50 Numerical Questions On Electricity Class 10

50 Numerical Questions

On Electricity Class 10 with Solutions

Source: omtexclasses.com

Numerical Problems based on Electric Current

This provides a comprehensive guide on solving numerical problems related to electricity, including electric current, potential difference, Ohm's law, resistance, and resistivity. It presents various formulas and examples to calculate charge, current, voltage, and resistance in different scenarios. Additionally, it covers the combination of resistances in series and parallel configurations.

Formulas to Remember

To solve the numericals on electricity, we will use the following formulas:

$$ I = \frac{Q}{t} $$

$$ I = \frac{ne}{t} $$

Where:
I = Electric current – Ampere
Q = charge – Coulomb
T = time - second
N = number of electrons
E = charge on an electron = \( 1.6 \times 10^{-19} \)

1. The filament of bulb draws a current of 0.5 ampere. Calculate the amount of charge if bulb glows for 2.5 hrs.

Solution

Given values Electric current (I) = 0.5 A
Time (t) = 2.5 Hrs = \( 2.5 \times 60 \times 60 = 9000 \) sec
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = It $$

$$ Q = 0.5 \times 9000 $$

Q = 4500 Coulomb

2. 10 Coulombs charge passing through a point in a circuit in 5 seconds. Find the electric current flowing in the circuit.

Solution

Given values Time (t) = 5 sec
Charge (Q) = 10 C
Electric current (I) = ?

$$ I = \frac{Q}{t} $$

$$ I = \frac{10}{5} $$

I = 2 Amp.

3. A current of 0.5 Ampere is drawn by a filament of an electric bulb for 15 minutes. Find the amount of electric charge.

Solution

Given values Electric current (I) = 0.5 A
Time (t) = 15 min = \( 15 \times 60 = 900 \) sec
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = I \times t $$

$$ Q = 0.5 \times 900 $$

Q = 450 Coulomb

4. A current of 1 Ampere is drawn by a filament of an electric bulb. Find the Number of electrons passing through a cross section of the filament in 10 seconds.

Solution

Given values Electric current (I) = 1 A
Time (t) = 15 s
Number of electrons (n) = ?
Charge on electrons (e) = \( 1.6 \times 10^{-19} \) C

$$ I = \frac{ne}{t} $$

After cross multiplication, we find:

$$ n = \frac{I \times t}{e} $$

Now put the given values in the formula:

$$ n = \frac{1 \times 10}{1.6 \times 10^{-19}} $$

$$ n = \frac{100}{16 \times 10^{-19}} $$

$$ n = \frac{6.25}{10^{-19}} $$

\( n = 6.25 \times 10^{19} \)

Number of electrons (n) has no unit required.

5. 150000 coulomb charge is required to deposit one mole of copper from the CuSO4 solution. Find the time to deposit 0.2 mole copper from the copper solution if a current of 5 Ampere is flowing the solution.

Solution

Given values Electric current (I) = 5 Amp
Charge (Q) = 150000 C

Charge required for 1 mole copper = 150000 C

Charge required for 0.2 mole copper = \( 150000 \times 0.2 = 30000 \) C

$$ I = \frac{Q}{t} $$

$$ t = \frac{Q}{I} = \frac{30000}{5} $$

t = 6000 S

6. An heater draws a current 10 A for 5 minutes. Calculate the charge flowing in heater.

Solutions

Given values Electric current (I) = 10 A
Time (t) = 5 min = \( 5 \times 60 = 300 \) s
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = I \times t $$

$$ Q = 10 \times 300 $$

Q = 3000 C

7. \( 1.25 \times 10^{18} \) electrons are passed from one end to another end of a conductor in 10 seconds. Find the current flowing through the conductor.

Solution

Given values Number of electrons (n) = \( 1.25 \times 10^{18} \)
Time (t) = 10 s
Electric current (I) = ?

$$ I = \frac{ne}{t} $$

[e = \( 1.6 \times 10^{-19} \)]

$$ I = \frac{1.25 \times 10^{18} \times 1.6 \times 10^{-19}}{10} $$

I = 0.2 A

8. Calculate the number of electrons present in 0.1 coulomb of charge.

Solution

Given values Charge (Q) = 0.1
Number of electrons (n) = ?

Number of elections in 1 Coulombs is = \( 6.25 \times 10^{18} \)

Number of electrons in 0.1 Coulomb charge is = \( 6.25 \times 10^{18} \times 0.1 \)

= \( 6.25 \times 10^{17} \)

Numerical Problems based on Potential difference and work

Formulas to Remember

$$ V = \frac{W}{Q} $$

Where:
V = Potential difference – volt
W = work done - Joule
Q = charge – Coulomb

9. Calculate the amount of work done in carrying 8C charge from a terminal of 150 volt to 200 volt.

Solution

Given values Charge (Q) = 8 C
Potential (V1) = 150 volt
Potential (V2) = 200 volt

Potential difference (V) = V2 - V1

200 – 150 = 50 volt

Work (W) = ?

$$ V = \frac{W}{Q} $$

$$ W = VQ $$

$$ W = 50 \times 8 = 400 J $$

10. How much energy is given to each coulomb of charge passing through a 10 Volt battery?

Solution

Given values Potential difference/voltage (V) = 10 volt
Charge (Q) = 1 C
Energy (E) = ?

$$ V = \frac{W}{Q} $$

Work is defined as energy transferred by force, so We can put E in the place of W

$$ E = VQ $$

= 10 × 1 = 10 J

11. Work done in moving 5 coulomb charge from point A to B is 80 Joule, if potential at point A is 25 volt then find the potential at B.

Solution

Given values Work (W) = 80 J
Charge (Q) = 5 C
Potential (VA) = 25 volt
Potential (VB) = ?

$$ V = V_B - V_A = \frac{W}{Q} $$

$$ V_B - 25 = \frac{80}{5} $$

$$ V_B - 25 = 16 $$

$$ V_B = 16 + 25 = 41 \text{ volt} $$

Numerical Problems based on Ohm’s law

Formulas to Remember

$$ V = IR $$

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

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

Where:
V = Voltage (Potential difference) = volt
R = Resistance – Ohm (Ω)
I = Electric current – Ampere

12. When a cell of 1.5 volt is applied in a circuit, a current of 0.5 ampere flows through it. Calculate the resistance of the circuit.

Solution

Given values Voltage (V) = 1.5 volt
Current (I) = 0.5 A
Resistance (R) = ?

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

Put the values in the formula of resistance:

$$ R = \frac{1.5}{0.5} = 3 \text{ Ohm} $$

13. How much current will an electric bulb draw from a 220-volt source. If the resistance of the filament is 1200 ohm?

Solution

Given values Voltage (V) = 220 volt
Resistance (R) = 1200 Ohm
Current (I) = ?

