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Physics Class 11th

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is:

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alok kumar singh

Contributor-Level 10

$η = \frac{W . D . b y g a s}{H e a t a b s o r b}$

$η = \frac{H e a t (N e t)}{H e a t (A b s o r b)}$

$η = \frac{A r e a u n d e r t h e c u r v e}{H e a t A b s o r b}$

$η = \frac{\frac{1}{2} * S_{0} T_{0}}{\frac{1}{2} (3 T_{0}) S_{0}} = \frac{1}{3}$

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2 months ago

Physics Waves

A sound source is located somewhere along the x-axis. Experiment shows that the same wavefront simultaneously reaches listeners at x = -10m and x = +4.0m. A third listener is positioned along the positive y-axis. What is y-coordinate (in m) if the same wavefront reaches at him at the same instant it does the first two listeners?

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alok kumar singh

Contributor-Level 10

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2 months ago

Physics Physics Current Electricity

In the circuit shown the readings of ammeter and voltmeter are $4 A$ and $20 V$ respectively. The meters are non-ideal, then R is:

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Raj Pandey

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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Physics Class 12th

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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alok kumar singh

Contributor-Level 10

Stopping potential increases and number of photons decreases.

$I = \frac{h c n_{p}}{λ t}$

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Physics Physics Ray Optics and Optical Instruments

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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alok kumar singh

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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Physics Physics Wave Optics

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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alok kumar singh

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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Physics 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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Physics Physics Wave Optics

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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Raj Pandey

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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2 months ago

Physics Work, Energy and Power

What minimum speed does a 100g particle need At point B to reach point A? The graph shows potential energy versus position.

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alok kumar singh

Contributor-Level 10

To reach at point A particle must cross the peak point.

Loss in KE = gain in PE

$\frac{1}{2} * \frac{100}{1000} * v^{2} = (5 - 0)$

v² = 100

v = 10 m/s

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2 months ago

Physics Class 12th

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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alok kumar singh

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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