Explore topic-wise InterviewSolutions in .

NCERT Solutions Class 11 Maths Chapter 14 Mathematical Reasoning

1. Which of the following sentences are statements? Give reasons for your answer.

(i) There are 35 days in a month.

(ii) Mathematics is difficult.

(iii) The sum of 5 and 7 is greater than 10.

(iv) The square of a number is an even number.

(v) The sides of a quadrilateral have equal length.

(vi) Answer this question.

(vii) The product of (–1) and 8 is 8.

(viii) The sum of all interior angles of a triangle is 180°.

(ix) Today is a windy day. (x) All real numbers are complex numbers

Answer :

(i) This sentence is incorrect because the maximum number of days in a month is 31. Hence, it is a statement.

(ii) This sentence is subjective in the sense that for some people, mathematics can be easy and for some others, it can be difficult. Hence, it is not a statement.

(iii) The sum of 5 and 7 is 12, which is greater than 10. Therefore, this sentence is always correct. Hence, it is a statement.

(iv) This sentence is sometimes correct and sometimes incorrect. For example, the square of 2 is an even number. However, the square of 3 is an odd number. Hence, it is not a statement.

(v) This sentence is sometimes correct and sometimes incorrect. For example, squares and rhombus have sides of equal lengths. However, trapezium and rectangles have sides of unequal lengths. Hence, it is not a statement.

(vi) It is an order. Therefore, it is not a statement.

(vii) The product of (–1) and 8 is (–8). Therefore, the given sentence is incorrect. Hence, it is a statement.

(viii) This sentence is correct and hence, it is a statement.

(ix) The day that is being referred to is not evident from the sentence. Hence, it is not a statement.

(x) All real numbers can be written as a × 1 + 0 × i. Therefore, the given sentence is always correct. Hence, it is a statement.

2. Give three examples of sentences which are not statements. Give reasons for the answers.

Answer :

The three examples of sentences, which are not statements, are as follows.

(i) He is a doctor.

It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.

(ii) Geometry is difficult. 

This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.

(iii) Where is she going?

This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.

3. Write the negation of the following statements:

(i) Chennai is the capital of Tamil Nadu.

(ii) √2 is not a complex number.

(iii) All triangles are not equilateral triangle.

(iv)The number 2 is greater than 7.

(v)Every natural number is an integer.

Answer :

(i) Chennai is not the capital of Tamil Nadu.

(ii) √2 is a complex number.

(iii) All triangles are equilateral triangles.

(iv) The number 2 is not greater than 7.

(v) (v) Every natural number is not an integer.

4. Are the following pairs of statements negations of each other?

(i) The number x is not a rational number.

The number x is not an irrational number.

(ii) The number x is a rational number.

The number x is an irrational number.

Answer :

(i) The negation of the first statement is “the number x is a rational number”. This is same as the second statement. This is because if a number is not an irrational number, then it is a rational number.

Therefore, the given statements are negations of each other.

(ii) The negation of the first statement is “the number x is not a rational number”. This means that the number x is an irrational number, which is the same as the second statement. Therefore, the given statements are negations of each other.

5. Find the component statements of the following compound statements and check whether they are true or false.

(i) Number 3 is prime or it is odd.

(ii) All integers are positive or negative.

(iii) 100 is divisible by 3, 11 and 5.

Answer :

(i) The component statements are as follows.

p: Number 3 is prime. q: Number 3 is odd. Both the statements are true.

(ii) The component statements are as follows. p: All integers are positive. q: All integers are negative. Both the statements are false.

(iii) The component statements are as follows.

p: 100 is divisible by 3. q: 100 is divisible by 11. r: 100 is divisible by 5.

Here, the statements, p and q, are false and statement r is true.

6. For each of the following compound statements first identify the connecting words and then break it into component statements.

(i) All rational numbers are real and all real numbers are not complex.

(ii) Square of an integer is positive or negative.

(iii) The sand heats up quickly in the Sun and does not cool down fast at night.

(iv) x = 2 and x = 3 are the roots of the equation 3x² – x – 10 = 0.

Answer :

(i) Here, the connecting word is ‘and’.The component statements are as follows.

p: All rational numbers are real.

q: All real numbers are not complex.

(ii) Here, the connecting word is ‘or’.

The component statements are as follows.

p: Square of an integer is positive.

q: Square of an integer is negative.

(iii) Here, the connecting word is ‘and’. The component statements are as follows.

p: The sand heats up quickly in the sun.

q: The sand does not cool down fast at night.

(iv) Here, the connecting word is ‘and’.

The component statements are as follows.

p: x = 2 is a root of the equation 3x² – x – 10 = 0

q: x = 3 is a root of the equation 3x² – x – 10 = 0

7. Identify the quantifier in the following statements and write the negation of the statements.

(i) There exists a number which is equal to its square.

(ii) For every real number x, x is less than x + 1.

(iii) There exists a capital for every state in India.

Answer :

(i) The quantifier is “There exists”.

The negation of this statement is as follows.

There does not exist a number which is equal to its square.

(ii) The quantifier is “For every”.

The negation of this statement is as follows. 

There exist a real number x such that x is not less than x + 1.

(iii) The quantifier is “There exists”.

The negation of this statement is as follows.

There exists a state in India which does not have a capital.

8. Check whether the following pair of statements is negation of each other. Give reasons for the answer.

(i) x + y = y + x is true for every real numbers x and y.

(ii) There exists real number x and y for which x + y = y + x.

Answer :

The negation of statement (i) is as follows.

There exists real number x and y for which x + y ≠ y + x. This is not the same as statement (ii). Thus, the given statements are not the negation of each other.

9. State whether the “Or” used in the following statements is “exclusive “or” inclusive. Give reasons for your answer.

(i) Sun rises or Moon sets.

(ii) To apply for a driving license, you should have a ration card or a passport.

(iii) All integers are positive or negative.

Answer :

(i) Here, “or” is exclusive because it is not possible for the Sun to rise and the moon to set together.

(ii) Here, “or” is inclusive since a person can have both a ration card and a passport to apply for a driving license.

(iii) Here, “or” is exclusive because all integers cannot be both positive and negative.

10. Rewrite the following statement with “if-then” in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd.

Answer :

The given statement can be written in five different ways as follows.

(i) A natural number is odd implies that its square is odd.

(ii) A natural number is odd only if its square is odd.

(iii) For a natural number to be odd, it is necessary that its square is odd.

(iv) For the square of a natural number to be odd, it is sufficient that the number is odd.

(v) If the square of a natural number is not odd, then the natural number is not odd.

11. Write the contrapositive and converse of the following statements.

(i) If x is a prime number, then x is odd.

(ii) It the two lines are parallel, then they do not intersect in the same plane.

(iii) Something is cold implies that it has low temperature.

(iv) You cannot comprehend geometry if you do not know how to reason deductively.

(v) x is an even number implies that x is divisible by 4

Answer :

(i) The contrapositive is as follows.

If a number x is not odd, then x is not a prime number.

The converse is as follows.

If a number x is odd, then it is a prime number.

(ii) The contrapositive is as follows.

If two lines intersect in the same plane, then they are not parallel.

The converse is as follows.

If two lines do not intersect in the same plane, then they are parallel.

(iii) The contrapositive is as follows. If something does not have low temperature, then it is not cold. The converse is as follows. If something is at low temperature, then it is cold.

(iv) The contrapositive is as follows. If you know how to reason deductively, then you can comprehend geometry. The converse is as follows. If you do not know how to reason deductively, then you cannot comprehend geometry.

(v) The given statement can be written as follows. If x is an even number, then x is divisible by 4. The contrapositive is as follows. If x is not divisible by 4, then x is not an even number. The converse is as follows. If x is divisible by 4, then x is an even number.

12. Write each of the following statement in the form “if-then”.

(i) You get a job implies that your credentials are good.

(ii) The Banana trees will bloom if it stays warm for a month.

(iii) A quadrilateral is a parallelogram if its diagonals bisect each other.

(iv) To get A⁺ in the class, it is necessary that you do the exercises of the book.

Answer :

(i) If you get a job, then your credentials are good.

(ii) If the Banana tree stays warm for a month, then it will bloom.

(iii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

(iv) If you want to get an A⁺ in the class, then you do all the exercises of the book.

13. Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other.

(a) If you live in Delhi, then you have winter clothes.

(i) If you do not have winter clothes, then you do not live in Delhi.

(ii) If you have winter clothes, then you live in Delhi.

(b) If a quadrilateral is a parallelogram, then its diagonals bisect each other.

(i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.

(ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Answer :

(a) (i) This is the contrapositive of the given statement (a).

(ii) This is the converse of the given statement (a).

(b) (i) This is the contrapositive of the given statement (b).

(ii) This is the converse of the given statement (b).

14. Show that the statement p: “If x is a real number such that x 3 + 4x = 0, then x is 0” is true by

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

Answer :

p: “If x is a real number such that x³ + 4x = 0, then x is 0”.

Let q: x is a real number such that x³ + 4x = 0 r: x is 0.

(i) To show that statement p is true, we assume that q is true and then show that r is true.

Therefore, let statement q be true.

∴ x³ + 4x = 0 x (x² + 4) = 0

⇒ x = 0 or x²+ 4 = 0

However, since x is real, it is 0.

Thus, statement r is true.

Therefore, the given statement is true.

(ii) To show statement p to be true by contradiction, we assume that p is not true.

Let x be a real number such that x³ + 4x = 0 and let x is not 0.

Therefore, x³ + 4x = 0 

x (x² + 4) = 0 

x = 0 or x² + 4 = 0 

x = 0 or x² = – 4

However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0.

Thus, the given statement p is true.

(iii) To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false.

Here, r is false implies that it is required to consider the negation of statement r.

This obtains the following statement.

∼r: x is not 0.

It can be seen that (x² + 4) will always be positive.

x ≠ 0 implies that the product of any positive real number with x is not zero.

Let us consider the product of x with (x² + 4). 

∴ x (x² + 4) ≠ 0

⇒ x³ + 4x ≠ 0

This shows that statement q is not true.

Thus, it has been proved that

∼r ⇒∼q

Therefore, the given statement p is true.

15. Show that the statement “For any real numbers a and b, a² = b² implies that a = b” is not true by giving a counter-example.

Answer :

The given statement can be written in the form of “if-then” as follows.

If a and b are real numbers such that a² = b² , then a = b.

Let p: a and b are real numbers such that a² = b² .

q: a = b

The given statement has to be proved false. For this purpose, it has to be proved that if p, then ∼q.

To show this, two real numbers, a and b, with a² = b² are required such that a ≠ b. Let a = 1 and b = –1 a² = (1)² = 1 and b² = (– 1)² = 1 

∴ a² = b²

However, a ≠ b

Thus, it can be concluded that the given statement is false.

16. Show that the following statement is true by the method of contrapositive. p: If x is an integer and x² is even, then x is also even.

Answer :

p: If x is an integer and x² is even, then x is also even.

Let q: x is an integer and x² is even. r: x is even.

To prove that p is true by contrapositive method, we assume that r is false, and prove that q is also false.

Let x is not even.

To prove that q is false, it has to be proved that x is not an integer or x² is not even.

x is not even implies that x² is also not even.

Therefore, statement q is false.

Thus, the given statement p is true.

17. By giving a counter example, show that the following statements are not true.

(i) p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.

(ii) q: The equation x² – 1 = 0 does not have a root lying between 0 and 2.

