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Rana Pratap Singh was defeated by the Mughal forces in the Battle of Haldighati.

Muslim clerics, Hindu scholars, Buddhist and Jain monks, Parsi priests, Christian missionaries, etc. assembled in the Ibadat Khanna.

The Ibadat Khana was built in 1575 at Fatehpur Sikri. At this hall, he used to call selected theologians of all religions, mystics and intellectuals and discuss religious and spiritual matters with them.

In 1575 ce, Akbar built the Ibadat Khana in his new capital for discussions on religious.

(b) Inverted red triangle

(i) 16562 

16562 = (7 × 7) × (13 × 13) × 2 

\(\frac{16562}{2}\)= (7 × 7) × (13 × 13) 

\(\frac{16562}{2}\)= (7 × 13) × (7 × 13) = 91 × 91 

= 91² 

Therefore, the resultant is the square of 91.

ii) 3698 

3698 = 2 × (43 × 43) 

\(\frac{3698}{2}\)= 43 × 43 

= 43² 

Therefore, the numbers must be divided by 2 and resultant is square of 43.

(iii) 5103 

5103 = (3 × 3) × (3 × 3) × (3 × 3) × 7 

\(\frac{5103}{7}\)= (3 × 3 × 3) × (3 × 3 × 3) 

= 27 × 27 

= 27² 

Therefore, the number must be divided by 7 and resultant is square of 27.

(iv) 3174 

3174 = 2 × 3 × (23 × 23) 

\(\frac{3174}{6}\)= 23 × 23 

= 23² 

Therefore, the number must be divided by 6 and the resultant is square of 23.

(v) 1575

1575 = 3 × 3 × 5 × 5 × 7

\(\frac{1575}{7}\) = 3 × 3 × 5 × 5

= 15 × 15 

= 15² 

Therefore, the number must be divided by 7 and the resultant is square of 15.

Given 78 X 82

We can write 78 as 80-2 and also 82 as 80+2

We know that (a + b) X (a – b) = a²– b²

Here a= 80 and b = 2

Applying the above formula we get

(80 + 2) X (80 – 2) 

= (80)² – (2)²

= 6400 – 4

= 6396

(i) False

Explanation: Because the number of digits present in perfect square also be odd.

(ii) False

Explanation: A prime number is one that is not divisible by any other, except by 1 and itself. So square of a prime number is not always prime.

(iii) False

Explanation: The sum of two perfect square can be any number.

(iv) False

Explanation: The difference of two perfect square can be any number.

(v) True

Explanation: According to the properties of perfect square we have,

The product of two perfect squares is a perfect square.

(i) 8820 

8820 = (2 × 2) × (3 × 3) × (7 × 7) × 5 

In the above factors only 5 is unpaired 

So, multiply the number with 5 to make it paired 

Again, 

8820 × 5 = 2 × 2 × 3 × 3 × 7 × 7 × 5 × 5 

= (2 × 2) × (3 × 3) × (7 × 7) (5 × 5) 

= (2 × 3 × 7 × 5) × (2 × 3 × 7 × 5) = 210 × 210 

= (210)² 

So, the product is the square of 210 

(ii) 3675 

3675 = (5 × 5) × (7 × 7) × 3 

In the above factors only 3 is unpaired 

So, multiply the number with 3 to make it paired

Again, 

3675 × 3 = 5 × 5 × 7 × 7 × 3 × 3 

= (5 × 5) × (7 × 7) × (3 × 3) 

= (3 × 5 × 7) × (3 × 5 × 7) 

= 105 × 105 

= (105)² 

So, the product is the square of 105 

(iii) 605 

605 = 5 × (11 × 11) 

In the above factors only 5 is unpaired 

So, multiply the number with 5 to make it paired

Again, 

605 × 5 = 5 × 5 × 11 × 11 

= (5 × 5) × (11 × 11) 

= (5 × 11) × (5 × 11) 

= 55 × 55 

= (55)² 

So, the product is the square of 55 

(iv) 2880 

2880 = 5 × (3 × 3) × (2 × 2) × (2 × 2) × (2 × 2) 

In the above factors only 5 is unpaired 

So, multiply the number with 5 to make it paired 

Again, 2880 × 5 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= (2 × 2) × (2 × 2) × (2 × 2) (3 × 3) × (5 × 5) 

= (2 × 2 × 2 × 3 × 5) × (2 × 2 × 2 × 3 × 5) 

