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Class 12th

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Class 12th NCERT Notes

How can I get full marks in the Class 12 Maths exam?

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nitesh singh

Contributor-Level 10

All students work hard to get a full score. Here are a few tips that can make your dream come true.

Master the concepts from the NCERT textbook before using any other notes and mark importnt points.
Consistentantly practice problems, highlight important formulas and theorems, and creating a personalized formula sheet.
Regular- daily and weekly revision of the important concepts and formulas.
Solve all examples and practice questions given in NCERT Textbooks.

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a year ago

Class 12th Dual Nature of Radiation and Matter

The temperature of an ideal gas in 3-dimensions is 300K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:

[m_e = mass of electron = 9 * 10^-³¹kg

h = Planck constant = 6.6 * 10^-³⁴ Js

k_B = Boltzmann constant = 1.38 * 10^-²³ JK^-¹]

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alok kumar singh

Contributor-Level 10

De Broglie wavelength

$λ = \frac{h}{m V} = \frac{h}{\sqrt[]{2 m E}} E \to K . E .$

$E = \frac{3}{2}$ KT for gas

$S o λ = \frac{h}{\sqrt[]{3 m K T}} = \frac{6.6 * 1 0^{- 34}}{\sqrt[]{3 * 9 * 1 0^{- 31} * 1.38 * 1 0^{- 23} * 300}}$

$\begin{array}{l} λ = 6.26 * 1 0^{- 9} m \\ λ = 6.26 n m \end{array}$

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Class 12th English

Directions for questions: A famous retail electronics showroom chain has six new mobile phone models – A, B, C, D, E and F – each equipped with at least one of the following three options: digital camera, music player, and office document viewer. No mobile has other option. The following condition apply: B features both a digital camera and an office document viewer. C has digital camera and music player. C and E have no options in common. D has more options as compared to C. B and F have exactly one option in common. A has fewer options as compared to F.

For exactly how many of the six mobile phones is it possible to determine exa

...more

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Payal Gupta

Contributor-Level 10

Here, DC = Digital Camera; ODV = Office Document Viewer; MP = Music Player

Number of options	Mobile	Options
1	A	DC/ODV/MP
2	B	DC, ODV
2	C	DC, MP
3	D	DC, MP, ODV
1	E	ODV
2	F	DC, MP/MP, ODV

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Class 12th English

Seven Trainers A, B, C, D, E, F and G – teach management courses at a ABC institute in west Delhi. Each instructor teaches during exactly one term: the first term, the second term, or the third term. The following conditions apply: B teaches during the third term. C and D teach during the same term. G teaches during either the first term or the second term. Exactly twice as many Trainers teach during the third term as teach during the first term. E and G teach during different terms. A and F teach during different terms.

Each of the following contains a list of Trainers who can all teach during the same term EXCEPT

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Payal Gupta

Contributor-Level 10

B, F, G can be two only, which leaves J and N same which is not possible.

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Class 12th English

If exactly one instructor teaches during the second term, which one of the following must be true?

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Payal Gupta

Contributor-Level 10

If D = 3 and C = 3

There are two teachers for first term and four for third term.

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a year ago

Class 12th Electromagnetic Waves

The magnetic field vector of an electromagnetic wave is given by B = B₀ = $\frac{\hat{i} + \hat{j}}{\sqrt[]{2}} c o s (k z - ω t);$ where $\hat{i}, \hat{j}$ represents unit vector along x and y-axis respectively. At t = 0 s, two electric charges q1 of 4π coulomb and q2 and 2p coulomb located at $(0, 0, \frac{π}{k}) a n d (0, 0, \frac{3 π}{k}),$ respectively, have the same velocity of 0.5 c $\hat{i}$ (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is:

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Vishal Baghel

Contributor-Level 10

Force on q₁ :

$F_{q_{1}} = q (V * B) = 4 π (0.5 c \hat{i}) * \frac{B_{0}}{\sqrt[]{2}} (c o s (k z - ω t) \hat{i} + c o s (k z - w t) \hat{j})$

$= 4 π (0.5 c) (\frac{B_{0}}{\sqrt[]{2}}) c o s (k \frac{π}{k}) \hat{k}$

For on q₂:

= $2 π (0.5 c) (\frac{B_{0}}{\sqrt[]{2}}) c o s (k \frac{3 π}{k}) \hat{k}$

So ratio will be 2 : 1

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New answer posted

a year ago

Class 12th Maths Determinants

If $a_{r} = c o s \frac{2 r π}{9} + i s i n \frac{2 r π}{9}, r = 1, 2, 3, . . . . ., i = \sqrt[]{- 1},$ then the determinant $| \begin{matrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{matrix} |$ is equal to:

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Raj Pandey

Contributor-Level 9

$? a_{r} = e^{i \frac{2 r π}{9}}$

$∴ a_{1}, a_{2}, a_{3}, . . . . . . . . . . .$ are in G.P.

$| \begin{matrix} a_{1} & a_{1} r & a_{1} r^{2} \\ a_{1} r^{3} & a_{1} r^{4} & a_{1} r^{5} \\ a_{1} r^{6} & a_{1} r^{7} & a_{1} r^{8} \end{matrix} | = {a_{1}}^{3} r^{9} | \begin{matrix} 1 & r & r^{2} \\ 1 & r & r^{2} \\ 1 & r & r^{2} \end{matrix} | = 0$

$(i) a_{2} a_{6} - a_{4} a_{8} = a_{1} r . a_{1} r^{5} - a_{1} r^{6} a_{1} r^{7} = {a_{1}}^{2} r^{3} - {a_{1}}^{2} r^{10} \neq 0$

$(i i) a_{9} = a_{1} r^{8} \neq 0$

$(i i i) a_{1} . a_{1} r^{8} - a_{1} r^{2} . a_{1} r^{6} = 0$

$(i v) a_{5} \neq 0$

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a year ago

Class 12th Physics Magnetism and Matter

Following plots show Magnetization (M) vs Magnetising field (H) and Magnetic susceptibility $(χ)$ vs temperature (T) graph:

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alok kumar singh

Contributor-Level 10

Magnetization (M) is directly proportional to magnetizing field and magnetic susceptibility does not depend on temperature so option 3 is correct.

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a year ago

Class 12th Physics

If VA and VB are the input voltages (either 5V or 0V) and Vo is the output voltage then the two gates represented in the following circuit (A) and (B) are:-

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Vishal Baghel

Contributor-Level 10

Using truth table

A B V₀

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0 1 1

1 0 1

1 1

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a year ago

Class 12th Physics

A capacitor is connected to a 20 V battery through a resistance of 10 $Ω .$ It is found that the potential difference across the capacitor rises to 2V in 1 $μ s .$ The capacitance of the capacitor is________ $μ F .$

Given $I n (\frac{10}{9}) = 0.105$

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alok kumar singh

Contributor-Level 10

V = 10 $(1 - e^{- t / R C})$

2 = 20 $(1 - e^{- t / R C})$

$\frac{1}{10} = 1 - e^{- t / R C}$

$e^{t / R C} = \frac{10}{9}$

$\frac{t}{R C} = l n (\frac{10}{9}) = 0.105$

$C = \frac{t}{R * 0.105} = \frac{1 0^{- 6}}{10 * 0.105} = 0.95 μ F$

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