Answer :

(i) The given statement is of the form “if q then r”.

q: All the angles of a triangle are equal. r: The triangle is an obtuse-angled triangle.

The given statement p has to be proved false.

For this purpose, it has to be proved that if q, then ∼r.

To show this, angles of a triangle are required such that none of them is an obtuse angle.

It is known that the sum of all angles of a triangle is 180°. Therefore, if all the three angles are equal, then each of them is of measure 60°, which is not an obtuse angle. In an equilateral triangle, the measure of all angles is equal. However, the triangle is not an obtuse-angled triangle.

Thus, it can be concluded that the given statement p is false.

(ii) The given statement is as follows. q:

The equation x² – 1 = 0 does not have a root lying between 0 and 2.

This statement has to be proved false. To show this, a counter example is required.

Consider x² – 1 = 0 x² = 1 x = ± 1

One root of the equation x² – 1 = 0, i.e. the root x = 1, lies between 0 and 2.

Thus, the given statement is false.

18. Which of the following statements are true and which are false? In each case give a valid reason for saying so.

(i) p: Each radius of a circle is a chord of the circle.

(ii) q: The centre of a circle bisects each chord of the circle.

(iii) r: Circle is a particular case of an ellipse.

(iv) s: If x and y are integers such that x > y, then –x < –y.

(v) t: √11is a rational number.

Answer :

(i) The given statement p is false. According to the definition of chord, it should intersect the circle at two distinct points.

(ii) The given statement q is false. If the chord is not the diameter of the circle, then the centre will not bisect that chord. In other words, the centre of a circle only bisects the diameter, which is the chord of the circle.

(iii) The equation of an ellipse is, \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) 

If we put a = b = 1, then we obtain x² + y² = 1, which is an equation of a circle

Therefore, circle is a particular case of an ellipse.

Thus, statement r is true.

(iv) x > y

⇒ –x < –y (By a rule of inequality)

Thus, the given statement s is true.

(v) 11 is a prime number and we know that the square root of any prime number is an irrational number. Therefore, √11 is an irrational number.

Thus, the given statement t is false.

19. Write the negation of the following statements:

(i) p: For every positive real number x, the number x – 1 is also positive.

(ii) q: All cats scratch.

(iii) r: For every real number x, either x > 1 or x < 1.

(iv) s: There exists a number x such that 0 < x < 1.

Answer :

(i) The negation of statement p is as follows.

There exists a positive real number x, such that x – 1 is not positive.

(ii) The negation of statement q is as follows.

There exists a cat that does not scratch.

(iii) The negation of statement r is as follows. There exists a real number x, such that neither x > 1 nor x < 1.

(iv) The negation of statement s is as follows. There does not exist a number x, such that 0 < x < 1.

20. State the converse and contrapositive of each of the following statements: 

(i) p: A positive integer is prime only if it has no divisors other than 1 and itself. 

(ii)q: I go to a beach whenever it is a sunny day. 

(iii) r: If it is hot outside, then you feel thirsty.

Answer :

(i) Statement p can be written as follows. 

If a positive integer is prime, then it has no divisors other than 1 and itself.

The converse of the statement is as follows. 

If a positive integer has no divisors other than 1 and itself, then it is prime.

The contrapositive of the statement is as follows. If positive integer has divisors other than 1 and itself, then it is not prime.

(ii) The given statement can be written as follows.

If it is a sunny day, then I go to a beach. 

The converse of the statement is as follows.

If I go to a beach, then it is a sunny day. The contrapositive of the statement is as follows. If I do not go to a beach, then it is not a sunny day.

(iii) The converse of statement r is as follows.

If you feel thirsty, then it is hot outside.

The contrapositive of statement r is as follows.

If you do not feel thirsty, then it is not hot outside.

21. Write each of the statements in the form “if p, then q”.

(i) p: It is necessary to have a password to log on to the server.

(ii) q: There is traffic jam whenever it rains.

(iii) r: You can access the website only if you pay a subscription fee.

Answer :

(i) Statement p can be written as follows. If you log on to the server, then you have a password.

(ii) Statement q can be written as follows. If it rains, then there is a traffic jam.

(iii) Statement r can be written as follows. If you can access the website, then you pay a subscription fee.

22. Re write each of the following statements in the form “p if and only if q”.

(i) p: If you watch television, then your mind is free and if your mind is free, then you watch television.

(ii) q: For you to get an A grade, it is necessary and sufficient that you do all the homework regularly.

(iii) r: If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular.

Answer :

(i) You watch television if and only if your mind is free.

(ii) You get an A grade if and only if you do all the homework regularly.

(iii) A quadrilateral is equiangular if and only if it is a rectangle.

23. Given below are two statements p: 25 is a multiple of 5. q: 25 is a multiple of 8.

Write the compound statements connecting these two statements with “And” and “Or”.

In both cases check the validity of the compound statement.

Answer :

The compound statement with ‘And’ is “25 is a multiple of 5 and 8”. This is a false statement, since 25 is not a multiple of 8.

The compound statement with ‘Or’ is “25 is a multiple of 5 or 8”. This is a true statement, since 25 is not a multiple of 8 but it is a multiple of 5.

24. Check the validity of the statements given below by the method given against it.

(i) p: The sum of an irrational number and a rational number is irrational (by contradiction method).

(ii) q: If n is a real number with n > 3, then n² > 9 (by contradiction method).

Answer :

(i) The given statement is as follows.

p: the sum of an irrational number and a rational number is irrational.

Let us assume that the given statement, p, is false. That is, we assume that the sum of an irrational number and a rational number is rational.

Therefore \(\sqrt a+\frac bc=\frac de\), where √a is irrational and b, c, d, e are integers.

\(\frac de-\frac bc\) is a rational number and √a is an irrational number. 

This is a contradiction. Therefore, our assumption is wrong.

Therefore, the sum of an irrational number and a rational number is rational.

Thus, the given statement is true.

(ii) The given statement, q, is as follows.

If n is a real number with n > 3, then n² > 9.

Let us assume that n is a real number with n > 3, but n² > 9 is not true.

That is, n² < 9 Then, n > 3 and n is a real number. Squaring both the sides, we obtain n² > (3)² 

⇒ n² > 9, which is a contradiction, since we have assumed that n² < 9.

Thus, the given statement is true. That is, if n is a real number with n > 3, then n² > 9.

25. Write the following statement in five different ways, conveying the same meaning. p: If triangle is equiangular, then it is an obtuse angled triangle.

Answer :

The given statement can be written in five different ways as follows.

(i) A triangle is equiangular implies that it is an obtuse-angled triangle.

(ii) A triangle is equiangular only if it is an obtuse-angled triangle.

(iii) For a triangle to be equiangular, it is necessary that the triangle is an obtuse-angled triangle.

(iv) For a triangle to be an obtuse-angled triangle, it is sufficient that the triangle is equiangular.

(v) If a triangle is not an obtuse-angled triangle, then the triangle is not equiangular.

NCERT Solutions Class 12 Biology Chapter 4 Reproductive Health

1. What do you think is the significance of reproductive health in a society?

Answer:

Reproductive health is the total well-being in all aspects of reproduction. It includes physical, emotional, behavioural, and social well-being. Sexually transmitted diseases such as AIDS, gonorrhoea, etc. are transferred from one individual to another through sexual contact. It can also lead to unwanted pregnancies. Hence, it is necessary to create awareness among people, especially the youth, regarding various reproduction related aspects as the young individuals are the future of the country and they are most susceptible of acquiring sexually transmitted diseases. Creating awareness about the available birth control methods, sexually transmitted diseases and their preventive measures, and gender equality will help in bringing up a socially conscious healthy family. Spreading awareness regarding uncontrolled population growth and social evils among young individuals will help in building up a reproductively healthy society.

2. Suggest the aspects of reproductive health which need to be given special attention in the present scenario.

Answer:

Reproductive health is the total well-being in all aspects of reproduction. The aspects which have to be given special attention in the present scenarios are 

♦ Counselling and creating awareness among people, especially the youth, about various aspects of reproductive health, such as sexually transmitted diseases, available contraceptive methods, case of pregnant mothers, adolescence, etc. 

♦ Providing support and facilities such as medical assistance to people during pregnancy, STDs, abortions, contraceptives, infertility, etc. for building a reproductively healthy society

3. Is sex education necessary in schools? Why?

Answer:

Yes, 

Introduction of sex education in schools is necessary. It would provide right information to young individuals at the right time about various aspects of reproductive health such as reproductive organs, puberty, and adolescence related changes, safe sexual practices, sexually transmitted diseases, etc. The young individual or adolescents are more susceptible in acquiring various sexually transmitted diseases. Hence, providing information to them at the right time would help them to lead a reproductively healthy life and also protect them from the myths and misconceptions about various sex related issues.

4. Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.

Answer:

Yes, the reproductive health has tremendously improved in India in the last 50 years. The areas of improvement are as follows. 

♦ Massive child immunization programme, which has lead to a decrease in the infant mortality rate 

♦ Maternal and infant mortality rate, which has been decreased drastically due to better post natal care 

♦ Family planning, which has motivated people to have smaller families 

♦ Use of contraceptive, which has resulted in a decrease in the rate of sexually transmitted diseases and unwanted pregnancies

5. What are the suggested reasons for population explosion?

Answer:

The human population is increasing day by day, leading to population explosion. It is because of the following two major reasons. 

♦ Decreased death rate 

♦ Increased birth rate and longevity 

The death rate has decreased in the past 50 years. The factor leading to decreased death rate and increased birth rate are control of diseases, awareness and spread of education, improvement in medical facilities, ensured food supply in emergency situation, etc. All this has also resulted in an increase in the longevity of an individual.

6. Is the use of contraceptives justified? Give reasons.

Answer:

Yes, the use of contraceptives is absolutely justified. The human population is increasing tremendously. Therefore, to regulate the population growth by regulating reproduction has become a necessary demand in the present times. Various contraceptive devices have been devised to reduce unwanted pregnancies, which help in bringing down the increased birth rate and hence, in checking population explosion.

7. Removal of gonads cannot be considered as a contraceptive option. Why?

Answer:

Contraceptive devices are used to prevent unwanted pregnancy and to prevent the spreading of STDs. There are many methods, such as natural, barrier, oral, and surgical methods, that prevent unwanted pregnancy. However, the complete removal of gonads cannot be a contraceptive option because it will lead to infertility and unavailability of certain hormones that are required for normal functioning of accessory reproductive parts. Therefore, only those contraceptive methods can be used that prevent the chances of fertilization rather than making the person infertile forever.

8. Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.

Answer:

Amniocentesis is a pre-natal diagnostic technique that is used to determine the sex and metabolic disorders of the developing foetus in the mother’s uterus through the observation of the chromosomal patterns. This method was developed so as to determine any kind of genetic disorder present in the foetus. However, unfortunately, this technique is being misused to detect the sex of the child before birth and the female foetus is then aborted. Thus, to prevent the increasing female foeticides, it is necessary to ban the usage of amniocentesis technique for determining the sex of a child.

9. Suggest some methods to assist infertile couples to have children.

Answer:

Infertility is the inability of a couple to produce a baby even after unprotected intercourse. It might be due to abnormalities present in either male or female, or might be even both the partners. The techniques used to assist infertile couples to have children are as follows. 