= 120 × 120 

= (120)² 

So, the product is the square of 120 

(v) 4056 

4056 = (2 × 2) × (13 × 13) × 2 × 3 

In the above factors only 2 and 3 are unpaired 

So, multiply the number with 6 to make it paired 

Again, 4056 × 6 = 2 × 2 × 13 × 13 × 2 × 2 × 3 × 3 

= (2 × 2) × (13 × 13) × (2 × 2) (3 × 3) 

= (2 × 2 × 3 × 13) × (2 × 2 × 3 × 13) 

= 156 × 156

 = (156)² 

So, the product is the square of 156 

(vi) 3468 

3468 = (2 × 2) × 3 × (17 × 17) 

In the above factors only 3 are unpaired 

So, multiply the number with 3 to make it paired 

3468 × 3 = (2 × 2) × (3 × 3) × (17 × 17) 

= (2 × 3 × 17) × (2 × 3 × 17) 

= 102 × 102 

= (102)² 

So, the product is the square of 102 

(vii) 7776 

7776 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2 × 3 

In the above factors only 2 and 3 are unpaired 

So, multiply the number with 6 to make it paired 

Again, 

7776 × 6 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 

= (2 × 2) × (2 × 2) × (2 × 2) (3 × 3) × (3 × 3) × (3 × 3) 

= (2 × 2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 3 × 3 × 3) 

= 216 × 216 

= (216)² 

So, the product is the square of 216.

Even 

Explanation: According to the property of perfect square i.e. the square of an Even number is always even.

Odd 

Explanation: According to the property of perfect square i.e. the square of an odd number is always odd.

Smaller

Explanation: According to the property of perfect square i.e. the square of a proper fraction is smaller than the fraction.

Odd

Explanation: According to the property of perfect square i.e. for ever natural number n, we have sum of first n odd natural numbers = n²

(i) Given 3675 Resolve 3675 into prime factors, we get 3675 = 3 X 5 X 5 X 7 X 7 Now to get perfect square we have to multiply the above equation by 3

Then we get, 3675 = 3 X 3 X 5 X 5 X 7 X 7

= (3 X 5² X 7²)

New number = (9 X 25 X 49)

= (3² X 5² X 7²)

Taking squares as common from the above equation we get

∴ New number = (3 X 5 X 7)²

= (105)²

Hence, the new number is square of 105

(ii) Given 2156 Resolve 2156 into prime factors, we get 2156 = 2 X 2 X 7 X 7 X 11 = (2² X 7² X 11) Now to get perfect square we have to multiply the above equation by 11

Then we get, 2156 = 2 X 2 X 7 X 7 X 11 X 11

New number = (4 X 49 X 121)

= (2² X 7² X 11²)

Taking squares as common from the above equation we get

∴ New number = (2 X 7 X 11)²

= (154)²

Hence, the new number is square of 154

(iii) Given 3332 Resolve 3332 into prime factors, we get 3332 = 2 X 2 X 7 X 7 X 17= (2² X 7² X 17) Now to get perfect square we have to multiply the above equation by 17

Then we get, 3332 = 2 X 2 X 7 X 7 X 17 X 17

New number = (4 X 49 X 289)

= (2² X 7² X 17²)

Taking squares as common from the above equation we get

∴ New number = (2 X 7 X 17)²

= (238)²

Hence, the new number is square of 238

(iv) Given 2925 Resolve 2925 into prime factors, we get 2925 = 3 X 3 X 5 X 5 X 13= (3² X 5² X 13)Now to get perfect square we have to multiply the above equation by 13

Then we get, 2156 = 3 X 3 X 5 X 5 X 13 X 13

New number = (9 X 25 X 169)

= (3² X 5² X 13²)

Taking squares as common from the above equation we get

∴ New number = (3 X 5 X 13)²

= (195)²

Hence, the new number is square of 195

(i) 8820

The prime factors for 8820

=8820 = 2×2×3×3×7×7×5  (grouping the prime factors in equal pairs we get)

= (2×2) × (3×3) × (7×7) × 5

Here,prime factor 5 is left out. So, multiply by 5 we get,

= 8820 × 5 

 = (2×2) × (3×3) × (7×7) × (5×5)

= (2×3×7×5) × (2×3×7×5)

= 210 × 210

= (210)²

(ii) 3675

The prime factors for 3675

3675 = 5×5×7×7×3 (grouping the prime factors in equal pairs we get)

= (5×5) × (7×7) × 3

Here,prime factor 3 is left out.So, multiply by 3 we get,

 = 3675 × 3 

= (5×5) × (7×7) × (3×3)

= (5×7×3) × (5×7×3)

= 105 × 105

= (105)²

(iii) 605

The prime factors for 605

605 = 5×11×11 (grouping the prime factors in equal pairs we get)

