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8 months ago

Physics Physics Magnetism and Matter

In a ferromagnetic region with zero magnetization.

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Payal Gupta

Contributor-Level 10

In ferromagnetic material, below Curie's temperature, a domain is defined as macroscopic region with zero magentisation.

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8 months ago

Physics Physics Atoms

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is

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Payal Gupta

Contributor-Level 10

Using Bohr's theory :

$λ = \frac{h c}{E_{2} - E_{1}} = \frac{1242 e V - n m}{(- 3.4) - (- 13.6)} = 121.8 n m$

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8 months ago

Physics Work, Energy and Power

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v₀. It encounters and inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

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Payal Gupta

Contributor-Level 10

Using conservation of energy between initial and final position shown :

$[\frac{1}{2} (\frac{2}{5} m a^{2}) {(\frac{V_{0}}{a})}^{2} + \frac{1}{2} m V_{0}^{2}] + 0 = m g l s i n θ$

$\Rightarrow \frac{7}{10} m V_{0}^{2} = m g l s i n θ$

$\Rightarrow l = \frac{7 v_{0}^{2}}{10 g s i n θ}$

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8 months ago

Physics Dual Nature of Radiation and Matter

An electron of mass m_e and a proton of mass m_p = 1836 m_e are moving with the same speed. The ratio of their de Broglie wavelength $\frac{λ_{e l e c t r o n}}{λ_{p r o t o n}}$ will be

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Payal Gupta

Contributor-Level 10

Using de-Broglie equation for wave nature of particle :

$λ = \frac{h c}{m v} \Rightarrow λ α \frac{1}{m}$

$\frac{λ_{e}}{λ_{P}} = \frac{\frac{1}{m_{e}}}{\frac{1}{m_{P}}} = \frac{m_{P}}{m_{e}} = 1836$

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8 months ago

Physics Physics Wave Optics

Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter $0.1 μ m .$ If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that

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Payal Gupta

Contributor-Level 10

Using diffraction formula of circular hole :

$r α \frac{1}{D}$

size will decease.

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8 months ago

Physics Class 12th

If a message signal of frequency 'f_m' is amplitude modulated with a carrier signal of frequency 'f_c' and radiated through an antenna, the wavelength of the corresponding signal in air is

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Payal Gupta

Contributor-Level 10

$λ_{a i r} = \frac{c}{f_{c}}$

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8 months ago

Physics Physics Electric Charge and Field

A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field $\vec{E}$ through the shaded area is

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Payal Gupta

Contributor-Level 10

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New answer posted

8 months ago

Physics Class 11th

Given below are two statements:

Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.

Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.

In the light of the above statements, choose the correct answer from the options given below

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Payal Gupta

Contributor-Level 10

Translational kinetic energy will be equal to rotational kinetic energy corresponds to each degree of freedom.

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8 months ago

Physics Physics Motion in Straight Line

A stone is dropped from the top of a building. When it crosses a point 5m below the top, another stone starts to fall from a point 25m below the top. Both stones reach the bottom of building simultaneously. The height of the building is

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Payal Gupta

Contributor-Level 10

When second stone is released

Equation for first ball:

$x + 20 = 10 t + \frac{1}{2} g t^{2}$

Equation for second ball :

$x = \frac{1}{2} g t^{2}$

Using these two equation

20 = 10t t = 2 sec

$x = \frac{1}{2} * 10 * 4 = 20 m$

H = 20 + 20 + 5 = 45 m.

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New answer posted

8 months ago

Physics Physics Alternating Current

An LCR circuit contains resistance of 110 $Ω$ and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45°. If on the other hand, only inductor is removed the current leads by 45° with the applied voltage. The rms current flowing in the circuit will be

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Payal Gupta

Contributor-Level 10

When capacitor is removed

tan 45° = $\frac{X_{L}}{X_{R}} \Rightarrow ω L = R$

When inductor is removed.

$t a n 45 ° = \frac{X_{C}}{X_{R}} \Rightarrow \frac{1}{ω C} = R$

$i_{0} = \frac{V_{0}}{Z} = \frac{V_{0}}{\sqrt[]{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}} = \frac{V_{0}}{R} = \frac{220}{110} = 2 A$

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