♦ Test tube babies This involves in-vitro fertilization where the sperms meet the egg outside the body of a female. The zygote, hence produced, is then transferred in the uterus or fallopian tube of a normal female. The babies produced from this method are known as test tube babies. 

♦ Gamete Intra fallopian transfer (GIFT) It is a technique that involves the transfer of gamete (ovum) from a donor into the fallopian tube of the recipient female who is unable to produce eggs, but has the ability to conceive and can provide right conditions for the development of an embryo. 

♦ Intra Cytoplasmic sperm injection (ICSI) It is a method of injecting sperm directly into the ovum to form an embryo in laboratory.

♦ Artificial insemination Artificial insemination is a method of transferring semen (sperm) from a healthy male donor into the vagina or uterus of the recipient female. It is employed when the male partner is not able to inseminate the female or has low sperm counts.

10. What are the measures one has to take to prevent from contracting STDs?

Answer:

Sexually transmitted diseases (STDs) get transferred from one individual to the other through sexual contact. Adolescents and young adults are at the greatest risk of acquiring these sexually transmitted diseases. Hence, creating awareness among the adolescents regarding its after-effects can prevent them from contracting STDs. The use of contraceptives, such as condoms, etc. while intercourse, can prevent the transfer of these diseases. Also, sex with unknown partners or multiple partners should be avoided as they may have such diseases. Specialists should be consulted immediately in case of doubt so as to assure early detection and cure of the disease.

11. State True/False with explanation 

(a) Abortions could happen spontaneously too. (True/False) 

(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False) 

(c) Complete lactation could help as a natural method of contraception. (True/False) 

(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people. (True/False)

Answer:

(a) Abortions could happen spontaneously too. True 

(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner.  False 

Infertility is defined as the inability of the couple to produce baby even after unprotected coitus. It might occur due to abnormalities/defects in either male or female or both.

(c) Complete lactation could help as a natural method of contraception. False 

Complete lactation or lactational amenorrhea is a natural method of contraception. However, it is limited till lactation period, which continues till six months after parturition. 

(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people. True

12. Correct the following statements: 

(a) Surgical methods of contraception prevent gamete formation. 

(b) All sexually transmitted diseases are completely curable. 

(c) Oral pills are very popular contraceptives among the rural women. 

(d) In E. T. techniques, embryos are always transferred into the uterus.

Answer:

(a) Surgical methods of contraception prevent gamete formation. 

Correction 

Surgical methods of contraception prevent the flow of gamete during intercourse. 

(b) All sexually transmitted diseases are completely curable. 

Correction 

Some of the sexually transmitted diseases are curable if they are detected early and treated properly. AIDS is still an incurable disease. 

(c) Oral pills are very popular contraceptives among the rural women. 

Correction 

Oral pills are very popular contraceptives among urban women. 

(d) In E. T. techniques, embryos are always transferred into the uterus. 

Correction 

In embryo transfer technique, 8 celled embryos are transferred into the fallopian tube while more than 8 celled embryos are transferred into the uterus.

NCERT Solutions Class 12 Biology Chapter 15 Biodiversity and Conservation

1. Name the three important components of biodiversity.

Answer:

Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Three important components of biodiversity are: 

• Genetic diversity 
• Species diversity
• Ecosystem diversity

2. How do ecologists estimate the total number of species present in the world?

Answer:

The diversity of living organisms present on the Earth is very vast. According to an estimate by researchers, it is about seven millions. The total number of species present in the world is calculated by ecologists by statistical comparison between a species richness of a well-studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the Earth.

3. Give three hypotheses for explaining why tropics show greatest levels of species richness.

Answer:

There are three different hypotheses proposed by scientists for explaining species richness in the tropics. 

• Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity. 
• Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness. 
• Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

4. What is the significance of the slope of regression in a species − area relationship?

Answer:

The slope of regression (z) has a great significance in order to find a species-area relationship. It has been found that in smaller areas (where the species-area relationship is analysed), the value of slopes of regression is similar regardless of the taxonomic group or the region. However, when a similar analysis is done in larger areas, then the slope of regression is much steeper.

5. What are the major causes of species losses in a geographical region?

Answer:

Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Biodiversity around the world is declining at a very fast pace. The following are the major causes for the loss of biodiversity around the world. 

♦ Habitat loss and fragmentation: Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanization. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species. 

♦ Over-exploitation: Due to over-hunting and over-exploitation of various plants and animals by humans, many species have become endangered or extinct (such as the tiger and the passenger pigeon). 

♦ Alien species Invasions: Accidental or intentional introduction of nonnative species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake. 

♦ Co−extinction: In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

6. How is biodiversity important for ecosystem functioning?

Answer:

An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods. 

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. As we all know, various trophic levels are connected through food chains. If any one organism or all organisms of any one trophic level is killed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food. If all deer’s are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. 

Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

7. What are sacred groves? What is their role in conservation?

Answer:

Sacred groves are tracts of forest which are regenerated around places of worship. Sacred groves are found in Rajasthan, Western Ghats of Karnataka and Maharashtra, Meghalaya and Madhya Pradesh. Sacred groves help in the protection of many rare, threatened, and endemic species of plants and animals found in an area. The process of deforestation is strictly prohibited in this region by tribals. Hence, the sacred grove biodiversity is a rich area.

8. Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?

Answer:

The biotic components of an ecosystem include the living organisms such as plants and animals. Plants play a very important role in controlling floods and soil erosion. The roots of plants hold the soil particles together, thereby preventing the top layer of the soil to get eroded by wind or running water. The roots also make the soil porous, thereby allowing ground water infiltration and preventing floods. Hence, plants are able to prevent soil erosion and natural calamities such as floods and droughts. They also increase the fertility of soil and biodiversity.

9. What measures, as an individual, you would take to reduce environmental pollution?

Answer:

The following initiatives can be taken to prevent environmental pollution: 

Measures for preventing Air pollution 

• Planting more trees 
• Use of clean and renewable energy sources such as CNG and bio-fuels 
• Reducing the use of fossil fuels 
• Use of catalytic converters in automobiles 

Measures for preventing water pollution 

• Optimizing the use of water 
• Using kitchen waste water in gardening and other household purposes

Measures for controlling Noise pollution 

• Avoid burning crackers on Diwali 
• Plantation of more trees 

Measures for decreasing solid waste generation 

• Segregation of waste 
• Recycling and reuse of plastic and paper 
• Composting of biodegradable kitchen waste 
• Reducing the use of plastics

10. Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?

Answer:

Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the Earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate small pox virus from the world through the use of vaccinations. This shows that humans deliberately want to make these species extinct. Several other eradication programmes such as polio and Hepatitis B vaccinations are aimed to eliminate these disease-causing microbes.

NCERT Solutions Class 11 Biology Chapter 12 Mineral Nutrition

1. ‘All elements that are present in a plant need not be essential to its survival’. Comment. 

Solution: 

Plants tend to absorb different kinds of nutrients from soil. However, a nutrient is inessential for a plant if it is not involved in the plant’s physiology and metabolism. For example, plants growing near radioactive sites tend to accumulate radioactive metals. Similarly, gold and selenium get accumulated in plants growing near mining sites. However, this does not mean that radioactive metals, gold, or selenium are essential nutrients for the survival of these plants. 

2. Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics? 

Solution: 

Hydroponics is the art of growing plants in a nutrient solution in the absence of soil. Since the plant roots are exposed to a limited amount of the solution, there are chances that the concentrations of oxygen and other minerals in the plant roots would reduce. Therefore, in studies involving mineral nutrition using hydroponics, purification of water and nutrient salts is essential so as to maintain an optimum growth of the plants. 

3. Explain with examples: macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements. 

Solution: 

Macronutrients: 

They are the nutrients required by plants in large amounts. They are present in plant tissues in amounts more than 10 m mole kg–1 of dry matter. Examples include hydrogen, oxygen, and nitrogen. 

Micronutrients: 

They are also called trace elements and are present in plant bodies in very small amounts, i.e., amounts less than 10 m mole kg– 1 of dry matter. Examples include cobalt, manganese, zinc, etc.

Beneficial nutrients: 

They are plant nutrients that may not be essential, but are beneficial to plants. Sodium, silicon, cobalt and selenium are beneficial to higher plants. 

Toxic elements: 

Micronutrients are required by plants in small quantities. An excess of these nutrients may induce toxicity in plants. For example, when manganese is present in large amounts, it induces deficiencies of iron, magnesium, and calcium by interfering with their metabolism. 

Essential elements: 

These elements are absolutely necessary for plant growth and reproduction. The requirement of these elements is specific and non-replaceable. They are further classified as macro and micro-nutrients. 

4. Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency. 

Solution: 

The five main deficiency symptoms arising in plants are: 

• Chlorosis. 
• Necrosis. 
• Inhibition of cell division. 
• Delayed flowering. 
• Stunted plant growth. 

Chlorosis or loss of chlorophyll leads to the yellowing of leaves. It is caused by the deficiencies of nitrogen, potassium, magnesium, sulphur, iron, manganese, zinc, and molybdenum. 

Necrosis is the death of plant tissues as a result of the deficiencies of calcium, magnesium, copper, and potassium. 

Inhibition of cell division is caused by the deficiencies of nitrogen, potassium, sulphur, and molybdenum. 

Delayed flowering is caused by the deficiencies of nitrogen, sulphur, and molybdenum. 

Stunted plant growth is a result of the deficiencies of copper and sulphur.

5. If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element? 

Solution: 

In plants, the deficiency of a nutrient can cause multiple symptoms. For example, the deficiency of nitrogen causes chlorosis and delayed flowering. 

In a similar way, the deficiency of a nutrient can cause the same symptom as that caused by the deficiency of another nutrient. For example, necrosis is caused by the deficiency of calcium, magnesium, copper, and potassium. 

Another point to be considered is that different plants respond in different ways to the deficiency of the same nutrient. 

Hence, to identify the nutrient deficient in a plant, all the symptoms developed in its different parts must be studied and compared with the available standard tables. 

6. Why is that in certain plants deficiency symptoms appear first in younger parts of the plant while in others they do so in mature organs? 

Solution: 

Deficiency symptoms are morphological changes in plants, indicating nutrient deficiency. Deficiency symptoms vary from one element to another. The plant part in which a deficiency symptom occurs depends on the mobility of the deficient element in the plant. Elements such as nitrogen, potassium, and magnesium are highly mobile. These elements move from the mature organs to the younger parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the older parts of the plant. Elements such as calcium and sulphur are relatively immobile. These elements are not transported out of the older parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the younger parts of the plant.

7. How are the minerals absorbed by the plants? 

Solution: 

The absorption of soil nutrients by the roots of plants occurs in two main phases – apoplast and symplast. 

During the initial phase or apoplast, there is a rapid uptake of nutrients from the soil into the free spaces of plant cells. This process is passive and it usually occurs through transmembrane proteins and ion-channels. 

In the second phase or symplast, the ions are taken slowly into the inner spaces of the cells. This pathway generally involves the expenditure of energy in the form of ATP. 

8. What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium. What is their role in N₂ -fixation? 

Solution: 

Rhizobium is a symbiotic bacteria present in the root nodules of leguminous plants. The basic requirements for Rhizobium to carry out nitrogen fixation are as follows: 

(a) Presence of the enzyme nitrogenase 

(b) Presence of leg-haemoglobin 

(c) Non-haem iron protein, ferrodoxin as the electron-carrier 

(d) Constant supply of ATP 

(e) Mg2+ions as co-factors 

Rhizobium contains the enzyme nitrogenase – a Mo-Fe protein – that helps in the conversion of atmospheric free nitrogen into ammonia. 

