1. Real Numbers

Exercise 1.2

Class X

Q.No. 1. Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution:

(i)                 140

Prime factors of 140 = 2x2x5x7 Ans.

(ii)               156

Prime factors of 156 = 2x2x3x13 Ans.

(iii)             3825

Prime factors of 3825 = 3x3x5x5x17 Ans.

(iv)             5005

Prime factors of 5005 = 5x7x11x13 Ans.

(v)               7429

Prime factors of 7429 = 17x19x23 Ans.

Q.No. 2. Find the LCM and HCF of the following pairs of integers and verify that LCMxHCF = Product of the two numbers.

(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Solution:

(i)                 26 and 91

Prime factors of 26 = 2x13

Prime factors of 91 = 13x7

LCM = 2x7x13 = 182

HCF = 13

LCM x HCF = 182x13=2366

Product of number = 26x91 = 2366

LCMxHCF = Product of number is verified.

(ii)               510 and 92

Prime factors of 510 = 2x3x5x17

Prime factors of 92 = 2x2x23

LCM = 2x2x3x5x17x23 = 23460

HCF = 2

LCMxHCF = 23460x2 = 46920

Product of number = 46920

Hence, LCMxHCF = Product of number is verified.

(iii)             336 and 54

Prime factors of 336 = 2x2x2x2x3x7

Prime factors of 54 = 2x3x3x3

LCM = 2x2x2x2x3x3x3x7 = 3024

HCF = 2x3 = 6

LCMxHCF = 18144

Product of number = 336x54 = 18144

Hence, LCMxHCF = Product of number is verified.

Q.No. 3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15and 21 (ii) 17, 23 and 29 (iii) 8,9 and 25

Solution:

(i)                 12, 15 and 21

Prime factorisation of 12 = 2x2x3

Prime factorisation of 15 = 3x5

Prime factorisation of 21 = 3x7

LCM = 2x2x3x5x7 = 420

HCF = 3

(ii)               17, 23 and 29

Prime factorisation of 17 = 17

Prime factorisation of 23 = 23

Prime factorisation of 29 = 29

LCM = 17x23x29 = 11339

HCF = 1

(iii)             8, 9 and 25

Prime factorisation of 8 = 2x2x2 = 8

Prime factorisation of 9 = 3x3

Prime factorisation of 25 = 5x5

LCM = 2x2x2x3x3x5x5 = 1800

HCF = 1

Q.No. 4.Given that HCF (306, 657) = 9 , find LCM (306, 657)

Solution:

LCMxHCF = Product of numbers

or, LCM = Product of numbers/HCF

or, LCM = 306x657/9

or, LCM = 22338 Ans.

Q.No. 5. Check whether  can end with the digit 0 for any natural number n.

Solution:

Prime factorisation of 6 = 2x3

So, the prime factorisation of 6 is not divisible by 5.

Hence, cannot end with 0 for any natural number n according to Fundamental Theorem of Arithmetic.

Q.No. 6. Explain why 7x11x13 + 13 and 7x6x5x4x3x2x1 + 5 are composite numbers.

Solution:

These numbers can be expressed as the form of the factorisation of prime numbers.

So, these numbers are composite numbers.

Q.No. 7. There is circular path around a sports field. Sonia takes 18 minutes to drive one round of the field. while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go the same direction. After how many minutes will they meet again at the starting point.

Solution:

Prime factorisation of 18 = 2x3x3

Prime factorisation of 18 = 2x2x3x3

LCM of 12 and 18 = 2x2x3x3 = 36

So, they will meet again after 36 minutes. Ans.

1. Real Numbers

Class X

NCERT Text Book Solution

Exercise 1.1

1. Use Euclid's division algorithm to find the HCF of 

(i)135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Solution:

(i)

From question it is clear that

225 > 135.

Apply Euclid's division lemma, to 135 and 225.

225 = 135x1 + 90

135 = 90x1 + 45

90 = 45x2 + 0

The remainder has now become zero.

The HCF of 225 and 135 is 45 Ans.

(ii)

From question it is clear that 

38220>196

Apply Euclid's division lemma, to 38220 and 196.

38220 = 196x195 + 0

The remainder has now become zero.

