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Solution: Circumference of first circle

= 2 πr = 2π x 19 = 38π cm

Circumference of second circle = 18π cm

Or, 2πr = 56π

Or, 2r = 56

Or, r = 28 cm

Circumference of the largest circle; as per question

= 38π + 18π = 56π cm

Atoms X and Y form bcc crystalline structure. Atom X is present at the comers of the cube Atom Y is present at die centre of the cube.

No of atoms of X in the unit cell = \(\frac{N_e}{8} = \frac{8}{8}\,=1\)

No of atoms of Y in the’unit cell = \(\frac{N_c}{1} = \frac{1}{1}\,=1\)

Ratio of atoms X : Y = 1 : 1 .

Hence formula of the compound = XY

OP in the chat

10^logsinxcotx log 10

Correct option is D) When

To funnel or restrict the respondent’s answer means to give the possible answer. It starts with open-ended questions but ends with closed questions. Answers are more restrictive in this criteria.

• In the rest of the options or questions, the respondent can answer anything of his choice and also in different forms.
• But in the question ‘how many books are there’, the answer to this question is restricted to the specific number of books present there.

Thus, it is concluded that an exemplar of a question to 'funnel' or restrict a respondent's answer is "How many books are there?"

Highest optical density = Diamond
Lowest optical density = Air
Optical density of a medium is directly related with the refractive index of that medium. A medium which has the highest refractive index will have the highest optical density and vice-versa.

It can be observed from table 10.3 that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical
density.

(d) 2Ag₂O (s) → 4Ag (s) + O₂ (g)

 2Pb₃O₄ (s) → 6PbO (s) + O₂ (g)

 2PbO₂ (s) → 2PbO (s) + O₂ (g)

 (c) S₂ is paramagnetic. It contains two unpaired electrons in the antibonding π * orbital

(a) Mn₂O₇ is an acidic oxide. BaO and Na₂O are basic oxides while N₂O is a neutral oxides

(c) Atomicity of sulphur in rhombic sulphur is 8

Chlorobenzene is less reactive than benzyl chloride In chlorobenzene the lone pairs present on Cl atom get involved in resonance with pi electrons of benzene due to which C-Cl bond acquires double bond character Hence, reactivity decreases.

Correct option is: (A) 1

Correct option is: (D) \(\overrightarrow{b}.\overrightarrow{a}\)

Correct option is: (D) \(\overrightarrow{a}\times\overrightarrow{c}\)

Correct option is: (B) 1

Correct option is: (D) e³ - e²

Correct option is: (D) \(\overrightarrow{k}\)

Correct option is: (D) \(\frac{19}{x}\)

Correct option is: (B) 1

Correct option is: (D) - tan x

Correct option is: (A) a^x log_e a

Correct option is: (C) \(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{c}\times\overrightarrow{a}\)

Correct option is: (B) \(\frac{\pi}{4}\)

Correct option is: (B) \(\sqrt{2}\)

Correct option is: (D) 90°

Correct option is: (C) e^x + e^-y = k

Correct option is: (A) 0

Correct option is: (C) 3

Solution to your question:

(i) f + g(x) = f(x) + g(x) = x² + 2x + 1 

(ii) (f – g)(x) = f (x) – g(x) = x² – 2x – 1 

(iii) (f × g)(x) = f(x) × g(x) = x²(2x + 1) = 2x³ + x² 

(iv) \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\)

\(= \frac{x^2}{2x+1}, x\neq -\frac{1}{2}\)

Correct option is: (D) 4

Correct option is: (B) \(\overrightarrow{i}\)

Let the quadratic  polynomial be

ax² + bx + c = 0 and its zeroes be 

α = 2 and β = −1

Sum of the zeroes = α + β

= 2 + (−1)

= 2 − 1

= 1

Product of the zeroes = αβ

= 2 × (−1)

= −2 

The quadratic polynomial  will be

ax² + bx + c is 

k[x² − (α + β)x + αβ]

k[x² − (1)x + (−2)]

If K = 1 , the quadratic polynomial will be

1[x² − (1)x + (−2)]

⇒ x² − x − 2

We have,

f(x) = x² −p(x+1)−c = 0

f(x) = x²−px−(p+c) = 0

Since,

α,β are the zeroes of the above polynomial.

So,

α + β = p

αβ = −(p + c) 

Since,

(α + 1)(β + 1) = 0

αβ + α + β + 1 = 0

−p − c + p + 1 = 0

−c + 1 = 0

c = 1