The reaction is as follows: 

N₂ + 8e^– + 8H⁺ + 16 ATP→ 2 NH₃ + H₂ + 16ADP + 16Pi 

The Rhizobium bacteria live as aerobes under free-living conditions, but require anaerobic conditions during nitrogen fixation. This is because the enzyme nitrogenase is highly sensitive to molecular oxygen. The nodules contain leghaemoglobin, which protects nitrogenase from oxygen.

9. What are the steps involved in formation of a root nodule? 

Solution: 

Multiple interactions are involved in the formation of root nodules. The Rhizobium bacteria divide and form colonies. These get attached to the root hairs and epidermal cells. The root hairs get curled and are invaded by the bacteria. This invasion is followed by the formation of an infection thread that carries the bacteria into the cortex of the root. The bacteria get modified into rod-shaped bacteroides. As a result, the cells in the cortex and pericycle undergo division, leading to the formation of root nodules. The nodules finally get connected with the vascular tissues of the roots for nutrient exchange. 

10. Which of the following statements are true? If false, correct them: 

(a) Boron deficiency leads to stout axis. 

(b) Every mineral element that is present in a cell is needed by the cell. 

(c) Nitrogen as a nutrient element, is highly immobile in the plants. 

(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities. 

Solution: 

(a) True. 

(b) All the mineral elements present in a cell are not needed by the cell. For example, plants growing near radioactive mining sites tend to accumulate large amounts of radioactive compounds. These compounds are not essential for the plants. 

(c) Nitrogen as a nutrient element is highly mobile in plants. It can be mobilised from the old and mature parts of a plant to its younger parts. 

(d) True.

NCERT Solutions Class 11 Biology Chapter 2 Biological Classification

1. Discuss how classification systems have undergone several changes over a period of time?

Solution:

The classification systems have undergone several changes with time. The first attempt of classification was made by Aristotle. He classified plants as herbs, shrubs, and trees. Animals, on the other hand, were classified on the basis of presence or absence of red blood cells. This system of classification failed to classify all the known organisms. Therefore, Linnaeus gave a two kingdom system of classification. It consists of kingdom Plantae and kingdom Animalia. However, this system did not differentiate between unicellular and multicellular organisms and between eukaryotes and prokaryotes. Therefore, there were large numbers of organisms that could not be classified under the two kingdoms. 

To solve these problems, a five kingdom system of classification was proposed by R.H Whittaker in 1969. On the basis of characteristics, such as cell structure, mode of nutrition, presence of cell wall, etc., five kingdoms, Monera, Protista, Fungi, Plantae, and Animalia were formed.

2. State two economically important uses of: 

(a) Heterotrophic bacteria 

(b) Archaebacteria

Solution:

(a) Heterotrophic bacteria 

i) They act as decomposers and help in the formation of humus. 

ii) They help in the production of curd from milk. 

iii) Many antibiotics are obtained from some species of bacteria. 

iv) Many soil bacteria help in fixation of atmospheric nitrogen. 

(b) Archaebacteria 

i) Methane gas is produced from the dung of ruminants by the methanogens. 

ii) Methanogens are also involved in the formation of biogas and sewage treatment.

3. What is the nature of cell-walls in diatoms?

Solution

The cell walls of diatoms are made of silica. Their cell wall construction is known as frustule. It consists of two thin overlapping shells that fit into each other such as a soap box. When the diatoms die, the silica in their cell walls gets deposited in the form of diatomaceous earth. This diatomaceous earth is very soft and quite inert. It is used in filtration of oils, sugars, and for other industrial purposes.

4. Find out what do the terms ‘algal bloom’ and ‘red-tides’ signify.

Solution:

Algal bloom

Algal bloom refers to an increase in the population of algae or blue-green algae in water, resulting in discoloration of the water body. This causes an increase in the biological oxygen demand (BOD), resulting in the death of fishes and other aquatic animals.

Red-tides

Red tides are caused by red dinoflagellates (Gonyaulax) that multiply rapidly. Due to their large numbers, the sea appears red in colour. They release large amounts of toxins in water that can cause death of a large number of fishes.

5. How are viroids different from viruses?

Solution:

Viroids were discovered in 1917 by T.O. Denier. They cause potato spindle tuber disease. They are smaller in size than viruses. They also lack the protein coat and contain free RNA of low molecular weight.

6. Describe briefly the four major groups of Protozoa.

Solution:

Protozoa are microscopic unicellular protists with heterotrophic mode of nutrition. They may be holozoic, saprobic, or parasitic. These are divided into four major groups.

(1) Amoeboid protozoa or sarcodines 

They are unicellular, jelly-like protozoa found in fresh or sea water and in moist soil. Their body lacks a periplast. Therefore, they may be naked or covered by a calcareous shell. They usually lack flagella and have temporary protoplasmic outgrowths called pseudopodia. These pseudopodia or false feet help in movement and capturing prey. They include free living forms such as Amoeba or parasitic forms such as Entamoeba. 

(2) Flagellated protozoa or zooflagellates 

They are free living, non-photosynthetic flagellates without a cell wall. They possess flagella for locomotion and capturing prey. They include parasitic forms such as Trypanosoma, which causes sleeping sickness in human beings. 

(3) Ciliated protozoa or ciliates 

They are aquatic individuals that form a large group of protozoa. Their characteristic features are the presence of numerous cilia on the entire body surface and the presence of two types of nuclei. All the cilia beat in the same direction to move the water laden food inside a cavity called gullet. They include organisms such as Paramoecium, Vorticella,etc.

(4) Sporozoans 

They include disease causing endoparasites and other pathogens. They are uninucleate and their body is covered by a pellicle. They do not possess cilia or flagella. They include the malaria causing parasite Plasmodium.

7. Plants are autotrophic. Can you think of some plants that are partially heterotrophic?

Solution:

Plants have autotrophic mode of nutrition as they contain chlorophyll pigment. Thus, they have the ability to prepare their own food by the process of photosynthesis. However, some insectivorous plants are partially heterotrophic. They have various means of capturing insects so as to supplement their diet with required nutrients derived from insects, causing proliferation of growth. The examples include pitcher plant (Nepenthes), Venus fly trap, bladderwort, and sundew plant.

8. What do the terms phycobiont and mycobiont signify?

Solution:

Phycobiont refers to the algal component of the lichens and mycobiont refers to the fungal component. Algae contain chlorophyll and prepare food for fungi whereas the fungus provides shelter to algae and absorbs water and nutrients from the soil. This type of relationship is referred to as symbiotic.

9. Give a comparative account of the classes of Kingdom Fungi under the following: 

(i) Mode of nutrition 

(ii) Mode of reproduction

Solution:

(A) Phycomycetes: 

This group of fungi includes members such as Rhizopus, Albugo, etc. 

(i) Mode of nutrition 

They are obligate parasites on plants or are found on decaying matter such as wood. 

(ii) Mode of reproduction 

Asexual reproduction takes place through motile zoospores or non-motile aplanospores that are produced endogenously in sporangium. Sexual reproduction may be of isogamous, anisogamous, or oogamous type. It results in the formation of thick-walled zygospore.

(B) Ascomycetes: 

This group of fungi includes members such as Penicillium, Aspergillus, Claviceps, and Neurospora. 

(i) Mode of nutrition 

They are sporophytic, decomposers, parasitic or coprophilous (growing on dung).

(ii) Mode of reproduction 

Asexual reproduction occurs through asexual spores produced exogenously, such as conidia produced on conidiophores. Sexual reproduction takes place through ascospores produced endogenously in saclike asci and arranged inside ascocarps.

(C) Basidiomycetes: 

This group of fungi includes members such as Ustilago, Agaricus and Puccinia. 

(i) Mode of nutrition 

They grow as decomposers in soil or on logs and tree stumps. They also occur as parasites in plants causing diseases such as rusts and smuts. 

(ii) Mode of reproduction 

Asexual reproduction takes place commonly through fragmentation. Asexual spores are absent. Sex organs are absent but sexual reproduction takes place through plasmogamy. It involves fusion of two different strains of hyphae. The resulting dikaryon gives rise to a basidium. Four basidiospores are produced inside a basidium.

(D) Deuteromycetes: 

This group of fungi includes members such as Alternaria, Trichoderma, and Colletotrichum. 

(i) Mode of nutrition 

Some members are saprophytes while others are parasites. However, a large number act as decomposers of leaf litter. 

(ii) Mode of reproduction 

Asexual reproduction is the only way of reproduction in deuteromycetes. It occurs through asexual spores called conidia. Sexual reproduction is absent in deuteromycetes.

10. What are the characteristic features of Euglenoids?

Solution:

Some characteristic features of Euglenoids are as follows. 

• Euglenoids (such as Euglena) are unicellular protists commonly found in fresh water. 
• Instead of cell wall, a protein-rich cell membrane known as pellicle is present. 
• They bear two flagella on the anterior end of the body. 
• A small light sensitive eye spot is present. 
• They contain photosynthetic pigments such as chlorophyll and can thus prepare their own food. However, in absence of light, they behave similar to heterotrophs by capturing other small aquatic organisms. 
• They have both plant and animal-like features, which makes them difficult to classify.

11. Give a brief account of viruses with respect to their structure and nature of genetic material. Also name four common viral diseases.

Solution:

Viruses are sub-microscopic infectious agents that can infect all living organisms. A virus consists of genetic material surrounded by a protein coat. The genetic material may be present in the form of DNA or RNA. 

Most of the viruses, infecting plants, have single stranded RNA as genetic material. On the other hand, the viruses infecting animals have single or double stranded RNA or double stranded DNA. Bacteriophages or viruses infecting bacteria mostly have double stranded DNA. 

Their protein coat called capsid is made up of capsomere subunits. These capsomeres are arranged in helical or polyhedral geometric forms. 

A.I.D.S, small pox, mumps, and influenza are some common examples of viral diseases.

12. Organise a discussion in your class on the topic- Are viruses living or non-living?

Solution:

Viruses are microscopic organisms that have characteristics of both living and nonliving. A virus consists of a strand of DNA or RNA covered by a protein coat. This presence of nucleic acid (DNA or RNA) suggests that viruses are alive. In addition, they can also respond to their environment (inside the host cell) in a limited manner. However, some other characters, such as their inability to reproduce without using the host cell machinery and their acellular nature, indicate that viruses are non-living. Therefore, classifying viruses has remained a mystery for modern systematics.

NCERT Solutions Class 12 Physics Chapter 12 Atoms

1. Choose the correct alternative from the clues given at the end of each statement:

(a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)

(b) In the ground state of ………. electrons are in stable equilibrium, while in ………. electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)

(c) A classical atom based on ………. is doomed to collapse. (Thomson’s model/ Rutherford’s model.)

(d) An atom has a nearly continuous mass distribution in a ……….but has a highly non-uniform mass distribution in ……….(Thomson’s model/ Rutherford’s model.)

(e) The positively charged part of the atom possesses most of the mass in ………. (Rutherford’s model/both the models.)

Answer:

(a)  The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.

(b) In the ground state of Thomson’s model electrons are in stable equilibrium, while in Rutherford’s model electrons always experience a net force.