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

$$ I = \frac{220}{1200} = 0.183 A $$

14. What is the potential difference between the ends of a conductor of 15 Ohm resistance when a current of 2 ampere flows through it.

Solution

Given values Resistance (R) = 15 Ohm
Electric current (I) = 2 A
Potential difference (V) = ?

$$ V = IR $$

$$ V = 2 \times 15 = 30 \text{ volt} $$

15. A heater draws a current of 5 A when it is connected to 110 volts. What current will the heater draw when it is connected to 220 volts.

Solution

Given values Current (I1) = 5 A
Voltage (V1) = 110 volts
Voltage (V2) = 220 volts
Current (I2) = ?

According to Ohm’s law, V=IR

$$ R = \frac{110}{5} = 22 \text{ Ohm} $$

When the voltage is 220 volt, The current will be:

$$ I_2 = \frac{220}{22} = 10 A $$

16. Calculate the potential difference required across a conductor of resistance 6 ohm to make a current of 2.5 A flow through it.

Solution

Given values Resistance (R) = 6 ohm
Current (I) = 2.5 A
Potential difference (V) = ?

$$ V = IR $$

$$ V = 2.5 \times 6 = 15 \text{ volt} $$

Numerical Problems based on Resistance and Resistivity

Formulas to Remember

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

Where:
R = Resistance - Ohm
\( \rho \) = Resisitivity - Ohm m
l = length – m
A = area of cross section - m²

17. Calculate the resistance of a copper wire of length 30 cm and area of cross section \( 3 \times 10^{-4} \text{ m}^2 \). The resistivity of copper is \( 1.7 \times 10^{-8} \text{ ohm m} \).

Solution

Given values Length of wire (l) = 30 cm = 0.3m
Area of cross section (A) = \( 3 \times 10^{-4} \text{ m}^2 \)
Resistivity of copper (\( \rho \)) = \( 1.7 \times 10^{-8} \text{ ohm m} \)
Resistance (R) = ?

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

$$ R = 1.7 \times 10^{-8} \times \frac{0.3}{3 \times 10^{-4}} $$

$$ R = \frac{0.51 \times 10^{-8}}{3 \times 10^{-4}} $$

$$ R = \frac{0.17 \times 10^{-8}}{10^{-4}} $$

$$ R = 0.17 \times 10^{-8} \times 10^4 $$

R = \( 0.17 \times 10^{-4} \text{ Ohm} \)

18. Calculate the resistivity of wire having length 1 m and area of cross section \( 1.20 \times 10^{-6} \text{ m}^2 \), if its resistance is 0.013 ohm.

Solution

Given values Area of cross section (A) = \( 1.20 \times 10^{-6} \text{ m}^2 \)
Length (l) = 1m
Resistance (R) = 0.013 ohm
Resistivity (\( \rho \)) = ?

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

Now put the values in the formula:

$$ 0.013 = \rho \frac{1}{1.20 \times 10^{-6}} $$

$$ \rho = 0.013 \times \frac{1.20 \times 10^{-6}}{1} $$

\( \rho = 1.56 \times 10^{-8} \text{ ohm m} \)

19. Calculate the area of cross section of a wire of 1 m and resistance 25 ohm, if the resistivity of material of the wire is \( 1.84 \times 10^{-6} \text{ ohm m} \).

Solution

Given values Length (l) = 1m
Resistance (R) = 25 ohm
Resistivity (\( \rho \)) = \( 1.84 \times 10^{-6} \text{ ohm m} \)
Area of cross section (A) = ?

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

$$ 25 = 1.84 \times 10^{-6} \times \frac{1}{A} $$

$$ A = 1.84 \times 10^{-6} \times \frac{1}{25} $$

A = \( 7.36 \times 10^{-8} \text{ m}^2 \)

20. 1 m long wire with resistance 0.85 ohm and diameter 0.2 mm, what will be the resistivity of the metal at 20°C.

Solution

Given values Length (l) = 1m
Resistance (R) = 0.85 ohm
Diameter (d) = 0.2 mm = \( 2 \times 10^{-4} \text{m} \)
Radius (r) = d/2 = \( 1 \times 10^{-4} \text{m} \)
Area of cross section (A) = ? (\( \pi r^2 \))

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

$$ R = \rho \frac{l}{\pi r^2} $$

$$ 0.85 = \rho \frac{7}{22 \times (1 \times 10^{-4})^2} \times 1 $$

$$ \rho = \frac{0.85 \times 22}{7 \times 10^{-8}} $$

\( \rho = 2.67 \times 10^{-8} \text{ ohm m} \)

Numerical Problems based on the Combination of Resistances

Formulas to Remember

$$ R = R_1 + R_2 + R_3 + \dots R_n $$

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \frac{1}{R_n} $$

Where:
\( R_1, R_2, R_3 \dots R_n \) are different resistances
R = total resistance of the combination

21. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in series, find the total resistance.

Solution

Given value R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
R = ?

$$ R = R_1 + R_2 + R_3 $$

$$ R = 5 + 10 + 15 = 30 \text{ ohm} $$

22. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in parallel, find the total resistance.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
R = ?

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$

$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{15} $$

$$ \frac{1}{R} = \frac{6+3+2}{30} = \frac{11}{30} $$

$$ \frac{1}{R} = \frac{11}{30} $$

$$ R = \frac{30}{11} = 2.72 \text{ ohm} $$

23. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in series, and system is connected with 90 volt battery. What will the current flowing in the circuit.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 90 volt
I = ?

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

We don’t have the value of ‘R’ so first we will find the value:

$$ R = 5 + 10 + 15 = 30 \text{ ohm} $$

$$ I = \frac{V}{R} = \frac{90}{30} = 3 A $$

24. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in parallel combination and the circuit is connected with 100 volt battery. Find electric current flowing in each of the resistance.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 100 volt
I1 = ?
I2 = ?
I3 = ?

$$ I_1 = \frac{V}{R_1} $$

$$ I_1 = \frac{100}{5} = 20 A $$

$$ I_2 = \frac{V}{R_2} $$

$$ I_2 = \frac{100}{10} = 10 A $$

$$ I_3 = \frac{V}{R_3} $$

$$ I_3 = \frac{100}{15} = 6.66 A $$

(Note: Current in all resistors is different in the parallel combination.)

25. Three resistors of 5ohm, 10 ohm and 15 ohm are connected in series combination and 10 ampere current is flowing in the system when connected with 110 Volt. Find the potential difference between two ends of each resistor.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 110 volt
I = 10 A
V1 = ?
V2 = ?
V3 = ?

$$ V_1 = I R_1 $$

$$ V_1 = 10 \times 5 = 50 \text{ volt} $$

$$ V_2 = I R_2 $$

$$ V_2 = 10 \times 10 = 100 \text{ volt} $$

$$ V_3 = I R_3 $$

$$ V_3 = 10 \times 15 = 150 \text{ volt} $$

26. Three resistances of 2 ohm, 3 ohm and 6 ohm are connected in series and then in parallel. Find the total resistance in both the arrangements.

