Sainavle

Electric Current and Ohm’s Law Combination Of Resistances

Several resistances or resistors such as light bulbs, fans, pumps, etc. may bo Joined In a network variously for various purposes.

A combination of resistances can ho of three types:

Aeries combination,

Parallel combination

Mixed combination.

Equivalent resistance:

If a single resistance is used Instead of a combination of resistances, to keep the current unchanged in the circuit, then that single resistance is called the equivalent resistance of the combination.

Ohm’s Law Combination Of Resistances Notes

Series Combination of Resistances:

A number of resistances are said to be connected In scries if they are connected end to end consecutively so that the same current flows through each resistance when a potential difference is applied across the combination.

Calculation of equivalent resistance: Three resistances R₁, R₂, and R₃ arc connected in series in between the two points A and D of on electrical circuit.

Let VA, VB, VC, and VD be the potentials at points A, B, C, and D respectively.

If I am the current flowing in the circuit, then according to Ohm’s law

V_A-V_B = IR₁….(1)

V_B – V_C = IR₂…..(2)

V_C – V_D = IR₃….(3)

Adding (1), (2) and (3) we have,

V_A-V_D = I(R₁+ R₂+R₃)…..(4)

If R be the equivalent resistance of the combination and if it is connected between die points A and D, the main current flowing in the circuit will remain die same. So

V_A-V_D = IR…..(5)

From (4) and (5) we have

R = R₁ + R₂ + R₃

Ohm’s Law Combination Of Resistances Notes

Similarly, if n number of resistances are connected in series instead of the three resistances then,

⇒ \(R=R_1+R_2+R_3^{\wedge}+\cdots+R_n=\sum_{i=1}^n R_i\)….(6)

So, the equivalent resistance of a series combination = the sum of the individual resistances.

A few characteristics of the series combination of resistances:

The same current flows through each resistance.

The total potential difference across the combination is equal to the sum of the individual potential difference across each resistance.

Since the current is constant, an individual potential difference is directly proportional to the individual resistance.

Unit 2 Current Electricity Chapter 1 Electric Current and Ohm’s Law Combination Of Resistances Numerical Examples

Example 1. Three resistances of magnitudes 20Ω, 30Ω, and 40Ω, are connected in series,

1. What is the equivalent resistance?
2. If the potential difference across the resistance 20 A is 1 V, calculate the potential difference across the other two resistances and also the total potential difference across the combination.

Solution:

1. Equivalent resistance, R = 20 + 30 + 40

= 90Ω

2. For the resistance of 20Ω

⇒ \(\text { current, } I=\frac{\text { potential difference }}{\text { resistance }}=\frac{1}{20} \mathrm{~A}\)

Since it is a series combination, the current everywhere is \(\frac{1}{20}\)A.

∴ The potential difference across 30Ω resistance

⇒ \(\frac{1}{20} \times 30\)

= 1.5 V

and across the 40Ω resistance = \(\frac{1}{20} \times 40\)

= 2.0 V

⇒ \(\frac{1}{20} \times 90\)

= 4.5 V

Ohm’s Law Combination Of Resistances Notes

Example 2. \(\rho_1 \text { and } \rho_2\) are the resistivities of the materials of two wires of the same dimensions. What will be the equivalent resistivity of the series combination of the two wires?
Solution:

Let l be the length and A be the cross-sectional area of each wire.

The equivalent resistance in series combination is

⇒ \(R=R_1+R_2=\rho_1 \frac{l}{A}+\rho_2 \frac{l}{A}=\left(\rho_1+\rho_2\right) \frac{l}{A}\)….(1)

In the series combination, the length of the conductor =21; cross-sectional area = A.

Let the equivalent resistivity be \(\rho\)

∴ \(R=\rho \frac{2 l}{A}\)…..(2)

Comparing (1) and (2) we have,

⇒ \(\rho \frac{2 l}{A}=\left(\rho_1+\rho_2\right) \frac{l}{A} \quad \text { or, } \rho=\frac{\rho_1+\rho_2}{2}\)

Parallel Combination of Resistances:

Two or more resistors are said to be connected in parallel if one end of each is connected to a common point, and the other end to another point.

When this combination is joined in a circuit, the main current is distributed among the resistors, but the potential difference across each is the same

Ohm’s Law Combination Of Resistances Notes

Calculation of equivalent resistance: Three resistances R₁, R₂, and R₃ are connected in parallel in between two common points A and B of an electrical circuit. Let VA and VB be the potentials at points A and B respectively.

A current I is divided into three parts through these resistances.

I = I₁ +I₂+ I₃…(1)

A and B are the common terminals of each of the three resistances. So according to Ohm’s law,

For the resistance \(R_1, I_1=\frac{V_A-V_B}{R_1}\)….(2)

For the resistance \(R_2, I_2=\frac{V_A-V_B}{R_2}\)…..(3)

For the resistance \(R_3, I_3=\frac{V_A-V_B}{R_3}\)….(4)

Adding (2), (3) and (4) we get,

⇒ \(I_1+I_2+I_3=\left(V_A-V_B\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)\)

or, \(I=\left(V_A-V_B\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)\)….(5)

If R is the equivalent resistance of the combination and if it is connected between points A and B, the main current flow in the circuit will remain the same. So

⇒ \(I_1+I_2+I_3=\left(V_A-V_B\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)\)…(6)

From (5) and (6) we get \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

Similarly, if n number of resistances are connected in parallel
instead of the three resistances, we have

⇒ \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots+\frac{1}{R_n}=\sum_{i=1}^n \frac{1}{R_i}\)…(7)

So, the reciprocal of the equivalent resistance of the parallel combination = the sum of the reciprocals of the individual resistances.

Ohm’s Law Combination Of Resistances Notes

Calculation of current in different resistances: If R be the equivalent resistance of a parallel combination connected between two points A and B of the circuit, the main current is given by,

⇒ \(I=\frac{V_A-V_B}{R} \text { or, } V_A-V_B=I R\)

The potential difference of each resistance is ( V_A– V_B).

So, the current flowing through Rl is

⇒ \(I_1=\frac{V_A-V_B}{R_1}=\frac{I \cdot R}{R_1}=I \cdot \frac{R}{R_1}\)

This rule is applicable to each resistance of the parallel combination.

So, the current flowing through any resistance

⇒ \(\text { main current } \times \frac{\text { equivalent resistance }}{\text { corresponding resistance }}\)

A few characteristics of parallel combinations of resistances:

1. The potential difference across each resistance is the same.

2. \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots ; \text { so } \frac{1}{R}\) is greater than each of \(\frac{1}{R_1}+\frac{1}{R_2}+….\) etc. Therefore, the value of R is less than each of R₁, R₂….etc i.e., the equivalent resistance is less than each of the resistances in the combination.

3. Total ament through the parallel combination is the sum of the individual currents through the resistances.

4. Since the potential difference across each resistance Is constant, individual currents are Inversely proportional to the individual resistances.

In domestic electrical circuits, appliances such as bulbs, electric fan, heaters, etc. operate with the same potential difference (220 V), So these are connected in parallel.

Special case (parallel combination of two resistances): Suppose two resistances R₁ and R₂, are connected in parallel. Let their equivalent resistance be R. Then,

⇒ \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \quad \text { or, } R=\frac{R_1 R_2}{R_1+R_2}\)

The potential difference across the two ends of the combination,

⇒ \(V=I R=I \frac{R_1 R_2}{R_1+R_2}\)

A potential difference of each resistance = V.

So, current flowing through \(R_1, I_1=\frac{V}{R_1}=I \cdot \frac{R_2}{R_1+R_2}\)

Similarly, current flowing through \(R_2, I_2=I \cdot \frac{R_1}{R_1+R_2}\)

So, current flowing through a resistance

⇒ \(=\text { main current } \times \frac{\text { other resistance }}{\text { sum of the two resistances }}\)

This formula will often be found handy in numerical calculations

Ohm’s Law Combination Of Resistances Notes

Electric Current and Ohm’s Law Numerical Examples

Example 1. ABC is a triangle formed by wires. The resistances of the sides AB, BC, and CA are respectively 40Ω, 60Ω and 100Ω. What is the equivalent resistance between points A and B?

