Mathematics Definition and Proof Based Problems

### Definition and Proof Based Problems

Q 3211191920

Define the term 'electrical conductivity' of a metallic wire. Write its SI unit.

Solution:

Electric conductivity ( sigma ) is defined as the reciprocal of electric resistivity.

i.e  sigma = 1/rho

Its SI unit is Omega^(-1) m^(-1) , mho m^(-1) or siemen m^(-1).
Q 3242512433

Define mobility of electron in a conductor.
How does electron mobility change when (i) temperature of conductor is decreased and (ii) applied potential difference is doubled at constant temperature ?

Solution:

Mobility is defined as the positive value of drift velocity per unit electric field applied.

:. mu = u_d/E = e/m tau

Therefore, the mobility (i) increases with the decrease in temperature and (ii) remains same.
Q 3282612537

Define relaxation time of the free electrons drifting in a conductor. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.

Solution:

Relaxation time is the duration for which an electron drifting through a conductor does not suffer any collision.

Relation between drift velocity of free electrons and relaxation time

 vec v_d = (- e vec E)/m tau

where E is the electric field across the conductor drift.ing the electrons.

We have the relation  v_d = (eE)/m tau

:. v_d = (eV)/(ml) tau ..........(i)

where V is the potential difference and l is the length of the conductor.

∵ I = v_d enA => v_d = I/(enA) .............(ii)

Substituting the value of v_d from equation (ii) in equation (i), we get

 I/(enA) = (eV)/(ml)tau => V = ((ml)/( n e^2 tau A) ) l

:. R = (ml)/( n e^2 tau A)

:. rho = m/(n e^2 tau)

rho is called the resistivity of the material of the conductor.
Q 3252812734

(a) Define the term 'drift velocity' of charge carriers in a conductor. Obtain the expression for the current density in terms of relaxation time.
(b) A 100 V battery is connected to the electric network as shown. If the power consumed in the 2 Omega resistor is 200 W, determine the power dissipated in the 5 Omega resistor.

Solution:

(a) Let A be the area of cross-section of a wire and n be the number of free electrons per unit volume. If v_d be the average drift speed of the electrons along the wire, then the number of electrons passing in time t through the cross-section of the wire will be given by nAv_d. If e is the charge on each electron, the charge q passing in time t through the cross-section of the wire will be given as

q = (nAv_d t) e

 ∵ I = n eAv_d

∵ j = I/A => j = n ev_d

where j is the current density

:. j = (n e^2 E tau)/m

(b) Power disspated in 2 Omega resistance = 200 W (given)

:. 200 = I^2 xx 2

 :. I = 10A (Total current in the circuit)

Voltage drop across 2 Omega, i.e. V = 2 xx 10 = 20 V

:. Remaining p.d. across the rest of the circuit = 80 V

Effective resistance of upper area  = ( 30 xx 6)/(30 + 6) + 5 = 5 + 5 = 10 Omega

:. R = (10 xx 40 )/(10 + 40) = 8 Omega

:. Current in 5 Omega resistance  = (80V)/(10 Omega) = 8A

:. Power across 5 Omega resistance  = 8^2 xx 5 = 64 xx 5 = 320 W
Q 3212512430

Explain the term 'drift velocity electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of drift velocity'.

Solution:

Consider a conductor of length l, area

of cross section A and having number

density of free electrons n. On establishing the potential difference across the conductor, suppose the · electrons drift from lower potential to higher potential side with velocity vec v_d . The volume of the conductor covered by an electron in unit time is

 V = v_d A ..............(i)

Electrons occupying the volume in unit time is

 N = nv_d A ............(ii)

Thus, the charge flow through any cross~section of the conductor in unit time is

q = env_d A ... (iii)

According to the definition, the electric current is the rate of flow of charge through any cross-section of the conductor. Hence,

 I = q/l

 I = env_d A ..............(iv)
Q 3282712637

State the underlying principle of a potentiometer. Describe briefly, giving the necessary circuit diagram, how a potentiometer is used to measure the internal resistance of a given cell.