Solution

Given values R1 = 2 ohm
R2 = 3 ohm
R3 = 6 ohm
R = Total resistance = ?

For series combination:

$$ R = R_1 + R_2 + R_3 $$

$$ R = 2 + 3 + 6 = 11 \text{ ohm} $$

For parallel combination:

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$

$$ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} $$

$$ \frac{1}{R} = \frac{6+4+2}{12} = \frac{12}{12} $$

R = 1 ohm

27. A 18 Volt battery is connected across a lamp whose resistance is 50 ohm, through a variable resistor. If the current flowing through the circuit is 0.3 A. Calculate the value of resistance used from the variable resistor.

Solution

Given values Voltage (V) = 18 volt
Current (I) = 0.3 A
Resistance of bulb (r) = 50 ohm
R1 = resistance form variable resistor = ?

Let R1 is the resistance from variable resistor that is used.

Both R1 and r are is series combination so the resultant resistance is:

$$ R = R_1 + r $$

According to Ohm’s law V=IR

$$ 15 = 0.3 \times (R_1 + r) $$

$$ R_1 + 50 = \frac{18}{0.3} = 60 $$

$$ R_1 = 60 - 50 = 10 \text{ ohm} $$

Numerical Problems based on the Circuit diagrams

28. Calculate the total resistance of the following circuit.

[Insert Image from Page 25 here: Series circuit with 5Ω, 10Ω, 20Ω]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 20 ohm
R = ?

All resistances in the given circuit are connected in the series so total resistance:

$$ R = R_1 + R_2 + R_3 $$

$$ R = 5 + 10 + 20 = 35 \text{ ohm} $$

29. Calculate the total resistance in the given circuit

[Insert Image from Page 26 here: Parallel circuit with 5Ω, 10Ω, 20Ω]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 20 ohm
R = ?

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$

$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{20} $$

$$ \frac{1}{R} = \frac{4+2+1}{20} = \frac{7}{20} $$

$$ R = \frac{20}{7} = 2.85 \text{ ohm} $$

30. You have been given the following circuit diagram. Find the following:
(i) Electric current through each resistor
(ii) Total resistance
(iii) Total current

[Insert Image from Page 27 here: Parallel circuit with 5Ω, 10Ω, 30Ω connected to 6V]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 30 ohm
V = 6 volt
R = ?
I1 = ?, I2 = ? and I3 = ?
I = ?

All the resistors are connected in parallel combination.

(i) Current through each resistor

$$ V = I_1 R_1 $$

$$ 6 = I_1 \times 5 $$

$$ I_1 = \frac{6}{5} = 1.2 A $$

$$ V = I_2 R_2 $$

$$ 6 = I_2 \times 10 $$

$$ I_2 = \frac{6}{10} = 0.6 A $$

$$ V = I_3 R_3 $$

$$ 6 = I_3 \times 30 $$

$$ I_3 = \frac{6}{30} = 0.2 A $$

(ii) Total resistance (R)

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$

$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} $$

$$ \frac{1}{R} = \frac{6 + 3 + 1}{30} = \frac{10}{30} $$

$$ R = \frac{30}{10} = 3 \text{ Ohm} $$

(iii) Total current (I)

$$ V = IR $$

$$ 6 = I \times 3 $$

$$ I = \frac{6}{3} = 2 A $$

31. Find the equivalent resistance in the given circuit and current flowing through the circuit.

[Insert Image from Page 29 here: 3Ω and 6Ω in parallel, connected to 4.5V]

Solution

Given values R1 = 3 ohm
R2 = 6 ohm
V = 4.5 volt
R = ?

We can see that the resistors of 3 ohm and 6 ohm are connected in parallel combination so the equivalent resistance R

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} $$

$$ \frac{1}{R} = \frac{1}{3} + \frac{1}{6} $$

$$ \frac{1}{R} = \frac{2 + 1}{6} = \frac{3}{6} $$

$$ R = \frac{6}{3} = 2 \text{ ohm} $$

According to Ohm’s law, V=IR

$$ 4.5 = I \times 2 $$

$$ I = \frac{4.5}{2} = 2.25 A $$

32. You have been given the following circuit of resistors connected with a battery.
Calculate:
(i) total resistance of the circuit
(ii) total current
(iii) voltage across 5 ohm resistor

[Insert Image from Page 31 here: 10Ω and 10Ω in parallel, connected in series with 5Ω, V=6V]

Solution

Given values R1 = 10 ohm
R2 = 10 ohm
R3 = 5 ohm
V = 6 volt

(i) total resistance

We can see R1 and R2 are in parallel so the total resistance RA

$$ \frac{1}{R_A} = \frac{1}{R_1} + \frac{1}{R_2} $$

$$ \frac{1}{R_A} = \frac{1}{10} + \frac{1}{10} $$

$$ \frac{1}{R_A} = \frac{1+1}{10} = \frac{2}{10} $$

$$ R_A = \frac{10}{2} = 5 \text{ Ohm} $$

Now RA and R3 in series so the total resistance R

R = RA + R3

R = 5 + 5 = 10 ohm

(ii) total current

$$ I = \frac{V}{R} = \frac{6}{10} = 0.6 A $$

(iii) Voltage across 5 ohm resistor

V1 = IR1

V1 = 0.6 × 5 = 3 volt

33. Find the equivalent resistance of the following circuit of resistors.

[Insert Image from Page 33 here: Triangle configuration. 4Ω on one side, 8Ω and 8Ω on others]

Solution

Given values R1 = 4 Ohm
R2 = 8 Ohm
R3 = 8 Ohm
R = ?

In the given circuit diagram we can see R2 and R3 are series so:

RA = R2 + R3

RA = 8 + 8 = 16 Ohm

Now the RA is parallel to R1 so the equivalent resistance R

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_A} $$

$$ \frac{1}{R} = \frac{1}{4} + \frac{1}{16} $$

$$ \frac{1}{R} = \frac{4+1}{16} = \frac{5}{16} $$

$$ R = \frac{16}{5} = 3.2 \text{ Ohm} $$

Numerical Problems based on Work, Power and Energy

Formulas to Remember

$$ V = \frac{W}{Q} $$

$$ W = V \times Q $$

$$ W = I^2Rt $$

$$ W = \frac{V^2t}{R} $$

Note = We can put E in place of W

$$ P = \frac{W}{t} $$

$$ P = \frac{V^2}{R} $$

$$ P = I^2R $$

Where:
P = power – Watt
W = work – Joule
V = voltage – Volt
I = current – Ampere
Q = charge = Coulomb

34. An electric bulb is connected to 110 volt electric source. The current is 0.5 A, then what is the power of the bulb?

Solution

Given values Voltage (V) = 110 volt
Current (I) = 0.5 A
Power (P) = ?

