Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

A,C D

Explanation:

checking option (A)

Resistance of the stop =

$\rho \frac{ \pi \: r }{tdr}$

R = net resistance

$\frac{1}{R} = \int \: \frac{tdr}{f \pi \: r} \frac{t}{ \pi \: p} \int \frac{dr}{r}$ limit R1 to R2

R = πr/tln(R2/R1)

Hence , current ,i = V/R = Vt/lnR2/R1/πp

_____option (A) correct

__________________________

for elementry dr

Checking option (D)

$di = \frac{v \circ}{ \rho \frac{ \pi \: r}{tdr} } = \frac{v \circ \: tdr}{ \rho \pi \:r}$

I = neav where ,

eE = mv²/r

$= m \frac{v \circ {}^{2} }{n {}^{2} l {}^{2} \rho { }^{2} \pi {}^{2}r {}^{2} } \alpha \frac{1}{r}$

•°•

$m \frac{v {}^{2} }{r} = m \frac{v \circ {}^{2} }{n {}^{2}l {}^{2} \rho {}^{2} \pi {}^{2} r {}^{2} } \alpha \frac{1}{r}$

$e \alpha \frac{1}{r {}^{3} }$

E = k ×1/r³

∆V =

$\int(Edr) = k \int \: \frac{dr}{v {}^{2} } \alpha v \circ \\ \\ \\ since \: \: \alpha \: i {}^{2}$

____________________________

Option (C) checking

For option (B) we already know that v is directly proportional to Resistance since,outer surface have lower voltage .

__________________________________

Answer 2

Shown in figure is a semicircular metallic strip that has thickness t and resistivity ρ. its inner radius is R₁ and outer radius R₂. if the voltage V₀ is applied between its two ends, a current I flows in it. In addition, it is observed that a transverse voltage ∆V develops between it's inner and outer surfaces due to purely kinetic effects of moving electrons ( ignore any role of the magnetic field due to the current ) Then...

solution : let cut an element dx at x distance from the origin.

we know R = ρ l/A

here here, R = dr , l = πx and A = t × dx [ element is rectangular when we stretch it so area, A = t dx ]

now 1/dr = t/ρπ ∫ dx/x

⇒1/r = t/πρ ln[R₂/R₁]

r = πρ/(t ln[R₂/R₁] )

resistance of metallic strip would be πρ/(t ln[R₂/R₁] )

Then current , I = V₀/R = V₀t ln[R₂/R₁]/πρ

option (A) is correct choice.

electric field (-eE) due to electrons will be inwards in order to provide centripetal acceleration. Therefore the electric field will be radially outward.

Therefore outer potential < inner potential

so option (C) is correct choice.

now, mV_d²/R = qE

⇒E = mV_d²/qR

⇒∆V = ∫ E dR

i.e., ∆V ∝ E ∝ V_d²

we know, drift velocity, V_d = i/neA ⇒V_d ∝ i²

so ∆V ∝ i²

so option (D) is correct choice.

Therefore options (A), (C) and (D) are correct choices.