The HCF of 38220 and 196 is 195 Ans.

(iii)

From question it is clear that

867>255

Apply Euclid's division lemma, to 867 and 255.

867 = 255x3 + 102

255 = 102x2 + 51

102 = 51x2 + 0 

The remainder has now become zero.

The HCF of 867 and 255 is 51 Ans

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:

Let a be any positive integer and b = 6.

Now, by Euclid's division algorithm,

a = 6q + r for some integer q ≥ 0,

r = 0, 1, 2, 3, 4, 5

Because 0 ≤ r < b

So, a can be

6q,or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5.

Since a is odd, a cannot be 6q, or 6q + 2, or 6q + 4 ( since they are all divisible by 2)

Therefore, any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5. Proved.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?

Solution:

For this, it will be find out the HCF of 616 and 32.

Apply Euclid's algorithm, to 616 and 32.

616 = 32x19 + 8

32 = 8x4 + 0

So, the HCF of 616 and 32 is 8.

Therefore, the maximum number of columns in which they can march = 8 Ans.

4. Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m, or 3m + 1 for some integer m.

Solution:

Let a be any positive integer and b = 3.

Now, Euclid's division lemma,

a = 3q + r for some integer q ≥ 0,

Since, 0 ≤ r < 3

a can be 3q, or 3q + 1, or 3q + 2

Therefore,

a² = 9q², or 9q² + 6q + 1, or 9q² + 12q + 1.

or, a² = 3x3q², or 3( 3q² + 2q) + 1, or 3(3q² + 4q) + 1

So, the square of any positive integer is either of the form 3m, or 3m + 1. Proved

5. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

Let a be any positive integer and b = 9.

Now, Euclid's division lemma,

a = 9q + r for some integer q ≥ 0,

Since, 0 ≤ r < 9

a can be 9q, or 9q + 1, or 9q + 2, 9q + 3, or ...... 9q + 8.

Therefore,

a³ = 9x81q³, or 9(81q³ + 27q² + 3q)+ 1, or 9(81q³ + 54q² + 6q) + 8 ..................

So, the cube of any positive integer is of the form 9m,or 9m + 1 or 9m + 8. Proved.

4.7 Ampere's Circuital Law

Subject: Physics

Class XII

Chapter: 4. Moving Charges and Magnetism

Ampere's Law

 "The integral of magnetic field around a closed path is equal to the product of absolute permeability and total current passing through it."

Mathematical form

∮B.dl = μ₀.I

Here,

B = Magnetic field

dl = Current element

μ₀ = Absolute Permeability

I = Current passing through it

Proof

Consider about an infinitesimal carrying conductor through which I current passing is passing. Magnetic field lines are produced around the conductor as the form of concentric circles.

Magnetic field at a distance due to this infinitesimal conductor. 

B = μ₀.2I/4πa ................... (i)

Now consider about a circle of radius a having a current element dl.

Both B and dl have same direction because B is along the the tangent of circle.

Hence,

B.dl = Bdlcosθ = Bdlcos0⁰ = Bdl

Now integrate the closed path

∮B.dl = ∮Bdl .............. (ii)

Now putting the value of B

∮B.dl = ∮μ₀.2Idl/4πa

∮B.dl  = μ₀.2I/4πa∮dl

For complete cirecle

∮dl = 2πa

∮B.dl  = μ₀.2I.2πa/4πa

∮B.dl = μ₀.I

Proved

4.6 Magnetic Field on the Axis of Circular Current Loop

Subject: Physics

Class : XII

Chapter 4. Moving Charges and Magnetism

4.6 Magnetic Field on the Axis of Circular Current Loop

Consider about a circular loop of radius R carrying the current I. The loop is placed in the y-z plane with its centre at the origin O. The x axis is the axis of the loop. There is a point P at a distance x from the centre of the loop at which magnetic field is to be determined.

Mathematical Calculation

dl is the conducting element of the loop.