Solution:

Along the path ACD the two resistances 100Ω and 60Ω are connected in series. So the equivalent resistance of the ACD

= 100 + 60

= 160Ω.

Again, the resistance 40Ω of the side All Is connected In parallel with the path ACB, So the equivalent resistance between the two points A and B is

⇒ \(R=\frac{160 \times 40}{160+40}=\frac{160 \times 40}{200}=32 \Omega\)

Example 2. Determine the equivalent resistance between points A and B

Solution:

The equivalent resistance of the path ADC = 3 + 7 = 10Ω.

With the path ADC, the 10Ω resistance of the diagonal AC Is connected in parallel.

So, the equivalent resistance of AD, DC, and AC

⇒ \(\frac{10 \times 10}{10+10}=5 \Omega \text {. }\)

With this 5Ω resistance, the 5Ω resistance of the side DC Is connected In series. So the equivalent resistance of 5 ft and DC

= 5 + 5

= 10Ω

Now, the 10Ω resistance of AD is also In, parallel with the above equivalent resistance 10Ω.

So, the equivalent resistance between A and B is

⇒ \(R=\frac{10 \times 10}{10+10}=5 \Omega\)

Ohm’s Law Combination Of Resistances Notes

Example 3. The two circuits draw equal currents from the battery. But the current through the resistance R in the second circuit is \(\frac{1}{10}\)th of that the first circuit. Determine the values of R₁ and R₂.

Solution:

Current drawn from the battery in first case,

⇒ \(I_1=\frac{E}{R}\)…..(1)

Current drawn from the battery in the second case,

⇒ \(I_2=\frac{\text { total emf }}{\text { total resistance }}=\frac{E}{R_1+\frac{R R_2}{R+R_2}}\) [∵ R1 is in series with the parallel combination of R and R2 ]

So, \(I_2=\frac{E\left(R+R_2\right)}{R_1 R+R_1 R_2+R R_2}\)….(2)

Accordingly, \(I_1=I_2 \quad \text { or, } \frac{E}{R}=\frac{E\left(R+R_2\right)}{R_1 R+R_1 R_2+R R_2}\)

or, R₁R + R₁R₂ + RR₂ = R₂ + RR₂

or, R₂-R₁R- R₁R₂ = 0 ….(3)

Now, current through the resistance R,

⇒ \(I_R=I_2 \frac{R_2}{R+R_2}\)…(4)

It is given that, \(I_R=\frac{I_1}{10} \quad \text { or, } I_R=\frac{I_2}{10} \quad\left(… I_1=I_2\right)\)

or, \(\frac{I_2 R_2}{R+R_2}=\frac{I_2}{10} \text { [from equation (4)] }\)

or, \(10 R_2=R+R_2 \quad \text { or, } R_2=\frac{R}{9}=0.11 R\)

Putting R2 in equation (3), we get

⇒ \(R^2-R_1 R-R_1 \frac{R}{9}=0 \quad \text { or, } R_1 \frac{10 R}{9}=R^2\)

⇒ \(R_1=\frac{9}{10} R=0.9 R\)

Required resistance, R₂ = 0.117R and 7R₁ = 0.9R

Ohm’s Law Combination Of Resistances Notes

Example 4. You are given several identical resistances, each of value R = 10Ω and each capable of carrying a maximum combination of these resistances to obtain a resistance of 5Ω which can carry a current of 4 A. Find the minimum number of resistances of the Type R that will be required.
Solution:

The resistances have to be connected in a series of parallel combinations. Since each resistance can carry a current of 1 A, to pass a 4 A current, we need four paths in parallel. Let r be the resistance of each path.

The equivalent resistance of 4 parallel paths will be \(\frac{r}{4}\).

According to the given problem,

⇒ \(\frac{r}{4}\) = 5

∴ r = 5 x 4

= 20Ω

In order to have 20Ω resistance in each path, two resistances each of resistance 10Ω have to be connected in series.

Since there are four paths, the total number of resistances required

=2 x 4

= 8.

Example 5. A wire of uniform cross-section and length l has a resistance of 16Ω. It is cut into four equal parts. Each part is stretched uniformly to length l and all four stretched parts are connected in parallel. Calculate the total resistance of the combination so formed. Assume that the stretching of the wire does not cause any change in the density of its material.
Solution:

Let the cross-section of the wire change from A to Aj when it is cut into four parts.

As density remains constant, so volume also remains constant.

Volume of each part before stretching = volume of each part after stretching

or, \(\frac{A l}{4}=A_1 l \quad \text { or, } A_1=\frac{A}{4}\)

So resistance of each part, \(R_1=\frac{\rho l}{\frac{A}{4}}=\frac{4 \rho l}{A}\)

Therefore, the equivalent resistance of the parallel combination,

⇒ \(R_{\mathrm{eq}}=\frac{R_1}{4}=\frac{\rho l}{A}=16 \Omega\) [∵ initial resistance = \(\frac{\rho l}{A}=16 \Omega\)]

Current Electricity

Electric Current and Ohm’s Law Shunt

For every electrical instrument, the current flowing through it has a maximum permissible limit.

If the current exceeds the limit, there is a possibility of damage to the instrument.

To protect sensitive instruments like galvanometers and ammeters against possible damages due to heavy current passing through them, an alternative passage is provided as an inbuilt device within these instruments such that a major part of the main current in the circuit passes through this alternative route and a very small part through the instrument.

This alternative passage which is nothing but a low resistance connected in parallel with the instrument is called a shunt.

The arrangement of a shunt where a low resistance S (shunt) is connected in parallel with a galvanometer of resistance G.

Let I be the main current, IG be the current through the galvanometer, and Is be the current through the shunt.

So, \(I=I_G+I_S\)

Now the equivalent resistance of the combination of the galvanometer and the shunt = \(\frac{G S}{G+S}\)

So, \(V_A-V_B=I \cdot \frac{G S}{G+S}\)

∴ \(I_G=\frac{V_A-V_B}{G}=I \cdot \frac{S}{S+G}\)…(2)

Class 12 Physics Electric Current and Ohm’s Law Shunt Notes

and \(I_S=\frac{V_A-V_B}{S}\)

= \(I \cdot \frac{G}{S+G}\) ….(3)

From equations (2) and (3) we get \(\frac{I_G}{I_S}=\frac{S}{G}\)

Particularly, if S << G, then IG << Is.

The ratio of the main current and the current passing through the galvanometer is known as the multiplying factor or power of a shunt, denoted by n

∴ \(\frac{I}{I_G}=n \quad \text { or, } \frac{S}{G+S}=\frac{1}{n} \text { [from equation (2)] }\)

or, \(\frac{S}{G}=\frac{1}{n-1} \quad \text { or, } S=\frac{G}{n-1}\)…(4)

Thus the shunt resistance can be adjusted according to the demand of n depending on G.

The shunt has another utility. With the help of the shunt, the galvanometer can almost accurately record the current flowing through the circuit. When a high-resistance galvanometer is connected to a circuit, the current drops considerably.

But with a shunt which is used in parallel, the equivalent resistance is reduced even below S. So, the main current of the circuit remains practically unchanged. Therefore, it becomes possible to measure the main current almost correctly.

Current Electricity Electric Current and Ohm’s Law Shunt Numerical Examples

Example 1. To reduce the action of a galvanometer by 25 times, a shunt is added to it. If the galvanometer resistance is 1000 ft, what is the resistance of the shunt? An ammeter is used to measure the current and a voltmeter
Solution:

⇒ \(I_G=I \cdot \frac{S}{S+G} \quad \text { or, } \frac{I_G}{I}=\frac{S}{S+G}\)

According to the question \(\frac{I_G}{I}=\frac{1}{25}\)

∴ \(\frac{1}{25}=\frac{S}{S+G} \quad \text { or, } 25 S=S+G \quad \text { or, } 24 S=1000\)

or, \(S=\frac{1000}{24}=41.67 \Omega\)

Example 2. If a shunt of 1Ω Is connected to a galvanometer of resistance 99Ω, what fraction of the main current will flow through the galvanometer?
Solution:

The galvanometer resistance G and the shunt resistance S are connected in parallel. So the galvanometer current

⇒ \(I_G=I \cdot \frac{S}{S+G} \quad[I=\text { main current }]\)

or, \(\frac{I_G}{I}=\frac{S}{S+G}=\frac{1}{1+99}=\frac{1}{100}=1 \%\)

i.e., 1% of the main current will flow through the galvanometer.

