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Chemical formula -CaOCl₂

Chemical equation -

Ca(OH)₂+Cl₂  →CaOCl₂+H₂O

Uses:

(i) For bleaching cotton and linen in the textile industry 

(ii) As an oxidising agent in a chemical industry.

(iii) For disinfecting water.

Baking powder

Baking soda and an edible acid like tartaric acid. 

Baking soda (NaHCO₃) is used to release CO₂ gas when heated.

Tartaric acid is used to avoid the bitter taste by reacting with the Na₂CO₃ formed

(i) Compound is NaHCO₃/baking soda/sodium hydrogen carbonate

Manufacture - NH₃ + NaCl + H₂O + CO₂ → NH₄Cl+NaHCO₃

(ii)2NaHCO₃ +Heat →  Na₂CO₃ + H₂O + CO₂  

Detailed Answer:

(i) The Compound is NaHCO₃ (Sodium hydrogen carbonate)(Commonly known as baking soda)

Manufacturing Sodium hydrogen carbonate :

NH₃ + NaCl + H₂O + CO₂ → NH₄Cl+NaHCO₃  

(ii) On heating, Baking Soda decomposes into Sodium carbonate, water and carbon diofde.

2NaHCO₃ +Heat →  Na₂CO₃ + H₂O + CO₂  

Sodium hydrogen carbonate is an essential ingredient in antacids because it neutralizes the effect HCl which is released in the stomach. So it is called as an antacid 

NaHCO ₃+ HCl -----------> NaCl + H₂O +CO₂

SI unit of energy is Joule (J).

Power is defined as the rate of doing work. It is the ratio of the work done (W) and the time taken (t).

Power (P) = \(\frac{Work(W)}{Time(t)}\)

The S.I unit of the power is Watt (W) in honour of the physicist James Watt.

SI unit of work is Joule. It is the scalar quantity. It is defined as the Work done by force of 1 N to move an object by 1 m. 

SI unit of Power is Watt denoted by W.

Following are the two examples where a body possess both kinetic energy as well as potential energy: 

a) a flying airplane 

b) A man climbing stairs

W = F × S

SI unit of F is N and that of S is m [N = Newton]

So, SI unit of work=N x m

1N m is defined as 1 joule.

i.e., 1 joule = 1 N m

So, SI unit of work is joule

• Work W = F × s
• Units of work ‘N-m’ (or) ‘Joule’.

• Some force must act on the object

• The point of application of force must move in the direction of force

The product of the force and the distance moved measures work done.

W = F × S

Where W is the work done, F is the force applied and S is the distance covered by the moving object. Work done is a scalar quantity.

Two conditions need to be satisfied in order to say that work has taken place. One is a force that should act on the object and another is the object must be displaced or there must be a change in the position of the object.

C) both A and B

Correct option is A) energy

D) potential energy

B) Seetha is pulling a toy car

Force exerted, F = 140N.
Displacement, S = 15m
Work done in ploughing the field
W = F × S
= 140 × 15
= 2100J
= 2.1 × 10³ J.

The magnitude of the relative velocity will remain same, i.e., no effect on its magnitude.

Magnet is fitted in the cap of a pin holder and in the door of a fridge. Iron objects stick to the magnet.

i) I observe that the uniformly spread iron filings come close and get concentrate at two points of the paper sheet. At some distance, I found some scattered iron filings between these two points.

ii) No. the ends of the bar magnet attract more iron filings than the middle part of the magnet.

iii) Yes, uniformly spread iron filings changed their pattern and concentrated more at endpoints of the bar magnet. Scattered iron filings between these two points are somehow in some lines from one point to other.

The material to which objects made from iron, nickel, cobalt get attracted is called as magnet.

i. True

ii. True

iii. True

iv. False: The north pole is indicated by Wand the south pole is indicated by ‘S’.

v. True

i. False: Material alnico is a mixture ofaluminium, nickel and cobal.

ii. False: Magnetism of electromagnet is temporary.

iii. True

iv. False: Like poles repel each other and unlike poles attract each other.

v. False: The magnetic force is concentrated at the poles of the magnet

No two boys and no two girls are together means the boys & girls must sit alternately. 

Four boys can be permuted in 4! ways & the 5 girls in 5! ways. 

∴ The total number of ways is 5! × 4! ways.

(c) 144000

First of all we arrange the 5 boys in 5 ! ways. 

Then we arrange the 3 girls in the remaining 6 places between the 5 boys and on the extreme in ⁶P₃ ways. 

