Class 12th

F₂₁ = QE₁ = λ₂l (2kλ₁/R)

R = mv/qB = √2mk/qB

M_p/M_d = ½ and q_p/q_d =1/1

R_p/r_d = 1/√2

In reverse biasing, due to collision of electrons and atom, avalanche breakdown occurs.

This negative sign is a result of the derivation of Lenz's law, which is used to signify the principle of conservation of energy. As per this law, the EMF always opposes the flow of magnetic flux in a circuit. And if this doesn't happen, the fundamental laws and theories of physics can be proved wrong by changing the magnetic flux.

Galvanometer is a device having low resistance which is used to detect tiny bits of electric current scattered among different directions in the circuit. Whereas, a voltmeter is used to calculate the potential difference between two different points in the circuit. A galvanometer can also be used as a voltmeter if connected to high resistance in a series.

They core difference between these two terms lies in the fact that self induction involves only one coil as compared to mutual induction involving two. In mutual induction, the change of electric current in one coil produces a magnetic field which results in generating EMF in the second coil. Whereas there is no concept of neighbouring coil in self induction and everything happens within the one coil itself.

kx + y + 2z = 1. (i)

3x – y – 2z = 2 . (ii)

2x – 2y – 4z = 3. (iii)

(ii) * 5 (i) $\equiv$ $$  (iii) * 3 (15 – k) = 6

K = 21

Let f (x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$    Non zero finite

So, d = e = f = 0

f (x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$ Non zero finite

f' (x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$

f' (1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 . (i)

f' (-1) = 0

-6 + 5a – 4b + 3 = 0 . (ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

$\therefore f\left(x\right)=x^6+\left(\frac{-3}{5}\right)x^5+\left(\frac{-3}{2}\right)x^4+x^3$

$\therefore 5.f\left(2\right)=144$

With the help of conservation of volume, we can write

  $27*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=3r.......\left(1\right)$                            

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U₁ =   $\frac{q^2}{8\pi {\epsilon }_{0}r}$

Potential energy of bigger drop =   $U_2=\frac{Q^2}{8\pi {\epsilon }_{0}R}=\frac{27*27*q^2}{8\pi {\epsilon }_0\left(3r\right)}=243\left(\frac{q^2}{8\pi {\epsilon }_{0}r}\right)=243U_1$

$\Rightarrow \frac{U_2}{U_1}=243$           

$l=\frac{90?30}{4000}=15mA$

$I_1=\frac{30}{5000}=6mA\&l_2=9mA$