Class 12th

Class 12th Physics Electromagnetic Waves

Contributor-Level 10

In forward bias diode act as a short circuit wire. Hence, the equivalent circuit is now.

So, $_{} \frac{}{}$ $V_{a b} = \frac{30}{5 + 10} * 5 = 10 V$

$50.00$

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The electric field of a plane electromagnetic wave is given by $\overset{?}{E} = E_{0} \frac{\hat{i} + \hat{j}}{\sqrt[]{2}} c o s ? (k z + ω t)$

At $t = 0$ , a positively charged particle is at the point $(x, y, z) = (0,0, \frac{π}{k})$ . If its instantaneous velocity at $(t = 0)$ is $v_{0} \hat{k}$ , the force acting on it due to the wave is:

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Contributor-Level 10

Electric field at $t = 0$ & $(x, y, z) = (0,0, \frac{π}{k})$ is

$\begin{array}{r} \overset{?}{E} = - (\frac{\hat{i} + \hat{j}}{\sqrt[]{2}}) E_{0} and \overset{?}{F} = \overset{?}{E} q \\ ∴ \overset{?}{F} = - (\frac{\hat{i} + \hat{j}}{\sqrt[]{2}}) q \end{array}$

Which is antiparallel to $(\frac{\hat{i} + \hat{j}}{\sqrt[]{2}})$

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Class 12th Nuclei

The activity of a radioactive sample falls from $700 s^{- 1}$ to $500 s^{- 1}$ in 30 minutes. Its half life is close to

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Contributor-Level 10

$A = A_{0} e^{- λ t}$

$\begin{array}{r} 500 = 700 e^{- λ t} \\ λ t = l n ? \frac{7}{5} \\ \frac{l n ? 2}{t_{1 / 2}} * 30 = l n ? \frac{7}{5} \\ ? t_{1 / 2} = \frac{l n ? 2 * 30}{l n ? \frac{7}{5}} \end{array}$

$t_{1 / 2} = 61.8 m i n \approx 62 m i n$

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The reaction of H₂O₂ with potassium permanganate in acidic medium leads to the formation of mainly.

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Contributor-Level 10

$2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} O_{2} \to 2 M n^{2 +} + 8 H_{2} O + 5 O_{2}$

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The compound(s) that is (are) removed as slag during the extraction of copper is:

(A) CaO

(B) FeO

(C) Al₂O_{3 &n}

...more

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Contributor-Level 10

$F e O + \underset{(A c i d i c f l u x)}{S i O_{2}} \to \underset{(S l a g)}{F e S i O_{3}}$

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The IUPAC nomenclature of an element with electronic configuration [Rn] 5f¹⁴6d¹7s² is:

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Contributor-Level 10

IUPAC nomenclature of element with atomic no. 103 is uniltrium.

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Class 12th Physics

The figure gives experimentally measured $B$ vs $H$ variation in a ferromagnetic material. The retentivity, co-ercivity and saturation, respectively, of the material are

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Contributor-Level 10

$x =$ retentivity

$y =$ coercivity

$z =$ saturation magnetization

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Match the List-I with List-II

List-I	List-II
(A) N₂(g) + 3H₂(g) -> 2NH₃(g)	(I) Cu
(B) CO(g) + 3H₂(g) -> CH₄(g) + H₂O(g)	(II) Cu/ZnO – Cr₂O₃
(C) CO(g) + H₂(g) -> HCOH(g)	(III) Fe_xO_y + K₂O + Al₂O₃
(D) CO(g) + 2H₂(g) -> CH₃OH(g)	(IV) Ni

Choose the correct answer from the options given below:

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Contributor-Level 10

Uses of catalyst is very specific for a particular reaction.

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Class 12th Wave Optics

In a Young's double slit experiment, the separation between the slits is $0.15 m m$ . In the experiment, a source of light of wavelength $589 n m$ is used and the interference pattern is observed on a screen kept $1.5 m$ away. The separation between the successive bright fringes on the screen is:

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Contributor-Level 10

$\frac{Energy}{Volume} = \frac{{M L}^{2} T^{- 2}}{L^{3}} = ({M L}^{- 1} T^{- 2})$

The distance between two successive bright fringes is fringe width $(β)$ .

$β = \frac{λ D}{d} = \frac{589 * 10^{- 9} * 1.5}{0.15 * 10^{- 3}} = 5.9 m m$

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Class 12th Solutions

The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L-1 is $0.0405 ° C .$ Density of formic acid is 1.05 g mL-1. The Van't Hoff factor of the formic acid solution is nearly: (Given for water $K_{f} = 1.86 k k g m o l^{- 1}$ $_{}^{}$ )

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