= (11×11) × 5

Here, prime factor 5 is left out.So, multiply by 5 we get,

= 605 × 5 

= (11×11) × (5×5)

= (11×5) × (11×5)

= 55 × 55

= (55)²

(iv) 2880

The prime factors for 2880

2880 = 5×3×3×2×2×2×2×2×2 (grouping the prime factors in equal pairs we get)

= (3×3) × (2×2) × (2×2) × (2×2) × 5

Here, prime factor 5 is left out. So, multiply by 5 we get,

= 2880 × 5 

= (3×3) × (2×2) × (2×2) × (2×2) × (5×5)

= (3×2×2×2×5) × (3×2×2×2×5)

= 120 × 120

= (120)²

(i) 11 it is a prime number.So it is not a perfect square.

(ii) 12 is not a perfect square.

(iii) 16= (4)² 

\(\therefore\) 16 is a perfect square.

(iv) 32 is not a perfect square.

(v) 36= (6)² 

\(\therefore\) 36 is a perfect square.

(vi) 50 is not a perfect square.

(vii) 64= (8)² 

\(\therefore\) 64 is a perfect square.

(viii) 79 it is a prime number. So it is not a perfect square.

(ix) 81= (9)² 

\(\therefore\) 81 is a perfect square.

(x) 111 it is a prime number. So it is not a perfect square.

(xi) 121= (11)² 

\(\therefore\) 121 is a perfect square.

Given,

180 = (2 × 2) × (3 × 3) × 5

=2² × 3² × 5

To make the unpaired 5 into paired, multiply the number with 5

180 × 5 = 2² × 3² × 5²

∴ Square root of √(180 × 5) = 2 × 3 × 5

= 30

(i) The square of an even number is even 

(ii) The square of an odd number is odd 

(iii) The square of a proper fraction is smaller than the given fraction 

(iv) n² = the sum of first n odd natural numbers

(i) 3675 

At first, We’ll resolve the given number into prime factors: 

Hence, 

3675 = 3 × 25 × 49 

= 7 × 7 × 3 × 5 × 5 

= (5 × 7) × (5 × 7) × 3 

In the above factors only 3 is unpaired 

So, in order to get a perfect square the given number should be multiplied by 3 

Hence, 

The number whose perfect square is the new number is as following: 

= (5 × 7) × (5 × 7) × 3 × 3 

= (5 × 7 × 3) × (5 × 7 × 3) 

= (5 × 7 × 3)² 

= (105)² 

(ii) 2156 

At first, We’ll resolve the given number into prime factors: 

Hence, 

2156 = 4 × 11 × 49 

= 7 × 7 × 2 × 2 × 11 

= (2 × 7) × (2 × 7) × 11 

In the above factors only 11 is unpaired 

So, 

in order to get a perfect square the given number should be multiplied by 11 

Hence, 

The number whose perfect square is the new number is as following: 

= (2 × 7) × (2 × 7) × 11 × 11 

= (2 × 7 × 11) × (2 × 7 × 11) 

= (5 × 7 × 11)² 

= (154)² 

(iii) 3332 

At first, We’ll resolve the given number into prime factors: 

Hence, 

3332 = 4 × 17 × 49 

= 7 × 7 × 2 × 2 × 17 

= (2 × 7) × (2 × 7) × 17 

In the above factors only 17 is unpaired 

So, in order to get a perfect square the given number should be multiplied by 17 

Hence, 

The number whose perfect square is the new number is as following: 

= (2 × 7) × (2 × 7) × 17 × 17 

= (2 × 7 × 17) × (2 × 7 × 17) 

= (2 × 7 × 17)² 

= (238)² 

(iv) 2925 

At first, We’ll resolve the given number into prime factors: 

Hence, 

2925 = 9 × 25 × 13 

= 3 × 3 × 13 × 5 × 5 

= (5 × 3) × (5 × 3) × 13 

In the above factors only 13 is unpaired 

So, in order to get a perfect square the given number should be multiplied by 13 

Hence, 

The number whose perfect square is the new number is as following: 

= (5 × 3) × (5 × 3) × 13 × 13 

= (5 × 3 × 13) × (5 × 3 × 13) 

= (5 × 3 × 13)² 

= (195)²

(i) 23805

The prime factors for 23805

23805 = 3×3×23×23×5  ( grouping the prime factors in equal pairs we get,)

= (3×3) × (23×23) × 5 

Here 5 is left out. So, multiply by 5 we get,

= 23805 × 5 

= (3×3) × (23×23) × (5×5)

= (3×5×23) × (3×5×23)