Class 12 Physics Electric Current and Ohm’s Law Shunt Notes

Example 3. A battery of internal resistance zero is connected to a galvanometer of resistance 80Ω and a resistance of 20Ω in series. A current flows through the galvanometer. If a shunt of 1Ω resistance is connected to the galvanometer, show that the current that will now flow through the galvanometer becomes \(\frac{1}{17}\) of the previous current.
Solution:

The main current before adding the shunt to the circuit is

⇒ \(I=\frac{E}{80+20}=\frac{E}{100}\)

After adding the shunt the main current is

⇒ \(I^{\prime}=\frac{E}{\frac{80 \times 1}{80+1}+20}=\frac{E \times 81}{80+20 \times 81}\)

So, \(I_G=I^{\prime} \cdot \frac{S}{S+G}=I^{\prime} \cdot \frac{1}{80+1}=I^{\prime} \times \frac{1}{81}=\frac{E}{80+20 \times 81}\)

∴ \(\frac{I_G}{I}=\frac{E}{80+20 \times 81} \times \frac{100}{E}=\frac{100}{20(4+81)}=\frac{100}{20 \times 85}=\frac{1}{17}\)

i.e., \(I_G=\frac{1}{17} I\)

Connection of Ammeter and Voltmeter in a Circuit:

An ammeter is used to measure the current and a voltmeter is used to measure the potential difference between any two points of a circuit.

Milliammeters and ammeters are used to measure small currents while millivoltmeters and microvoltmeters are used to measure small potential differences.

Connection of ammeter:

An ammeter is connected an electrical circuit, so that it may give the reading of the current when the circuit current passes through it. The main current decreases a little due to the resistance of the ammeter. To overcome this disadvantage a low-resistance ammeter should be used

Connection of voltmeter: A voltmeter of resistance Rv is inserted in a parallel connection between the two points of the circuit across which the potentia difference is to be measured. The potential difference across the voltmeter is recorded as that between the two points. But there is a difficulty in this arrangement. Before inserting the voltmeter the resistance between the points A and B was R.

After joining the voltmeter, the resistance between points A and B is equal to the equivalent resistance of R and Rv, which is always less than R.

Class 12 Physics Electric Current and Ohm’s Law Shunt Notes

So, the main current increases i.e., the potential difference between A and B also increases. Hence it is desirable that Rv should be much greater than R.

In that case, the equivalent resistance becomes almost equal to R and the slight increase of the potential difference can be ignored. So, the voltmeter is an instrument having a very high resistance and is connected in parallel in an electrical circuit.

Current Electricity

Electric Current and Ohm’s Law Numerical Examples

Example 1. The internal resistance of a battery of 100 V is 5fl. j When the emf of the battery is measured by a voltmeter 20% error is found. What is the resistance of the voltmeter?
Solution:

Suppose, the resistance of the voltmeter is R. Its reading is the potential difference of the external circuit.

According to the question, voltmeter reading =IR = 80% of 100 V = 80 V

∴ Lost volt =Ir = 100-80 = 20 V

∴ \(\frac{I R}{I r}=\frac{80}{20} \quad \text { or, } \frac{R}{r}=4\)

or, R = 4r

= 4 x 5

= 20Ω

Class 12 Physics Electric Current and Ohm’s Law Shunt Notes

Example 2. In a supply line of 100 V, there is a resistance of 1000Ω. In between one terminal of the resistance and its point, a voltmeter is connected which gives a reading of 40 V. Determine the resistance of the voltmeter
Solution:

Suppose, the resistance of the voltmeter is R. I C is the midpoint of the resistance AB.

So, resistance of each portion AC and BC = \(\frac{1000}{2}\)

= 500Ω

Since the reading of the voltmeter = 40 V,

∴ V_A – V_C = 40V,

SO V_C– V_B = 100 – 40

= 60V

Now, the main current of the circuit,

⇒ \(I=\frac{V_C-V_B}{\text { resistance of } B C}=\frac{60}{500}=0.12 \mathrm{~A}\)

Again, current in the resistance AC,

⇒ \(I^{\prime}=\frac{V_A-V_C}{\text { resistance of } A C}\)

= \(\frac{40}{500}\)

= 0.08A

So, current in the voltmeter

IV = I-V

= 0.12 – 0.08

= 0.04 A

Therefore, the resistance of the voltmeter,

⇒ \(R=\frac{V_A-V_C}{I_V}=\frac{40}{0.04}=1000 \Omega\)

Class 12 Physics Electric Current and Ohm’s Law Shunt Notes

Example 3. When a voltmeter of resistance 100Ω is connected. with an electric cell, the reading of the voltmeter Is 2 V. When the cell is connected with a resistance of 15Ω, an ammeter of resistance 1Ω gives the reading of 0.1 A. Determine the emf of the cell.
Solution:

In the first circuit

⇒ \(\text { current, } I_1=\frac{\text { reading of the voltmeter }}{\text { resistance of the voltmeter }}=\frac{2}{100}=0.02 \mathrm{~A}\)

If the die emf of the cell is E and the internal resistance is r then,

lost volt = I₁r

= 0.02r

i.e., E = 2 + 0.02r…(1)

Now, in the second circuit

⇒ \(\text { current, } I_2=\frac{E}{r+15+1}\)

or, \(0.1=\frac{E}{r+16}\)

or, E = 0.1r + 1.6…(2)

From the equations (1) and (2) we have,

2 + 0.02r = 0.1r+ 1.6

or, 0.08r = 0.4

or, \(r=\frac{0.4}{0.08}=5 \Omega\)

So, from (1), E = 2 + 0.02 x 5

= 2 + 0.1

= 2.1V

Class 12 Physics Ohm’s Law Multiple Choice Question And Answers

Question 1. Two copper wires have a ratio of 1: 4 between their diameters. If the same current passes through both of them, the drift velocity of the electrons will be in the ratio of

1. 16:1
2. 4: 1
3. 1:4
4. 1:16

Answer: 1. 16:1

Question 2. A conductor of uniform cross-section is carrying a current of 1 ampere. The number of free electrons flowing across the cross-section of the conductor per second is

1. 6.25 x 10¹⁸
2. 6.25 x 10¹⁷
3. 6.25 X 10¹⁶
4. 6.025 x 10²³

Answer: 1. 6.25 x 10¹⁸

1 A electric current = the flow of 1 C charge through the cross-sectional area of the conductor in 1 second.

∴ Number of free electrons flowing per second

⇒ \(\frac{1 \mathrm{C}}{\text { charge of an electron }}\)

= \(\frac{1 \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}\)

= \(6.25 \times 10^{18}\)

The option 1 is correct.

Question 3. In the circuit,

1. I = 3A
2. I₁ = 2A
3. I₂ = 1A
4. V_AB = 8V

Answer:

1. I = 3A

2. I₁ = 2A

3. I₂ = 1A

Question 4. In the circuit,

1. I = 1 A
2. I = \(\frac{4}{3}\)A
3. V_AB = 4 V
4. V_AB = 3 V

Answer:

1. I = 1 A

4. V_AB = 3 V

Class 12 Physics Ohm’s Law MCQs Question 5. If I = 2 A in the circuit

1. E = 5 V
2. I₁ = 1.5 A
3. I₂ = 0.5 A
4. V_AB = 3 V

Answer:

1. E = 5 V

2. I₁ = 1.5 A

3. I₂ = 0.5 A

4. V_AB = 3 V

Question 6. If the resistance of the rheostat Rh is gradually increased in the circuit,

1. I will rise gradually
2. I will fall gradually
3. I₁ will rise gradually
4. I₁ will fall gradually

Answer:

2. I will fall gradually

3. I₁ will rise gradually

Question 7. Which of the following observations is correct if the galvanometer resistance is 200Ω in the circuit

1. S = 5Ω, I = 1.5 A,IG = 36.6 mA
2. S = 1Ω, I = 1.5 A, IG = 14.9 mA
3. S = 2Ω, I = 2 A, IG = 19.8 mA
4. S = 3Ω, I = 2 A, IG = 29.6 mA

Answer:

1. S = 5Ω, I = 1.5 A,IG = 36.6 mA

3. S = 2Ω, I = 2 A, IG = 19.8 mA

4. S = 3Ω, I = 2 A, IG = 29.6 mA

Question 8. Brown, black, orange, and gold are the respective colors of the characteristic rings on a carbon resistor. Which of the following values of its resistance are definitely wrong?

