Class 12th

Contributor-Level 9

Voltmeter readings

$\begin{array}{l} \Rightarrow & 20 = I_{1} R \\ \Rightarrow & 20 = (\frac{R_{v}}{R + R_{v}}) 4 R \\ \Rightarrow & R_{v} R = 5 R + 5 R_{v} \Rightarrow R = \frac{5 R_{v}}{R_{v} - 5} \\ ∴ & R = \frac{5}{1 - \frac{5}{R_{v}}} \end{array}$

For $5 < R_{v} < \infty, R > 5 Ω$

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Figure shows the anode potential vs photo-current graph when photons of 6 eV are incident on cathode in a photoelectric experiment set-up. In the same experimental set-up, if photons of 8 eV and same intensity are incident on cathode, then anode-potential vs photo-current graph will be (assume 100% efficiency of photons to eject photo-electrons in both cases)

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Class 12th Physics Ray Optics and Optical Instruments

Contributor-Level 10

Stopping potential increases and number of photons decreases.

$I = \frac{h c n_{p}}{λ t}$

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The figure shows the initial position of a point source of light s, a detector D and lens L. Now at t = 0, all three starts moving towards right with different velocity as shown in figure. The times at which detector receives the maximum light, is (Assume that the detector is initially just touching the lens).

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Class 12th Physics Wave Optics

Contributor-Level 10

$\frac{1}{0.9} = + \frac{1}{(10 ? 2 t)} + \frac{1}{2 t}$

t = 0.5 sec and t = 4.5 sec.

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A parallel beam of monochromatic light is incident on a narrow rectangular slit of width 1mm. When the diffraction pattern in seen on a screen placed at a distance of 2m, the width of principal maxima is found to be 2.5mm. The wavelength of light is

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Class 12th Physics Wave Optics

Contributor-Level 10

Here the width of principal maxima is 2.5 mm

Then its half width is = $\frac{β}{2}$

$= \frac{2.5}{2} = 1.25 * 1 0^{- 3} m$

Diffraction angle q = $\frac{(\frac{β}{2})}{D} = \frac{1.25 * 1 0^{- 3}}{2}$ …(i)

$∴ a θ = λ ∴ θ = \frac{λ}{a}$ …(ii)

From (i) and (ii)

$λ = θ a$

$\frac{1.25 * 1 0^{- 3}}{2} * 1 0^{- 3}$

$∴ λ = 6250$ = 6.25 * 10^-7 m

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Class 12th NCERT Notes

What are key characteristics of Shiksha's NCERT Notes?

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Piyush Vimal

Beginner-Level 5

Shiksha provides NCERT Notes for all class 11 and 12 important topics. We have covered all concepts in detail. You can check the key characteristics below;

We have covered the topic-wise NCERT Notes of all NCERT textbook chapters, including deleted topics and chapters.
These NCERT Notes are prepared keeping in mind the CBSE Exam pattern and CBSE syllabus 2025-2026.
We have designed these NCERT Notes as revision notes along with a concise summary.
You will also get the NCERT-based definition and derivation along with basic problem-solving techniques.
We have also included solved JEE and NEET questions for holis

...more

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Light of wavelength $λ$ in air enters a medium of refractive index $μ$ . Two points in this medium, lying along the path of this light are at a distance $x$ apart. The phase difference between these points is:

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Contributor-Level 9

Optical path $= μ x$ =μx

Path-difference $= μ x$ =μx

Phase-difference $= \frac{2 π}{λ} *$ =2πλ* path-difference $= \frac{2 π}{λ} μ x$ =2πλμx

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In the figure, there is a conducting wire having current i and which has a shape of closed half ellipse $[\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1]$ is kept in a uniform magnetic field B as shown. The magnitude of magnetic dipole moment of loop and torque acting on it are-

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Class 12th Physics Semiconductor Devices

Contributor-Level 10

Area of ellipse = $π a b$

Magnetic dipole moment = $\frac{π a b i}{2}$

Torque $\vec{τ} = \vec{M} * \vec{B}$

$τ = M B s i n θ$

$? \vec{M} | | \vec{B}$ $τ = 0$

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Suppose a pure Si crystal has $5 * 10^{28}$ atoms $/ m^{3}$ . It is doped by $1 p p m$ concentration of pentavalent As. The number of holes $(n_{i} = 1.5 * 10^{16} / m^{3})$ is:

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Contributor-Level 9

${(n_{i})}^{2} = n_{e} * n_{h}$ where $n_{e} = \frac{5 * 10^{28}}{10^{6}}$

$\begin{array}{r} ∴ n_{h} & = \frac{{(n_{i})}^{2}}{n_{e}} = \frac{(1.5 * 10^{16}) (1.5 * 10^{16}) * 10^{6}}{5 * 10^{28}} \\ n_{h} & = 4.5 * 10^{9} / m^{3} \end{array}$

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The binding energy (in MeV) of a nitrogen nucleus $(_{7}^{14} N)$ is :

(Given mass of nucleus $m (_{7}^{14} N) = 14.00307 u$ , mass of proton $m_{p} = 1.00783 u$ , mass of neutron $m_{n} = 1.00867 u$ and $1 u = 931.5 M e V / c^{2}$ )

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Contributor-Level 9

$_{7}^{14} N$ contains 7 protons and 7 neutrons.

Mass defect, $(_{}_{}) (\begin{matrix} \end{matrix})$ $Δ m = (7 m_{H} + 7 m_{n}) - m (\begin{matrix} 14 \\ 7 \end{matrix}) = (7 * 1.00783 u) + (7 * 1.00867 u) - 14.00307 u$

$= 0.11243 u$

Binding energy $= 0.11243 * 931.5 M e V = 104.7 M e V$

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