Physics NCERT Exemplar Solutions Class 12th Chapter Three

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

Explanation- Metallic strips have negligible resistance and need not to be counted in the length l₁, of the null point of potentiometer. That's why the thick metallic strips are used in potentiometer. It is for the convenience of experimenter as he measures only their lengths along the straight segments each of lengths 1 m.

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

Explanation- relaxation time =mean free path/rms velocity of electron

Also $\rho$ = $\frac{1}{\sigma }$ = $\frac{m}{ne2t}$ relaxation time is inversely proportional to velocities

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

Explanation – In the circuit when an electron approaches a junction, in addition to the uniform E that faces it normally (which keep the drift velocity fixed), as drift velocity (v_d) is directly proportional to Electric field (E). That's why there are accumulation of charges on the surface of wires at the junction.
These produce additional electric fields. These fields alter the direction of momentum. Thus, the motion of a charge across junction is not momentum conserving

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

Explanation- applying kirchhoff's junction rule I₁ = I+I₂

Applying kirchhoff's rule in outer loop containing 10V cell

10= IR+10I₁……………. (1)

Applying kirchhoff's rule in outer loop containing 2V cell

2= 5I₂-RI= 5 (I₁-I)-RI

4= 10I₁-10I-RI………… (2)

From 1 and 2

6=3RI+10I

2=I (R+10/3)

V= I (R+R_eff)

After comparing V=2V, R_eff= 10/3 ohm

Since effective internal resistance R_eff of two cells 10/3 ohm, being the parallel combination 5 ohm and 10 ohm . the equivalent circuit is

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

Explanation- according to ohm's law V= IR

I= 6/6 = 1A

I= AneV_d  or V_d= i/neA

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

Explanation – power consumption in a day i.e in 5  = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $*$ I

2000W= 220V $*$ l or l= 9A approx.

R= $\frac{\rho l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{\rho l}{A}$ l²= 1.7 $*$ 10^{-8 $*\frac{10}{\pi *{10}^{-6}}*81$} j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000}*$ 100=0.2%

Power loss in Al wire =1.6 $*$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000}*$ 100= 0.32%