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Firstly we have to find the prime factors of 64

64 = 2 × 2 × 2 × 2 × 2 × 2

And the prime factors of 343

343 = 7 × 7 × 7

= \(\sqrt[3]{64/343}\)

= \(\sqrt[3]{(4 × 4 × 4/7 × 7 × 7)}\)

= 4/7

∴ 4/7 is the correct answer.

Firstly we have to find the prime factors of 125

125 = 5 × 5 × 5

And the prime factors of 64

64 = 2 × 2 × 2 × 2 × 2 × 2

\(\sqrt[3]{(125 × 64) }\)

= \(\sqrt[3]{(5 × 5 × 5 × 4 × 4 × 4) }\)

= \(\sqrt[3]{ (5 × 5 × 5) }\)× \(\sqrt[3]{ (4 × 4 × 4) }\)

= 5 × 4 = 20

∴ 20 is the correct answer.

The cube root of 8,57,375 is 95.

The cube root is denoted by the ³√ sign.

(i) ∛(0.027/0.008) ÷ √(0.09/0.04) – 1

= ∛(0.3×0.3×0.3/0.2×0.2×0.2) ÷ √(0.3×0.3/0.2×0.2)

= (∛(0.3)³ / ∛(0.2)³) ÷ (√(0.3)² / √(0.2)²)

= (0.3/0.2) ÷ (0.3/0.2) -1

= (0.3/0.2 × 0.2/0.3) -1

= 1 – 1

= 0

(ii) ∛(0.1×0.1×0.1×13×13×13)

= ∛(0.1³×13³)

= 0.1 × 13

= 1.3

(c) gravitational potential energy of the block

Explanation: 

In this case velocity of the block will change, hence (a) and (b) will change. Some part of the K.E. of the bullet will be converted into heat energy due to friction between bullet and the block during the embedment. Hence (d) will also change. Since height of the block does not change hence (c) will not change.

Cube root – 216 is -6.

(b) the kinetic energy but not the linear momentum

Explanation: 

If there is no external force the linear momentum is conserved because the velocity of CoM is not changed, so internal forces cannot change the linear momentum of the system. Since internal forces can change the velocity of the constituent particles of the system, K.E. of the system can change because even for negative velocity K.E. =½mv² will be positive.

The least natural number by 392 should be multiplied to get a number which is a perfect cube is 7.

(i) If only we know the mass of the particle, it's kinetic energy can be found out. Because both the K.E. and the magnitude of the momentum involve mass m and magnitude of the velocity v. 

(ii) Even if we know the mass of the particle, only the magnitude of the momentum can be calculated. The direction of the momentum is unknown. Momentum, being a vector thus cannot be found out.

-(27/343)

= 3√((3 x 3 x 3)/(7 x 7 x 7)) = -(3/7)

(i) 35

We have, a = 3 and b = 5

∴ The cube of 35 is 42875

(ii) 56

We have, a = 5 and b = 6

∴ The cube of  56 is 175616

(i) No. 

Consider the mechanical energy of a particle under gravity. The total mechanical energy = K.E. + P.E. 

K.E. is always positive because either v is positive or negative, v² will always be positive and so will ½mv² = K.E. but P.E. = mgh may be positive or negative because h is positive or negative depending upon its position with respect to the reference line. Suppose we take ground as the reference line. If the particle falls into a dry well, at some depth below ground level the magnitude of negative P.E. may be equal to K.E. thus making the total mechanical energy of the particle zero. But due to its velocity, the linear momentum is not zero. 

(ii) No. It is not necessarily nonzero. Suppose the particle is on the ground at rest. K.E. = 0, and P.E. = 0. Total mechanical energy = 0. Since velocity is zero, the linear momentum is also zero.

-(512/343) = - 3√(512/343)

- 3√(8 x 8 x 8)/(7 x 7 x 7)

= -(8/7)

i) 44447

Unit digit of 44447 is = 7

Cube of 7, 

= 7³ 

= 343

∴ Unit digit of cube of 44447 is always 3

ii) 125125125

Unit digit of 125125125 is = 5

Cube of 5. 

= 5³ 

= 125

∴ Unit digit of cube of 125125125 is always 5

Property : abc = 100a + 106 + c ………(i)

bca = 1006 + 10c + a ……..(ii)

and cab = 100c + 10a + b ……….(iii)

Adding, (i), (ii) and (iii), we get abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 3 x 37(a + b + c) 

Now, let us try this method on 

527 + 752 + 275 to check is it exactly divisible by 14 

Here, a = 5, 6 = 2, c = 7 

527 + 752 + 275 = 3 x 37(5 + 2 + 7) = 3 x 37 x 14 

Hence, it shows that 527 + 752 + 275 is exactly divisible by 14

The cube root of 343/512 is 7/8.