(c) A classical atom based on Rutherford’s model is doomed to collapse.

(d) An atom has a nearly continuous mass distribution in a Thomson’s model but has a highly non-uniform mass distribution in Rutherford’s model.

(e) The positively charged part of the atom possesses most of the mass in both the models.

2. Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Answer:

We know that the mass of the incident alpha particle (6.64 × 10^-27kg) is more than the mass of hydrogen (1.67 × 10^-27Kg). Hence, the target nucleus is lighter, from which we can conclude that the alpha particle would not rebound. Implying to the fact that solid hydrogen isn’t a suitable replacement to gold foil for the alpha particle scattering experiment.

3. What is the shortest wavelength present in the Paschen series of spectral lines?

Answer:

We know the Rydberg’s formula is written as:

\(\frac{hc}{\lambda }=21.76\times 10^{-19}[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}]\)

Where,

h = Planck’s constant = 6.6 × 10^-34

c = Speed of light = 3 × 10⁸ m/s

n₁ and n₂ are integers

So the shortest wavelength present in the Paschen series of the spectral lines is given for n₁ = 3 and n₂ = ∞

\(\frac{hc}{\lambda }=21.76\times 10^{-19}[\frac{1}{3^{2}}-\frac{1}{\infty^{2}}]\)

\(\lambda =\frac{6.6\times 10^{-34}\times 3\times 10^{8}\times 9}{21.76\times 10^{-19}}\)

= 8.189 × 10⁷ m

= 818.9 nm

4. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Answer:

It is given that

Separation of two energy levels in an atom,

E = 2.3 eV = 2.3 × 1.6 × 10^-19

= 3.68 × 10^-19 J

Consider v as the frequency of radiation emitted when the atom transits from the upper level to the lower level.

So the relation for energy can be written as:

E = hv

Where,

h = Planck’s constant = 6.62 × 10^-34 Js

v = E/h

= \(\frac{3.68 \times 10^{-19}}{6.62\times 10^{-32}}\)

= 5.55 × 10¹⁴ Hz

= 5.6 × 10¹⁴ Hz

Therefore, the frequency of radiation is 5.6 × 10¹⁴ Hz.

5. The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Answer:

Ground state energy of hydrogen atom, E = − 13.6 eV

The total energy of hydrogen atom is -13.6 eV. The kinetic energy is equal to the negative of the total energy.

Kinetic energy = − E = − (− 13.6) = 13.6 eV

Potential energy = negative of two times of kinetic energy.

Potential energy = − 2 × ( 13.6 ) = − 27.2 eV

6. A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

Answer:

For ground level, n₁= 1

Let, E₁ be the energy of this level. It is known that E₁ is related with n₁ as:

\(E_{1}=\frac{-13.6}{{(n_{1})^{2}}}\)​ eV \(= \frac{-13.6}{1^{2}}=\)​  -13.6 eV

The atom is excited to a higher level, n₂= 4

Let, E₂ be the energy of this level:

\(E_{2}=\frac{-13.6}{{(n_{2})^{2}}}\) eV  \(=\frac{-13.6}{4^{2}}\)​ \(=\frac{-13.6}{16}\)​ eV

Following is the amount of energy absorbed by the photon:

E = E₂ − E₁

= \(\frac{-13.6}{16}-(\frac{-13.6}{1})\)

= \(\frac{-13.6\times 15}{16}\)​ eV

= \(\frac{-13.6\times 15}{16}\times 1.6\times 10^{-19}\)

= 2.04 × 10^-18 J

For a photon of wavelength λ, the expression of energy is written as:

\(E = \frac{hc}{\lambda }\)

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

c = Speed of light = 3 × 10⁸ m/s

\(\lambda = \frac{hc}{E}\)​

\(=\frac{6.6 \times 10 ^{-34}\times 3\times 10^{8}}{2.04 \times 10^{-18}}\)

\(\lambda = 97nm\)

\(c = \lambda v\)

v = c/λ

= \(\frac{3\times 10^{8}}{9.7\times 10^{-8}}\)

= 3.1 × 10¹⁵ Hz

Therefore, 97nm and 3.1×10¹⁵Hz is the wavelength and frequency of the photon.

7. (a) Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

Answer:

(a) Let v₁ be the orbital speed of the electron in a hydrogen atom in the ground state level, n₁ = 1. For charge (e) of an electron, v₁ is given by the relation,

\(v_1 = \frac{e^{2}}{n_1 4\;\pi\;\epsilon_0\; (\frac{h}{2\pi}) } = \frac{e^{2}}{2\;\epsilon_0 \;h}\)

Where, e = 1.6 × 10⁻¹⁹ C

ϵ₀​ = Permittivity of free space = 8.85 × 10⁻¹² N⁻¹ C ² m⁻²

h = Planck’s constant = 6.62 × 10⁻³⁴ J s

\(v_1 = \frac{(1.6\times10^{-19})^{2}}{2\times8.85\times10^{-12}\times6.62\times10^{-34}}=\) 0.0218 x 10⁸ = 2.18 × 10⁶m/s

For level n₂ = 2, we can write the relation for the corresponding orbital speed as:

\(v_2 = \frac{e^2}{n_2 2 \epsilon_0 h}= \frac{(1.6 \times 10^{-19})^2}{2 \times 2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}=​ \)1.09 × 10⁶ m/s

For level n₃=  3, we can write the relation for the corresponding orbital speed as:

\(v_3 = \frac{e^2}{n_3 2 \epsilon_0 h}​ = \frac{(1.6 \times 10^{-19})^2}{2 \times 3 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}=\)​ 7.27 × 10⁵ m/s

Therefore, in a hydrogen atom, the speed of the electron at different levels that is n = 1, n = 2, and n = 3 is 2.18 × 10⁶m/s, 1.09×10⁶m/s and 7.27 × 10⁵ m/s.

(b) Let, T₁ be the orbital period of the electron when it is in level n₁= 1.

Orbital period is related to orbital speed as:

\(T_1 = \frac{ 2 \pi r_1 } { v_1 }​​ \) [Where, r₁ = Radius of the orbit] \(=\frac{ { n _ 1 } ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)

h = Planck’s constant = 6.62 × 10⁻³⁴ J s

e = Charge on an electron = 1.6 × 10⁻¹⁹ C

ε₀ = Permittivity of free space = 8.85 × 10⁻¹² N⁻¹ C² m⁻²

m = Mass of an electron = 9.1 × 10⁻³¹ kg

\(T_1 = \frac{ 2 \pi r_1 } { v_1 }​​ =\frac{2 \;\pi \;\times ( 1 ) ^ { 2 } \;\times \;( 6.62 \;\times \;10 ^ { -34 } ) ^ { 2 } \;\times \;8.85 \;\times \;10 ^{ -12 }}{ 2.18 \;\times \;10 ^ { 6 } \;\times \;\pi \;\times 9.1 \;\times \;10 ^ { -31 } \;\times \;( 1.6 \;\times \;10 ^ { -19 } ) ^ { 2 } }=\) 15.27 × 10^-17 = 1.527 × 10^-16 s

For level n₂ = 2, we can write the period as:

\(T_2 = \frac{ 2 \pi r_2 } { v_2 }​​  \) [Where, r₂ = Radius of the electron in n₂ = 2] \(=\frac{ {( n _ 2 )} ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)

\(=\frac{2 \;\pi \;\times \;( 2 ) ^ { 2 } \;\times \;( 6.62 \;\times \;10 ^ { -34 } ) ^ { 2 }\;\times \;8.85 \;\times 10 ^{ -12 }}{ 1.09 \;\times \;10 ^ { 6 } \;\times \;\pi \;\times \;9.1 \;\times \;10 ^ { -31 } \;\times \;( 1.6 \;\times \;10 ^ { -19 } ) ^ { 2 } }=\)​ 1.22 × 10^-15s

For level n₃ = 3, we can write the period as:

\(T_3= \frac{ 2 \pi r_3 } { v_3 }\) [Where, r₃ = Radius of the electron in n₃ = 3] \(=\frac{ {( n _ 3 )} ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)

\(=\frac{2 \;\pi \;\times \;( 3 ) ^ { 2 } \;\times \;( 6.62 \;\times \;10 ^ { -34 } ) ^ { 2 }\;\times \;8.85 \;\times 10 ^{ -12 }}{ 7.27 \;\times \;10 ^ { 5 } \;\times \;\pi \;\times \;9.1 \;\times \;10 ^ { -31 } \;\times \;( 1.6 \;\times \;10 ^ { -19 } ) ^ { 2 } }=\)= 4.12 × 10^-15s

Therefore, 1.52 × 10^-16s, 1.22 × 10^-15s and 4.12 × 10^-15s are the orbital periods in each levels.

8. The radius of the innermost electron orbit of a hydrogen atom is 5.3×10^–11 m. What are the radii of the n = 2 and n = 3 orbits?

Answer:

The radius of the innermost orbit of a hydrogen atom, r₁ = 5.3 × 10⁻¹¹ m.

Let r₂ be the radius of the orbit at n = 2. It is related to the radius of the inner most orbit as:

r₂ ​= (n)²r₁​ = 4 × 5.3 × 10^-11 = 2.12 × 10^-10 m

For n = 3, we can write the corresponding electron radius as:

r₃​ = (n)²r₁​ = 9 × 5.3 x 10^-11 = 4.77 × 10^-10 m

Therefore, 2.12 × 10⁻¹⁰ m and 4.77 × 10⁻¹⁰ m are the radii of an electron for n = 2 and n = 3 orbits respectively.

9. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer:

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.

Orbital energy is related to orbit level (n) as:

\(E = \frac {-13.6 } { n ^ { 2 } } \; eV\)

For n = 3, E = -13.6 / 9 = – 1.5 eV

This energy is approximately equal to the energy of gaseous hydrogen.

It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

\(\frac { 1 } { \lambda } = R_y \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { n ^ { 2 } } \right )\)

Where, R_y = Rydberg constant = 1.097 × 10⁷ m⁻¹

λ = Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λ as:

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { 3 ^ { 2 } } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( 1 -\frac { 1 } { 9 } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 8 } { 9 } \right )\)

\(\lambda = \frac { 9 }{8 \times 1.097 \times 10 ^ {7 }}= 102.55 \,nm\) 

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { 2 ^ { 2 } } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( 1 -\frac { 1 } { 4 } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 3 } { 4 } \right )\)

\(\lambda = \frac { 4 }{3 \times 1.097 \times 10 ^ {7 }} = 121.54 \;nm\)

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 2 ^ { 2 } } -\frac { 1 } { 3 ^ { 2 } } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 4 } -\frac { 1 } { 9 } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 5 } { 36 } \right )\)

\(\lambda = \frac { 36 }{ 5 \times 1.097 \times 10 ^ {7 }}= 656.33 \,nm\)

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Therefore, there are two wavelengths that are emitted in Lyman series and they are approximately 102.5 nm and 121.5 nm and one wavelength in the Balmer series which is 656.33 nm.

10. In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 3 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6.0 × 10²⁴ kg.)