1. 10.6 kΩ
2. 10.2 kΩ
3. 9.8 kΩ
4. 9.4 kΩ

Answer:

1. 10.6 kΩ

4. 9.4 kΩ

Question 9. Three 4Ω resistances can be connected in different combinations. The probable values of the equivalent resistance are

1. 12Ω
2. 6Ω
3. \(\frac{10}{3}\)Ω
4. \(\frac{4}{3}\)Ω

Answer:

1. 12Ω

2. 6Ω

4. \(\frac{4}{3}\)Ω

Class 12 Physics Ohm’s Law MCQs  Question 10. E₁, E₂, and r₁,r₂ are respectively, the emf’s and internal resistances of two cells. The current through an external resistance R, when it is connected to the first cell, is equal to that when it is connected to the second. Here, the probable relations are

1. E₁ = E₂,r₁ = r₂
2. E₁ > E₂, r₁ > r₂
3. E₁ < E₂, r₁ < r₂
4. E₁ > E₂, r₁ < r₂

Answer:

1. E₁ = E₂,r₁ = r₂

2. E₁ > E₂, r₁ > r₂

3. E₁ < E₂,r₁ < r₂

Question 11. A voltmeter and an ammeter are connected in series to an ideal cell of emf E. The voltmeter reading is V and the ammeter reading is I. Choose the correct options.

1. The voltmeter resistance is \(\frac{V}{I}\)
2. The potential difference across the ammeter is (E- V)
3. V<E
4. Voltmeter resistance + ammeter resistance = \(\frac{E}{I}\)

Answer:

1. The voltmeter resistance is \(\frac{V}{I}\)

2. The potential difference across the ammeter is (E- V)

3. V

Question 12. Part of a circuit. Which points have the potential same as that of point m?

1. p
2. r
3. t
4. u

Answer:

2. r

3. t

Question 13. A galvanometer has a resistance of 100Ω and a full-scale range of 50μA. It can be used as a voltmeter or as a higher range ammeter, provided a resistance is added to it. Pick the correct range and resistance combination(s).

1. 50 V range with 10kΩ resistance in series
2. 10 V range with 200 kΩ .resistance in series
3. 5 mA range with 1Ω resistance in parallel
4. 10 mA range with 1Ω resistance in parallel

Answer:

2. 10 V range with 200 kΩ .resistance in series

3. 5 mA range with 1Ω resistance in parallel

Question 14. A straight conductor AB lies along the axis of a hollow metal cylinder, which is connected to the earth through a conductor C. A quantity of charge will flow through C if

1. A Current begins to flow through AB
2. The current through AB is reversed
3. AB is removed and a beam of protons flows in its place
4. AB is removed, and a beam of electrons flows in its place

Answer:

3. AB is removed and a beam of protons flows in its place

4. AB is removed, and a beam of electrons flows in its place

Class 12 Physics Ohm’s Law MCQs  Question 15. To double the full-scale voltage reading of any galvanometer turned into a voltmeter, you must

1. Increase the resistance to 3R
2. Half the resistance R
3. Increase the resistance to 4R
4. None of the above

Answer: 4. None of the above

Question 16. The resistance Ω of a conducting wire depends on its material, length l, and area of cross-section A. The resistivity of the material of the wire is p = \(\frac{RA}{l}\) the value of p is different for different materials. It is very low for conducting materials, like metals. Besides, the resistance of a conductor also depends on its temperature. If the resistance of a conductor is R0 at 0° C, and Rt at t° C, then Rt = R₀(1 + αt), where a is called the temperature coefficient of resistance. The resistance increases with temperature for metallic conductors but decreases for graphite, a few metal alloys, and for semiconductors like silicon and germanium.

1. The resistance of a metal wire increases by 10% when its temperature rises from 10° C to 110° C. The temperature coefficient of resistance of the metal is

1. 0.02 °C^-1
2. 0.01 °C^-1
3. 0.002 °C^-1
4. 0.001 °C^-1

Answer: 4. 0.001 °C^-1

2. The length of this metal wire is doubled by stretching. What will be the change in its resistance?

1. 100% increase
2. 200% increase
3. 300% increase
4. 500% decrease

Answer: 3. 300% increase

3. The temperature of this new wire is again raised from A B 10°C to 110°C. The percentage increase of its resistance would be

1. 5%
2. 10%
3. 20%
4. 40%

Answer: 2. 10%

4. The temperature coefficient of resistance of a semiconductor is

1. Zero
2. Positive
3. Negative
4. Positive or negative depending on the material

Answer: 3. Negative

5. The graphs of the relations between current (I) and potential difference (V) of a metal wire at two different temperatures t₁ and t₂. The relation between t₁ and t₂ is

1. t₁ = t₂
2. t₁ < t₂
3. t₁ > t₂
4. Insufficient data

Answer: 2. t₁ < t₂

Question 17. If a current passes through a metal conducting wire of area of cross-section A, the drift velocity of free electrons inside the metal is vd = \(\frac{1}{neA}\) where the amount of electric charge of an electron = e, and the number of free electrons per unit volume of the metal =n. The applied electric field on the wire is E = \(\frac{V}{l}\) y, where a potential difference V exists between two points,l apart, along the length of the wire. If R is the resistance of the wire between those two points, then the resistivity of its material is \(\rho=\frac{R A}{l}\). Besides, the mobility (μ) of the free electrons inside a wire is defined as their drift velocity for a unit-applied electric field.

1. Two copper wires have both lengths and radii in the ratio 1: 2. If the ratio between the electric currents flowing through them is also 1: 2, what would be the ratio between the drift velocities of free electrons?

1. 1:1
2. 1:2
3. 2:1
4. 4:1

Answer: 3. 2:1

2. The radii of two wires of the same metal are in the ratio 1: 2. The same potential difference is applied between two points at a distance on each of the wires. The ratio between the drift velocities of the free electrons in two wires is

1. 1:1
2. 1:2
3. 2:1
4. 1:4

Answer: 2. 1:2

Class 12 Physics Ohm’s Law MCQs  3. The radii of two wires, made of two different metals, are in the ratio 1: 2. The number density of free electrons in the first metal is double that in the second metal. If the current in the first wire is 1 A, then the current In the second wire producing the same drift velocity is

1. 1 A
2. 2 A
3. 4 A
4. 8 A

Answer: 3. 4 A

4. Tire current through unit cross-section of a conductor, culled tire electric current density J, Is related to the applied electric field E as

1. J = \(\rho\)E
2. \(J=\frac{1}{\rho} E\)
3. J = μE
4. \(J=\frac{1}{\mu} E\)

Answer: 2. \(J=\frac{1}{\rho} E\)

Question 18. Measurements and Interpretations of voltage and electric current signals are common In modern medicine. Occasionally, a situation arises In which a voltmeter or an ammeter Is needed but it Is not available. A galvanometer Is an instrument dial that can be used to construct an ammeter (for measuring electric currents). It can also be used to construct a voltmeter (to measure voltages). In both cases, a resistor R must be connected to the galvanometer to effect the change. To turn the galvanometer into an ammeter, the resistor R is connected in parallel. The resistor R is connected in series with the galvanometer in order to turn it into a voltmeter. The current required to produce a full-scale deflection in a galvanometer is 10mA. The internal resistance of the galvanometer is 100Ω. Let Vr be the voltage across r and VR the voltage across R.

1. Which of the following relations correctly applies to the ammeter?

1. V_r>V_R
2. V_r < V_R
3. V_r=V_R
4. More information is required

Answer: 1. V_r>V_R

2. What resistance must be connected in parallel to the galvanometer to turn It into an ammeter capable of reading electric currents up to 10.01 A?