The magnitude of dB of the magnetic field due to dl is according to Biot-Savart Law

dB = 𝝁₀I|dlxr|/4𝜋r³
Now 
r² = x² + R²
Any element to loop will be perpendicular to the displacement vector r from dl to the axial point P is in the x-y plane.
Hence,
|dlxr| = rdl
∴dB = 𝝁₀Idlr/4𝜋r³ ............... (i)
Now putting the value of r
dB = 𝝁₀Idl/4𝜋(x² + R²)
The direction of dB is perpendicular to the plane formed by dl and r.
It has an x- component dBₓ and a component perpendicular to x-axis dB⫡.
When the components perpendicular to the x- axis are summed over, they cancel out.
Only the x component survives.
∴dBₓ = dBcos𝜃
From figure 
cos𝜃 = R/√(x² + R²) ............. (ii)
Now from eqn. no. (i) and (ii)
dBₓ = 𝝁₀IdlR/4𝜋(x² + R²)³/²
For whole circular loop
dl = 2𝜋R
Magnetic field at P due to entire circular loop
= B = Bₓi = 𝝁₀IR²/2𝜋(x² + R²)³/²i
Field at the centre of the loop. Here x = 0
B₀ = 𝝁₀I/2R

The direction of the magnetic field is determined by the help of 
Fleming Right Hand Thumb rule.

11. Alcohols, Phenols and Ethers

Alcohols, Phenols and Ethers

Class XII

Subject : Chemistry

Alcohols

Alcohols are compounds in which a hydroxyl(-OH) group attached to the saturated carbon atom.

General formula of alcohol - R-OH

The hydroxyl group is the functional group of alcohols.

Alcohols containing one hydroxyl are called Monohydric Alcohols.

Alcohols with two, three, or more hydroxyl groups are known as Dihydric Alcohols, Trihydric Alcohols, and Polyhydric Alcohols respectively.

For example,

Primary, Secondary and Tertiary Alcohols

Monohydric alcohols are classified as primary, secondary or tertiary depending upon whether the -OH group is attached to primary, secondary or a tertiary carbon.

Structure

Consider about the structure of Methyl Alcohol (CH₃OH). In methyl alcohol both oxygen and carbon are sp³ hybridized. Two of the sp³ orbitals of oxygen are completely filled and cannot take part in bond formation.

The C-O bond methyl alcohol is formed by overlap of an sp³ orbital of carbon and an sp³ orbital of oxygen. The O-H bond is formed by overlap of an sp³ orbital of oxygen and s orbital of hydrogen. The C-O-H bond angle is 105⁰. It is less than the normal tetrahedral angle. This is because the two completely filled sp³ orbitals of oxygen repel each other. This result in reduction of the bond angle.

Structure of Methyl Alcohol is given below

Continue ................ 

4.4.2 Cyclotron Class XII.

The Cyclotron is a machine which is used to accelerate the charged particles.

This machine was invented by E.O. Lowrence and M.S Livingston. 

Construction

It has two hollow chambers of D shaped is called dees. There is some gap between the dees in which source of positive particles are kept. These dees are connected by high frequent oscillator which provides high frequent electric field between the gap of dees. This machine is placed between the two powerful electromagnetic poles.

Working Principle

Positive charged particles accelerated from D1 to D2 when D1 and D2 are positive and negative respectively. In D2 charged particles accelerated perpendicular to the magnetic field.

The schematic figure of Cyclotron is drawn below.

The charged particles move in a semi circular path perpendicular to the magnetic field.

Let the time taken to complete one revolution is T.

T = 2πm/qB
Frequency = 1/T
= qB/2πm

During the circular motion of charged particles.
mv²/r = qvB
v = qBr/m

Kinetic Energy = mv²/2
Now, putting the value of v.
Kinetic Energy = q²B²r²/2m.

Time of revolution is independent of speed of particles and radius of trajectory.

Uses of Cyclotron are as follow

1. Uses in the nuclear reactor plant
2. To implant ions into solids
3. To synthesis materials
4. To produce radio active substances in the hospital.

3. Atoms and Molecules IX Chapter 3

Page 32

Q.No. 1:

In a reaction 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g

of carbon dioxide, 0.9g water and 8.2 g of sodium ethanoate. Show that these observations are in

agreement with the law of conservation of mass.

Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Answer:

In a reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate,

carbondioxide, and water.

