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Class 11th Maths

A point P moves so that the sum of squares of its distances from the points (1, 2) and (-2, 1) is 14. Let f(x, y) = 0 be the locus of P, which intersects the x-axis at the points A , B and the y-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to:

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Raj Pandey

Contributor-Level 9

Let point P : (h, k)

Therefore according to question, ${(h - 1)}^{2} + {(k - 2)}^{2} + {(h + 2)}^{2} + {(k - 1)}^{2} = 14$

$∴$ locus of P(h, k) is $x^{2} + y^{2} + x - 3 y - 2 = 0$

Now intersection with x – axis are $x^{2} + x - 2 = 0 \Rightarrow x = - 2, 1$

Now intersection with y – axis are $y^{2} - 3 y - 2 = 0 \Rightarrow y = \frac{3 \pm \sqrt[]{17}}{2}$

Therefore are of the quadrilateral ABCD is = $\frac{1}{2} (| x_{1} | + | x_{2} |) (| y_{1} | + | y_{2} |) = \frac{1}{2} * 3 * \sqrt[]{17} = \frac{3 \sqrt[]{17}}{2}$

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Class 11th Physics

In an experiment to verigy Newton's law of cooling, a graph is plotted between, the temperature difference $(Δ T)$ of the water and surrounding and time as shown in figure. The initial temperature of water is taken as $80 ° C .$ The value of t₂ as mentioned in the graph will be _______.

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alok kumar singh

Contributor-Level 10

$\frac{d T}{d t} = - k (T - T_{0})$

$\frac{d T}{(T - T_{0})} = - k d t$

$\Rightarrow k = - \frac{1}{6} l n (\frac{2}{3})$ - (1)

Now $\frac{d T}{d t} = - k (T - T_{0})$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

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Class 11th Maths

Consider two G.P.'s. 2, 2², 2³, …… and 4, 4², 4³, …… of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is ${(2)}^{\frac{225}{8}}$ , then $\sum_{k = 1}^{n} k (n - k)$ is equal to:

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Raj Pandey

Contributor-Level 9

Given G.P's 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. = ${(2)}^{\frac{225}{8}} \Rightarrow {(2, 2^{2}, 2^{3}, . . . .)}^{\frac{1}{60 + n}} = {(2)}^{\frac{225}{8}} \Rightarrow n = \frac{57}{8}, 20 s o n = 20$

$∴ \sum_{k = 1}^{n} k (n - k) 20 * \frac{20 * 21}{2} - \frac{20 * 21 * 41}{6} = 1330$

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Class 11th Maths

Let O be the origin and A be the point z₁ = 1 + 2i. If B is the point z₂, Re(z₂) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

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Raj Pandey

Contributor-Level 9

$\frac{z_{2} - 0}{(1 + 2 i) - 0} = | \frac{O B}{O A} | e^{\frac{i π}{4}} \Rightarrow z_{2} = (1 + 2 i) (1 + i) = - 1 + 3 i ∴ a r g z_{2} = π - t a n^{- 1} 3 a n d | z_{2} | = \sqrt[]{10}$

$z_{1} - 2 z_{2} = 3 - 4 i ∴ a r g (z_{1} - 2 z_{2}) = - t a n^{- 1} \frac{4}{3} \Rightarrow | z_{1} - 2 z_{2} | = | 2 + 4 i + 1 - 3 i | = \sqrt[]{10}$

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2 months ago

Class 11th Ncert Solutions Chemistry Class 11th

On complete combustion of 0.492g of an organic compound containing C, H and O, 0.7938 g of CO₂ and 0.4428 g of H₂O was produced. The % composition of oxygen in the compounds is____________.

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Raj Pandey

Contributor-Level 9

% of C in organic compound

$= \frac{12}{14} * \frac{W C O_{2}}{W . O . C .} * 100$

$= \frac{952.56}{21.648} = 44 %$

$% o f H = \frac{2}{18} * \frac{W H_{2} O}{W . O . C .} * 100$

= $\frac{20}{18} * \frac{0.4428}{0.492} * 100$

= $\frac{88.56}{8.856} = 10 %$

= 100 – 54 = 46%

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Class 11th Redox Reactions

The disproportionation of $M n O_{4}^{2 -}$ in acidic medium resultant in the formation of two manganese compounds A and B. If the oxidation state of Mn is B is smaller than that of A, then the spin-only magnetic moment $(μ)$ value of B in BM is___________.