Answer:

Radius of the orbit of the Earth around the Sun, r = 1.5 × 10¹¹ m

Orbital speed of the Earth, ν = 3 × 10⁴ m/s

Mass of the Earth, m = 6.0 × 10²⁴ kg

According to Bohr’s model, angular momentum is quantized and given as:

\(m v r = \frac { n h } { 2 \pi }\)

Where,

h = Planck’s constant = 6.62 × 10⁻³⁴ J s

n = Quantum number

\(n = \frac { m v r 2 \pi } { h }​ = \frac{ 2 \pi \times 6 \times 10^{24} \times 3 \times 10 ^ {4} \times 1.5 \times 10^{11} }{ 6.62 \times 10^{-34} }\)​ = 25.61 ×10⁷³ = 2.6 × 10⁷⁴

Therefore, the earth revolution can be characterized by the quantum number 2.6 × 10⁷⁴.

11. Choose a suitable solution to the given statements which justify the difference between Thomson’s model and Rutherford’s model

(a) In the case of scattering of alpha particles by a gold foil, average angle of deflection of alpha particles stated by Rutherford’s model is (less than, almost the same as, much greater than) stated by Thomson’s model.

(b) Is the likelihood of reverse scattering (i.e., dispersing of α-particles at points more prominent than 90°) anticipated by Thomson’s model ( considerably less, about the same, or much  more prominent ) than that anticipated by Rutherford’s model?

(c) For a small thickness T, keeping other factors constant, it has been found that amount of alpha particles scattered at direct angles is proportional to T. This linear dependence implies?

(d) To calculate average angle of scattering of alpha particles by thin gold foil, which model states its wrong to skip multiple scattering?

Answer:

(a) almost the same

The normal point of diversion of alpha particles by a thin gold film anticipated by Thomson’s model is about the same as from anticipated by Rutherford’s model. This is on the grounds that the average angle was taken in both models.

(b) much less

The likelihood of scattering of alpha particles at points more than 90° anticipated by Thomson’s model is considerably less than that anticipated by Rutherford’s model.

(c) Dispersing is predominantly because of single collisions. The odds of a single collision increment linearly with the amount of target molecules. Since the number of target particles increment with an expansion in thickness, the impact likelihood depends straightly on the thickness of the objective.

(d) Thomson’s model

It isn’t right to disregard multiple scattering in Thomson’s model for figuring out the average angle of scattering of alpha particles by a thin gold film. This is on the grounds that a solitary collision causes almost no deflection in this model. Subsequently, the watched normal scattering edge can be clarified just by considering multiple scattering.

12. The gravitational attraction amongst proton and electron in a hydrogen atom is weaker than the coulomb attraction by a component of around 10⁻⁴⁰. Another option method for taking a gander at this case is to assess the span of the first Bohr circle of a hydrogen particle if the electron and proton were bound by gravitational attraction. You will discover the appropriate response fascinating.

Answer:

Radius of the first Bohr orbit is given by the relation:

\(r_1 = \cfrac{4 \pi \epsilon_0 \left ( \frac{h}{2 \pi} \right )^2}{m_e e^2 }\)     ........(1)

Where,

∈₀ = Permittivity of free space

h = Planck’s constant = 6.63 × 10⁻³⁴ J s

m_e = Mass of an electron = 9.1 × 10⁻³¹ kg

e = Charge of an electron = 1.9 × 10⁻¹⁹ C

m_p = Mass of a proton = 1.67 × 10⁻²⁷ kg

r = Distance between the electron and the proton

Coulomb attraction between an electron and a proton is:

\(F_C = \frac{e^{2}}{4 \pi \epsilon_0 r^{2}}\)       ........(2)

Gravitational force of attraction between an electron and a proton is:

\(F_G = \frac{G m_e m_p}{r^{2}}\)     .........(3)

Where,

G = Gravitational constant = 6.67 × 10⁻¹¹ N m²/kg²

Considering electrostatic force and the gravitational force between an electron and a proton to be equal, we get:

∴ F_G = F_C

\(\frac{G m_e m_p}{r^{2}}=\frac{e^{2}}{4 \pi \epsilon_0 r^{2}}\)

\({G m_e m_p}=\frac{e^{2}}{4 \pi \epsilon_0 }\)     ^.........(4)

Putting the value of equation (4) in equation (1), we get:

\(r_{1} = \cfrac{\left ( \frac{h}{2 \pi} \right )^2}{G m_e m_p}\)

\(r_{1} = \cfrac{\left ( \frac{6.63\times10^{-34}}{2 \times 3.14} \right )^2}{6.67\times 10^{-11}\times 1.67 \times 10^{-27}\times(9.1\times 10^{-31})^{2} }= 1.21\times10^{29}\,m\)​ 

It is known that the universe is 156 billion light years wide or 1.5 × 10²⁷ m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

13. Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Answer:

It is given that the hydrogen atom de-excites from the level n to level (n-1).

The equation for energy (E₁) of the radiation at the level n is

\(E_{1}=h v_{1}=\cfrac{h m e^{4}}{(4 \pi)^{3} \epsilon_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3}} \times\left(\frac{1}{n^{2}}\right)\)  ..........(1)

Here,

ν₁ is the frequency at level n

h = Planck’s constant

m = mass of the hydrogen atom

e = Electron charge

ε₀ = Permittivity of free space

The energy (E₂) of the radiation at level (n - 1) is

\(E_{2}=h v_{2}=\cfrac{h m e^{4}}{(4 \pi)^{3} \epsilon_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3}} \times\left(\frac{1}{(n-1)^{2}}\right)\)   .........(2)

Here,

ν₂ is the frequency at level (n – 1)-

Energy (E) is released as a result of de-excitation

E = E₂ – E₁

hν = E₂ – E₁  ..........(3)

Here,

ν = Frequency of the radiation emitted

Substituting (1) and (2) in (3) we get

\(\begin{aligned} v &=\frac{m e^{4}}{(4 \pi)^{3} \epsilon_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3}}\left[\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}\right] \\ &=\frac{m e^{4}(2 n-1)}{(4 \pi)^{3} \epsilon_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3} n^{2}(n-1)^{2}} \end{aligned}\)

For large n, we can write (2n -1 ) ≈ 2n and (n-1) ≈ n

Therefore, \(v=\cfrac{m e^{4}}{32 \pi^{3} \epsilon_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3} n^{3}}\)     ........(4)

Classical  relation of the frequency of revolution of an electron is given as

ν_c = v/2πr   .........(5)

In the n^th orbit, the velocity  of the electron is

\(v=\cfrac{e^{2}}{4 \pi \epsilon_{0}\left(\frac{h}{2 \pi}\right) n}\)   ........(6)

The radius of the n^th orbit r is given as

\(r=\cfrac{4 \pi \epsilon_{0}\left(\frac{h}{2 \pi}\right)^{2}}{m e^{2}} n^{2}\)     .........(7)

Putting the equation (6) and equation (7) in equation (5) we get

\(v=\cfrac{m e^{4}}{32 \pi^{3} \epsilon_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3} n^{3}}\)   ........(8)

Therefore, the frequency of radiation emitted by the hydrogen atom is equal to the classical orbital frequency.

14. Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10^–10m).

(a) Construct a quantity with the dimensions of length from the fundamental constants e, m_e , and c. Determine its numerical value.

(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in a non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, m_e, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, m_e, and e and confirm that its numerical value has indeed the correct order of magnitude.

Answer:

(a) Charge of an electron, e = 1.6 x 10^-19 C

Mass of the electron, m_e = 9.1 x 10^-31 kg

Speed of light, c = 3 x 10⁸ m/s

The equation from the given quantities is given as,

\(\frac{e^{2}}{4 \pi \epsilon_{0} m_{e}c^{2}}\)

\(\frac{1}{4 \pi \epsilon_{0} } = 9 \times 10^{9}Nm^{2}C^{-2}\)

ε₀ is the permittivity of free space

The numerical value of the quantity is

= 9 x 10⁹ x [ (1.6 x 10 ^-19)²/ 9.1 x 10^-31 x (3 x 10⁸)²]

= 2.81 x 10^-15 m

The numerical value of the equation taken is much lesser than the size of the atom.

(b) Charge of an electron, e = 1.6 x 10^-19 C

Mass of the electron, m_e = 9.1 x 10^-31 kg

Planck’s constant , h = 6.63 x 10^-34  Js

Considering a quantity involving all these values as

\(\cfrac{4 \pi \epsilon_{0} \left [ \frac{h}{2\pi } \right ]^{2}}{m_{e}e^{2}}\)

Where, ε₀ = Permittivity of free space

\(\frac{1}{4 \pi \epsilon_{0} } = 9 \times 10^{9}Nm^{2}C^{-2}\)

The numerical value of the above equation is

\(4\Pi \epsilon _{0}\times \cfrac{(\frac{h}{2\Pi })^{2}}{m_{e}e^{2}}\)

\(=\cfrac{1}{9\times 10^{9}}\times \cfrac{\left ( \frac{6.63\times 10^{-34}}{2\times 3.14} \right )^{2}}{9.1\times 10^{-31}\times (1.6\times 10^{-19})^{2}}\)

= 0.53 x 10^-10 m

The numerical value of the quantity is of the order of the atomic size.

15. The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.

(a) What is the kinetic energy of the electron in this state?

(b) What is the potential energy of the electron in this state?

(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Answer:

(a) Total energy of the electron, E = – 3.4 eV

The kinetic energy of the electron is equal to the negative of the total energy.

K.E = – E

= – (- 3.4 ) = + 3.4 eV

Kinetic energy =  + 3.4 eV

(b) Potential energy (U) of the electron is equal to the negative of twice of the kinetic energy,

P.E = – 2 (K.E)

= – 2 x 3.4 = – 6.8 eV

Potential energy = – 6.8 eV

(c) The potential energy of the system depends on the reference point. If the reference point is changed from zero then the potential energy will change. The total energy is given by the sum of the potential energy and kinetic energy. Therefore, the total energy will also change.

16. If Bohr’s quantization postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion as well. Why then do we never speak of quantization of orbits of planets around the sun?

Answer:

The quantum level for a planetary motion is considered to be continuous. This is because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). 10⁷⁰h is the order of the angular momentum of the Earth in its orbit. As the values of n increase, the angular momenta decreases. So, the planetary motion is considered to be continuous.

17. Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ⁻) of mass about 207m_e orbits around a proton].

Answer:

Mass of a negatively charged muon,  m_μ = 207 m_e

According to Bohr’s model:

Bohr radius, \(r_e \alpha \left ( \frac{1}{m_e} \right )\)

And, energy of a ground state electronic hydrogen atom, \(Ee  \;\alpha\; m_e\)

Also, the energy of a ground state muonic hydrogen atom, E_μ α m_μ

We have the value of the first Bohr orbit, r_e  = 0.53 A = 0.53 × 10^-10 m

Let, r_μ be the radius of muonic hydrogen atom.

Following is the relation at equilibrium:

m_μr_μ= m_e r_e

207 m_e × r_μ = m_e r_e

r_μ= ( 0.53 × 10^-10 )/ 207 = 2.56 × 10^-13 m

Hence, for a muonic hydrogen atom 2.56 × 10⁻¹³ m is the value of first Bohr radius.

We have,

E_e= − 13.6 eV

Considering the ratio of the energies we get:

\(\frac{E_e}{E_μ}=\frac{m_e}{m_μ}=\frac{m_e}{207\;m_e}\)

E_μ= 207 E_e = 207 × ( – 13.6 ) = – 2.81 k eV

−2.81 k eV is the ground state energy of a muonic hydrogen atom.

NCERT Solutions Class 12 Physics Chapter 5 Magnetism And Matter

1. Answer the following:

(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.

(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?