= 345 × 345

= (345)²

(ii) 12150

The prime factors for 12150

12150 = 2×2×2×2×3×3×5×5×2  (grouping the prime factors in equal pairs we get,)

= (2×2) × (2×2) × (3×3) × (5×5) × 2

Here, prime factor 2 is left out. So, multiply by 2 we get,

 = 12150 × 2 

= (2×2) × (2×2) × (3×3) × (5×5) × (2×2)

= (2×2×3×5×2) × (2×2×3×5×2)

= 120 × 120

= (120)²

(iii) 7688

The prime factors for 7688

7688 = 2×2×31×31×2 (grouping the prime factors in equal pairs we get,)

= (2×2) × (31×31) × 2

Here, prime factor 2 is left out. So, multiply by 2 we get,

= 7688 × 2 

= (2×2) × (31×31)× (2×2)

= (2×31×2) × (2×31×2)

= 124 × 124

= (124)²

As we know that, 

Square of an odd number is always an odd number. 

Hence, 

1369 is the square of an odd number. 

Therefore, 

Option (C) is the correct option.

(i) 1156

The prime factors for 1156

1156 = 2×2×17×17

= (2×2) × (17×17)  (none of the prime factors are left out..)

Hence, 1156 is a perfect square.

Now,

1156 = (2×17) × (2×17)

= 34 × 34

= (34)²

∴ 1156 is a square of 34

(ii) 2025

The prime factors for 2025

2025 = 3×3×3×3×5×5

= (3×3) × (3×3) × (5×5)  (none of the prime factors are left out..)

Hence, 2025 is a perfect square.

Now,

2025 = (3×3×5) × (3×3×5)

= 45 × 45

= (45)²

∴ 2025 is a square of 45.

(iii) 14641

 The prime factors for 14641

14641 = 11×11×11×11

= (11×11) × (11×11)  (none of the prime factors are left out..)

Hence, 14641 is a perfect square.

Now,

14641 = (11×11) × (11×11)

= 121 × 121

= (121)²

∴ 14641 is a square of 121.

(c) 81000

Explanation:

According to the property of a square a number ends in an odd number of zeros is not a perfect square.

(b) 4787

Explanation:

According to the property of square a number ending with 2, 3, 7 and 8 is not a perfect square.

From the question it is given that total number of seats = 2704

Let us assume number of seats in each row be ‘a’

As per the condition given in the question,

Number of rows is equal to the number of seats in each row i.e. ‘a’

Then,

Total number of seats = a × a

2704 = a²

By taking square root on both side,

a = √(2704)

a = √(52 × 52)

a = √(52)²

a = 52

∴The number of seats in each row is 52.

(b) 105

Let the number of rows = number of trees = x 

\(\therefore\) Total number of trees in the garden 

= x × x = x² = 10914 + 111 = 11025 

\(\therefore\) No. of rows of trees = \(\sqrt{11025}\) = 105

Let the number of rows be x

Therefore,

The number of plants in each row is also x

Hence,

Total number of plants = (x × x) = x² = 1225

x² = 1225 = 5 × 5 × 7 × 7

x = \(\sqrt{1225}\) = 5 × 7 = 35

Thus, 

The total number of rows is 35 and the number of plants in each row is also 35

Number of squares in the shaded portion = 15 Number of squares in the unshaded portion = 33 So, the ratio of the shaded portion to the unshaded portion = 15 : 33 = 15/33 = 5x3/11x3 = 5/11 = 5:11

(i) Ratio = 1 : 2

(ii) Ratio = 5 : 4

Teacher to Student Ratio of 6th Std = 6 : 12 = 1 : 2

Teacher to Student Ratio of 7th Std = 9 : 27 = 1 : 3

Teacher to Student Ratio of 8th Std = 4 : 16 = 1 : 4

In standard 8th the ratio of teacher to student is 1 : 4 and which is the least.

Further capital introduced during the year is deducted from closing capital in order to find out the correct profit.

Explanation: Under single-entry system, profit or loss is calculated by comparing capital at two dates, i.e. opening capital and closing capital (net worth method). The profit is calculated as closing capital less opening capital and also, the following adjustments are made:

a. Drawings: If drawings are made during the year, they should be added to the amount of closing capital.

b. Additional capital: If additional capital is introduced in the business during the year, it should be deducted from the amount of closing capital.

c. Interest on capital: If interest is provided on capital, it should be deducted from the amount of closing capital.

i. Balance Sheet

ii. Opening Capital

iii. Added

iv. Double Entry System

v. Capital

vi. Sole Trader

vii. Deducted

viii. Small

Correct option is (d) Profit or Loss

(i) Added

(ii) Deducted

(iii) Single Entry System

(iv) the closing capital balance

(i) Disagree

(ii) Disagree

(iii) Disagree

(iv) Disagree

(v) Disagree

(i) Bank overdraft

(ii) Interest on Loan

(iii) Reserve for a discount on creditors

(iv) Prepaid Expenses

(i) Single Entry

(ii) Assets, Liabilities

(iii) Conventional Accounting

(iv) Arithmetical

Here, Marked price = Rs 1700, selling price = Rs 1540 

Selling price = Marked price – Discount 

∴ 1540 = 1700 – Discount 

∴ Discount = 1700 – 1540 = Rs 160 

∴ The amount of discount is Rs 160.