× B × B × B × B × B × 

∴ Required number of ways = 5 ! × ⁶P₃ 

= 120 × 6 × 5 × 4 = 144000

There is a total of 13 marbles in a bag.

Out of these 5 are Red, 4 Blue, and 4 are Green marbles.

All balls of the same colour are taken to be identical.

∴ Required number of arrangements = \(\frac{13!}{5!4!4!}\)

The seating arrangement may be done as desired in two operations. 

(i) First we fix the positions of 6 boys. Their positions are indicated by B₁, B₂,...., B₆. x B₁ x B₂ x B₃ x B₄ x B₅ x B₆ x 

This can be done in 6 ! ways. 

(ii) Now if the positions of girls are fixed at places (including those at the two ends) shown by the crosses, the four girls will never come together. In any one of these arrangements there are 7 places for 4 girls and so the girls can sit in ⁷P₄ ways. 

Hence the required number of ways of seating 6 boys and 4 girls under the given condition 

= ⁷P₄ × 6! = 7 × 6 × 5 × 4 × 6 × 5 × 4 × 3 × 2 × 1 = 604800.

There are 6 letters in the word INDIAN in which I and N are repeated twice. Number of different words that can be formed using the letters of the word INDIAN = \(\frac{6!}{2!2!}\) = \(\frac{6\times5\times4\times3\times2!}{2\times2!}\) =180

When two N’s are together. Let us consider the two N’s as one unit. They can be arranged with 4 other letters in \(\frac{5!}{2!}=\frac{5\times4\times3\times2!}{2!}\) = 60 ways.

∴ 2 N can be arranged in \(\frac{2!}{2!}\) = 1 way.

∴ Required number of words = 60 x 1 = 60

(i) 5 books arranged in ⁵P₅ = 5! = 120 ways.

(ii) 2 books are together.

Let us consider two books as one unit. This unit with the other 3 books can be arranged in ⁴P₄ = 4! = 24 ways.

Also, two books can be arranged among themselves in ²P₂ = 2 ways.

∴ Required number of arrangements = 24 × 2 = 48

(iii) Say books are B₁ , B₂ , B₃ , B₄ , B₅ are to be arranged with B₁ , B₂ never together.

B₃ , B₄ , B₅ can be arranged among themselves in ³P₃ = 3! = 6 ways.

B₃ , B₄ , B₄ create 4 gaps in which B₁ , B₂ are arranged in ⁴P₂ = 4 × 3 = 12 ways.

∴ Required number of arrangements = 6 × 12 = 72

(d) 3! × 5! × 4! × 3!

Let us consider all the books of same size as one entity. Now there are three different entities which have to be arranged on the shelf and this can be done is 3! ways. 

Also the five large books can be arranged amongst themselves in 5! ways, four medium books can be arranged amongst themselves in 4! ways and the three small books can be arranged amongst themselves in 3! ways. So,

Required number of ways of arranging the books so that all same sized books are together = 3! × 5! × 4! × 3!

Regarding the two particular books as one book, there are (n – 1) books now which can be arranged in ^{n – 1}P_{n –1}, i.e., (n – 1)! ways. Now, these two books can be arranged amongst themselves in 2 ! ways. Hence the total number of permutations in which these two books are placed together is 2 ! . (n – 1)!. The number of permutations of n books without any restriction is n !. 

Therefore, the number of arrangements in which these two books never occur together 

= n !–2 ! .(n – 1) ! = n . (n – 1) ! – 2 . (n – 1) ! = (n – 2) . (n – 1) !

Total number of books = 3 × 4 = 12 

Each of the 4 different titles has 3 copies each 

∴ Required number of ways of arranging them on a shelf = \(\frac{12!}{3!3!3!3}\) = \(\frac{12!}{(3!)^4}\) = 369600.

There are 12 letters in the word MATHEMATICAL in which ‘M’ is repeated 2 times, ‘A’ repeated 3 times and ‘T’ repeated 2 times.

∴ Required number of arrangements = \(\frac{12!}{2!3!2!}\) 

When all the vowels,

i.e., ‘A’, ‘A’, ‘A’, ‘E’, ‘I’ are to be kept together.

Let us consider them as one unit.

Number of arrangements of these vowels among themselves = \(\frac {5!}{3!}\) ways.

This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.

∴ Number of such arrangements = \(\frac{8!}{2!2!}\) 

∴ Required number of arrangements = \(\frac{8!\times5!}{2!3!2!}\)

The required number of ways = \(\frac{(18)!}{(6!)^3}\)

A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places.

There are 4 odd digits, i.e. 3, 5, 7, 9.

