Physics

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- eeq= $\frac{e_2{r}_{1+}{e}_2{r}_1}{r_1+r_2}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation- as we know that J=E, and current density is directly proportional to electric field, so electric field produced by charges accumulated on the surface of wire.

This is a long answer type question as classified in NCERT Exemplar

Given volume V = 1m³

area = 0.01mm²

= 8.01 $*{10}^{-6}$ m²= ${10}^{-8}$ m²

Temperature both inside and outside are equal

So initial pressure inside the box = 1.50atm

Final pressure inside the box= 0.1atm

Assuming V_x= speed of nitrogen molecule in x direction

n_i = number of molecules per unit volume in a time interval of $?T$

Let area of the wall, number of particles colliding in time

$?t$ = $\frac{1}{2}n$ _i (v_{x $?t$} )A , here we use ½ because particle moves both in positive and negative direction.

V_x²+ V_y²+ V_z²= V_rms²

V_x²= V_rms²/3 if all three velocities are equal.

½ mv_rms²= 3/2K_BT/m

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- according to ohms law

I= $\frac{V_{eff}}{R_{eff}+R}$

If voltage and resistance increase

V'= nVeff, R'= nReff

I'= $\frac{\mathrm{n}\mathrm{V}\mathrm{e}\mathrm{f}\mathrm{f}}{\mathrm{n}\mathrm{R}\mathrm{e}\mathrm{f}\mathrm{f}+\mathrm{n}\mathrm{R}}=$ $\frac{V_{eff}}{R_{eff}+R}$ =

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- R= $\rho l$ /A

Resistance of first conductor, RA= $\frac{\rho l}{\pi {(10}^{-3}*0.5)}$ 2

Resistance of second conductor, RB= $\frac{\rho l}{\pi \left. {{(10}^{-6}\right)-(0.5*0.5*{10}^{-6})\}}$

Now ratio $\frac{R_A}{R_B}=\frac{\frac{\rho l}{\pi {(10}^{-3}*0.5)}2}{\frac{\rho l}{\pi \left. {{(10}^{-6}\right)-(0.5*0.5*{10}^{-6})\}}}$ = 3:1

This is a long answer type question as classified in NCERT Exemplar

Time require to avoid the collision T= l/v where l = mean free path =1/ $\sqrt[]{2}\pi d^{2}n$

Where n = N/V

n=number of aeroplanes/volume

= $\frac{10}{20*20*1.5}=0.0167km$ ^-3

T= $\frac{1V}{\sqrt[]{2}\pi d^{2}N}*\frac{1}{v}$

T= $\frac{1}{\sqrt[]{2}*3.14*{20}^2*0.0167*{10}^{-6}*150}$ = $\frac{{10}^6}{1776.25*2.505}$

$=\frac{{10}^6}{4449.5}=224.74h\approx 225h$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- according to ohms law

I= $\frac{E+E}{R+r1+r2}$

The potential difference across terminal is

V=e-Ir= E- $\frac{2E}{R+r1+r2}$ r1=0

E= $\frac{2Er1}{R+r1+r2}$

1= $\frac{2r1}{R+r1+r2}$

R+r1+r2 = 2r1

R= r1-r2

This is a long answer type question as classified in NCERT Exemplar

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- when all resistance are in parallel

$\frac{1}{R_p}=\frac{1}{R_1}+\dots \dots \dots \dots ..+\frac{1}{R_n}$ by multiplying this equation by R_min we have

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\dots \dots \dots \dots ..+\frac{R_{min}}{R_n}$

But there exist a term in RHS $\frac{R_{min}}{R_{min}}=1$ and other terms are positive so we have

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\dots \dots \dots \dots ..+\frac{R_{min}}{R_n}$ >1

This shows that equivalent resistance is less than maximum resistance available.

But when all resistance are in series

R_s =R₁+R₂………R_n

here must be R_max value in RHS

R_s= R₁+……R_max+….R_n

And R_s> R_max

So equivalent resistance is less than R_max