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9 months ago

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

The velocity of sound in a gas, in which two wavelengths 4.08m and 4.16m produce 40 beats in 12s, will be

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Payal Gupta

Contributor-Level 10

$| f_{1} - f_{2} | = \frac{40}{12} = \frac{10}{3}$

$\Rightarrow v (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = \frac{10}{3}$

$\Rightarrow$ $V$ $=$ $\frac{10}{3}$ $*$ $\frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}}$ $=$ $\frac{10}{3}$ $*$ $\frac{4.08 * 4.16}{. 08}$ $=$ $\frac{10}{3}$ $*$ $\frac{408 * 4.16}{8}$ $=$ $7$ $0$ $7$ $.$ $2$ $m$ $/$

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9 months ago

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

Let $\vec{a} = 3 \hat{i} + \hat{j} a n d \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} .$ Let $\vec{c}$ be vector satisfying $\vec{a} * (\vec{b} * \vec{c}) = \vec{b} + λ \vec{c} .$ If $\vec{b} a n d \vec{c}$ are non-parallel, then the value of $λ$ is:

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Vishal Baghel

Contributor-Level 10

$(\overset{?}{a} . \overset{?}{c}) \overset{?}{b} ? (\overset{?}{a} . \overset{?}{b}) \overset{?}{c}$

$= \overset{?}{b} + ? \overset{?}{c} (g i v e n)$

$\overset{?}{a} . \overset{?}{c} = 1, ? = ? \overset{?}{a} . \overset{?}{b}$

$? = ? (3, 1, 0) . (1, 2, 1)$

= (3 + 2)= 5

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9 months ago

Class 12th Communication Systems

Match List – I with List – II

	List – I		List – II
A.	Television signal	I.	03 KHz
B.	Radio signal	II.	20 KHz
C.	High Quality Music	III.	02 MHz
D.	Human speech	IV.	06 MHz

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

Using factual data.

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9 months ago

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

IF the foot of the perpendicular from the point A(1, 4, 3) on the plane P : 2x + my + nz = 4, is $(- 2, \frac{7}{2}, \frac{3}{2}),$ then the distance of the point A form the plane P, measured parallel to a line with direction ratios 3, 1, 4, is equal to:

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Vishal Baghel

Contributor-Level 10

A (1, 4, 3)

2x + my + nz = 4

$M \Rightarrow - 4 + \frac{7 m}{2} + \frac{3 n}{2} = 4$

7m + 3n = 16

$\vec{A M} = (- 1, \frac{- 1}{2}, \frac{- 3}{2})$

$\frac{2}{- 1} = \frac{m}{\frac{- 1}{2}} = \frac{n}{\frac{- 3}{2}}$

m = 1, n = 3

Plane : 2x + y + 3z = 4

$c o s θ = \frac{| 6 - 1 - 12 |}{\sqrt[]{26} \sqrt[]{14}} = \frac{7}{2 \sqrt[]{91}}$

AM = $\frac{| - 2 + 4 + 9 - 4 |}{\sqrt[]{14}} = \frac{7}{\sqrt[]{14}}$

$c$ $o$ $s$ $θ$ $=$ $\frac{A M}{A B}$ $\Rightarrow$ $A$ $B$ $=$ $\frac{7 / 14}{7 / 291}$ $=$ $\frac{291}{14}$ $=$ $2$ $\frac{13 * 7}{2 * 7}$ $=$

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9 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : n-p-n transistor permits more current that a p-n-p transistor.

Reason R : Electrons have greater mobility as a change carrier.

Choose the correct answer form the options given below

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Payal Gupta

Contributor-Level 10

Using factual data.

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9 months ago

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

Let a line L pass through the point of intersection of the lines bx + 10y – 8 = 0 and $2 x - 3 y = 0, b \in R - {\frac{4}{3}} .$ If the line L also passes through the point (1, 1) and touches the circle 17(x2 + y2) = 16, then the eccentricity of the ellipse $\frac{x^{2}}{5} + \frac{y^{2}}{b^{2}} = 1$ is:

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Vishal Baghel

Contributor-Level 10

y – 1 = m(x – 1)

mx – y + 1 – m = 0

$\frac{| 1 - m |}{\sqrt[]{m^{2} + 1}} = \frac{4}{\sqrt[]{17}}$

$17 {(1 - m)}^{2} = 16 (m^{2} + 1)$

m² – 3m + 1 = 0

$\begin{array}{l} b x + 10 y - 8 = 0 \\ y = \frac{2}{3} x \end{array}} \Rightarrow (b + \frac{20}{3}) x = 8$

$x = \frac{24}{3 b + 20}, y = \frac{16}{3 b + 20}$

b = $\frac{2 (3 - 2 \sqrt[]{2}) (4 + 3 \sqrt[]{2})}{- 2}$

$= \frac{12 - 8 \sqrt[]{2} + 9 \sqrt[]{2}}{- 1} = - \sqrt[]{2}$

b² = 2

$\frac{x^{2}}{5} + \frac{y^{2}}{2} = 1$

2 = 5 (1 – e²)

$e^{2} = \frac{3}{5}$

$e$ $=$

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9 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

For using a multimeter to identify diode from electrical compounds, choose the correct statement out of the following about the diode

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Payal Gupta

Contributor-Level 10

Diode conducts current only in one direction.

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9 months ago

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

Let the solution curve y = y(x) of the differential equation $(1 + e^{2 x}) (\frac{d y}{d x} + y) = 1$ pass through the point $(0, \frac{π}{2}) .$ Then, $\underset{x \to \infty}{l i m} e^{x} y (x)$ is equal to:

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Vishal Baghel

Contributor-Level 10

IF =e^x

$y e^{x} = \int \frac{e^{x}}{1 + e^{2 x}} d x$

$\int \frac{d t}{1 + t^{2}}$

= tan^-1 (t) + c

$\underset{x \to \infty}{l i m} e^{x} y = \underset{x \to \infty}{l i m} t a n^{- 1} e^{x} + \frac{π}{4} = \frac{3 π}{4}$

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9 months ago

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

The integral $\int_{0}^{\frac{π}{2}} \frac{1}{3 + 2 s i n x + c o s x} d x$ is equal to:

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Vishal Baghel

Contributor-Level 10

$\int_{0}^{\frac{π}{2}} \frac{d x}{3 + 2 s i n x + c o s x} P u t t a n \frac{x}{2} = t \Rightarrow \frac{1}{2} s e c^{2} \frac{x}{2} d x = d t$

$= \int_{0}^{1} \frac{2 d t}{2 t^{2} + 4 t + 4} = \int_{0}^{1} \frac{d t}{{(t + 1)}^{2} + 1} = {[t a n^{- 1} (t + 1)]}_{0}^{1} = t a n^{- 1} 2 - t a n^{- 1} 1$

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9 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

In the following circuit, the correct relation between output (Y) and inputs A and B will be

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Payal Gupta

Contributor-Level 10

From Truth table $Y = \bar{A B}$ .

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