
\(B = -\frac{p}{\frac{\Delta V}{V}}\)
\(\Rightarrow \frac{\Delta V}{V} = -\frac{p}{B}~~~…(1)\)
The volume of the sphere is given by;
\(V = \frac{4}{3}\pi R^3\)
\(\Rightarrow \frac{\Delta V}{V} = 3\frac{\Delta R}{R}\)
\(\Rightarrow \frac{\Delta R}{R} = \frac{1}{3}\frac{\Delta V}{V}~~~…(2)\)

Step 2: Find the change in the cross-sectional area of the sphere.
The surface area of the sphere is given by;
\(A= 4\pi R^2\)
\(\Rightarrow \frac{\Delta A}{A} = 2\frac{\Delta R}{R}~~~…(3)\)
From the equation \((1), (2),\) and \((3)\) we get;
\(\frac{\Delta A}{A} = -\frac{2p}{3B}\)
\(\Rightarrow \Delta A = -\frac{2pA}{3B}\)
Therefore, the magnitude of the cross-sectional area of the sphere is \(\frac{2pA}{3B}\).
Hence, option (3) is the correct answer. 
 

Step: Find the wavelength of the transmitted wave.
The frequency of the wave pulse remains unchanged since it depends only on the source. The wavelength of the transmitted wave is given by;
\(\frac{\lambda_1}{\lambda_2}=\frac{v_2}{v_1}\)
\(\Rightarrow \frac{\lambda_1}{\lambda_2}= \sqrt{\frac{\mu_2}{\mu_1}}~~~~~~~~~~~~\left[v=\sqrt{\frac{T}{\mu}}\right]\)
\(\Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{1}{2}\)
\(\Rightarrow \lambda_2=2\lambda _1=80~\text{cm}\)
Hence, option (2) is the correct answer.
 

Step: Find the inductance of the combination.
If two or more inductors are connected in series, then the current flowing through them will be the same, but the voltage difference across them will be different. Therefore, the inductance of the combination is \(L_{eff}=L_1+L_2.\)
Hence, option (1) is the correct answer.