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11 months ago

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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11 months ago

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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11 months ago

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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11 months ago

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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11 months ago

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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11 months ago

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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11 months ago

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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11 months ago

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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11 months ago

Class 12th Physics

For an object placed at a distance 2.4m from a lens, a sharp focused image is observed on a screen placed at a distance 12cm from the lens. A glass plate of refractive index 1.5 and thickness 1cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

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Payal Gupta

Contributor-Level 10

When glass slab is not put in between lens and screen

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{12} - \frac{1}{- 240} = \frac{1}{f} . . . . . . (1)$

When glass slab is put in between lens and screen

Shift = $t (1 - \frac{1}{μ}) = 1 (1 - \frac{1}{1.5}) = \frac{1}{3} c m$

$\Rightarrow \frac{3}{35} - \frac{1}{- x} = \frac{1}{f} . . . . . . . . . (2)$

With the help of equations (1) and (2), we can write

$\frac{3}{35} - \frac{1}{- x} = \frac{1}{12} - \frac{1}{- 240}$

$\Rightarrow | Δ u | = 560 - 240 = 320 c m = 3.2 m$

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11 months ago

Class 12th Dual Nature of Radiation and Matter

The ratio of wavelengths of proton and deuteron accelerated by potential V_p and V_d is $1 : \sqrt[]{2}$ . Then, the ratio of V_p to V_d will be

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Payal Gupta

Contributor-Level 10

According to de-Broglie's hypothesis, we can write

$λ = \frac{h}{p} = \frac{h}{\sqrt[]{2 m e V}}$

$\Rightarrow \frac{λ_{P}}{λ_{D}} = \frac{\sqrt[]{2 m_{D} e V_{D}}}{\sqrt[]{2 m_{P} e V_{P}}} = \frac{1}{\sqrt[]{2}} \Rightarrow \frac{2 m_{P} V_{D}}{m_{P} V_{P}} = \frac{1}{2} \Rightarrow \frac{V_{P}}{V_{D}} = 4 : 1$

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