∴ They can be arranged at 4 odd places among themselves in 4! = 24 ways.

There are 3 even digits, i.e. 4, 6, 8.

∴ They can be arranged at 3 even places among themselves in 3! = 6 ways.

∴ Required number of numbers formed = 24 × 6 = 144

(d) 313

As there are 4 digits 1, 2, 3, 4 and repetition of digits is allowed. 

Total number of 1-digit numbers = 4 

Total number of 2-digit numbers = 4 × 4 = 16 

Total number of 3-digit numbers = 4 × 4 × 4 = 64

Number of 4-digit numbers beginning with 1 = 4 × 4 × 4 = 64 

(∵ The first place is occupied by 1) 

Number of 4-digit numbers beginning with 2 = 4 × 4 × 4 = 64 

Number of 4-digit numbers beginning with 3 = 4 × 4 × 4 = 64 

Number of 4-digit numbers beginning with 41 = 4 × 4 = 16 

Number of 4-digit numbers beginning with 42 = 4 × 4 = 16 

Number of 4-digit numbers beginning with 431 = 4 

Number of 4-digit numbers beginning with 432 = 1 (4321 only) 

∴ Total number of 4-digit numbers = 64 + 64 + 64 + 16 + 16 + 4 + 1 = 229 

∴ Total number of natural numbers not exceeding 4321 = 4 + 16 + 64 + 229 = 313.

There are 7 letters in the word ARRANGE in which ‘A’ and ‘R’ repeat 2 times each.

∴ Number of ways to arrange the letters of word ARRANGE = \(\frac{7!}{2!2!}\) = 1260

Consider the words in which 2A are together and 2R are together.

Let us consider 2A as one unit and 2R as one unit.

These two units with remaining 3 letters can be arranged in = \(\frac{5!}{2!2!}\) = 30 ways.

Number of arrangements in which neither 2A together nor 2R are together = 1260 – 30 = 1230

1st representative can be selected from first section in 30 ways. 

2nd representative can be selected from second section in 30 ways. 

3rd representative can be selected from third section in 30 ways. 

∴ Required number of ways = 30 × 30 × 30 = 27000.

5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5.

Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5

Number of such numbers = \(\frac{5!}{2!}+\frac{5!}{2!}=5!=120\) 

Case II: Numbers are formed from 2, 2, 3, 3 and any one of 4 or 5

Number of such numbers = \(\frac{5!}{2!2!}+\frac{5!}{2!2!}=60\)

Required number of numbers = 120 + 60 = 180

Since the required numbers are less than 1000 therefore, they are 1-digit, 2-digit or 3-digit numbers. 

One-digit numbers. Only two odd one-digit numbers are possible, namely, 5 and 7. 

Two-digit numbers. For two-digit odd numbers the unit place can be filled up by 5 or 7 i.e., in two ways and ten’s place can be filled up by 2, 5 or 7 (not 0) in 3 ways. 

∴ No. of possible 2-digit odd numbers = 2 × 3 = 6. 

Three-digit numbers. For three-digit odd numbers, the unit place can be filled up by 5 or 7 in 2 ways. The ten’s place can be filled up by any one of the digits 0, 2, 5, 7 in 4 ways. The hundred’s place can be filled up by 2, 5 or 7 (not 0) in 3 ways. 

∴ No. of possible 3-digit numbers = 2 × 4 × 3 = 24 

Hence total number of odd numbers = 2 + 6 + 24 = 32.

Case I: Number of three-digit numbers formed from 2, 3, 4, 5, 6, greater than 400.

100’s place can be filled by any one of the numbers 4, 5, 6. 100’s place digit can be selected in 3 ways.

Since repetition is not allowed, 10’s place can be filled by any one of the remaining four numbers.

∴ 10’s place digit can be selected in 4 ways.

Unit’s place digit can be selected in 3 ways.

∴ Total number of three-digit numbers formed

= 3 × 4 × 3 = 36

Case II: Number of four-digit numbers formed from 2, 3, 4, 5, 6.

Since repetition of digits is not allowed,

1000’s place digit can be selected in 5 ways.

100’s place digit can be selected in 4 ways. 10’s place digit can be selected in 3 ways.

Unit’s place digit can be selected in 2 ways.

∴ Total number of four-digit numbers formed

= 5 × 4 × 3 × 2 = 120

Case III: Number of five-digit numbers formed from 2, 3, 4, 5, 6

 Similarly, since repetition of digits is not allowed,

Total number of five digit numbers formed

= 5 × 4 × 3 × 2 × 1 = 120.

