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Class 12th Chemistry

In liquation process used for tin (Sn), the metal

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alok kumar singh

Contributor-Level 10

Liquation refining process is applicable for the metal having low m.p, but containing impurities have higher m.p

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Class 12th Chemistry

Given below are the critical temperature of some of the gases:

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alok kumar singh

Contributor-Level 10

Higher adsorbtion of gas is corresponds to higher liquefaction and higher liquefaction is directly proportional to the higher critical temperature.

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Class 12th Maths Determinants

The number of real values of $λ$ , such that the system of linear equations

2x – 3y + 5z = 9

x + 3y – z = 18

3x – y $(λ^{2} - | λ |) z = 16$ has no solutions, is

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Payal Gupta

Contributor-Level 10

System of equation can be written as

$(\begin{matrix} 2 & - 3 & 5 \\ 1 & 3 & - 1 \\ 3 & - 1 & λ^{2} - | λ | \end{matrix}) (\begin{array}{l} x \\ y \\ z \end{array}) = (\begin{array}{l} 9 \\ - 18 \\ 16 \end{array})$

for no solution

$\frac{9}{10} (λ^{2} - | λ | + 3) - 7 = 0$

$\Rightarrow 9 λ^{2} - 9 | λ | - 43 = 0$

$\Rightarrow | λ | = \frac{9 + \sqrt[]{81 + 36 * 43}}{18} > 0$

$∴$ only two possible value of $λ$ are there.

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Class 12th Maths

For $z \in C$ if the minimum value of $(| z - 3 \sqrt[]{2} | + | z - p \sqrt[]{2} i |) i s 5 \sqrt[]{2}$ , then a value of p is

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Payal Gupta

Contributor-Level 10

$| z - 3 \sqrt[]{2} | + | z - p \sqrt[]{2} i |$ is minimum for z, $3 \sqrt[]{2} & p \sqrt[]{2} i$ are collinear.

$\Leftrightarrow {(3 \sqrt[]{2})}^{2} + {(p \sqrt[]{2})}^{2} = {(5 \sqrt[]{2})}^{2}$

$\Rightarrow 18 + 2 p^{2} = 50$

$2 p^{2} = 32$

$p = \pm 4$

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Class 12th Physics Atoms

Hydrogen atom from exited state comes to the ground state by emitting a photon of wavelength $λ .$ The value of principal quantum number 'n' of the excited state will be

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Payal Gupta

Contributor-Level 10

According to definition of wave number, we can write

$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{n^{2}}) \Rightarrow 1 - \frac{1}{n^{2}} = \frac{1}{λ R}$

$\Rightarrow \frac{1}{n^{2}} = 1 - \frac{1}{λ R} = \frac{λ R - 1}{λ R} \Rightarrow n = \sqrt[]{\frac{λ R}{λ R - 1}}$

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Class 12th Physics

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1%, 2% and 3% respectively. The maximum percentage error in the detection of the dissipated heat will be

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Payal Gupta

Contributor-Level 10

$H = l^{2} R t$

$% e r r o r i n H = \frac{Δ H}{H} * 100 = 2 (\frac{Δ l}{l}) * 100 + (\frac{Δ R}{R}) * 100 + (\frac{Δ t}{t}) * 100 = 2 * 2 + 1 + 3 = 8 %$

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Class 12th Physics

A current of 15 mA flow in the circuit as shown in figure. The value of potential difference between the pointes A and B will be

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Payal Gupta

Contributor-Level 10

$R_{A B} = 5 + (10 / / 5) + 10 = 15 + \frac{10 * 5}{10 + 5} = 15 + \frac{10}{3} = \frac{55}{3} k Ω$

$V_{A B} = R_{A B} l = (\frac{55}{3} * 1 0^{3}) * (15 * 1 0^{- 3}) = 275 V$

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Class 12th Physics Moving Charges and Magnetism

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 * 10^-4 Wbm^-2. The frequency of revolution of the electron will be

(Take mass of electron = 9.0 * 10^-31kg)

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Payal Gupta

Contributor-Level 10

f = $\frac{q B}{2 π m} = \frac{1.6 * 1 0^{- 19} * 1.0 * 1 0^{- 4}}{2 * 3.14 * 9 * 1 0^{- 31}} = 0.028 * 1 0^{8} H z = 2.8 * 1 0^{6} H z$

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Class 12th Physics

When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on

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Payal Gupta

Contributor-Level 10

This phenomenon works on resonance in AC circuit.

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Class 12th Physics Electromagnetic Waves

Light wave travelling in air along x-direction is given by Ey = $540 s i n π * 1 0^{4} (x - c t) V m^{- 1} .$ The, the peak value of magnetic field of wave will be (Given c = 3 * 10⁸ms^-1)

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Payal Gupta

Contributor-Level 10

$c = \frac{E_{0}}{B_{0}} \Rightarrow B_{0} = \frac{E_{0}}{c} = \frac{540}{3 * 1 0^{8}} = 18 * 1 0^{- 8} T$

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