 Sodium + Ethanoic → Sodium + Carbon + Water

Carbonate acid ethanoate dioxide

Mass of sodium carbonate = 5.3g (Given)

Mass of ethanoic acid = 6g (Given)

Mass of sodium ethanoate = 8.2g (Given)

Mass of carbon dioxide = 2.2 (Given)

Mass of water = 0.9g (Given)

Now, total mass before the reaction = (5.3 + 6)g

= 11. 3g

and total mass after the reaction = (8.2 + 2.2 + 0.9)g

= 11.3g

Therefore, Total mass before the reaction = Total mass after the reaction

Hence, the given observations are in agreement with the law of conservation of ma

Page 33

Q.No. 2:

Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen

gas would be required to react completely with 3g of hydrogen gas?

Answer:

It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8. Then, the mass of

oxygen gas required to react completely with 1g of hydrogen gas is 8g. Therefore, the mass of

oxygen gas required to react completely with 3g of hydrogen gas is 8 × 3g = 24 g.

Q.No. 3:

Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?

Answer:

The postulate of Dalton's atomic theory which is a result of the law of conservation of mass is

“Atoms are indivisible particles, which can neither be created nor destroyed in a chemical

reaction”.

Q.No. 4:

Which postulate of Dalton's atomic theory can explain the law of definite proportions?

Answer:

The postulate of Dalton's atomic theory which can explain the law of definite proportion is “The

relative number and kind of atoms in a given compound remains constant”.

Page 35

Q.No. 1:

Define atomic mass unit.

Answer:

Mass unit equal to exactly one- twelfth the mass of one atom of carbon - 12 is called one atomic

mass unit. It is written as 'u'.

Q.No. 2:

Why is it not possible to see an atom with naked eyes?

Answer:

The size of an atom is so small that it is not possible to see it with naked eyes. Also, atom of an

element does not exist independently.

Page 39

Q.No. 1:

Write down the formula of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium suphide

(iv) magnesium hydroxide

Answer:

(i) Sodium oxide → Na2O

(ii) Aluminium chloride → AlCl3

(iii) Sodium suphide → Na2S

(iv) Magnesium hydroxide → Mg(OH)2

Q.No. 2:

Write down the names of compounds represented by the following formula:

(i) Al2(SO4)3

(ii) CaCl2

(iii) K2SO4

(iv) KNO3

(v) CaCO3

Answer:

(i) Al(SO4)3→ Aluminium sulphate

(ii) CaCl2→ Calcium chloride

(iii) K2SO4→ Potassium sulphate

(iv) CaCO3→ Calcium carbonate

Q.No. 3:

What is meant by the term chemical formula?

Answer:

The chemical formula of a compound means the symbolic representation of the composition of a

compound. From the chemical formula of a compound, we can know the number and kinds of

atoms of different elements that constitute the compound. For example, from the

chemical formula CO2 of carbon dioxide, we come to know that one carbon atom and two oxygen

atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

Question 4:

How many atoms are present in a

(i) H2S molecule and

(ii) PO4

3-

ion?

Answer :

(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.

(ii) In a PO4

3-

ion, five atoms are present; one of phosphorus and four of oxygen.

Page 40

Question 1:

Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Answer:

Molecular mass of H2 = 2 × Atomic mass of H

= 2 × 1 = 2u

Molecular mass of O2 = 2 × Atomic mass of O

= 2 × 16 = 32u

Molecular mass of Cl2 = 2 × Atomic mass of Cl

= 2 × 35.5 = 71 u

Molecular mass of CO2 = Atomic mass of C + 2 × Atomic mass of O

= 12 + 2 × 16 = 44 u

Molecular mass of CH4 = Atomic mass of C + 4 × Atomic mass of H

= 12 + 4 × 1 = 16 u

Molecular mass of C2H6 = 2 × Atomic mass of C + 6 × Atomic mass of H

= 2 × 12 + 6 × 1 = 30u

Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H

= 2 × 12 + 4 × 1 = 28u

Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H

= 14 + 3 × 1 =17 u

Molecular mass of CH3OH Atomic mass of C+4 ×Atomic mass of H+Atomic mass of O

= 12 + 4 × 1 + 16 = 32 u

Q.No. 2:

Calculate the formula unit masses of ZnO, Na2O, K2CO3, given masses of Zn = 65u, Na = 23u, K

= 39u, C = 12u, and O = 16u.