P = VI

P = 110 × 0.5

P = 55 Watt

35. A bulb rated 5 volt – 100mA, calculate its (i) power (ii) resistance

Solution

Given values Voltage (V) = 5 volt
Current (I) = 100mA = \( 100 \times 10^{-3} \) A

P = VI

P = \( 5 \times 100 \times 10^{-3} \)

P = 0.5 Watt

36. Two electric bulbs rated 60W, 220V and 100W, 220 V. which one of them has higher resistance?

Solution

Given values (i) P1 = 60 W, V = 220 volt
(ii) P2 = 110 W , 220 volt
R1 = ?
R2 = ?

$$ R_1 = \frac{V^2}{P_1} $$

$$ R_1 = \frac{(220)^2}{60} $$

$$ R_1 = 806.67 \text{ Ohm} $$

$$ R_2 = \frac{V^2}{P_2} $$

$$ R_2 = \frac{(220)^2}{100} $$

$$ R_2 = 484 \text{ Ohm} $$

60 watt bulb has higher resistance than 100 watt bulb.

37. Two lamps rated 100 W, 220 Volt and 60W, 220 V connected in parallel. Calculate the current drawn by the circuit.

Solution

Given values P1 = 100W
P2 = 60 W
V = 220volt
I = ?

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

In the above formula we don’t have value of R so we will have to find the value of R

$$ R_1 = \frac{V^2}{P_1} = \frac{(220)^2}{100} \text{ Ohm} $$

$$ R_2 = \frac{V^2}{P_2} = \frac{(220)^2}{60} \text{ Ohm} $$

Both the lamps are in parallel, so the resultant resistance is

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} $$

$$ \frac{1}{R} = \frac{100}{220 \times 220} + \frac{60}{220 \times 220} $$

$$ \frac{1}{R} = \frac{100+60}{220 \times 220} $$

$$ \frac{1}{R} = \frac{160}{220 \times 220} \text{ Ohm} $$

Now we can find the value of current (I)

$$ I = \frac{220}{\frac{220 \times 220}{160}} $$

$$ I = \frac{220 \times 160}{220 \times 220} $$

$$ I = \frac{160}{220} = 0.72 A $$

38. An electric lamp is rated 100W, 220V. it is used for 10 hours daily then calculate energy consumed in KWh per day.

Solution

Given values P = 100 W
V = 220 Volt

$$ \text{Energy consumed per day} = \frac{P \times t}{1000} $$

$$ = \frac{100 \times 10}{1000} $$

Energy consumed per day = 1KWh or 1 Unit

39. A bulb rated 2.5 V and 650mA. Calculate (i) power (ii) resistance (iii) energy consumed for 5 hours.

Solution

Given values V = 2.5 Volt
I = 650 mA = 0.65A
T = 5 hours

(i) P=VI

P = 2.5 × 0.65 = 1.625 W

(ii) Resistance

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

$$ R = \frac{2.5}{0.65} = 3.84 \text{ Ohm} $$

(iii) Energy consumed in 5 hours

$$ = \frac{P \times t}{1000} $$

$$ = \frac{1.625 \times 5}{1000} = 0.008125 \text{ KWh} $$

40. An immersion rod of 750W is used for one hour. Find the energy consumed in (i) KWh (ii) Joule

Solution

Given values P = 750 W
T = 1 hour

(i) Energy consumed in KWh

$$ \text{Energy consumed for 1 hour} = \frac{P \times t}{1000} $$

$$ = \frac{750 \times 1}{1000} = 0.75 \text{ KWh} $$

(iii) Energy consumed in Joule

1 KWh = \( 3.6 \times 10^6 \) Joule

So \( 0.75 \times 3.6 \times 10^6 \) Joule

= \( 2.7 \times 10^6 \) Joule

41. One heater is rated 220 V- 5 A. calculate the energy consumed if it is used for 5 hours.

Solution

Given values V = 220 volt
I = 5 A

$$ \text{Energy consumed for 5 hour} = \frac{P \times t}{1000} $$

But in this formula we need value of Power (P)

P = VI

P = 220 × 5

P = 1100W

Now we can put the value of P in the formula

$$ \text{Energy consumed for 5 hour} = \frac{1100 \times 5}{1000} $$

= 5.5 KWh

42. Which uses more energy a 250W computer in 1 hour or a 1200W heater in 10 minutes.

Solution

Given values P1 = 250W
T1 = 1hour
P2 = 1200W
T2 = 10 min = 10/60 hour

(i) A 250 W computer is used for 1 hour

= P1 × t1 = 250 × 1

= 250Wh

(ii) A 1200W heater is used for 10 min

= P2 × t2 = 1200 × 10/60

= 200Wh

So, computer uses more energy.

43. An electric heater is rated 1kW, 220V. calculate its resistance.

Solution

Given values P = 1kW = 1000W
V = 220 volt
R = ?

$$ R = \frac{V^2}{P} $$

$$ R = \frac{220^2}{1000} $$

R = 48.4 Ohm

44. An electric bulb rated 220 V- 100 W. calculate its resistance.

Solution

Given values Power (P) = 100W
Voltage (V) = 220volt
Resistance (R) = ?

$$ P = \frac{V^2}{R} $$

$$ 100 = \frac{220^2}{R} $$

$$ R = \frac{220^2}{100} $$

R = 484 Ohm

Numerical Problems based on Heating effect of current

45. How much heat will a device of 15 W produces in one minute if it is connected to a battery of 15 Volt.

Solution

Given values P = 15W
V = 15 volt
T = 1 min = 60 s

H = P × t

H = 15 × 60 = 900 J

46. 1 kJ hat is produced each second in 50 Ohm resistor. Calculate the potential difference across the resistor.

Solution

Given values H = 1kJ = 1000J
Time (t) = 1 s
Resistance (R) = 50 Ohm
Voltage (V) = ?

$$ H = I^2Rt $$

$$ 1000 = I^2 \times 50 \times 1 $$

$$ I^2 = \frac{1000}{50} = 20 $$

$$ I = \sqrt{20} = 4.47A $$

So, potential difference across the resistor

V = IR

V = 4.47 × 50 = 223.5 volt

47. A heater of resistance 10 Ohm draws 10 A current from electric source for 2 hours. Calculate the heat produced in the heater.