(i) 91125

Unit digit of 91125 in 5. 

Therefore the units digit in its cube root is 5. 

Let us split 91125 as 91 and 125. We find that 4³ = 64 < 91 < 125 = 5³ 

Hence 40³ = 64000 <91125 < 125000 = 50³ 

∴ Cube root of 91125 lies between 40 and 50 and units digit is 5 the only such number is 45.

∴ \(3\sqrt{91125} = 45\)

(ii) 166375

Units digit of 166375 is 5. 

Therefore the units digit of its cube root is 5. 

Let us split 166375 as 166 and 375. 

5³ = 125 < 166 < 216 = 6³ 

Hence 50³ = 125000, < 166375 < 216000 = 60³ 

∴ \(3\sqrt{166375}\) lies between 50 and 60.

Since the units digit is 5 the only such number in 55

 ∴ \(3\sqrt{166375}\) = 55

(iii) 704969

The units digit of 704969 is 9. Therefore the units digits of its cube root are 9. 

Let us split 704969 as 704 and 969 

8³ = 512 < 704 < 729 – 9³ 

Hence = 80³= 512000 < 704969 < 72900 = 90³

∴ \(3\sqrt{704969}\) lies between 80 and 90. 

Since the units digits is 9 the only such number is 89.

∴ \(3\sqrt{704969}\)  = 89

(i) Prime factorisation of 175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2  x 7 x 7 x 7 

∴  3√175616 = 2 x 2 x 2 x 7 = 56

(ii) Prime factorisation of 91125 = 3 x 3 x 3 x 3 x 3 x 3 x5 x 5 x 5

∴ 3√91125 = 3 x 3 x 5 = 45

(i) Prime factorisation of 13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2x 3 x 3 x 3

∴ 3√13824 = 2 x 2 x 2 x 3 = 24

(ii) Prime factorisation of 110592 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

∴ 3√110592 = 2 x 2 x 2 x 2 x 3 = 48

(iii) Prime factorisation of 46656 = 2 x 2 x 2 x 2 x 2 x 2  x 3 x 3 x 3 x 3 x3 x 3

∴ 3√46656 = 2 x 2 x 3 x 3 = 36

(i) -2744

2744 = 2 × 2 × 2 × 7 × 7 × 7

= 2³ × 7³

= 14³

2744 is a perfect cube.

∴ -2744 is a cube of negative integer – 14.

(ii) -42875

42875 = 5 × 5 × 5 × 7 × 7 × 7

= 5³ × 7³

= 35³

42875 is a perfect cube.

∴ -42875 is a cube of negative integer – 35.

(i) 226981

Unit digit of 226981 = 1

Cube root of 226981 = 1

(ii) 13824

Unit digit of 13824 = 4

 Cube root of 13824 = 4

(iii) 571787

Unit digit of 571787 = 7

Cube root of 571787 = 7

(iv) 175616

Unit digit of 175616 = 6

Cube root of 175616 = 6

(i) 64

64 – 1 = 63

63 – 7 = 56

56 – 19 =37

37 – 37 = 0

Here, subtraction is performed 4 times.

Hence cube root of 64 is 4.

(ii) 512

Let’s perform subtraction

512 – 1 = 511

511 – 7 = 504

504 – 19 = 485

485 – 37 = 448

448 – 61 = 387

387 – 91 = 296

296 – 127 = 169

169 – 169 = 0

Here, subtraction performed 8 times.

Hence cube root of 512 is 8.

(i) 343

∛343 = ∛(7×7×7) 

= 7

(ii) 2744

∛2744 = ∛ (2×2×2×7×7×7) 

= ∛ (23×73) 

= 2×7 

= 14

(iii) 4913

∛4913 = ∛(17×17×17) 

= 17

(iv) 1728

∛1728 = ∛(2×2×2×2×2×2×3×3×3) 

= ∛(23×23×33) 

= 2×2×3 

= 12

(v) 35937

∛35937 = ∛(3×3×3×11×11×11) 

= ∛ (33×113) 

= 3×11 

= 33

(vi) 17576

∛17576 = ∛(2×2×2×13×13×13) 

= ∛ (23×133) 

= 2×13 

= 26

(i) 792

792 – 1 = 791

791 – 7 = 784

784 – 19 = 765

765 – 37 = 728

728 – 61 = 667

667 – 91 = 576

576 – 127 = 449

449 – 169 = 280

280 – 217 = 63

Other number to be subtracted is 271, which is greater than 63

∴ 792 is not a perfect cube

(ii) 1331

1331 – 1 = 1330

1330 – 7 = 1323

1323 – 19 = 1304

1304 – 37 = 1267

1267 – 61 = 1206

1206 – 91 = 1115

1115 – 127 = 988

988 – 169 = 819

819 – 217 = 602

602 – 271 = 331

331 – 331 = 0

Here ,subtraction is performed 11 times

Cube root of 1331 is 11

∴ 1331 is a perfect cube.