Correct option is (D) density

Data : Activity = 10.0 mCi = 10.0 × 10^-3 Ci

= (10.0 × 10^-3 )(3.7 × 10¹⁰) dis/s = 3.7 × 10⁸ dis/s

T_1/2 = 5.3 years = (5.3)(3.156 × 10⁷ ) s

= 1.673 × 10⁸ s

Decay constant, \(\lambda\) = \(\cfrac{0.693}{T_{1/2}}\) = \(\cfrac{0.693}{1.673\times10^8}s-1\)

=4.142 × 10^-9 s^-1

Activity = Nλ

∴ N = \(\cfrac{activity}{\lambda}\) = \(\cfrac{3.7\times10^8}{4.142\times10^{-9}}\) atoms

= 8.933 × 10¹⁶ atoms 

= 60 grams of \(^{60}_{27}Co\) contain 6.02 × 10²³ atoms

Mass of 8.933 × 10¹⁶ atoms of \(^{60}_{27}Co\)

= \(\cfrac{8.933\times10^{16}}{6.02\times10^{23}}\) x 60 g

= 8.903 × 10^-6 g = 8.903 µg 

This is the required amount.

Correct option is  (B) \(\frac{N}4\)

Correct answer is (a) body

The author says that people think soul, spirit or mind to be the essence of life. They compare the soul, spirit or mind with the wine and the body to its bottle. They see one’s death in the sense that time has consumed the essence of life and the bottle i.e. body is of no use now.

लेखक कहता है कि लोग आत्मा, भावना या सोच को जीवन का सार मानते हैं। वे आत्मा, भावना या सोच की तुलना शराब से करते हैं और शरीर की तुलना इसकी बोतल से। वे व्यक्ति की मृत्यु को इस रूप में देखते हैं कि समय ने जीवन के सार का उपभोग कर लिया है और अब बोतल अर्थात् शरीर व्यर्थ है।

The momentum is transferred to the metal. At the microscopic level, atoms absorb the photon and its momentum is transferred mainly to the nucleus and electrons. The excited electron is emitted. Conservation of momentum needs to be accounted for the momentum transferred to the nucleus and electrons.

(a) momentum

As p = h/λ, so electrons and photons having the same wavelength λ will have the same momentum p.

Power (P) = 1/f

where f is focal length of lens.

Data: y = 5 × 10¹⁴ Hz, h = 6.63 × 10^-34 Js,

1eV =1.6 × 10^-19 J

The energy of each photon,

E = hv = (6.63 × 10^-34 J.s)(5 × 10¹⁴ Hz)

= 3.315 × 10^-19 J

= \(\cfrac{3.315\times10^{-19}J}{1.6\times10^{-19}{\frac{j}{eV}}}\) = 2.072 eV

Lens maker’s formula:

1/f = (n - 1) (\(\frac {1}{R_1} - \frac {1}{R_2}\))

n = Refractive index of the medium 

R₁ = Radius of curvature of 1 st surface 

R₂ = Radius of curvature of 2nd surface 

f = Focal length

(i) dB 

(ii) MHz 

(iii) 20 m 

(iv) Kg.m.s^-2

(a) \(\frac{hv}{c}\)

p = \(\frac{E}{c^2} =\frac{hv}{c}\)

hv = (6.63 × 10^-34)(6 × 10¹⁴)

= 3.978 × 10^-19 J is the energy of the photon.

Data : h = 6.63 × 10^-34 J∙s, 

v = 100 MHz = 100 × 10⁶ Hz, 

λ = 10000 Å = 10⁶ m, 

c = 3 × 10⁸ m/s

(i) The energy of a photon, E = hv

= (6.63 × 10^-34)(100 × 10⁶ ) = 6.63 × 10^-26J

(ii) The energy of a photon, E = \(\cfrac{hc}λ\)

= \(\cfrac{(6.63\times10^{-34})(3\times10^8)}{10^{-6}}\) = 1.989 × 10^-19 J