Solution

Given values Resistance (R) = 10 Ohm
Current (I) = 10 A
Time (t) = 2 hours = 2 × 60 × 60 = 7200 s
Heat (H) = ?

$$ H = I^2Rt $$

$$ H = (10)^2 \times 10 \times 7200 $$

H = \( 72 \times 10^5 \) J

48. 100 Joule of heat is produced each second in 10 Ohm resistance. find the potential difference across the resistor.

Solution

Given values Heat (H) = 100J
Time (t) = 1s
Resistance (R) = 10 Ohm
Potential difference (V) = ?

V = IR

In the above formula we need the value of current (I) so at first we must find the value of I form the given values

$$ H = I^2Rt $$

$$ I = \sqrt{\frac{H}{Rt}} $$

$$ I = \sqrt{\frac{100}{10 \times 1}} $$

$$ I = \sqrt{10} $$

$$ I = 3.16 A $$

Now we can find the value of potential difference

V = IR

V = 3.16 × 10

V = 31.6 volt

49. Find the heat produced in the following combination of resistors if current drawn for 10 seconds.

[Insert Image from Page 47 here: Two resistors in series (2Ω and 2Ω) connected to 10V]

Solution

Given values R1 = 2 Ohm
R2 = 2 Ohm
Potential difference (V) = 10 volt
Time (t) = 10 s
Heat (H) = ?

H = V I t

We will have to find the value of current (I)

R = R1 + R2

R = 2 + 2 = 4 Ohm

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

$$ I = \frac{10}{4} = 2.5 A $$

H = V I t

H = 10 × 2.5 × 10

H = 250 J

50. How much energy is given to 10 coulomb charge passing through 15 volt battery?

Solution

Given values Charge (Q) = 10 Coulomb
Voltage (V) = 15 volt
Energy (E) = ?

E = VQ

E = 15 × 10 = 150 J

10th Science Acoustics: Comprehensive Book Back Questions and Answers

Book Back Questions with Answers - Acoustics | Science

I. Choose the correct answer

1. When a sound wave travels through air, the air particles

• a) vibrate along the direction of the wave motion
• b) vibrate but not in any fixed direction
• c) vibrate perpendicular to the direction of the wave motion
• d) do not vibrate

Answer: a) vibrate along the direction of the wave motion

2. Velocity of sound in a gaseous medium is 330 m s^–1. If the pressure is increased by 4 times without causing a change in the temperature, the velocity of sound in the gas is

• a) 330 m s^–1
• b) 660 m s¹
• c) 156 m s^–1
• d) 990 m s^–1

Answer: a) 330 m s^–1

3. The frequency, which is audible to the human ear is

• a) 50 kHz
• b) 20 kHz
• c) 15000 kHz
• d) 10000 kHz

Answer: b) 20 kHz

4. The velocity of sound in air at a particular temperature is 330 m s^–1. What will be its value when temperature is doubled and the pressure is halved?

• a) 330 m s^–1
• b) 165 m s^–1
• c) 330 × √2 m s^–1
• d) 320 / √ 2 m s^–1

Answer: c) 330 × √2 m s^–1

5. If a sound wave travels with a frequency of 1.25 × 10⁴ Hz at 344 m s^–1, the wavelength will be

• a) 27.52 m
• b) 275.2 m
• c) 0.02752 m
• d) 2.752 m

Answer: c) 0.02752 m

6. The sound waves are reflected from an obstacle into the same medium from which they were incident. Which of the following changes?

• a) speed
• b) frequency
• c) wavelength
• d) none of these

Answer: d) none of these

7. Velocity of sound in the atmosphere of a planet is 500 m s^–1. The minimum distance between the sources of sound and the obstacle to hear the echo, should be

• a) 17 m
• b) 20 m
• c) 25 m
• d) 50 m

Answer: c) 25 m

II. Fill up the blanks

1. Rapid back and forth motion of a particle about its mean position is called Vibration.
2. If the energy in a longitudinal wave travels from south to north, the particles of the medium would be vibrating in both north and south.
3. A whistle giving out a sound of frequency 450 Hz, approaches a stationary observer at a speed of 33 m s^–1. The frequency heard by the observer is (speed of sound = 330 m s^–1) 500 Hz.
4. A source of sound is travelling with a velocity 40 km/h towards an observer and emits a sound of frequency 2000 Hz. If the velocity of sound is 1220 km/h, then the apparent frequency heard by the observer is 2068 Hz.

III. True or false (If false give the reason)

1. Sound can travel through solids, gases, liquids and even vacuum. - False

   Reason: Sound waves cannot travel through vacuum.

2. Waves created by Earth Quake are Infrasonic. - True

3. The velocity of sound is independent of temperature. - False

   Reason: The velocity of sound is dependent on temperature.

4. The Velocity of sound is high in gases than liquids. - False

   Reason: The velocity of sound is high in liquids than gases.

IV. Match the following

Answers:

1. Infrasonic10 Hz

2. EchoUltrasonography

3. Ultrasonic22 kHz

4. High pressure regionCompressions

V. Assertion and Reason Questions

Mark the correct choice as
a. If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
b. If both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
c. Assertion is true, but the reason is false.
d. Assertion is false, but the reason is true.

1. Assertion: The change in air pressure affects the speed of sound.
Reason: The speed of sound in a gas is proportional to the square of the pressure.

Answer: c. Assertion is true, but the reason is false.

2. Assertion: Sound travels faster in solids than in gases.
Reason: Solid posses a greater density than that of gases.

Answer: b. If both the assertion and the reason are true but the reason is not the correct explanation of the assertion.

VI. Answer very briefly

1. What is a longitudinal wave?

It is a wave in which particles are vibrating along the direction of wave motion.

2. What is the audible range of frequency?

The audible range of frequency is 20 to 20,000 Hz.

3. What is the minimum distance needed for an echo?

The minimum distance needed for an echo 17.2 m.

4. What will be the frequency sound having 0.20 m as its wavelength, when it travels with a speed of 331 m s^–1?

Frequency Sound = n
Wavelength λ = 0.20 m
Frequency of sound, $$n = \frac{V}{\lambda}$$ $$n = \frac{331}{0.20}$$ $$n = 1655 \text{ Hz}$$

5. Name three animals, which can hear ultrasonic vibrations.

Dogs, Bats and Mosquitoes.

VII. Answer briefly

1. Why does sound travel faster on a rainy day than on a dry day?

During rainy days, the humidity is more in the atmosphere. The speed of the sound generally increases with humidity and speed of sound in water is more than 4X then speed in air. So sound travels faster on a rainy day.

2. Why does an empty vessel produce more sound than a filled one?

In an empty vessel, only air is present inside it. Whenever sound is produced in an empty vessel, the vibration of air molecule will be more due to multiple reflections. But a filled vessel has very less number of air molecules than the contents in it, so it does not produce more sound.

