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Class 12th Physics

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kW. The voltage gain of the transistor will be ________.

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alok kumar singh

Contributor-Level 10

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_{B} = \frac{V_{B}}{I_{B}} = \frac{10 * 1 0^{- 3}}{10 * 1 0^{- 6}} = 1 0^{3} Ω$

$A_{v} = (\frac{Δ l_{C}}{Δ l_{B}}) * (\frac{R_{C}}{R_{B}})$

= $\frac{1.5 * 1 0^{- 3}}{10 * 1 0^{- 6}} * \frac{5 * 1 0^{3}}{1 0^{3}}$

A_v = 750

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Class 12th Physics Ray Optics and Optical Instruments

Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is ________ cm.

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alok kumar singh

Contributor-Level 10

f₁ = 15 cm

$P_{1} = \frac{1}{15} * 100 = \frac{100}{15} D$

$P_{3} = \frac{100}{15} D$

$\frac{1}{f^{2}} = (μ_{2} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$= (1.25 - 1) * \frac{- 2}{R}$

$\frac{1}{f_{2}} = \frac{0.25 * - 2}{15}$

$\frac{1}{f_{2}} = \frac{- 1}{30} c m^{- 1}$

$\begin{array}{l} p_{2} = \frac{1}{f_{2}} = - \frac{100}{30} \\ P_{R} = P_{1} + P_{2} + P_{3} \\ P_{R} = \frac{100}{15} - \frac{100}{30} + \frac{100}{15} \\ P_{R} = 10 D \end{array}$

$P_{R} = \frac{1}{f_{R}}$

$f_{R} = \frac{1}{10} * 100 c m$

$f_{R} = 10 c m$

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Class 12th Physics Wave Optics

Consider the given arrangement. The two slits $S_{1}$ and $S_{2}$ are illuminated by monochromatic light of wavelength $λ$ . Slits $S_{3}$ and are at separation $d_{1} = \frac{λ D}{3 d}$ , then ratio of maximum and minimum intensity on the screen will be

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

Answer (12.00)

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Class 12th Ncert Solutions Chemistry Class 12th

The highest industrial consumption of molecular hydrogen is to produce compounds of element :

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Raj Pandey

Contributor-Level 9

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having manufacturing of N-based fertizers)

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2 months ago

Class 12th Chemistry Chemical Kinetics

For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is

(Given : In 10 = 2.303 and log 2 = 0.3010)

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Raj Pandey

Contributor-Level 9

$T_{90 %} = \frac{2.303}{k} l o g \frac{100}{10}$

$T_{50 %} = \frac{2.303}{k} l o g \frac{100}{50}$

$\frac{T_{90 %}}{T_{50 %}} = \frac{l o g 10}{l o g 2} = \frac{1}{0.301} = 3.32$

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Class 12th Physics Nuclei

A radioactive $β -$ emission. A detector records $n β -$ particles in $2 s$ and by next $2 s$ (accumulatively) it records $1.1 n β -$ particle. Number of $β -$ particles recorded by detector after a long time, is

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alok kumar singh

Contributor-Level 10

$I_{S_{4}} = 4 I_{0}$

$I_{S_{3}} = I_{0}$

$∴ \frac{I_{m a x}}{I_{m i n}} = \frac{{(\sqrt[]{I_{S_{4}}} + \sqrt[]{I_{S_{3}}})}^{2}}{{(\sqrt[]{I_{S_{4}}} - \sqrt[]{I_{S_{3}}})}^{2}} = {(\frac{2 + 1}{2 - 1})}^{2} = \frac{9}{1}$

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Class 12th Dual Nature of Radiation and Matter

A free particle having one electronic charge with initial kinetic energy $9 e V$ and de Broglie wavelength $1 m m$ enters a region of $V_{0}$ potential difference such that new de Broglie wavelength is now $1.5 m m$ . Then $e V_{0}$ is

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alok kumar singh

Contributor-Level 10

$λ_{1} = \frac{h}{\sqrt[]{2 m E_{1}}} = λ_{2} = \frac{h}{\sqrt[]{2 m E_{2}}}$

$= E_{2} = \frac{4}{9} E_{1} = 4 e V$

$E_{2} = E_{1} - e V_{0} = V_{0} = 5 V$

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2 months ago

Class 12th Ncert Solutions Chemistry Class 12th

The energy of one mole of photons of radiation of wavelength 300nm is (Given: h = 6.63 * 10³⁴ Js,N_A = 6.02 * 10²³ mol^-1 c = 3 * 10⁸ ms^-1)

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Raj Pandey

Contributor-Level 9

Energy of 1 photon = $\frac{h c}{λ}$

Energy of 1 mole of photon = $N_{A} * \frac{h c}{λ}$

$= \frac{6.022 * 1 0^{23} * 6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{300 * 1 0^{- 9}}$

=0.399 * 10⁶ J = 399KJ

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Class 12th Physics

A simple $L R$ circuit is connected to a battery at $t = 0$ . The time instant at which rate of energy storage in inductor is half of power delivered by battery

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alok kumar singh

Contributor-Level 10

$i = \frac{v_{0}}{R} (1 - e^{\frac{t R}{L}})$

$U = \frac{1}{2} L i^{2}$

$\frac{d U}{d t} = L i \frac{d i}{d t}$

$2 L * \frac{R}{L} * \frac{v_{0}^{2}}{R^{2}} (1 - e^{\frac{t R}{L}}) e^{\frac{t R}{L}} = \frac{v_{0}^{2}}{R} (1 - e^{\frac{t R}{L}})$

$t = \frac{L}{R} I n 2$

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Class 12th Sets

If f(A, B) = $\frac{1 + s i n A}{c o s A} + \frac{c o s B}{1 - s i n B}$ and g(A, B) = $\frac{(s i n A - s i n B)}{s i n (A - B) + c o s A - c o s B}$ and h(A,B) = $\frac{f (A, B)}{g (A, B)}$ then the value of $h (18 °, 36 °) .$ (wherever defined)

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Vishal Baghel

Contributor-Level 10

$f (A, B) = \frac{1 - s i n B + s i n A - s i n A s i n B + c o s A c o s B}{c o s A (1 - s i n B)}$

$\begin{array}{l} \Rightarrow f (A, B) (c o s A - c o s A s i n B - c o s B + s i n A c o s B) \\ = 2 s i n A - 2 s i n B \end{array}$

$\Rightarrow f (A, B) = \frac{2 (s i n A - s i n B)}{c o s A - c o s B + s i n (A - B)} = 2 g (A, B)$

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