1. 0.1Ω
2. 1Ω
3. 10Ω
4. None

Answer: 1. 0.1Ω

3. What resistance R must be connected in series to the galvanometer in order to convert it to a 100 V voltmeter?

1. 900Ω
2. 1000Ω
3. 9900Ω
4. 10000Ω

Answer: 3. 9900Ω

4. In the voltmeter circuit, the current in the resistor R must be

1. Negligible, so that it has only a small effect on the voltage reading
2. Substantial, but does not have any effect on the voltage reading
3. Substantial, but does have some effect on the voltage reading
4. None of the above

Answer: 1. Negligible, so that it has only a small effect on the voltage reading

Class 12 Physics Ohm’s Law MCQs  5. Which of the following relations correctly applies to the voltmeter circuit?

1. V_r>V_R
2. V_r<V_R
3. V_r=V_R
4. V_r = 2V_R

Answer: 2. V_r<V_R

Question 19. Two cells each of emf e but internal resistances r₁ and r₂ are connected in series through an external resistance R. If the potential difference across the first cell is zero while current flows, the relation of R in terms of r₁ and r₂ is

1. R = r₁ + r₂
2. R = r₁-r₂
3. R = \(\frac{1}{2}\) (rj + r2)
4. R = \(\frac{1}{2}\)(r1 – r2)

Answer: 2. R = r₁-r₂

Question 20. Resistance of the thinner wire is 10Ω, then the resistance of the other wire will be

1. 40Ω
2. 20Ω
3. 10Ω
4. 5Ω

Answer: 3. 10Ω

⇒ \(R=\rho \frac{l}{A}=\frac{\rho l}{\pi r^2}\)

∴ \(\frac{R_1}{R_2}=\frac{\rho_1}{\rho_2} \cdot \frac{l_1}{l_2} \cdot\left(\frac{r_2}{r_1}\right)^2=\frac{1}{3} \times \frac{1}{3} \times\left(\frac{3}{1}\right)^2=1\)

R₂ = R₁ = 10Ω

The option 3 is correct

Question 21. Four cells, each of emf E . and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then the current in the external circuit is

1. \(\frac{2 E}{4 r+R}\)
2. \(\frac{3 E}{4 r+R}\)
3. \(\frac{3 E}{3 r+R}\)
4. \(\frac{2 E}{3 r+R}\)

Answer: 1. \(\frac{2 E}{4 r+R}\)

Currently the circuit,

⇒ \(I=\frac{3 E-E}{4 r+R}=\frac{2 E}{4 r+R}\)

The option 1 is correct.

Class 12 Physics Ohm’s Law MCQs  Question 22. A circuit consists of three batteries of emf E₁ = 1V, E₂ = 2 V, and E₃ = 3V and internal resistances 1Ω, 2Ω, and 1Ω respectively which are connected in parallel. The potential difference between points P and Q is

1. 1.0V
2. 2.0V
3. 2.2V
4. 3V

Answer: 2. 2.0V

Equivalent resistance of the internal resistances connected in parallel

= \(\frac{2}{5}\)Ω

Total current = \(\frac{1}{1}+\frac{2}{2}+\frac{3}{1}\)

= 5V

∴ The potential difference between points P and

Q = 5 x \(\frac{2}{5}\)

= 2V

The option 2 is correct

Question 23. A metal wire of a circular cross-section has a resistance. The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes R₂ then R₂: R₁ is

1. 1:1
2. 1:2
3. 4:1
4. 1:4

Answer: 3. 4: 1

If the length of the wife is l and its cross-sectional area is A, the volume of the wire, V = lA = constant.

Then, A = \(\frac{V}{l}\)

Now, from the relation \(R=\rho \frac{l}{A}\)

⇒ \(\frac{R_1}{R_2}=\frac{l_1}{l_2} \cdot \frac{A_2}{A_1}=\frac{l_1}{l_2} \cdot \frac{V / l_2}{V / l_1}=\left(\frac{l_1}{l_2}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

or, \(\frac{R_2}{R_1}=\frac{4}{1}=4: 1\)

The option 3 is correct

Question 24. Two equal resistances, 400Ω each, are connected in series with an 8 V battery. If the resistance of the first one increases by 0.5%, the charge required in the resistance of the second one in order to keep the potential difference across it unaltered is to

1. Increase it by 1Ω
2. Increase it by 2Ω
3. Increase it by 4Ω
4. Decrease it by 4Ω

Answer: 2. Increase it by 2Ω

Increase in first resistance =400 x \(\frac{0.5}{100}\) = 2Ω

Initially, the emf 8 V will be divided equally between the two resistances. So the voltage across each resistance will be 4 V. When the first resistance is increased, the second resistance should also be increased by 2Ω to keep the voltage across it unchanged.

The option 2 is correct.

Class 12 Physics Ohm’s Law MCQs  Question 25. Two wires of the same radius having lengths l₁ and l₂ and resistivities p₁ and p₂ are connected in series. The equivalent resistivity will be

1. \(\frac{\rho_1 l_2+\rho_2 l_1}{\rho_1+\rho_2}\)
2. \(\frac{\rho_1 l_1+\rho_2 l_2}{l_1+l_2}\)
3. \(\frac{\rho_1 l_1-\rho_2 l_2}{l_1-l_2}\)
4. \(\frac{\rho_1 l_2+\rho_2 l_1}{l_1+l_2}\)

Answer: 2. \(\frac{\rho_1 l_1+\rho_2 l_2}{l_1+l_2}\)

⇒ \(\frac{\rho_1 l_1}{A}+\frac{\rho_2 l_2}{A}=\rho_{\mathrm{eq}} \frac{\left(l_1+l_2\right)}{A}\)

∴ \(\rho_{\mathrm{eq}}=\frac{\rho_1 l_1+\rho_2 l_2}{l_1+l_2}\)

The option 2 is correct.

Question 26. The effective resistance between A and B is \(\frac{7}{12}\)Ω if each side of the cube has 1Ω resistance. The effective resistance between the same two points when the link AB is removed is

1. \(\frac{7}{12}\)Ω
2. \(\frac{5}{12}\)Ω
3. \(\frac{7}{5}\)Ω
4. \(\frac{5}{7}\)Ω

Answer: 3. \(\frac{7}{5}\)Ω

If x is the effective resistance between A and B of the remaining cube, then

⇒ \(\frac{7}{12}=\frac{1 \times x}{1+x}\)

Solving we get, x = \(\frac{7}{5}\)Ω

The option 3 is correct

Question 27. Four resistors 100Ω, 200Ω, 300Ω, and 400Ω are connected to form four sides of a square. The resistors can be connected in any; order. What is the maximum possible equivalent resistance across the diagonal of the square?

1. 210Ω
2. 240Ω
3. 300Ω
4. 250Ω

Answer: 4. 250Ω

The resistance across the diagonal of the square formed by the four resistors is equal to the equivalent resistance of the resistors in parallel combination.

The equivalent resistance of the parallel combination is maximum when the resistance on the two sides of the diagonal is equal.

∴ Equivalent resistance,

⇒ \(R=\frac{500 \times 500}{500+500}=250 \Omega\)

The option 4 is correct.

Class 12 Physics Ohm’s Law MCQs  Question 28. What will be the current through the 200Ω resistor in the given circuit a long time after the switch K is made on?

1. 0
2. 100mA
3. 10mA
4. 1mA

Answer: 3. 10mA

A long time after the switch K is turned on, the 1μF and 2μF capacitors will be open-circuited and no current will flow through them.

∴ Current through, the 200Ω resistor

⇒ \(\frac{6}{200+400}=\frac{1}{100} \mathrm{~A}\)

= 10mA

The option 3 is correct.

Question 29. When the 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 x 10^-4 m s^-1. If the electron density in the wire is 8 x 10²⁸ m^-3, the resistivity of the material is close to

1. 1.6 x 10^-8Ω.m
2. 1.6 x 10^-7Ω.m
3. 1.6 x 10^-6Ω.m
4. 1.6 x 10^-5Ω.m

Answer: 4. 1.6 x 10^-5Ω.m

Drift velocity, \(v=\frac{I}{n e A}\)

⇒ \(I=\frac{V}{R}=\frac{V}{\rho \frac{l}{A}} \quad \text { or, } \rho=\frac{V}{\frac{I}{A} l}=\frac{V}{n e v l}\)

∴ \(\rho=\frac{5}{\left(8 \times 10^{28}\right) \times\left(1.6 \times 10^{-19}\right) \times\left(2.5 \times 10^{-4}\right) \times 0.1}\)

= 1.56 X 10^-5Ω.m

= 1.6 X 10^-5Ω m

The option 4 is correct.