3. Air temperature in the Rajasthan desert can reach 46°C. What is the velocity of sound in air at that temperature? (V₀ = 331 m s^–1)

Velocity of sound in gas at 0° C
V₀ =331 ms^-1
Air temperature in Rajasthan, T = 46° C
The formula for velocity at temperature T is:
$$V_T = (V_o + 0.61T)$$ $$V_T = 331 + (0.61 \times 46)$$ $$V_T = 331 + 28.06$$ $$V_T = 359.06 \text{ m/s}$$

4. Explain why, the ceilings of concert halls are curved.

They are made curved so that the sound after reflecting from the ceiling reaches every corner of the concert hall and the audience listen the sound clearly.

5. Mention two cases in which there is no Doppler effect in sound?

(i) When source(S) and listener (L) both are at rest.
(ii) When S and L move in such a way that distance between them remains constant.
(iii) When source S and L are moving in mutually perpendicular directions.

VIII. Problem Corner

1. A sound wave has a frequency of 200 Hz and a speed of 400 m s^–1 in a medium. Find the wavelength of the sound wave.

Given

Frequency of a wave, n = 200 Hz
Speed of sound, V = 400 ms^-1

To find

Wavelength, λ = ?

Solution

Velocity of light, $$V = n\lambda$$ $$\lambda = \frac{V}{n} = \frac{400}{200}$$ $$\lambda = 2 \text{ m}$$

2. The thunder of cloud is heard 9.8 seconds later than the flash of lightning. If the speed of sound in air is 330 m s^–1, what will be the height of the cloud?

Given

Time, t = 9.8 s
Speed of sound, V = 330 ms^-1

To find

Height of the cloud, d = ?

Solution

We know, $$V = \frac{d}{t}$$ Height of the cloud, $$d = V \times t$$ $$d = 330 \times 9.8$$ $$d = 3234 \text{ m}$$

3. A person who is sitting at a distance of 400 m from a source of sound is listening to a sound of 600 Hz. Find the time period between successive compressions from the source?

Given

Frequency of sound, n = 600 Hz

To find

Time period between successive compressions, T = ?

Solution

$$T = \frac{1}{n} = \frac{1}{600} = 0.00166 \text{ s}$$ Time period, T ≈ 0.0017 s

4. An ultrasonic wave is sent from a ship towards the bottom of the sea. It is found that the time interval between the transmission and reception of the wave is 1.6 seconds. What is the depth of the sea, if the velocity of sound in the seawater is 1400 m s^–1?

Given

Time interval between sending and receiving of the wave, t = 1.6 s
Velocity of sound in sea wave, V = 1400 ms^-1

To find

Depth of the sea, d = ?

Solution

$$2d = V \times t$$ Depth of the sea, $$d = \frac{1400 \times 1.6}{2}$$ $$d = \frac{2240}{2}$$ $$d = 1120 \text{ m}$$

5. A man is standing between two vertical walls 680 m apart. He claps his hands and hears two distinct echoes after 0.9 seconds and 1.1 second respectively. What is the speed of sound in the air?

Given

Time of first echo, t₁ = 0.9 s
Time of second echo, t₂ = 1.1 s
Total distance between walls = 680 m. The total distance travelled by sound for both echoes is 2d, where d is the distance between the walls.

To find

Speed of sound in air, V = ?

Solution

The total time for sound to travel to both walls and back to the man is t = t₁ + t₂. The total distance covered is 2d.
Speed of sound in air, $$V = \frac{2d}{t_1+t_2}$$ $$V = \frac{2 \times 680}{0.9 + 1.1}$$ $$V = \frac{1360}{2}$$ Speed of sound in air, V = 680 m/s

6. Two observers are stationed in two boats 4.5 km apart. A sound signal sent by one, under water, reaches the other after 3 seconds. What is the speed of sound in the water?

Given

Distance between two observers, d = 4.5 km = 4500 m
Time taken to reach underwater, t = 3 s

To find

Speed of sound, V = ?

Solution

Speed of sound in water, $$V = \frac{d}{t}$$ $$V = \frac{4.5 \text{ km}}{3 \text{ s}} = 1.5 \text{ km/s}$$ or $$V = \frac{4500 \text{ m}}{3 \text{ s}} = 1500 \text{ m/s}$$

7. A strong sound signal is sent from a ship towards the bottom of the sea. It is received back after 1s.What is the depth of sea given that the speed of sound in water 1450 m s^–1?

Solution

Total time for signal to go and come back = 1 s.
Time taken by the signal to reach the bottom of the sea, t = 1/2 s = 0.5 s.
Speed of sound in water, V = 1450 m/s
Depth of the sea (distance travelled by signal), d = Speed × Time
$$d = V \times t$$ $$d = 1450 \times 0.5$$ Depth of the sea, d = 725 m

IX. Answer in Detail

1. What are the factors that affect the speed of sound in gases?

(i) Effect of density: Velocity of sound in gas is inversely proportional to the square root of density of the gas. Hence, the velocity decreases as the density of the gas increases.
$$v \propto \sqrt{\frac{1}{d}}$$

(ii) Effect of temperature: Velocity of sound in a gas is directly proportional to the square root of its temperature. Velocity of sound in a gas increases with increase in temperature, $$v \propto \sqrt{T}$$. Velocity at temperature T is given by the following equation.
$$v_T = (v_0 + 0.61 T) \text{ ms}^{-1}$$ Here, v₀ is the velocity of sound in the gas at 0° C. For air, v₀ = 331 ms^-1. Hence, the velocity of sound changes by 0.61 ms^-1 when temperature changes by each degree celcius.

(iii) Effect of relative humidity: When humidity increases, the speed of sound increases. That is why we can hear sound from long distances clearly during rainy seasons.

2. What is mean by reflection of sound? Explain:
a) reflection at the boundary of a rarer medium
b) reflection at the boundary of a denser medium
c) Reflection at curved surfaces

When sound waves travel in a given medium and strike the surface of another medium, it can be bounced back into the first medium is called as reflection.

a) Reflection in rarer medium:

(i) Consider a wave travelling in a solid medium striking on the interface between the solid and the air.
(ii) The compression exerts a force F on the surface of the rarer medium.
(iii) As a rarer medium has smaller resistance for any deformation, the surface of separation is pushed backwards.
(iv) As the particles of the rarer medium are free to move, a rarefaction is produced at the interface.
(v) Thus, a compression is reflected as a rarefaction and a rarefaction travels from right to left.

b) Reflection in denser medium:

(i) A longitudinal wave travels in a medium in the form of compressions and rarefactions.
(ii) Suppose a compression travelling in air from left to right reaches a rigid wall. The compression exerts a force F on the rigid wall.
(iii) In turn, the wall exerts an equal and opposite reaction R = - F on the air molecules. This results in a compression near the rigid wall.
(iv) Thus, a compression travelling towards the rigid wall is reflected back as a compression. That is, the direction of compression is reversed.

c) Reflection in curved surfaces:

(i) When sound waves are reflected from curved surfaces, the intensity of reflected waves is changed.
(ii) When reflected from a convex surface, the reflected waves are diverged out and the intensity is decreased.
(iii) When sound is reflected from a concave surface, the reflected waves are converged and focused at a point. So the intensity of reflected waves is concentrated at a point.
(iv) Parabolic surfaces are used when it is required to focus the sound at a particular point.
(v) Hence, many halls are designed with parabolic reflecting surfaces.
(vi) In elliptical surfaces, sound from one focus will always be reflected to the other focus, no matter where it strikes the wall.