Solution:

The principle of working of a potentiometer: The potential drop along a wire is directly proportional to its corresponding length provided the current is constant and the wire is of uniform area of cross-section and physical conditions remain the same.

i.e. V alpha l

To measure the internal resistance of a cell: With an open key K_2, the balancing length is obtained. Let it be length l_1 We have relation

epsilon = phi l_1 ... (i)

where phi = potential gradient When key K_2 is closed, the cell passes a current I through the resistance box having resistance R. If V is the terminal potential difference of the cell and l_2 is the balancing length. We have relation

V = phi l_2 ........(ii)

From equations (i) and (ii), we have epsilon/V = l_1/l_2

We know that  r = ( epsilon/V - 1) R => r = (l_1/l_2 - 1) R
Q 3222612531

State the principle on which. the working of a meter bridge is based. Under what condition is the error in determining the unknown resistance minimised ?

Solution:

Meter bridge is based on the principle of Wheatstone bridge. An error, in determination of resistance, can be minimised by adjusting the balance point near the middle of the meter bridge.
Q 3252712634

(i) State the principle of working of a meter bridge.
(ii) In a meter bridge balance point is found at a distance l_1 with resistances R and S as shown in the figure. When an unknown resistance X is connected in parallel with the resistance S, the balance point shifts to a distance l_2 . Find the expression for X in terms of l_1 , l_2 and S.

Solution:

With R and S alone, we have

R/S = l_1/(100 - l_1) => R ( 100 - l_1) = Sl_1 ..............(i)

With S and X in parallel with R on the left gap,

 R/((SX)/(S + X)) = l_2/(100 - l_2) => R ( 100 - l_2) = (SXl_2)/(S + X) ...........(iii)

Dividing equations (i) and (ii), we get

(100 - l_1)/(100 - l_2) = (l_1 (S + X))/(Xl_2)

=> 100 X l_2 - l_1 l_2 X = 100 l_1 S + 100 l_1 X - l_1 l_2 S - l_2 l_1 X

:. X = (100 l_1 S - l_1 l_2 X)/( 100 (l_2 - l_1))
Q 3282712637

State the underlying principle of a potentiometer. Describe briefly, giving the necessary circuit diagram, how a potentiometer is used to measure the internal resistance of a given cell.

Solution:

The principle of working of a potentiometer: The potential drop along a wire is directly proportional to its corresponding length provided the current is constant and the wire is of uniform area of cross-section and physical conditions remain the same.

i.e. V alpha l

To measure the internal resistance of a cell: With an open key K_2, the balancing length is obtained. Let it be length l_1 We have relation

epsilon = phi l_1 ... (i)

where phi = potential gradient When key K_2 is closed, the cell passes a current I through the resistance box having resistance R. If V is the terminal potential difference of the cell and l_2 is the balancing length. We have relation

V = phi l_2 ........(ii)

From equations (i) and (ii), we have epsilon/V = l_1/l_2

We know that  r = ( epsilon/V - 1) R => r = (l_1/l_2 - 1) R
Q 3262812735

(a) State, with the help of circuit diagram, the working principle of a meter bridge. Obtain the expression used for determining the unknown resistance.
(b) What happens if the galvanometer and cell are interchanged at the balance point of the bridge?
(c) Why is it considered important to obtain the balance point near the midpoint of the wire?

Solution:

(a) The metre bridge is based on the Wheatstone bridge principle. The connections are made as shown in the figure. A resistance R is introduced from the resistance box and the key K is closed. The jockey is moved on the wire to the point where there is no deflection in the galvanometer. In such as case points B and D are at the same potential. The point D is called the null point.

Let, in this position, AB = L cm and BC = (100 - L)  cm. Therefore, the

resistance of AB, i.e.  P alpha L and resistance of BC, i.e. Q alpha (100 - L) hence

 P/Q = L/(100 - L) ...........(i)

In a balanced state, by the Wheatstone bridge principle, we have

 P/Q = R/X ..........(ii)

Substituting equation (i) in equation (ii), we have

 R/X = L/(100 - L) ...............(iii)

Rewriting equation (iii), we have

 X = ((100 - L)/L ) R

(b) There is no change in the position of the balance point if the galvanometer and the cell are interchanged.