Question 30. The temperature dependence of resistances of Cu and undoped Si in the temperature range of 300-400 K, is best described by

1. Linear increase for Cu, linear increase for Si
2. Linear increase for Cu, exponential increase for Si
3. Linear increase for Cu, exponential decrease for Si
4. Linear decrease for Cu, linear decrease for Si

Answer: 3. Linear increase for Cu, exponential increase for Si

Question In the given circuit the current in each resistance is

1. 1A
2. 0.25A
3. 0.5A
4. zero

Answer: 4. zero

There are two cells of equal electromotive force in opposite directions with each other in each loop. So, the electromotive force in each loop is zero. Hence, the current is also zero.

Option 4 is correct

Class 12 Physics Ohm’s Law MCQs  Question 31. In the given circuit diagram when the current reaches a steady state in the circuit, the charge on the capacitor of capacitance C will be

1. CE
2. \(\frac{C E r_1}{r_2+r}\)
3. \(\frac{C E r_2}{r+r_2}\)
4. \(\frac{C E r_1}{r_1+r}\)

Answer: 3. \(\frac{C E r_2}{r+r_2}\)

The current reaches a steady state in the circuit means that the current through the capacitor is zero

∴ Current in the circuit, \(i=\frac{E}{r+r_2}\)

Potential differences across the capacitor,

⇒ \(V_{A D}=V_{A B}=i r_2=\frac{E r_2}{r+r_2}\)

Therefore, the charge storedin the capacitor C,

⇒ \(Q=C V_{A D}=C E \frac{r_2}{r+r_2}\)

The option 3 is correct

Question 32. A, B, and C are voltmeters of resistance R, 1.5R, and 3R respectively. When some potential difference is applied between X and Y, the voltmeter readings are V_A, V_B,  and V_C respectively. Then

1. V_A=V_B=V_C
2. V_A ≠ V_B = V_C
3. V_A = V_B ≠ V_C
4. V_A ≠ V_B ≠ V_C

Answer: 1. V_A=V_B=V_C

Clearly, V_B = V_C

Again, equivalent resistance for B and C = \(\frac{1.5 \times 3}{1.5+3}=1 \Omega\)

= resistance of A.

Hence, V_A = V_B = V_C

The option 1 is correct.

Question 33. Across a metallic conductor of a non-uniform cross-section, a constant potential difference is applied. The quantity which remains constant along the conductor is

1. Current density
2. Current
3. Drift velocity
4. Electric field

Current remains constant along the conductor.

The option 2 is correct.

Class 12 Physics Ohm’s Law MCQs  Question 34. In the electrical circuit, the current I through the side AB is

1. \(\frac{6}{2}\)A
2. \(\frac{10}{33}\)A
3. \(\frac{1}{5}\)A
4. \(\frac{10}{63}\)A

Answer: 1. \(\frac{6}{2}\)A

E=IR

or, \(10=I\left(10+\frac{20 \times 30}{20+30}+3\right)=0.25 I\)

Here, \(I=\frac{10}{25}=\frac{2}{5} \mathrm{~A}\)

∴ \(I_{A B}=\frac{2}{5} \times \frac{30}{20+30}=\frac{6}{25} \mathrm{~A}\)

The option 1 is correct

Question 35. A cell of emf E and internal resistance r is connected to a variable external resistor R. The graph which gives the terminal voltage of cell V with respect to R is

Answer: 2.

⇒ \(V=E-I r=E-\frac{E r}{R+r}\)

or, \(\frac{d V}{d R}=\frac{E r}{(R+r)^2}\)

Therefore, the slope of the V-R graph is positive and it decreases within resistance R

The option 2 is correct.

Question 36. A carbon resistor of (47 ± 4.7)kΩ is to be marked with rings of different colors for its identification. The color code sequence will be

1. Yellow—Green—Violet—Gold
2. Yellow—Violet—Orange—Silver
3. Violet—Yellow—Orange—Silver
4. Green—Orange—Violet—Gold

Answer: 2. Yellow—Violet—Orange—Silver

⇒ \((47 \pm 4.7)=47 \pm\left(\frac{4.7}{47} \times 100\right) \%=47 \pm 10 \%\)

∴ \((47 \pm 4.7) \mathrm{k} \Omega=47 \times 10^3 \pm 10 \% \Omega\)

Therefore, the color code sequence will be yellow—Violet—Orange—Silver

The option 2 is correct

Class 12 Physics Ohm’s Law MCQs  Question 37. A set of n equal resistors, of value R each, are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then the current drawn from the battery becomes 10I. The value of n is

1. 20
2. 11
3. 10
4. 9

Answer: 3. 10

Equivalent resistance in series,

R₁ = nR + R = (n + 1)R

∴ \(I=\frac{E}{R_1}=\frac{E}{(n+1) R}\)…..(1)

Equivalent resistance in parallel

⇒ \(R_2=\left(\frac{R}{n}+R\right)=\left(1+\frac{1}{n}\right) R\)

∴ \(10 I=\frac{E}{R_2}=\frac{E}{\left(1+\frac{1}{n}\right) R}\)…..(2)

From equations (1) and (2)

⇒ \(\frac{10 I}{I}=\frac{(n+1) R}{\left(1+\frac{1}{n}\right) R} \text { or, } 10\left(\frac{n+1}{n}\right)=(n+1)\)

or, n = 10

The option 3 is correct.

Question 38. A battery consists of a variable number n of identical cells (having internal resistance r each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?

Answer: 3.

Current in the circuit containing n identical cells connected in series,

I = \(\frac{ne}{nr}\) [ e = emf of each identical cell]

or, I = \(\frac{e}{r}\)

So, I remain the same with any change of n.

The option is correct.

Question 1. The Q -the value of a nuclear reaction A + b → C + d is defined by Q = [m_A + m_b-m_C-m_d]c² , where the masses refer to nuclear rest masses. Determine from the given data whether the following reactions are exothermic or endothermic

1. \({ }_1^1 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H}\)

2. \({ }_6^{12} \mathrm{C}+{ }_6^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_2^4 \mathrm{He}\)

Atomic masses are given to be

\(m\left({ }_1^1 \mathrm{H}\right)=1.007825 \mathrm{u}, m\left({ }_1^2 \mathrm{H}\right)=2.014102 \mathrm{u}\)

\(m\left({ }_1^3 \mathrm{H}\right)=3.016049 \mathrm{u}, m\left({ }_1^{12} \mathrm{C}\right)=12.000000 \mathrm{u}\)

\(m\left({ }_{10}^{20} \mathrm{Ne}\right)=19.992439 \mathrm{u}, \quad m\left({ }_2^4 \mathrm{He}\right)=4.002603 \mathrm{u}\)

Answer:

1. \({ }_1^1 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H}\)

Q= \(m_{\mathrm{N}}\left({}_1^1\mathrm{H}\right)+m_{\mathrm{N}}\left({ }_1^3 \mathrm{H}\right)m_{\mathrm{N}}\left({ }_1^2 \mathrm{H}\right)-m_{\mathrm{N}}\left({ }_1^2 \mathrm{H}\right)\)

= \(m\left({ }_1^1 \mathrm{H}\right)-m_e+m\left({ }_1^3 \mathrm{H}\right)-m_e-m\left({ }_1^2 \mathrm{H}\right)\) + \(m_em\left({}_1^2\mathrm{H}\right)+m_e\)

= \(m\left({ }_1^1 \mathrm{H}\right)+m\left({ }_1^3 \mathrm{H}\right)-m\left({ }_1^2 \mathrm{H}\right)-m\left({ }_1^2 \mathrm{H}\right)\)

1.007825 + 3.016049-2 × 2.014102

= -0.00433 ×  931.5 MeV = -4.03 MeV

∴ Q < 0, the reaction is endothermic

2. \({ }_6^{12} \mathrm{C}+{ }_6^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_2^4 \mathrm{He}\)

Q = \(2 m_{\mathrm{N}}\left({ }_6^{12} \mathrm{C}\right)-m_{\mathrm{N}}\left({ }_{10}^{20} \mathrm{Ne}\right)-m_{\mathrm{N}}\left({ }_2^4 \mathrm{He}\right)\)

= \(=2 m\left({ }_6^{12} \mathrm{C}\right)-12 m_e-m\left({ }_{10}^{20} \mathrm{Ne}\right)+10 m_e\) \(-m\left({ }_2^4 \mathrm{He}\right)+2 m_e\)

= [2 × 12.000000-19.992439-4.002603] × 931.5 MeV

= 0.004958 × 931.5 MeV = 4.62 MeV

Q > 0; so, the reaction is exothermic.