3. a) What do you understand by the term ‘ultrasonic vibration’?
b) State three uses of ultrasonic vibrations.
c) Name three animals which can hear ultrasonic vibrations.

a) Ultrasonic vibrations: These are high- frequency sound waves beyond the range of human hearing. The frequency, for these sound waves, is greater than 20 kHz.

b) Uses:
(i) To kill micro organisms.
(ii) To find direction and range of submarines.
(iii) To clean dental plates, jewellery and coins.
(iv) For welding.

c) Animals which can hear Ultrasonic vibrations:
(i) Dogs,
(ii) Bats,
(iii) Dolphins.

4. What is an echo?
a) State two conditions necessary for hearing an echo.
b) What are the medical applications of echo?
c) How can you calculate the speed of sound using echo?

An echo is the sound reproduced due to the reflection of original sound from various rigid surfaces such as walls, ceilings, surfaces of mountains, etc.

a) Conditions necessary for hearing an echo:

(i) The minimum time gap between the original sound and echo must be 0.1s.
(ii) The minimum distance required to hear an echo is 17.2 m.

b) The medical applications of echo:

The principle of echo is used in obstetric ultrasonography, which is used to create real-time visual images of the developing embryo or fetus in the mother’s uterus.

c) Calculate the speed of sound using echo:

(i) The sound pulse emitted by the source travels a total distance of 2d while traveling from the source to the wall and then back to the receiver.
(ii) The time taken for this has been observed to be “t”. Hence, the speed of sound wave is given by:
$$\text{Speed of sound} = \frac{\text{Distance travelled}}{\text{Time taken}} = \frac{2d}{t}$$

X. HOT Questions

1. Suppose that a sound wave and a light wave have the same frequency, then which one has a longer wavelength?

• a) Sound
• b) Light
• c) both a and b
• d) data not sufficient

Answer: b) Light

2. When sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound remain the same. Do you hear an echo sound on a hotter day? Justify your answer.

(i) An echo can only be heard if it reaches the ear after 0.1s. Time taken = Total distance / Velocity.
(ii) If the temperature rises (i.e. on a hotter day), the velocity of sound will increase. This in turn will decrease the time required for the sound to travel the same distance. If this time becomes less than 0.1s, an echo will not be heard.

Concept Map

Electric Power: Definition, Formula, SI Unit, and Energy Consumption Explained

Electric Power - Definition, Formula, Unit, Consumption

ELECTRIC POWER

In general, power is defined as the rate of doing work or rate of spending energy. Similarly, the electric power is defined as the rate of consumption of electrical energy. It represents the rate at which the electrical energy is converted into some other form of energy.

Suppose a current ‘I’ flows through a conductor of resistance ‘R’ for a time ‘t’, then the potential difference across the two ends of the conductor is ‘V’. The work done ‘W’ to move the charge across the ends of the conductor is given by the equation (4.19) as follows:

$$ W = VIt $$

$$ P = \frac{\text{Work}}{\text{Time}} = \frac{VIt}{t} $$

$$ P = V I \quad (4.21) $$

Thus, the electric power is the product of the electric current and the potential difference due to which the current passes in a circuit.

1. Unit of Electric Power

The SI unit of electric power is watt. When a current of 1 ampere passes across the ends of a conductor, which is at a potential difference of 1 volt, then the electric power is

\( P = 1 \text{ volt} \times 1 \text{ ampere} = 1 \text{ watt} \)

Thus, one watt is the power consumed when an electric device is operated at a potential difference of one volt and it carries a current of one ampere. A larger unit of power, which is more commonly used is kilowatt.

2. Consumption of electrical energy

Electricity is consumed both in houses and industries. Consumption of electricity is based on two factors: (i) Amount of electric power and (ii) Duration of usage. Electrical energy consumed is taken as the product of electric power and time of usage. For example, if 100 watt of electric power is consumed for two hours, then the power consumed is 100 × 2 = 200 watt hour. Consumption of electrical energy is measured and expressed in watt hour, though its SI unit is watt second. In practice, a larger unit of electrical energy is needed. This larger unit is kilowatt hour (kWh) . One kilowatt hour is otherwise known as one unit of electrical energy. One kilowatt hour means that an electric power of 1000 watt has been utilized for an hour. Hence,

$$ 1 \text{ kWh} = 1000 \text{ watt hour} = 1000 \times (60 \times 60) \text{ watt second} = 3.6 \times 10^6 \text{ J} $$

Understanding Electrical Resistivity and Conductivity: Definitions, Formulas, and Solved Problems

Electrical Resistivity & Conductivity

1. Electrical Resistivity

You can verify by doing an experiment that the resistance of any conductor ‘R’ is directly proportional to the length of the conductor ‘L’ and is inversely proportional to its area of cross section ‘A’.

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

Where, ρ (rho) is a constant, called as electrical resistivity or specific resistance of the material of the conductor.

From the above equation, we can write:

$$ \rho = \frac{RA}{L} $$

If L = 1 m, A = 1 m² then, from the above equation ρ = R

Hence, the electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section. Its unit is ohm metre (Ω m).

Electrical resistivity of a conductor is a measure of the resisting power of a specified material to the passage of an electric current. It is a constant for a given material.

2. Conductance and Conductivity

Conductance of a material is the property of a material to aid the flow of charges and hence, the passage of current in it. The conductance of a material is mathematically defined as the reciprocal of its resistance (R). Hence, the conductance ‘G’ of a conductor is given by:

$$ G = \frac{1}{R} $$

Its unit is ohm^–1. It is also represented as ‘mho’.

The reciprocal of electrical resistivity of a material is called its electrical conductivity ($\sigma$).

$$ \sigma = \frac{1}{\rho} $$

Its unit is ohm^–1 metre^–1. It is also represented as mho metre^–1. The conductivity is a constant for a given material. Electrical conductivity of a conductor is a measure of its ability to pass the current through it. Some materials are good conductors of electric current. Example: copper, aluminium, etc. While some other materials are non-conductors of electric current (insulators). Example: glass, wood, rubber, etc.

