Class 12th

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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Class 12th Physics Wave Optics

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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4 months ago

Visible light of wavelength $6000 * 10^{- 8} c m$ $^{}$ falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $60^{?}$ $^{}$ from the central maximum. If the first minimum is produced at $θ_{1}$ $_{}$ then $θ_{1}$ $_{}$ is close to:

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Class 12th Physics Magnetism and Matter

Contributor-Level 10

For $2^{nd}$ $^{}$ minima

$d s i n ? θ = 2 λ$

$s i n ? θ = \frac{\sqrt[]{3}}{2}$ (given)

$\Rightarrow \frac{λ}{d} = \frac{\sqrt[]{3}}{4}$

So for $1^{st}$ $^{}$ minima is

$d s i n ? θ = λ$

$s i n ? θ = \frac{λ}{d} = \frac{\sqrt[]{3}}{4}$ (from equation (i)

$θ = {25.65}^{?}$ (from sin table)

$^{}$ $θ \approx 25^{?}$ .

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4 months ago

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

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4 months ago

A long solenoid of radius $R$ carries a time ( $t)$ dependent current $_{}$ $I (t) = I_{0} t (1 - t)$ . A ring of radius $2 R$ is placed cordially near its middle. During the time internal $0 \leq t \leq 1$ , the induced current $(I_{R})$ $(_{})$ and the induced $E M F ? (V_{R})$ $(_{})$ in the ring changes as:

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Contributor-Level 10

$I = I_{0} t - I_{0} t^{2}$ $_{}_{}^{}$

$? = B A$

$? = μ_{0} n I A$

$V_{R} = \frac{d ?}{d t} = - μ_{0} {n A l}_{0} (1 - 2 t)$

$V_{R} = 0$ at $\frac{}{}$ $t = \frac{1}{2}$ and $I_{R} = \frac{V_{R}}{Resistance of loop}$ $_{} \frac{_{}}{}$

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Class 12th Physics Wave Optics

In finding the electric field using Gauss law the formula $| \overset{?}{E} | = \frac{q_{enc}}{ε_{0} | A |}$ $\frac{_{}}{_{}}$ is applicable. In the formula $ε_{0}$ $_{}$ is permittivity of free space, $A$ is the area of Gaussain surface and $q_{enc}$ $_{}$ is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation? Only when the Gaussian surface is an

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Raj Pandey

Contributor-Level 9

Magnitude of electric field is constant & the surface is equipotential.

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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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4 months ago

When photon of energy $4.0 e V$ strikes the surface of a metal $A$ , the ejected photoelectrons have maximum kinetic energy $T_{A}$ $_{}$ eV and de-Broglie wavelength $λ_{A}$ $_{}$ . The maximum kinetic energy of photoelectrons liberated form another metal B by photon of energy 4.50 $e V$ is $T_{B} = (T_{A} - 1.5) e V$ $_{} (_{})$ . If the de-Broglie wavelength of these photoelectrons $λ_{B} = 2 λ_{A}$ $_{}_{}$ , then the work function of metal $B$ is

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Raj Pandey

Contributor-Level 9

Relation between De-Broglie wavelength and K.E. is

$λ = \frac{h}{\sqrt[]{2 (K E) m_{e}}} \Rightarrow λ \propto \frac{1}{\sqrt[]{K E}}$

$\begin{array}{r} \frac{λ_{A}}{λ_{B}} = \frac{\sqrt[]{{K E}_{B}}}{\sqrt[]{K E}} \Rightarrow \frac{1}{2} = \sqrt[]{\frac{T_{A} - 1.5}{T_{A}}} \\ \Rightarrow T_{A} 2 e V \\ ∴ {K E}_{B} = 2 - 1.5 = 0.5 e V \\ ?_{B} = 4.5 - 0.5 = 4 e V \end{array}$

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4 months ago

Two infinite planes each with uniform surface charge density $+ σ$ are kept in such a way that the angle between them is $30^{?}$ $^{}$ . The electric field in the region shown between them is given by:

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Class 12th Physics Wave Optics

Contributor-Level 10

$\overset{?}{E} = \frac{σ}{2 ε_{0}} c o s ? 60^{?} (- \hat{x}) + [\frac{σ}{2 ε_{0}} - \frac{σ}{2 ε_{0}} s i n ? 60^{?}] (\hat{y})$

$\overset{?}{E} = \frac{σ}{2 ε_{0}} [(1 - \frac{\sqrt[]{3}}{2}) \hat{y} - \frac{1}{2} \hat{x}]$

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4 months ago

A polarizer-analyser set is adjusted such that the intensity of light coming out of the analyser is just $10 %$ of the original intensity. Assuming that the polarizer-analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero is

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Class 12th Electromagnetic Induction

Contributor-Level 10

$I = I_{0} {c o s}^{2} ? θ$

$\frac{I_{0}}{10} = I_{0} {c o s}^{2} ? θ$

$c o s ? θ = \frac{1}{\sqrt[]{10}} = 0.31 < \frac{1}{\sqrt[]{2}}$ which is 0.707

So, $θ > 45^{?}$ $^{}$ and $90 - θ < 45^{?}$ $^{}$ so only one option is correct i.e. ${18.4}^{?}$ $^{}$

Angle rotated should be $= 90^{?} - {71.6}^{?} = {18.4}^{?}$ $^{}^{}^{}$ .

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4 months ago

A planar loop of wire rotates in a uniform magnetic field. Initially, at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

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