(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

(d) In which direction would a compass free to move in the vertical plane point to, is located right on the geomagnetic north or south pole?

(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of the magnetic moment 8 × 10²² JT^-1 located at its centre. Check the order of magnitude of this number in some way.

(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

Answer:

(a) The three independent conventional quantities used for determining the earth’s magnetic field are:

(i) Magnetic declination,

(ii) Angle of dip

(iii) The horizontal component of the earth’s magnetic field

(b) The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. Hence, as the location of Britain on the globe is closer to the magnetic North pole, the angle of dip would be greater in Britain (About 70°70°) than in southern India.

(c) It is assumed that a huge bar magnet is submerged inside the earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole.

Magnetic field lines originate from the magnetic north pole and terminate at the magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to move away from the ground.

(d) If a compass is placed in the geomagnetic North Pole or the South Pole, then the compass will be free to move in the horizontal plane while the earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.

(e) Magnetic moment, M = 8 × 10²² J T⁻¹

Radius of earth, r = 6.4 × 10⁶ m

Magnetic field strength, B = \(\frac{\mu _{0}M}{4\pi r^{3}}\)

Where,

μ₀​ = Permeability of free space = 4π × 10⁻⁷ TmA⁻¹

Therefore, B = \(\frac{4\pi\times 10^{-7}\times 8\times 10^{22}}{4\pi \times (6.4\times 10^{6})^{3}} = 0.3\; G\)

This quantity is of the order of magnitude of the observed field on earth.

(f) Yes, there are several local poles on earth’s surface oriented in different directions. A magnetized mineral deposit is an example of a local N-S pole.

2. Answer the following:

(a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?

 (b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?

(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such a distant past?

(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

(f) Interstellar space has an extremely weak magnetic field of the order of 10^–12 T. Can such a weak field be of any significant consequence? Explain.

Answer:

(a) Earth’s magnetic field varies with time and it takes a couple of hundred years to change by an obvious sum. The variation in the Earth’s magnetic field with respect to time can’t be ignored.

(b) The Iron core at the Earth’s centre cannot be considered as a source of Earth’s magnetism because it is in its molten form and is non-ferromagnetic.

(c) The radioactivity in the earth’s interior is the source of energy that sustains the currents in the outer conducting regions of the earth’s core. These charged currents are considered to be responsible for the earth’s magnetism.

(d) The Earth’s magnetic field reversal has been recorded several times in the past about 4 to 5 billion years ago. These changing magnetic fields were weakly recorded in rocks during their solidification. One can obtain clues about the geomagnetic history from the analysis of this rock magnetism.

(e) Due to the presence of ionosphere, the Earth’s field deviates from its dipole shape substantially at large distances. The Earth’s field is slightly modified in this region because of the field of single ions. The magnetic field associated with them is produced while in motion.

(f) A remarkably weak magnetic field can deflect charged particles moving in a circle. This may not be detectable for a large radius path. With reference to the gigantic interstellar space, the deflection can alter the passage of charged particles.

3. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. What is the magnitude of the magnetic moment of the magnet?

Answer:

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, T = 4.5 × 10⁻²J

The angle between the bar magnet and the external magnetic field, θ = 30°

Torque is related to magnetic moment (M) as:

T = MBsinθ

\(∴ M = \frac{T}{B sin\theta }\)

\(= \frac{4.5\times 10^{-2}}{0.25\times sin 30°} = 0.36\; J\; T^{-1}\)

Hence, the magnetic moment of the magnet is 0.36 J T⁻¹.

4. A short bar magnet of magnetic moment m = 0.32 JT^–1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Answer:

Moment of the bar magnet, M = 0.32 J T⁻¹

External magnetic field, B = 0.15 T

(a) The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle θ, between the bar magnet and the magnetic field is 0°.

Potential energy of the system = −MBcosθ

= −0.32 × 0.15 cos0°

= −4.8 × 10⁻² J

(b) The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium. θ=180°

Potential energy = −MBcosθ

= −0.32 × 0.15cos180°

= 4.8 × 10⁻² J

5. A closely wound solenoid of 800 turns and area of cross-section 2.5 × 10^–4 m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Answer:

Number of turns in the solenoid, n = 800

Area of cross-section, A = 2.5 × 10⁻⁴m²

Current in the solenoid, I = 3.0 A

A current-carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis, i.e., along with its length.

The magnetic moment associated with the given current-carrying solenoid is calculated as:

M = n I A

= 800 × 3 × 2.5 × 10⁻⁴

= 0.6JT⁻¹

6. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Answer:

Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6  T⁻¹

The angle θ, between the axis of the solenoid and the direction of the applied field, is 30°.

Therefore, the torque acting on the solenoid is given as:

τ = MBsinθ

= 0.6 × 0.25 sin30°

= 7.5 × 10⁻² J

7. A bar magnet of magnetic moment 1.5 J T^–1 lies aligned with the direction of a uniform magnetic field of 0.22 T.

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?

Answer:

(a) Magnetic moment, M = 1.5 J T⁻¹

Magnetic field strength, B = 0.22 T

(i) Initial angle between the axis and the magnetic field, θ₁ ​= 0°

Final angle between the axis and the magnetic field, θ₂ ​= 90°

The work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = - MB(cosθ₂​ – cosθ₁​)

= −1.5 × 0.22(cos90° – cos0°)

= – 0.33 (0 – 1)

= 0.33 J

(ii) Initial angle between the axis and the magnetic field, θ₁​ = 0°

Final angle between the axis and the magnetic field, θ₂​ = 180°

The work required to make the magnetic moment opposite to the direction of the magnetic field is given as:

W = - MB(cosθ₂ ​– cosθ₁​)

= - 1.5 × 0.22(cos180° – cos0°)

= – 0.33 (– 1 – 1)

= 0.66 J

(b) For case (i):

θ = θ₂​ = 90°

∴ Torque, τ = MBsinθ

= MB sin90°

= 1.5 × 0.22 sin 90°

= 0.33 J

The torque tends to align the magnitude moment vector along B.

For case (ii):

θ = θ_2​ = 180°

∴ Torque, τ = MBsinθ

= MB sin180°

= 0 J

8. A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴m², carrying 4.0 A current, is suspended through its centre, thereby allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid?

(b) What is the force and torque on the solenoid if a uniform the horizontal magnetic field of 7.5 × 10^–2 T is set up at an angle of 30° with the axis of the solenoid?

Answer:

Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A = 1.6 × 10⁻⁴ m²

Current in the solenoid, I = 4.0 A

(a) The magnetic moment along the axis of the solenoid is calculated as:

M = nAI

= 2000 × 4 × 1.6 × 10⁻⁴

= 1.28 Am²

(b) Magnetic field, B = 7.5 × 10⁻²T

The angle between the magnetic field and the axis of the solenoid, θ = 30°

Torque, τ = MBsinθ

= 1.28 × 7.5 × 10⁻²sin30°

= 0.048 J

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 0.048 J.

9. A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10^–2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^–1. What is the moment of inertia of the coil about its axis of rotation?

Answer:

Number of turns in the circular coil, N = 16

Radius of the coil, r = 10 cm = 0.1 m

Cross-section of the coil, A = πr² = π × (0.1)² m²

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 × 10⁻² T

Frequency of oscillations of the coil, v = 2.0s⁻¹

∴ Magnetic moment, M = NIA = NIπr²

16 × 0.75 × n × (0.1)²

= 0.377 J T⁻¹

Frequency is given by the relation:

\(v = \frac{1}{2\pi }\sqrt{\frac{MB}{I}}\)

Where,

I = Moment of inertia of the coil

Rearranging the above formula, we get:

\(∴ I = \frac{MB}{4 \pi^{2} v^{2}}\)

= \(\frac{0.377\times 5\times 10^{-2}}{4\pi^{2}\times (2)^{2}}\)

= 1.2 × 10⁻⁴ kg m²

Hence, the moment of inertia of the coil about its axis of rotation is 1.19 × 10⁻⁴ kgm².

10. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Answer:

Horizontal component of earth’s magnetic field, B_H​ = 0.35G

Angle made by the needle with the horizontal plane = Angle of dip = δ = 22°

Earth’s magnetic field strength = B

We can relate B and B_H​ as:

B_H ​= Bcosδ

\(∴ B = \frac{B_{H}}{cos\delta}\)

= \(\frac{0.35}{cos\; 22°}\)​ = 0.38 G

Hence, the strength of the earth’s magnetic field at the given location is 0.38 G.

11. At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

Answer:

Angle of declination, θ = 12°

Angle of dip, δ = 60°

Horizontal component of earth’s magnetic field, BH​ = 0.16 G

Earth’s magnetic field at the given location = B

We can relate B and B_H​ as:

BH ​= Bcosδ

\(∴ B = \frac{B_{H}}{cos\delta}\)

\(= \frac{0.16}{cos\; 60°}\)​ = 0.32 G

Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.

12. A short bar magnet has a magnetic moment of 0.48 J T^–1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Answer:

Magnetic moment of the bar magnet, M = 0.48 J T⁻¹

(a) Distance, d = 10 cm = 0.1 m

The magnetic field at distance d, from the centre of the magnet on the axis, is given by the relation:

\(B = \frac{\mu_{0}\;2M}{4\pi d^{3}}\)

Where,

μ_0​ = Permeability of free space = 4π × 10⁻⁷ TmA⁻¹

\(∴ B = \frac{4\pi \times 10^{-7}\times 2\times 0.48}{4\pi \times (0.1)^{3}}\)

= 0.96 × 10⁻⁴ 

T = 0.96 G

The magnetic field is along the S-N direction.

(b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:

\(B = \frac{\mu_{0}\times M}{4\pi \times d^{3}}\)

= \(\frac{4\pi \times 10^{-7}\times 0.48}{4\pi (0.1)^{3}}\)

= 0.48 G

The magnetic field is along with the N – S direction.

13. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer:

Earth’s magnetic field at the given place, H = 0.36 G

The magnetic field at a distance d, on the axis of the magnet, is given as:

\(B_{1} = \frac{\mu_{0}2M}{4\pi d^{3}} = H\)     ......(i)

Where,

μ₀​ = Permeability of free space

M = Magnetic moment

The magnetic field at the same distance d, on the equatorial line of the magnet, is given as:

\(B_{2} = \frac{\mu_{0}M}{4\pi d^{3}} = \frac{H}2\)​       [Using equation (i)]

Total magnetic field, B = B₁​ + B_2​

= H + \(\frac{H}{2}\)

= 0.36 + 0.18 = 0.54 G

Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.

14. If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?

Answer:

The magnetic field on the axis of the magnet at a distance d₁​ = 14 cm, can be written as:

\(B_{1} = \frac{\mu_{0}2M}{4\pi (d_{1})^{3}} = H\)        …(1)

Where,

M = Magnetic moment

μ₀​ = Permeability of free space

H = Horizontal component of the magnetic field at d_1​

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.

Hence, the magnetic field at a distance d₂​, on the equatorial line of the magnet can be written as:

\(B_{1} = \frac{\mu_{0}2M}{4\pi (d_{2})^{3}} = H\)      …(2)

Equating equations (1) and (2), we get:

\(\frac{2}{(d_{1})^{3}} = \frac{1}{(d_{2})^{3}}\)

\(\left [ \frac{d_{2}}{d_{1}} \right ]^{3} = \frac{1}{2}\)

\(∴ d_{2} = d_{1}\times \left ( \frac{1}{2} \right )^{\frac{1}{3}}\)

= 14 × 0.794 = 11.1 cm

The new null points will be located 11.1 cm on the normal bisector.