Answer :

Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O

= 65 + 16 = 81 u

Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O

= 2 × 23 + 16 = 62u

Formula unit mass of K2CO3

= 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O

= 2 × 39 + 12 + 3 × 16 = 138u

Page 42

Question 1:

If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?

Answer:

One mole of carbon atoms weighs 12g (Given)

i.e., mass of 1 mole of carbon atoms = 12g

Then, mass of 6.022× 1023 number of carbon atoms = 12g

Therefore, mass of 1 atom of carbon =

12

6.022× 10– 23 g g

= 1.9926 × 10– 23 g

Q.No. 2:

Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass

of Na = 23u, Fe =56 u)?

Answer 2:

Atomic mass of Na = 23u (Given)

Then, gram atomic mass of Na = 23g

Now, 23g of Na contains = 6.022×1023 number of atoms

Thus, 100g of Na contains =

6.022×1023 ×10⁻²³

number of atoms

= 2.6182 × 1024 number of atoms

Again, atomic mass of Fe = 56u (Given)

Then, gram atomic mass of Fe = 56g

Now, 56 g of Fe contains = 6.022×1023 number of atoms

Thus, 100 g of Fe 6.022×1023 ×100

56

number of atoms

= 1.0753 × 1024 number of atoms

Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Q.No. 5:

Give the names of the elements present in the following compounds:

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

Answer:

Compound Chemical formula Elements present

Quick lime CaO Calcium, oxygen

Hydrogen bromide HBr Hydrogen, bromine

Baking powder NaHCO3 Sodium, hydrogen, carbon,

oxygen

Potassium sulphate K2SO4 Potassium, sulphur, oxygen

Q.No. 6:

Calculate the molar mass of the following substances:

(a) Ethyne, C2H2

(b) Sulphur molecule, S8

(c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO3

Answer:

(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28g

(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256g

(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124g

(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g

(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63g

Q.No. 7:

What is the mass of

(a) 1 mole of nitrogen atoms?

(b) 4 mole of aluminium atoms (Atomic mass of aluminium = 27)?

(c) 10 moles of sodium sulphite (Na2SO3)?

Answer:

(a) The mass of 1 mole of nitrogen atoms is 14g.

(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g

(c) The mass of 10 moles of sodium sulphite (Na2SO3) is 10 × [2 × 23 + 32 + 3 × 16]g

= 10 × 126g = 1260g

Q.No. 8:

Convert into mole.

(a) 12g of oxygen gas

(b) 12g of water

(c) 22g of carbon dioxide

Answer:

(a) 32 g of oxygen gas = 1 mole

Then, 12g of oxygen gas = 12/32 mole = 0.375 mole

(b) 18g of water = 1 mole

Then, 20 g of water = 20/18 mole = 1.11 moles (approx.)

(c) 44g of carbon dioxide = 1 mole

Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole

Q.No. 9:

What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Answer:

(a) Mass of one mole of oxygen atoms = 16g

Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g

(b) Mass of one mole of water molecule = 18g

Then, mass of 0.5 mole of water molecules = 0.5 × 18g = 9g

Question 10:

Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.

Answer 10:

 1 mole of solid sulphur (S8) = 8 × 32g = 256g

i.e., 256g of solid sulphur contains = 6.022 × 1023 molecules

Then, 16g of solid sulpur contains 6.022×1023

256

× 16 molecules

= 3.76 × 1022 molecules (approx)

Q.No. 11:

Calculate the number of aluminium ions present in 0.051g of aluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al

= 27u)

Answer:

1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g

i.e., 102g of Al2O3 = 6.022 × 10⁻²³ molecules of Al2O3

Then, 0.051 g of Al2O3 contains =

6.022 × 10⁻²³ 

102

× 0.051 molecules

= 3.011 × 1020

 molecules of Al2O3

The number of aluminium ions (Al3+

) present in one molecules of aluminium oxide is 2.

Therefore, The number of aluminium ions (Al3+

) present in

3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020

= 6.022 × 1020

Write for the Answers of other Chapter.

For any error report to me