## Shown in the figure is a semicircular metallic strip that has thickness t and resistivity ρ . Its inner radius is R1 and outer radius R2 ….

Q: Shown in the figure is a semicircular metallic strip that has thickness t and resistivity ρ . Its inner radius is R1 and outer radius R2 . If a voltage Vo is applied between its two ends , a current I flows in it . In addition , it is observed that a transverse voltage ΔV develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignore any role of the magnetic field due to the current ) . Then

(a) $I = \frac{V_o}{\pi \rho} ln(\frac{R_2}{R_1})$

(b) the outer surface is at a higher voltage than the inner surface

(c) the outer surface is at a lower voltage than the inner surface

(d) $\Delta V \propto I^2$

Click to See Solution :
Ans: (a,c,d)
Sol: Since all the elements are in parallel

$\int \frac{1}{dr} = \int_{R_1}^{R_2} \frac{t dx}{\rho \pi x}$

$\frac{1}{r} = \frac{t}{\pi \rho}ln(\frac{R_2}{R_1})$

$Resistance = \frac{\pi \rho}{t ln(\frac{R_2}{R_1})}$

$I = \frac{V}{Resistance}$

$I = \frac{V t}{\pi \rho}ln(\frac{R_2}{R_1})$

$eE = \frac{m v^2}{r}$

$I = n e A v_d$

$\frac{V_o t dr}{\rho \pi r} = n e (dr t)v$

$v = \frac{V_o }{\rho \pi n e r}$

$E = \frac{m}{e r} \frac{V_o^2}{\rho^2 \pi^2 n^2 e^2 r^2}$

$E = \frac{m V_o^2}{e^3 r^2 \rho^2\pi^2 n^2}$

$dV = -E dr$

$dV = – \frac{K}{r^3}dr$

$V = K \int_{R_1}^{R_2}r^{-3} dr$

$\Delta V = \frac{m V_o^2}{e^3 r^2 \rho^2\pi^2 n^2} (\frac{1}{r_1^2} – \frac{1}{r_2^2})$

$\Delta V \propto V_o^2$

$I \propto V_o$

$\Delta V \propto I^2$

## 12 cells each having same emf , are connected in series and are kept in a closed Box.Some of cells are wrongly connected…

Q: 12 cells each having same emf , are connected in series and are kept in a closed Box. Some of cells are wrongly connected. This battery is connected is series with an ammeter and two cells identical with the others. The current is 3 A with the cells and battery aid with each other and 2 A when the cells and battery oppose each other. How many cells are wrongly connected.

(a) 1

(b) 2

(c) 3

(d) 4

Ans: (a)
Sol: Let n cells be wrongly connected. Net emf of battery = (12-2 n)E when 2 other identical cells are connected so that these aid the battery

Net emf = (12 – 2 n)E + 2 E and I = 3 A

so , ((12-2n)E +2E )/R = 3 …(i)

When 2 cells oppose the battery, net emf

=(12-2 n)E – 2 E

As per the question

Current = ((12-2n)E -2E )/R = 2 …(ii)

Dividing (i) and (ii),

we get , n = 1

## A student measures the terminal potential difference (V) of a cell (of emf E and internal resistance r) …

Q: A student measures the terminal potential difference (V) of a cell (of emf E and internal resistance r) as a function of current (I) flowing through it. The slope and the intercept, of the graph between V and I, then respectively, equal

(a) -r and E

(b) r and -E

(c) -E and r

(d) E and -r

Ans: (a)

Sol: Terminal potential difference ,

V = E – I r

V = -I r + E ….(i)

It shows as I increases, V decreases.

The variation of V and I will be a straight line with negative slope .

On comparing with (i) , y = m x + c

Slope , m = -r and Intercept = E

## The internal resistance of primary cell is 4 Ω. It generates a current of 0.2 A in an external resistance of 21 Ω…

Q: The internal resistance of primary cell is 4 Ω. It generates a current of 0.2 A in an external resistance of 21 Ω. The rate at which chemical energy is consumed in providing the current is

(a) 0.42 J/s

(b) 0.24 J/s

(c) 5 J/s

(d) 1 J/s