Reverse biased

At t = T ½ N = No / 2 Using N = No e- λt
No / 2 = No e -λt ^½
Solving we get, T1/2 = ln2/λ = 0.693/ λ

No Apparent depth is maximum for that part of the bottom of the tank which is observed normally. Apparent depth decreases with increasing obliquity. Due to this unequal refraction, the flat bottom of the tank appears concave.

doubled

Self - inductance is defined as double the work done against the induced emf in producing unit current in the coil itself \(L=\frac{2 W}{I^2}\)

(i) Decreases From X to Z 
(ii) Decreases From X to Z
In case of  X Quality factor Is More , Impedance, Decreases From X to Z.

M = nIA = 100 x 5 x 10^-4 = 500 x 10^-4 J/T
Net force = 0
Torque = MB sin \(\theta\) = 5 x 10² x 0.1 x sin30  = 25 x 10^-4 Nm
H = R cos \(\eth \) = 4 x 1/2 = 0.2 G

M1 = M/2, M = M2

The magnetic Field due to AB and EF is as the direction of length vector is along the radius vector. Also the magnetic field due to BCE and BDE are equal opposite and equal so they cancel the effect of each other. So the net magnetic field at O is 0.

Now, 1/R = 1R₁ + 1R₂
Because 1 = 48/240 = 0.2 m

\(\text { Toral } \phi=\mathrm{q} 0 / \varepsilon_0=2 / \varepsilon_0, \rightarrow \text { flux through one face }=\phi / 6=1 / 3 \varepsilon_0\)

On removing one electron from He⁴ and He³, the energy levels, as worked out on the basis of Bohr's model will be very close as both the nuclei are very heavy as compared to electron mass. Also, after removing one electron, He⁴ and He³ atoms contain one electron and are hydrogen like atoms.

Total energy of an electron E_e = mc²
Total energy of a photon, \(E_p=\frac{h c}{\lambda}\)
de-Broglie wavelength of electron of mass m moving with velocity V
\( \lambda=\frac{h}{m v} \\ \Rightarrow m=\frac{h}{\lambda v} \)
\(E_e=m c^2=\frac{h c^2}{\lambda v}\)
\(\therefore \frac{E_e}{E_p}=\frac{\frac{h_0^2}{\lambda v}}{\frac{h o}{\lambda}}=\frac{c}{v}\)
As c > > v, therefore the total energy of electron is more than the total energy of photon.

As, P(1, 0, 4) and (2,-1, 5)
\(\therefore \ 2l=PQ=[(2-1)\hat { i } +(-1-0)\hat { j } +(5-4)\hat { k] }\\ =(\hat { i } -\hat { j } +\hat { k } )\)
and \(\ q=\pm 9\times { 10 }^{ -6 }C,E=0.20\hat { i } V/cm.\tau =?\)
Since, \(\tau =\rho \times E=q(2l)\times E\)
\(\therefore \ \tau =9\times { 10 }^{ -6 }(\hat { i } -\hat { j } +\hat { k } )\times 0.20=18\times { 10 }^{ -7 }(\hat { k } -\hat { j) } \)
Magnitude of torque,
\(\tau =18\times { 10 }^{ -7 }[\sqrt { { (1) }^{ 2 }+{ (1) }^{ 2 } } ]=25.45\times { 10 }^{ -7 }N-m\)

As electrons are negatively charged, they have tendency to go to regions of high potential.