∴ Total number of numbers that exceed 400

= 36 + 120 + 120 = 276

Any number between 100 and 1000 is of 3 digits. The unit’s place can be filled by 2 or 9 in 2 ways. 

Similarly ten’s place can be filled in 2 ways. 

The hundred’s place can also be filled in 2 ways.

∴ Required no. of numbers = 2 × 2 × 2 = 8.

Correct option  (d)  6666600

Explanation:

Sum of n digit numbers

= (Sum of digits) (10ⁿ  - 1)/(10 - 1)(n -n - 1)!

=  (1 + 3 + 5 + 7 + 9)(10⁵ - 1)/10 - 1(5 - 1)!

= 25 × 11111 x 24

= 6666600

Between any set of digits, the greatest number is possible when digits are arranged in descending order.

∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.

Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed,

1000’s place digit can be selected in 4 ways,

100’s place digit can be selected in 3 ways,

10’s place digit can be selected in 2 ways,

Unit’s place digit can be selected in 1 way.

∴ Total number of numbers not exceeding

7432 that can be formed with the digits 2, 3, 4, 7

= Total number of four-digit numbers possible from the digits 2, 3, 4, 7

= 4 × 3 × 2 × 1

= 24

A number of 6 different digits is to be formed from the digits 3, 4, 5, 6, 7, 8 which can be done in ⁶P₆ = 6! = 720 ways.

(a) If the number is to be divisible by 5, the unit’s place digit can be 5 only.

∴ it can be arranged in 1 way only.
The other 5 digits can be arranged among themselves in ⁵P₅ = 5! = 120 ways.

∴ Required number of numbers divisible by 5 = 1 × 120 = 120

(b) If the number is not divisible by 5, unit’s place can be any digit from 3, 4, 6, 7, 8. ∴ it can be arranged in 5 ways.

Other 5 digits can be arranged in ⁵P₅ = 5! = 120 ways.

∴ Required number of numbers not divisible by 5 = 5 × 120 = 600

Numbers between 100 and 1000 are 3-digit numbers.

A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where the unit place digit is 4.

Unit’s place digit is 4.

∴ it can be selected in 1 way only.

10’s place digit can be selected in 10 ways. For a 3-digit number, 100’s place digit should be a non-zero number.

∴ 100’s place digit can be selected in 9 ways.

∴ By using the fundamental principle of multiplication,

total numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90.

Numbers between 100 and 1000 are 3-digit numbers.

A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.

When 7 is in unit’s place:

Unit’s place digit is 7.

∴ it can be selected in 1 way only.

10’s place digit can be selected in 9 ways.

100’s place digit can be selected in 8 ways.

Total numbers which have 7 in unit’s place = 1 × 9 × 8 = 72.

When 7 is in 10’s place:

Unit’s place digit can be selected in 9 ways.

10’s place digit is 7.

∴ it can be selected in 1 way only. 100’s place digit can be selected in 8 ways.

∴ A total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72.

When 7 is in 100’s place:

Unit’s place digit can be selected in 9 ways. 10’s place digit can be selected in 9 ways.

100’s place digit is 7.

∴ it can be selected in 1 way only.

∴ Total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81.

∴ Total numbers between 100 and 1000

having digit 7 exactly once = 72 + 72 + 81 = 225

33! = 33 × 32 × 31 × 30 × 29 … × 16 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 

= 33 × (2)⁵ × 31 × 30 × 29 × … × (2)⁴ … × (2)³ × 7 × 6 × 5 × (2)² × 3 × (2)¹ × 1 

Now take all 2’s out we get 

33! = (2)⁵ ∙ (2)⁴ ∙ (2)³ ∙ (2)² ∙ (2)¹ [33 × 31 × 30 × 29 × … × 1] [33 × 31 × 30 × 29 × … × 1] 

contains numbers 6, 10, 12, 14, 18, 20, 22, 24, 26, 28, 30, 

which have 2 as one of their factors. 

So, the product of 2 present in these numbers 

= 2 × 2 × 2² × 2 × 2 × 2² × 2 × 2³ × 2 × 2² × 2 

= 2¹⁶ 

Therefore, we have 2¹⁵ × 2¹⁶ = 2³¹ 

Thus, 31 is the largest integer such that 33! Is divisible by 2n.

(i) ⁶C₃ = \(\frac{6\times5\times4}{1\times2\times3}\) = 20

^{(ii) 30}C_{28 =} ³⁰C_{28 - 20} = ³⁰C₂ = \(\frac{30\times29}{1\times2}\) = 435.