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Class 12th Physics

Which one of the following chemicals is responsible for the production of HCl in the stomach leading to irritation and pain?

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alok kumar singh

Contributor-Level 10

Histamine stimulate the secretion of HCl

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Class 12th Electromagnetic Induction

A small square look offside 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop offside 'b', then the coefficient of mutual inductance between the two loops is:

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Raj Pandey

Contributor-Level 9

$? = M I$

Also

$? = \vec{B} . \vec{A} = B A c o s 0 °$

= 4 {\frac{μ_{0} l}{4 π (\frac{b}{2})} [\frac{1}{\sqrt[]{2}} + \frac{1}{\sqrt[]{2}}]} . a^{2}

? = 4 * \frac{μ_{0} l}{2 π b} \sqrt[]{2} a^{2}

And on comparing $?$ with

$? = M I$

$\frac{μ_{0}}{4 π} 8 \sqrt[]{2} \frac{a^{2}}{b} = M$

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Class 12th Physics Alternating Current

In an ac circuit, an inductor, a capacitor and a resistor are connected in series with $X_{L} = R = X_{C} .$ Impedance of this circuit is:

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Raj Pandey

Contributor-Level 9

$Z = \sqrt[]{R^{2} + {(X_{C} - X_{L})}^{2}}$

$Z = \sqrt[]{R^{2} + {(R - R)}^{2}} [? X_{L} = X_{C} = R \to G i v e n]$

Z = R

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Class 12th Physics Semiconductor Devices

Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one:

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Raj Pandey

Contributor-Level 9

Diode, in forward biased condition only, will allow current to flow through it.

Pot. different across resistor is

Δ V = (10 s i n ω t - 3) v o l t

But in reverse biased condition of diode,

Δ V = 0

(across diode)

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11 months ago

Class 12th Chemistry

Which one of the following is the major product of the given reaction?

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alok kumar singh

Contributor-Level 10

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11 months ago

Class 12th Chemistry

Given below are two statements:

Statement I: Ethyl pent -4-yn-oate on reaction with CH₃MgBr gives a 3° -alcohol.

Statement II: In this reaction one mole of ethyl pent-4-yn-oate utilizes two moles of CH₃MgBr.

In the light of the above statements, choose the most appropriate answer from the options given below.

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alok kumar singh

Contributor-Level 10

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New answer posted

11 months ago

Class 12th Physics Ncert Solutions Class 12th

If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160m, then the maximum range of LOS communication is_______ km.

(Take radius of Earth = 6400km)

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Raj Pandey

Contributor-Level 9

$d M = \sqrt[]{2 R h_{T}} + \sqrt[]{2 R h_{R}}$

$? h_{T} + h_{R} = 160$

[Given]

So, hR = 160 - hT

$d_{M} = \sqrt[]{2 R h_{T}} + \sqrt[]{2 R (160 - h_{T})}$

$∴ d_{M} = \sqrt[]{2 * 6400000 * 80} + \sqrt[]{2 * 6400000 * 80}$

$= 2 * 4 * 8 * 1 0^{3}$

dM = 64 km

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11 months ago

Class 12th Chemistry

Which one of the following is formed (mainly) when red phosphorus is heated in a sealed tube at 803K?

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alok kumar singh

Contributor-Level 10

When red phosphorus is heated in a sealed tube at 803 K, a - black phosphorus is formed.

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11 months ago

Class 12th Coordination Compounds

The number of miles of NH₃, that must be added to 2L of 0.80 M AgNO₃ in order to reduce the concentration of Ag⁺ions to 5.0 * 10^-8 M (K_formation for [Ag(NH₃)₂]⁺ = 1.0 * 10⁸) is_________.

(Nearest integer)

[Assume no volume change on adding NH₃]

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Vishal Baghel

Contributor-Level 10

$A g^{+} (a q) + 2 N H_{3} (a q) ? A g {(N H_{3})}_{2}^{+} (a q)$

t = 00.8 M $(\frac{a}{2}) M$

0 $= \infty$ 5 * 10^-8 M $(\frac{a}{2} - 1.6) M$ 0.8M

$k_{f} = \frac{[A g {(N H_{3})}_{2}^{+}]}{[A g^{+}] {[N H_{3}]}^{2}}$

$1 0^{8} = \frac{0.8}{(5 * 1 0^{- 8}) {(\frac{a}{2} - 1.6)}^{2}}$

$\frac{a}{2} = 2$

Concentration of NH₃ added = $\frac{a}{2} = 2 M$

Volume of solution = 2L

Moles of NH₃ added = 2 * 2 mol = 4 mol.

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Class 12th Physics Electromagnetic Waves

The electric field in an electromagnetic wave is given by E = $(50 N C^{- 1}) s i n ω (t - x ? / ? c)$ The energy contains in a cylinder of volume V is 5.5 * 10^-12 J. The value of V is________cm³.

(given by $\in_{0} = 8.8 * 1 0^{- 12} C^{2} N^{- 1} m^{- 2})$

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Raj Pandey

Contributor-Level 9

E = (50 N/s) sin (t – x/c)

Energy 5.5 * 10^-12 J., vol v =?

$u = \frac{1}{2} ε_{0} E^{2} \to$ energy density

$? u = \frac{U}{v} \Rightarrow U = u v$

= $\frac{1}{2} ε_{0} E^{2} . v$

$v = \frac{5.5 * 1 0^{- 12} * 2}{8.55 * 2500 * 1 0^{- 12}} * 1 0^{6} c m^{3}$

= 497

$≅ 500 c m^{3}$

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