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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

If the variance of the following frequency distribution :

$\begin{array}{l} Class & : & 10 - 20 & 20 - 30 & 30 - 40 \\ Frequency & : & 2 & x & 2 \end{array}$ is 50 , then $x$ is equal to:

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alok kumar singh

Contributor-Level 10

Since, ${l i m}_{x \to 0} ? \frac{f (x)}{x}$ exist $\Rightarrow f (0) = 0$

Now, $f^{'} (x) = {l i m}_{h \to 0} ? \frac{f (x + h) - f (x)}{h} = {l i m}_{h \to 0} ? \frac{f (h) + x h^{2} + x^{2} h}{h} ($ take $y = h)$

$= {l i m}_{h \to 0} ? \frac{f (h)}{h} + {l i m}_{h \to . 0} ? (x h) + x^{2}$

$\Rightarrow f^{'} (x) = 1 + 0 + x^{2}$

$\Rightarrow f^{'} (3) = 10$

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11 months ago

Class 12th physics ncert exemplar solution class 12th chapter one

The three charges q/2, q and q/2 are placed at the corners A, B and C of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the corner D of the square, is

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Payal Gupta

Contributor-Level 10

$E = \frac{k q}{2 a^{2}}$

$E' = \frac{k q}{{(\sqrt[]{2} a)}^{2}} = \frac{k q}{2 a^{2}}$

$E_{N e t} = 2 E c o s 45 ° + E'$

$= \frac{k q}{a^{2}} (\frac{1}{\sqrt[]{2}} + \frac{1}{2})$

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11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Fifteen

In a linear tetrapeptide (Constituted with different amino acids), (number of amino acids) (number of peptide bonds) is_______________.

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Vishal Baghel

Contributor-Level 10

No. of Amino acid units are 4 and

No. of peptide bonds are 3

In tetrapeptide

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11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

The reaction between X and Y is first order respect to X and zero order with respect to Y.

Examine the data of table and calculate ratio of numerical values of M and L (Nearest integer)

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Vishal Baghel

Contributor-Level 10

According to condition

Rate $\propto {[X]}^{1} {[Y]}^{0}$

By using experimental data I and II

$\frac{4 * 1 0^{- 3}}{2 * 1 0^{- 3}} = \frac{L}{0.1} ∴ L = 0.2$

And by using experimental data I and II

$\frac{M * 1 0^{- 3}}{2 * 1 0^{- 3}} = \frac{0.4}{0.1}$

$∴$ M = 8

Ratio of $(\frac{M}{L}) = \frac{8}{0.2} = 40$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Let $P Q$ be a diameter of the circle $x^{2} + y^{2} = 9$ . If $α$ and $β$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line, $x + y = 2$ respectively, then the maximum value of $α β$ is.

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alok kumar singh

Contributor-Level 10

$D = |\begin{matrix} 1 & - 2 & 3 \\ 2 & 1 & 1 \\ 1 & - 7 & a \end{matrix}| = 0 \Rightarrow a = 8$

also, $D_{1} = |\begin{matrix} 9 & - 2 & 3 \\ b & 1 & 1 \\ 24 & - 7 & 8 \end{matrix}| = 0 \Rightarrow b = 3$

hence, $a - b = 8 - 3 = 5$

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11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

If the solubility product of PbS is 8 * 10^-²⁸, then the solubility of PbS in pure water at 298 K is $x * 1 0^{- 16} m o l L^{- 1} .$ The value of x is__________. (Nearest integer)

[Given: $\sqrt[]{2} = 1.41]$

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Vishal Baghel

Contributor-Level 10

$S = \sqrt[]{K s p} = \sqrt[]{8 * 1 0^{- 28}} = 2.82 * 1 0^{- 14}$

Or solubility of PbS = 282 * 10^-16

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11 months ago

Class 12th physics ncert exemplar solution class 12th chapter one

Given below are two statements:

Statement – I : A point change is brought in an electric field. The value of electric field at a point near to the change may increase if the charge is positive.

Statement – I : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

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New answer posted

11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

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alok kumar singh

Contributor-Level 10

$\begin{array}{r} D_{1} = |\begin{matrix} - 7 & 4 & - 1 \\ 8 & 1 & 5 \\ 15 & b & 6 \end{matrix}| = 0 \Rightarrow b = - 3 \\ D = |\begin{matrix} 1 & 4 & - 1 \\ 3 & 1 & 5 \\ a & b & 6 \end{matrix}| = 0 \Rightarrow 21 a - 8 b - 66 = 0 \\ P : 2 x - 3 y + 6 z = 15 \end{array}$

so required distance $= \frac{21}{7} = 3$

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11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

If O₂ gas is bubbled through water at 303 K, the number of millimoles of O₂ gas that dissolve in 1 litre of water is______________. (Nearest integer)

(Given: Henry's Law constant for O₂ at 303 K is 46.82 k bar and partial pressure of O₂ = 0.920 bar)

(Assume solubility of O₂ in water in too small, nearly negligible)

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Vishal Baghel

Contributor-Level 10

According to Henry's law

P = K_H * mole fraction of solute

$0.92 = 46.82 * 1000 * \frac{m o l o f O_{2}}{55.5}$

$∴$ Mole of O₂ = 1.09 * 10^-3

$∴ M i l l i m o l e = 1.09 \approx 1$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ , then the value of $| \hat{i} * (\vec{a} * \hat{i}) |^{2} + | \hat{j} * (\vec{a} * \hat{j}) |^{2} + | \hat{k} * (\vec{a} * \hat{k}) |^{2}$ is equal to

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alok kumar singh

Contributor-Level 10

Given ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} ? a_{r} x^{r}$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{r} \frac{2^{10} {(2 x^{2} + 3 x + 4)}^{10}}{x^{20}} = \sum_{r = 0}^{20} ? ? \frac{a_{r} 2^{r}}{x^{r}} \\ \Rightarrow 2^{10} \sum_{r = 0}^{20} ? ? a_{r} x^{r} = \sum_{r = 0}^{20} ? ? a_{r} 2^{r} x^{(20 - r)} (from (i)) \end{array}$

now, comparing coefficient of $x^{7}$ from both sides

(take $r = 7$ in L.H.S. and $r = 13$ in R.H.S.)

$2^{10} a_{7} = a_{13} 2^{13} \Rightarrow \frac{a_{7}}{a_{13}} = 2^{3} = 8$

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