Class 12 Physics Atomic Nucleus Question And Answers

Question 2. Is the fission of  ⁵⁶Fe₂₆ nucleus into two equal fragments,  ²⁸Al₁₃ energetically possible? Argue by working out the Q of the process. Given, m ( ⁵⁶Fe₂₆ ) = 55.93494 u and m(²⁸Al₁₃) = 27.98191 u
Answer:

If possible, let the reaction be \({ }_{26}^{56} \mathrm{Fe} \rightarrow{ }_{13}^{28} \mathrm{Al}+{ }_{13}^{28} \mathrm{Al}\)

Q -value of the process = \(m\left({ }_{26}^{56} \mathrm{Fe}\right)-2 m\left({ }_{13}^{28} \mathrm{Al}\right)\)

55.934944-2 × 27.98191

= -0.02888 × 931.5 MeV

= -26.90 MeV

Since the Q -value is negative, fission is not possible

Question 3. The fission properties of ²³⁹Pu₉₄ are very similar to those of ²³⁵U₉₂ The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in ²³⁹Pu₉₄1 kg of pure undergo fission?
Answer:

Number of nuclei in 1 kg of ²³⁹Pu₉₄

= \(\frac{6.023 \times 10^{23}}{239} \times 1000.0\)

= 2.52 × 10 ²⁴

The energy released per fission = 180 MeV

Total energy released = 2.52 x 10²⁴ × 180 MeV

= 4.54 × 10²⁶ MeV

Class 12 Physics Atomic Nucleus Question And Answers

Question 4. A 1000 MW fission reactor consumes half of its fuel in 5y. How much ²³⁵U₉₂ did it contain initially? Assume that the reactor was active 80% of the time and all the energy generated arises from the fission of ²³⁵U₉₂ and that this nuclide is consumed by the fission process.
Answer:

Energy generated per gram of ²³⁵U₉₂

= \(\frac{6.023 \times 10^{23}}{235} \times 200 \times 1.6 \times 10^{-13} \mathrm{~J} \cdot \mathrm{g}^{-1}\)

Energy generated in 5 y

Power × Time × 80%

(1000 × 10⁶) × (5 × 365 × 24 × 60 × 60) × 80% J

Amount of spent

= \(\frac{\left(1000 \times 10^6\right) \times(5 \times 365 \times 24 \times 60 \times 60)}{6.023 \times 10^{23} \times 200 \times 1.6 \times 10^{-13}} \mathrm{x} \times 235\)

= 1538 kg

Initial mass of ²³⁵U₉₂  = 2 × 1538 kg = 3076 kg

Question 5. How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction can be taken as = \({ }_1^2 \mathrm{H}+{ }_1^2\mathrm{H}\rightarrow{}_2^3\mathrm{He}+\mathrm{n}+3.2 \mathrm{MeV}\)
Answer:

Number of nuclei in 2 kg of ²H₁

= \(\frac{6.023 \times 10^{23} \times 2000}{2}=6.023 \times 10^{26}\)

Energy generated by the fusion of these nuclei

E = \(\frac{3.2 \times 6.023 \times 10^{26}}{2} \mathrm{MeV}\)

Power of the bulb = 100 W

Let the bulb = 100 W

Energy spent = \(\frac{100 \times t}{1.6 \times 10^{-13}} \mathrm{MeV}\)

∴ \(\frac{100 t}{1.6 \times 10^{-13}}=\frac{3.2 \times 6.023}{2} \times 10^{23}\)

t = \(\frac{3.2 \times 6.023 \times 10^{23} \times 1.6 \times 10^{-13}}{2 \times 100}\)

– 1.54 × 10 12 s

= 4.89 × 104 y

Class 12 Physics Atomic Nucleus Question And Answers

Question 6. A source contains two phosphorus radionuclides  ³²P₁₅ (T_1/2 = 14.3 d) and ³³P₁₅ (T_1/2 = 25.3 d). Initially, 10% of the decay comes from ³³P₁₅. How long one must wait until 90% do so. 33 32
Answer:

Let i R₀₁ and R₀₂ be the initial activities of ³³P₁₅ and ³²P₁₅ respectively and R₁ and R₂ be their activities at any instant t. According to the first observation

R₀₁ = 10% (R₀₁ +R₀₂)

R₀₂ =  9 R₀₁ ……………………………. (1)

Again, R₁ = 90% (R₁ + R₂)

Or, \(\frac{R_2}{R_{02}}=\frac{1}{81} \frac{R_1}{R_{01}}\) ……………………………. (1)

Combining equation (1) and (2)

⇒ \(\frac{R_2}{R_{02}}=\frac{1}{81} \frac{R_1}{R_{01}}\)

Or, \(\frac{R_{02} e^{-\lambda_2 t}}{R_{02}}=\frac{1}{81} \times \frac{R_{01} e^{-\lambda_1 t}}{R_{01}}\)

Or, \(81 e^{-\lambda_2 t}=e^{-\lambda_1 t}\)

⇒ \(\left(\lambda_2-\lambda_1\right) t=\dot{2} .303 \log 81\)

t = \(\frac{2.303 \log 81}{\frac{0.693}{14.3}-\frac{0.693}{25.3}}\)

Since Or, λ = 0.693T_½

= 208 . 5 d

Question 7. Under certain circumstances, a nucleus can decay by = 231.1 MeV emitting a particle more massive than an a -particle. Consider the following decay processes

1. \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_6^{14} \mathrm{C}\)

2. \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_2^4 \mathrm{He}\)

Calculate the Q -values for these two decays and determine that both are energetically possible. 

m( ²²³Ra₈₈ ) = 223.01850 u, m( ²⁰⁹Ra₈₂ ) = 208.98107 u,

m(²¹⁹Ra₈₆ ) = 219.00948 u, m(¹⁴C₆) = 14.00324 u and m(⁴He₂) = 4.00260 u

Answer:

1. \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_6^{14} \mathrm{C}+Q\)

= \(\left[m_{\mathrm{N}}\left({ }_{88}^{223} \mathrm{Ra}\right)-m_{\mathrm{N}}\left({ }_{82}^{209} \mathrm{~Pb}\right)-m_{\mathrm{N}}\left({ }_6^{14} \mathrm{C}\right)\right]\) x 931.2 MeV

= \(\left[m_{88}^{223} \mathrm{Ra}-m\left({ }_{82}^{209} \mathrm{~Pb}\right)-m\left({ }_6^{14} \mathrm{C}\right)\right] \times 931.2 \mathrm{MeV}\)

31. 8 MeV

∴ Q > 0: so, the decay is possible

2. \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_2^4 \mathrm{He}+Q^{\prime}\)

Q’ = 5.98 MeV [by similar calculation as above]

∴ Q’ > 0; so, this decay is also possible.

