## 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

Ans: (d)

Sol: Here, r = 4 Ω, I = 0.2 A, R = 21 Ω

If E is emf of the cell, then

E = I(R + r) = 0.2(21+4) = 5 volt.

Rate at which chemical energy is consumed

P = EI = 5 × 0.2 = 1 J/s

## Two metal wires of identical dimensions are connected in series. If σ_1 and σ_2 are the conductivities …

Q: Two metal wires of identical dimensions are connected in series. If σ1 and σ2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is

$\displaystyle (a) \frac{\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

$\displaystyle (b) \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

$\displaystyle (c) \frac{\sigma_1 + \sigma_2}{2 \sigma_1 \sigma_2}$

$\displaystyle (d) \frac{\sigma_1 + \sigma_2}{ \sigma_1 \sigma_2}$

Ans: (b)
Sol: When two resistances are connected in series then ,

$\displaystyle R_s = R_1 + R_2$

$\displaystyle R_s = \rho_1 \frac{l}{A} + \rho_2 \frac{l}{A}$

$\displaystyle R_s = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A}$

$\displaystyle \frac{2 l}{\sigma_{eq} A} = \frac{l}{A}[ \frac{1}{\sigma_1} + \frac{1}{\sigma_2} ]$

$\displaystyle \frac{2 l}{\sigma_{eq} A} = \frac{l}{A}[ \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} ]$

$\displaystyle \sigma_{eq} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

## The temperature dependence of resistances of Cu and Si in the temperature range 300-400 K is best described by :

Q: The temperature dependence of resistances of Cu and Si in the temperature range 300-400 K is best described by :

(a) linear increase for Cu, linear increase for Si

(b) linear increase for Cu, exponential increase for Si

(c) linear increase for Cu, exponential decreases for Si

(d) linear decrease for Cu, linear decrease for Si