15. A short bar magnet of magnetic moment 5.25 × 10^–2 J T^–1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. The magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Answer:

Magnetic moment of the bar magnet, M = 5.25 × 10⁻² J T⁻¹

Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10⁻⁴ T

(a) The magnetic field at a distance R from the centre of the magnet on the ordinary bisector is given by:

\(B = \frac{\mu_{0}M}{4\pi R^{3}}\)

Where,

μ₀​ = Permeability of free space = 4π × 10⁻⁷ TmA⁻¹

When the resultant field is inclined at 45° with earth’s field, B = H

\(∴ \frac{\mu_{0}M}{4\pi R^{3}} = H = 0.42\times 10^{-4}\)

\(R^{3} = \frac{\mu_{0}M}{0.42\times 10^{-4}\times 4\pi}\)

\(= R^{3} = \frac{4\pi \times 10^{-7}\times 5.25\times 10^{-2}}{4\pi \times 0.42\times 10^{-4}}\)

= 12.5 × 10⁻⁵

∴ R = 0.05 m = 5 cm

(b) The magnetic field at a distanced ‘R’ from the centre of the magnet on its axis is given as:

\(B’ = \frac{\mu_{0}2M}{4\pi R^{3}}\)

The resultant field is inclined at 45° with the earth’s field.

∴ B’ = H

\(\frac{\mu_{0}2M}{4\pi (R’)^{3}} = H\)

\((R’)^{3} = \frac{\mu_{0}\;2M}{4\pi \times H}\)

\(= \frac{4\pi \times 10^{-7}\times 2\times 5.25\times 10^{-2}}{4\pi\times 0.42\times 10^{-4}} = 25\times 10^{-5}\)

∴ R = 0.063 m = 6.3 cm

16. Answer the following questions:

(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?

(b) Why is diamagnetism, in contrast, almost independent of temperature?

(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?

(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?

(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

Answer:

(a) The thermal motion reduces at a lower temperature and the tendency to disrupt the alignment of the dipoles decreases.

(b) The dipole moment induced is always opposite to the magnetising field. Therefore, the internal motion of the atoms due to the temperature will not affect the magnetism of the material.

(c) Bismuth is diamagnetic substance. Therefore,  a toroid with bismith core will have a field slightly less than when the core is empty.

(d) Permeability of the ferromagnetic material depends on the magnetic field. Permeability is greater for lower fields.

(e) Proof of this important fact (of much practical use) is based on boundary conditions of magnetic fields (B and H) at the interface of two media. (When one of the media has µ >> 1, the field lines meet this medium nearly normally.)

(f) Yes. Apart from minor differences in strength of the individual atomic dipoles of two different materials, a paramagnetic sample with saturated magnetisation will have the same order of magnetisation. But of course, saturation requires impractically high magnetising fields.

17. Answer the following questions:

(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.

(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy? 

(c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.

Answer:

(a) The domain aligns in the direction of the magnetic field when the substance is placed in an external magnetic field. Some energy is spent in the process of alignment. When the external field is removed, the substance retains some magnetisation. The energy spent in the process of magnetisation is not fully recovered. It is lost in the form of heat. This is the basic cause for the irreversibility of the magnetisation curve of a ferromagnet substance.

(b) Carbon steel piece, because heat lost per cycle is proportional to the area of the hysteresis loop.

(c) Magnetisation of a ferromagnet is not a single-valued function of the magnetising field. Its value for a particular field depends both on the field and also on the history of magnetisation (i.e., how many cycles of magnetisation it has gone through, etc.). In other words, the value of magnetisation is a record or memory of its cycles of magnetisation. If information bits can be made to correspond to these cycles, the system displaying such a hysteresis loop can act as a device for storing information.

(d) Ceramics (specially treated barium iron oxides) also called ferrites.

(e) Surrounding the region with soft iron rings. Magnetic field lines will be drawn into the rings, and the enclosed space will be free of a magnetic field. But this shielding is only approximate, unlike the perfect electric shielding of a cavity in a conductor placed in an external electric field.

18. A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? 

(At neutral points, magnetic field due to a current-carrying the cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer:

Current in the wire = 2.5 A

The earth’s magnetic field at a location, R= 0.33 G = 0.33 x 10^-4 T

Angle of dip is zero, δ = 0

Horizontal component of earth’s magnetic field, B_H= R cosδ = 0.33 x 10^-4 Cos 0 = 0.33 x 10^-4 T

Magnetic field due to a current carrying conductor, B_c = (μ₀/2π) x (I/r)

B_c = (4π x 10^-7/2π) x (2.5/r) = (5 x 10^-7/r)

B_H = B_c

0.33 x 10^-4  = 5 x 10^-7/r

r =  5 x 10^-7/0.33 x 10^-4 

= 0.015 m = 1.5 cm

Hence neutral points lie on a straight line parallel to the cable at a perpendicular distance of 1.5cm.

19. A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Answer:

First let us decide the direction which would best represent the situation.

We know that

B_H = B cos δ

= 0.39 × cos 35^o G

B_H = 0.32G

Here

B_V = B sin δ

= 0.39 × sin 35^o G

B_V = 0.22G

It is given that the telephone cable carry a total current of 4.0 A in the direction east to west. So the resultant magnetic field 4.0 cm below.

B_wire = \(\frac{\mu _{0}}{4\pi }\frac{2I}{r}\)

= \(10^{-7}\times \frac{2\times 4}{4\times 10^{-2}}\)

= 2 × 10^-5 T

= 0.2G

Net magnetic field

\(B_{net}=\sqrt{(B_{H}-B_{wire})^{2}+B_{V}^{2}}\)

\(=\sqrt{(0.12)^{2}+(0.22)^{2}}\)

\(=\sqrt{0.0144+0.0484}\)

= 0.25G

The resultant magnetic field at points 4 cm below the cable 0.25 G.

20. A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

(a) Determine the horizontal component of the earth’s magnetic field at the location.

(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Answer:

Number of turns = 30

Radius of the coil = 12 cm

Current in the coil = 0.35 A

Angle of dip, δ = 45⁰

(a) Horizontal component of earth’s magnetic field,

B_H = B sinδ

B is the magnetic field strength due to the current in the coil

B = (μ₀/4π) (2πnI/r)

= (4π x 10^-7/4π) (2π x 30 x 0.35/0.12)

= 5.49 x 10^-5 T

Therefore, B_H = B sinδ

=  (5.49 x 10^-5 ) sin 45°

= 3.88 x 10^-5 T = 0.388 G

(b)  The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise direction. The needle will point from east to west.

21. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10^–2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?

Answer:

Magnitude of one of the magnetic field, B₁ = 1.2 × 10^–2 T

Let the magnitude of the other field is B₂

Angle between the field, θ = 60°

At stable equilibrium, the angle between the dipole and the field B₁, θ₁ = 15°

Angle between the dipole and the field B₂, θ₂ =θ – θ₁ = 45°

Torque due to the field B₁ = Torque due to the field B₂

MB₁ sin θ₁ = MB₂ sin θ₂

here , M is the magnetic moment of the dipole

B₂  = MB₁ sin θ₁/M sin θ₂

= (1.2 × 10^–2 ) x sin 15°/sin 45° = 4.39 x 10^-3 T

Magnetic field due to the other magnetic field is 4.39 x 10^-3 T

22. A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (m_e = 9.11 × 10^–31 kg). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

Answer:

Magnetic field, B = 0.04 G

Mass of the electron, m_e = 9.11 × 10^–31 kg

Distance to which the beam travels, d = 30 cm = 0.3 m

Kinetic energy of the electron beam is

E = (1/2) mv²

\(v=\sqrt{\frac{2\times 18\times 10^{3}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}= 0.795\times 10^{8} m/s\)

The electron beam deflects along the circular path of radius r.

The centripetal force is balanced by the force due to the magnetic field

Bev = mv²/r

r = mv/Be

= (9.11 x 10^-31 x 0.795 x 10⁸)/(0.04 x 10^-4 x 1.6 x 10^-19)

= (7.24 x 10^-23)/(0.064 x 10^-23)

= 113.125

Let the up and down deflection of the beam be x = r (1 – cosθ)

here, θ is the angle of deflection

sin θ = d/r

= 0.3/113.12 = 0.0026

θ = sin ^-1 (0.0026)= 0.1489°

x = r (1 – cosθ) = 113.12 (1 – cos 0.1489°)

= 113.12 (1- 0.999) 

= 113.12 x 0.01 

= 1.13 mm

23. A sample of paramagnetic salt contains 2.0 × 10²⁴ atomic dipoles each of dipole moment 1.5 × 10^–23 J T^–1. The sample is placed under a homogeneous magnetic field of 0.64 T and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)

Answer:

Number of diatomic dipoles =  2.0 × 10²⁴

Dipole moment of each dipole, M’= 1.5 × 10^–23 J T^–1

Magnetic field strength, B₁= 0.64 T

Cooled to a temperature, T₁= 4.2 K

Total dipole moment of the sample = n x M’ = 2.0 × 10²⁴  x 1.5 × 10^–23 = 30

Degree of magnetic saturation = 15 %

Therefore, M₁ = (15/100) x 30 = 4.5 J/T

Magnetic field strength, B₂ = 0.98 T

Temperature, T₂ = 2.8 K

The ratio of magnetic dipole from Curie temperature,

\(\frac{M_{2}}{M_{1}}=\frac{B_{2}}{B_{1}}\times \frac{T_{1}}{T_{2}}\)

\(M_{2}=M_{1}\frac{B_{2}}{B_{1}}\times \frac{T_{1}}{T_{2}}\)

\(M_{2}=4.5\frac{0.98}{0.64}\times \frac{4.2}{2.8}\)

= 18.52/1.79 

= 10.34 J/T

24. A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

Answer:

Mean radius of the Rowland ring = 15 cm

Number of turns = 3500

Relative permeability of the core, μ_r = 800

Magnetising current, I = 1.2 A

Magnetic field at the core, \(B = \frac{\mu _{r}\mu _{0}IN}{2\pi r}\)

\(B = \frac{800\times 4\pi \times 10^{-7}\times 1.2\times 3500}{2\pi \times 0.15}= 4.48 T\)

The magnetic field in the core is 4.48 T

25. The magnetic moment vectors µ_s and µ_l associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron, are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:

µ_s= –(e/m) S,

µ_l= –(e/2m)l

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Answer:

Of the two, the relation µ_l = – (e/2m)l is in accordance with classical physics. It follows easily from the definitions of µ_l and l:
µ_l =  IA =  (e/T) πr²    ........(1)

l = mvr = m (2πr²/T)   .......(2)

where r is the radius of the circular orbit which the electron of mass m and charge (–e) completes in time T. Dividing (1) by (2).

Clearly, µ_I/l = [ (e/T) πr²  ]/[m (2πr²/T) ] = –(e/2m)

Therefore, µ_I= (−e /2 m) l

Since the charge of the electron is negative (–e), it is easily seen that µ_I and l are antiparallel, both normal to the plane of the orbit.

Note µ_s/S in contrast to µ_l /l is e/m, i.e., twice the classically expected value. This latter result (verified experimentally) is an outstanding consequence of modern quantum theory and cannot be obtained classically.