Class 12 Physics Atomic Nucleus Question And Answers

Question 8. Consider the fission of ²³⁸U₉₂ by fast neutrons. In one fission event, no neutrons are emitted and the final stable end products, after the beta decay of the primary fragments, are¹⁴⁰Ce₅₈ and ⁹⁹Ru₄₄ Calculate Q for this fission process
Given

\(m\left({ }_{92}^{238} \mathrm{U}\right)=238.05079 \mathrm{u}, m\left({ }_{58}^{140} \mathrm{Ce}\right)=139.90543 \mathrm{u}\)

\(m\left({ }_{44}^{99} \dot{\mathrm{Ru}}\right)=98.90594 \mathrm{u}, \dot{m}_n=1.008667 \mathrm{u}\)

Answer:

The fission reaction is

⇒ \({ }_{92}^{238} \mathrm{U}+{ }_0^1 \mathrm{n}{\beta}{ }_{58}^{140} \mathrm{Ce}+{ }_{44}^{99}\mathrm{Ru}+Q\)

∴ Q = \(\left[m\left({}_{92}^{238}\mathrm{U}\right)+m\left({ }_0^1\mathrm{n}\right)m\left({}_{58}^{140}\mathrm{Ce}\right)-m\left({ }_{44}^{99} \mathrm{Ru}\right)\right] \mathrm{u}\)

= \([238.05079+1.00867-139.90543-98.90594]\)

= 23 1.1 MeV

Question 9. Suppose India had a target of producing by 2020 AD, 200, 000 MW of electric power, 10% of which was to be obtained from nuclear power plants. Suppose we are given that, on average, the efficiency of utilization (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium did our country need per year by 2020? Take the heat per fission of ²³⁵U₉₂ to be about 200 MeV. Avogadro’s number = 6.023 × 10 mol^-1. Nuclear power target = 10% of total generation
Answer:

Nuclear power target = 10% of total generation

= 10% × 2 × 10¹¹ W

= 2 × 10¹⁰ W

Efficiency, η = 25%

∴ Total power generated by the nuclear reactor

= \(\frac{2 \times 10^{10}}{25 \%} \mathrm{~W}=8 \times 10^{10} \mathrm{~W}\)

∴ Generated heat by the reactor in 2020

H = 8 × 10¹⁰ × 366 × 24 × 60 ×60 J

∴ Number of fission required for generation of this heat,

N = \(\frac{H}{200 \times 1.6 \times 10^{-13}}\)

If m g of  ²³⁵U₉₂ contains this number of nuclei, then,

m = \(\frac{N \times 235}{6.023 \times 10^{23}} \mathrm{~g}\)

= \(\frac{8 \times 10^{10} \times 366 \times 24 \times 60 \times 60 \times 235}{200 \times 1.6 \times 10^{-13} \times 6.023 \times 10^{23}} \mathrm{~g}\)

= 3.084 × 10⁴  kg

Class 12 Physics Atomic Nucleus Question And Answers

Question 10. Calculate and compare the energy released by

1. Fusion of: 1.0 kg of hydrogen deep within the sun and
2. The fission oqc of 1.0 kg of ²³⁵U in a fission reactor.

Answer: 

1. Equation of fusion reaction,

4 ¹H → ⁴He₂ + 2 ⁰e₊₁ + 26 MeV

∴ 26 MeV of energy is released on a combination of four H nuclei.

Number of nuclei in 1 kg of hydrogen

= 6.023 × 1023 × 1000

Energy released,

E_H = 6.023  × 1026 ×  26 MeV

2. Fission of one  nucleus releases 200 MeV of energy

Number of nuclei in 1 kg of ²³⁵U₉₂ = \(\frac{6.023 \times 10^{23} \times 1000}{235}\)

∴ Energy released,

E_U = \(\frac{6.023 \times 10^{23} \times 1000 \times 20}{235}\)

= 5.12 × 10²⁶ MeV

∴ \(\frac{E_{\mathrm{H}}}{E_{\mathrm{U}}}=\frac{3.913 \times 10^{27}}{5.12 \times 10^{26}}\)

= 7.6

Question 11. The half-life of a radioactive substance is 30 days. The number of atoms in the substance is 10¹². How many disintegrations of atoms per second does occur?
Answer:

T = 30 d =30 × 24 × 60 × 60 s

⇒ \(\lambda=\frac{0.693}{T}\)

= \(=\frac{0.693}{30 \times 24 \times 60 \times 60}\)

= 2.67 × 10^-7(approx)

If t = 1 s, t = 2.67 × 10^-7

Hence, e^λt  = 1.000000267(approx)

N = N₀e^λt  = \(\)

= 9.99999733 × 10¹¹ (approx)

∴ Number of atoms disintegrated per second

= N₀ – N = 2.67 ×  10 5 (approx)

Question 12. Draw a plot of the potential energy of a pair of nucleons as a function of their separations. Mark the region where the nuclear force is (1) attractive and (il) repulsive. Write any two characteristic features of nuclear forces
Answer:

The required plot

In region AB, nuclear force is attractive.

The nuclear force is not repulsive. The repulsive force corresponding to the region DF is the repulsive coulomb force between protons

Class 12 Physics Atomic Nucleus Question And Answers

Question 13.

1. Draw the Plot of binding energy per nucleon (B.E./A ) as a function of mass number A . Write two important conclusions that can be drawn regarding the nature of
   nuclear force
   
2. Use this graph to explain the release of energy in both the processes of nuclear fusion and fission.
3. Write the basic nuclear process of neutrons undergoing β-decay. Why is the detection of neutrinos found very difficult?

Answer:

1.

1. Two conclusions from the plot: the nuclear force is
2. Short-range and
3. Charge independent

2. Mass of a nucleus (M) = mass of its nucleons- binding energy (B)

So, M decreases with an increase in B.

Now, we consider nuclear fission: 1 → 2 + 3.

From mass-energy equivalence,

M₁ = M₂ + M₃+ energy release (Q)

or, Q = M₁ – M₂ – M₃

Q is positive, i.e., energy is released if

M₁ > M₂ – M₃,i.e., B₁ < B₂ – BM₃

or A₁e₁<A₂e₂ + A₃e₃

Where A₁A₂, A₃ are mass numbers A₁ = A₂ + A₃) and e₁ , e₂ , e₃ are binding energy per nucleon. From the plot, we see that this condition is satisfied for high A₁, where both e₂ and e₃ are higher than e₁ of the large nucleus 1.

The fission of a large nucleus releases energy. On the other hand, for low A nuclei, e_2, and e₃ will be less than the e₁ of the larger nucleus 1. So, energy will be released rather in the opposite process: 2 + 3 → 1. Therefore, a fusion of small nuclei releases energy

3. It is very difficult to detect neutrinos or antineutrinos experimentally because they have neither any charge nor any mass.

Question 14. Define the activity of a radioactive sample. Write its SI unit. A radioactive sample has the activity of 10000 disintegrations per second (DPS) after 20 hours. After the next 10 hours its activity reduces to 5000 dps. Find out its half-life and initial

The activity of a radioactive sample is defined as the rate of disintegration of the sample. It is also called the count rate. Its SI unit is becquerel (Bq).

1 Bq = 1 decays/s

We know A = A₈ e^-λt

5000 = 10000e^-λt

Or, e^λt = \(\frac{10000}{5000}\)

= 2

Taking logs on both sides we get

λt = log 2

Or, λ = \(\frac{\log 2}{t}\)

Now half-life T = \(=\frac{0.693}{\lambda}=\frac{0.693 \times t}{\log 2}=\frac{0.693 \times 600}{\log 2}\)

= 1381. 26 min 23 h

In this table, the last two values are the given values. Prom these values, the first two values have been calculated,

Class 12 Physics Atomic Nucleus Question And Answers

Question 15.

1. A radioactive nucleus ‘A ‘ undergoes a series of decays as given below:

The mass number and atomic number of A₂ are 176 and 71 respectively. Determine the mass and atomic numbers of A4 and A.

2. Write the basic nuclear processes underlying α and β decays

Answer:

1. When β^– decay occurs

When β⁺  decay occurs

In the case of β^–   decay, the mass number of A is 180 and its atomic number is 72. In the case of β⁺ decay, the mass number of A is 100 and its atomic number is 74. In both cases, the mass number of A₄   Is 172 and its atomic number is 69.

Question 16. Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A

Nuclear fission:

The binding energy per nucleon for heavier nuclei is approximately 7.6 MeV, but for lighter nuclei, it is roughly 8.4 MeV. The heavier nuclei exhibit reduced stability, leading to the fission of the heavier nucleus into lighter nuclei, which therefore releases energy. This process is referred to as nuclear fission.

Nuclear fusion:

The binding energy per nucleon for nuclei with mass number A < 12 is minimal, rendering them less stable. Consequently, two such nuclei can amalgamate to create a somewhat heavier nucleus, which possesses a greater binding energy per nucleon.

Consequently, energy is liberated in this process, referred to as nuclear fusion.

Class 12 Physics Atomic Nucleus Question And Answers

Question 17. A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to 3.125%?
Answer:

Activity A = λN

Given A = 3. 125 % of \(\frac{3.125 A_0}{100}=\frac{A_0}{32}\)

Or, \(\frac{A}{A_0}=\left(\frac{1}{2}\right)^5\)

∴ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^5\)

So the activity will reduce to 3.125% after 5 half-lives. Hence required time =5 × 10 = 50 years