(i) 64 

We have, 

64 – 1 = 63 

63 – 7 = 56 

56 – 19 =37 

37 – 37 = 0 

∵ Subtraction is performed 4 times. 

Hence, cube root of 64 is 4. 

(ii) 512 

We have, 

512 – 1 = 511 

511 – 7 = 504 

504 – 19 = 485 

485 – 37 = 448 

448 – 61 = 387 

387 – 91 = 296 

296 – 127 = 169 

169 – 169 = 0 

∵ Subtraction is performed 8 times. 

Hence, cube root of 512 is 8. 

(iii) 1728 

We have, 

1728 – 1 = 1727 

1727 – 7 = 1720 

1720 – 19 = 1701 

1701 – 37 = 1664 

1664 – 91 = 1512 

1512 – 127 = 1385 

1385 – 169 = 1216 

1216 – 217 = 999 

999 – 271 = 728 

728 – 331 = 397 

397 – 397 = 0 

∵ Subtraction is performed 12 times. 

Hence, cube root of 1728 is 12.

(i) 1728

1728 – 1 = 1727

1727 – 7 = 1720

1720 – 19 = 1701

1701 – 37 = 1664

1664 – 91 = 1512

1512 – 127 = 1385

1385 – 169 = 1216

1216 – 217 = 999

999 – 271 = 728

728 – 331 = 397

397 – 397 = 0

Hence ,subtraction is performed 12 times.

∴ Cube root of 1728 is 12.

(a) lower to higher

Domain Name:

1. com

2. edu

3. gov

4. mil

Meaning: 

1. Commercial Organisation

2. Educational Institution

3. Government (US)

4. Military groups

(d)  a and b

There are 3 important components in the Domain Name System

(I) protocol is needed to form a connection between the resolver and the user program

The number of address that can be formed in IPV6 is 2¹²⁸

128 bits are used in the IPv6 addresses.

(c) (i), (ii) & (iii)

(b) (i)-3 (ii)-4 (iii)-2 (iv)-1

(b) (i)-4 (ii)-2 (iii)-1 (iv)-3

(I) Relative URL is used when the file is on the different server.

(b) A is false and R is true.

In IPV6 Addressing, hexadecimal numbers are separated by :

Give Example

Generic Top-Level Domain names: Top level domain is the last part of a domain name. Generic top level domains are used for generic purpose and maintained by IANA.

Generic Domian Names:

Generic top-level domains (gTLDs) are one of the categories of top-level domains (TLDs) maintained by the Internet Assigned Numbers Authority (IANA) for use in the Domain Name System of the Internet. A top-level domain is the last level of every fully qualified domain name.

Absloute and Relative

1. It is the area up to which the server has access.

2. Zone is defined as a group of contiguous domains and sub domains. If the zone has a single domain, then zone and domain are the same.

There are 4 parts in the URL

There are 2 IP addressing methods

1. Domain name is the sequence of labels. In domain name the sequence of labels are separated ‘ by dot (.).

2. The domain name is always read from the lower level to higher level i.e., from the leaf node to root node.

3. Since the root node always represent NULL string, all the domain name ending with dot.

Basic rules of Domain names:

1. Domain can consists of Alphabets a through z, and the digits 0 through 9.

2. Hyphens are allowed, but hyphens can not be used as first character of a domain name. 

3. Spaces are not allowed

4. Special symbols (such as !,$,&, _ and so on) are not permitted, length of 2, and the maximum length of 63 characters.

5. The entire name may be at most 253 characters long.

6. Domain names are not case-sensitive.(It may be upper, lower or mixing of both case letters)

1. A domain is a single node of the Domain Namespace.

2. In the domain name space (DNS) tree structure domain is a sub structure tree. The domain can be further divided into sub domains.

Absolute URL contains all the four necessary and fundamental parts of URL.

ICANN, Internet Corporation for Assigned Name and Numbers is the Non-profit Organization which assigns names and numbers for all Internet resources like domain names and IP addresses.