(Nearest integer)

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Raj Pandey

Contributor-Level 9

$M n O_{4}^{- 2} \to A + B$

Oxidation state of Mn in B < A

$M n O_{4}^{- 2} + H^{+} \to M n O_{4}^{-} + M n O_{2}$

B is MnO₂

$∴$ Oxidation state of Mn = +4

$∴_{25} M n^{+ 4} = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{3}$

$∴$ unpaired electron = 3

Spin only magnetic moment $(μ)$

...more

$M n O_{4}^{- 2} \to A + B$

Oxidation state of Mn in B < A

$M n O_{4}^{- 2} + H^{+} \to M n O_{4}^{-} + M n O_{2}$

B is MnO₂

$∴$ Oxidation state of Mn = +4

$∴_{25} M n^{+ 4} = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{3}$

$∴$ unpaired electron = 3

Spin only magnetic moment $(μ)$

$= \sqrt[]{n (n + 2)} = \sqrt[]{3 (3 + 2)} = \sqrt[]{15}$

$= 3.87 \approx 4$

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Class 11th Ncert Solutions Chemistry Class 11th

The number of interhalogens from the following having square pyramidal structure is:

CIF₃, IF₇, BrF₅, BrF₃, I₂Cl₆, IF₅, ClF, ClF₅

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Raj Pandey

Contributor-Level 9

$C l F_{3}$ ® T-shaped (sp³d)

IF₇ ® Pentagonal bipyramidal (sp³d³)

BrF₅ ® Square pyramidal (sp³d²)

BrF₃ ® T-Shaped (sp³d)

I₂Cl₆ ® Triangular bipyramidal (sp³d)

$I F_{5} \to$ Square Pyramidal (sp³d²)

ClF ® (sp³)

ClF₅ ® Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

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Class 11th Chemistry Chemical Bonding and Molecular Structure

The number of paramagnetic species among the following is

B₂, Li₂, C₂, $C_{2}^{-}$ , $O_{2}^{- 2}$ , $O_{2}^{+}$ and $H e_{2}^{+}$ .

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Raj Pandey

Contributor-Level 9

$B_{2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{1} = π_{2 p y}^{1} \to$ Paramagnetic

$\begin{array}{l} L i_{2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} \to D i a m a g n e t i c \\ C_{2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} \to D i a m a g n e t i c \\ C_{2}^{-} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{1} \to P a r a m a g n e t i c \end{array}$

$O_{2}^{- 2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p z}^{2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} π_{2 p x}^{* 2} \equiv π_{2 p y}^{* 2}$ Diamagnetic

$O_{2}^{+} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p z}^{2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{0} \to$ Paramagnetic

$H e_{2}^{+} = σ_{1 s}^{2} σ_{1 s}^{* 1} \to$ Paramagnetic

$∴$ Paramagnetic molecules are $= B_{2}, C_{2}^{-}, O_{2}^{+}, H e_{2}^{+}$

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Class 11th Chemistry

In the given reaction,

X + Y + 3Z XYZ₃

If one mole of each of X and Y with 0.05 mol of Z gives compound XYZ₃. (Given: Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively.) The yield of XYZ₃ is_____________g.

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Raj Pandey

Contributor-Level 9

x + y + 3z = xyz₃

1 mole 1 mole 0.05 mole

$∴ n_{x} = 1$

$n_{y} = 1$

$n_{z} = \frac{0.05}{3} = 0.0167$ here z is limiting reagent.

$? \frac{0.05}{3}$ mole z gives 1 mole xyz₃

mass of xyz₃ = n * molecular mass

= $\frac{0.05}{3} * (10 + 20 + 3 * 30) a . m . u .$

= 0.5 * 4 = 2g

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2 months ago

Class 11th Ncert Solutions Chemistry Class 11th

The geometry around boron in the product 'B' formed from the following reaction is

$B F_{3} + N a H \overset{450 K}{\to} A + N a F$

A + NMe₃ $\overset{}{\to} B$

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Raj Pandey

Contributor-Level 9

$2 B F_{3} + 6 N a H \to B_{2} H_{6} + 6 N a F$

$B_{2} H_{6} + 2 \overset{?}{N} {(M e)}_{3} \to 2 H_{3} B \leftarrow : N {(M e)}_{3}$

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