Conductivity is more for conductors than for insulators. But, the resistivity is less for conductors than for insulators. The resistivity of some commonly used materials is given in Table 4.2.

Table 4.2: Resistivity of some materials

Solved Problem

The resistance of a wire of length 10 m is 2 ohm. If the area of cross section of the wire is 2 × 10^–7 m², determine its (i) resistivity (ii) conductance and (iii) conductivity.

Solution:

Given: Length, L = 10 m, Resistance, R = 2 ohm and Area, A = 2 × 10^–7 m²

(i) Resistivity ($\rho$):

$$ \rho = \frac{R \times A}{L} = \frac{2 \times (2 \times 10^{-7})}{10} = \frac{4 \times 10^{-7}}{10} = 4 \times 10^{-8} \, \Omega \text{ m} $$

(ii) Conductance (G):

$$ G = \frac{1}{R} = \frac{1}{2} = 0.5 \, \text{mho} $$

(iii) Conductivity ($\sigma$):

$$ \sigma = \frac{1}{\rho} = \frac{1}{4 \times 10^{-8}} = 0.25 \times 10^{8} \, \text{mho m}^{-1} $$

Understanding Electrical Resistance of a Material: Definition, Unit (Ohm), and Solved Problems

Resistance of a Material

RESISTANCE OF A MATERIAL

In Figure 4.4, a Nichrome wire was connected between X and Y. If you replace the Nichrome wire with a copper wire and conduct the same experiment, you will notice a different current for the same value of the potential difference across the wire. If you again replace the copper wire with an aluminium wire, you will get another value for the current passing through it. From equation (4.3), you have learnt that V/I must be equal to the resistance of the conductor used. The variations in the current for the same values of potential difference indicate that the resistance of different materials is different. Now, the primary question is, “what is resistance?”

Figure 4.4: Electric circuit to understand Ohm's law

Resistance of a material is its property to oppose the flow of charges and hence the passage of current through it. It is different for different materials.

Defining Resistance with Ohm's Law

From Ohm’s Law, V / I = R.

\( R = \frac{V}{I} \)

The resistance of a conductor can be defined as the ratio between the potential difference across the ends of the conductor and the current flowing through it.

Unit of Resistance

The SI unit of resistance is ohm and it is represented by the symbol Ω.

Resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its ends.

1 ohm = 1 volt / 1 ampere

\( 1 \, \text{ohm} = \frac{1 \, \text{volt}}{1 \, \text{ampere}} \)

Solved Problem

Calculate the resistance of a conductor through which a current of 2 A passes, when the potential difference between its ends is 30 V.

Solution:

Current through the conductor I = 2 A

Potential Difference V = 30 V

From Ohm’s Law: \( R = \frac{V}{I} \).

Therefore, \( R = \frac{30}{2} = 15 \, \Omega \)

Understanding Electric Potential and Potential Difference in Physics

Electric Potential and Potential Difference

ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE

You are now familiar with the water current and air current. You also know that there must be a difference in temperature between two points in a solid for the heat to flow in it. Similarly, a difference in electric potential is needed for the flow of electric charges in a conductor. In the conductor, the charges will flow from a point in it, which is at a higher electric potential to a point, which is at a lower electric potential.

1. Electric Potential

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.

2. Electric Potential Difference

The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.

Suppose, you have moved a charge Q from a point A to another point B. Let ‘W’ be the work done to move the charge from A to B. Then, the potential difference between the points A and B is given by the following expression:

Potential difference is also equal to the difference in the electric potential of these two points. If \(V_A\) and \(V_B\) represent the electric potential at the points A and B respectively, then, the potential difference between the points A and B is given by:

\(V = V_A – V_B\) (if \(V_A\) is more than \(V_B\))

\(V = V_B – V_A\) (if \(V_B\) is more than \(V_A\))

3. Volt

The SI unit of electric potential or potential difference is volt (V).

The potential difference between two points is one volt, if one joule of work is done in moving one coulomb of charge from one point to another against the electric force.

1 volt = 1 Joule / 1 coulomb

Solved Problem

The work done in moving a charge of 10 C across two points in a circuit is 100 J. What is the potential difference between the points?

Solution:

Charge, Q = 10 C
Work Done, W = 100 J

Potential Difference \(V = \frac{W}{Q} = \frac{100}{10}\)

Therefore, V = 10 volt

Understanding Electric Circuits and Their Components

Electric Circuit

What is an Electric Circuit?

ELECTRIC CIRCUIT

An electric circuit is a closed conducting loop (or) path, which has a network of electrical components through which electrons are able to flow. This path is made using electrical wires so as to connect an electric appliance to a source of electric charges (battery). A schematic diagram of an electric circuit comprising of a battery, an electric bulb, and a switch is given in Figure 4.2.

Figure 4.2 A simple electric circuit

In this circuit, if the switch is ‘on’, the bulb glows. If it is switched off, the bulb does not glow. Therefore, the circuit must be closed in order that the current passes through it. The potential difference required for the flow of charges is provided by the battery. The electrons flow from the negative terminal to the positive terminal of the battery.

By convention, the direction of current is taken as the direction of flow of positive charge (or) opposite to the direction of flow of electrons. Thus, electric current passes in the circuit from the positive terminal to the negative terminal.

Electrical Components

The electric circuit given in Figure 4.2 consists of different components, such as a battery, a switch and a bulb. All these components can be represented by using certain symbols. It is easier to represent the components of a circuit using their respective symbols.

The symbols that are used to represent some commonly used components are given in Table 4.1. The uses of these components are also summarized in the table.

Understanding Refraction Through a Concave Lens: Image Formation Explained

Refraction Through a Concave Lens

REFRACTION THROUGH A CONCAVE LENS

Let us discuss the formation of images by a concave lens when the object is placed at two possible positions.

Object at Infinity

When an object is placed at infinity, a virtual image is formed at the focus. The size of the image is much smaller than that of the object (Figure 2.12).

Figure 2.12 Concave lens-Object at infinity

Object anywhere on the principal axis at a finite distance

When an object is placed at a finite distance from the lens, a virtual image is formed between optical center and focus of the concave lens. The size of the image is smaller than that of the object (Figure 2.13).

Figure 2.13 Concave lens-Object at a finite distance

But, as the distance between the object and the lens is decreased, the distance between the image and the lens also keeps decreasing. Further, the size of the image formed increases as the distance between the object and the lens is decreased. This is shown in (figure 2.14).

Figure 2.14 Concave lens- Variation in position and size of image with object distance