(c) It is important to get the balance point near the mid-point of the wire because then the resistances in the four arms of the bridge are of the same order. The sensitivity of the bridge is maximum and the resistance is determined move accurately.
Q 3282812737

(a) State the principle of potentiometer. Define potential gradient. Obtain an expression for potential gradient in terms of resistivity of the potentiometer wire.
(b) Figure shows a long potentiometer wire AB having a constant potential gradient. The null points for the two primary cells of emfs epsilon_1 and epsilon_2 connected in the manner shown a:re obtained at A a distance of l_1 = 120 cm and l_2 = 300 cm from the end A. Determine (i) epsilon_1// epsilon_2 and (ii) position of null point for the cell epsilon_1 only.

Solution:

(a) Principle of potentiometer: The principle of working of a potentiometer: The potential drop along a wire is directly proportional to its corresponding length provided the current is constant and the wire is of uniform area of cross-section and physical conditions remain the same.

i.e. V alpha l

To measure the internal resistance of a cell: With an open key K_2, the balancing length is obtained. Let it be length l_1 We have relation

epsilon = phi l_1 ... (i)

where phi = potential gradient When key K_2 is closed, the cell passes a current I through the resistance box having resistance R. If V is the terminal potential difference of the cell and l_2 is the balancing length. We have relation

V = phi l_2 ........(ii)

From equations (i) and (ii), we have epsilon/V = l_1/l_2

We know that  r = ( epsilon/V - 1) R => r = (l_1/l_2 - 1) R

Potential gradient: The fall of potential per unit length of the potentiometer wire is called potential gradient.

 k = V/L

where L is the length of potentiometer wire and V is the potential difference across it.

 k = (IR)/L = (I rho l)/(LA)

(b) (i) Given: l_1 = 120 cm, l_2 = 300 cm

epsilon_1 + epsilon_2 = k 300 ; epsilon_1 - epsilon_2 = k 120

Dividing both the equations, we get

 :. ( epsilon_1 + epsilon_2)/(epsilon_1 - epsilon_2) = (300)/(120) = (10)/4 => 4 epsilon_1 + 4 epsilon_2 = 10 epsilon_1 - 10 epsilon_2 ......(i)

:. 6 epsilon_1 = 14 epsilon_2 => epsilon_1/epsilon_2 = (14)/6 = 7/3 ........(ii)

(ii) epsilon_1 + epsilon_2 = k 300

=> epsilon_1 - epsilon_2 = k 120

Adding both the equations, we get

:. 2 epsilon_ 1 = k 420

=> epsilon_1 = k 210

i.e. balancing length is 210 cm.
Q 3202812738

(a) State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emfs of two primary cells. Obtain the :required expression used for comparing the emfs.
(b) Write two possible causes for one sided deflection in a potentiometer experiment.

Solution:

(a) Principle of potentiometer : The principle of working of a potentiometer: The potential drop along a wire is directly proportional to its corresponding length provided the current is constant and the wire is of uniform area of cross-section and physical conditions remain the same.

i.e. V alpha l

To measure the internal resistance of a cell: With an open key K_2, the balancing length is obtained. Let it be length l_1 We have relation

epsilon = phi l_1 ... (i)

where phi = potential gradient When key K_2 is closed, the cell passes a current I through the resistance box having resistance R. If V is the terminal potential difference of the cell and l_2 is the balancing length. We have relation

V = phi l_2 ........(ii)

From equations (i) and (ii), we have epsilon/V = l_1/l_2

We know that  r = ( epsilon/V - 1) R => r = (l_1/l_2 - 1) R

Comparison of emf's of two primary cells: The circuit diagram for comparing the emfs of two cells is given below. B First, the key point 1 of two way key joined with point 2 is inserted. This brings the cell of emf E_1 in the circuit. The jockey is moved on the wire to obtain a balance point, i.e. a point on the wire where the galvanometer gives zero deflection. Let the balancing length be l_1. Then, by the potentiometer principle, we have

E_1 alpha l_1 ... (i)

Now, the key point 2 of two way key joined with point 1 is inserted. This brings the cell of emf E_ 2 in the circuit. The jockey is again moved on the wire to obtain the balance point. Let the balancing length be l_2. Then, by potentiometer principle, we have

 E_2 alpha l_2 .............(iii)

Dividing equation (i) by (ii), we have

 E_1/E_2 = l_1/l_2 ...............(iii)

Knowing the values of l_1 and l_2, the emfs can be compared.

(b) (i) The connections are not correct, i.e. at point A, the positive terminals of driver cell and battery are not connected.

(ii) The emf of a driver cell is lesser than the emfs of auxiliary cells.