A metallic loop is placed in a nonuniform magnetic field. Will an…

Question

A metallic loop is placed in a nonuniform magnetic field. Will an emf be induced in the loop?

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Answer

As the magnetic field is non uniform thus it will induce only small electric field in different directions so there would be no net current in the loop.

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When light of sufficiently high frequency is incident on a metallic surface, electrons are emitted from the metallic surface. TI1is phenomenon is called photoelectric emission. Kinetic energy of the emitted photoelectrons depends on the wavelength of incident tight and is independent of the intensity of light. Number of emitted photoelectrons depends on intensity. $(\text{h}\upsilon-\phi)$ is the maximum kinetic energy of emitted photoelectrons (where $\phi$ is the work function of metallic surface). Reverse effect of photo emission produces X-ray. X-ray is not deflected by electric and magnetic fields. Wavelength of a continuous X-ray depends on potential difference across the tube. Wavelength of characteristic X-ray depends on the atomic number.

Einstein's photoelectric equation is:

$\text{E}_\text{max}=\text{h}\upsilon-\phi$
$\text{E}=\text{mc}^2$
$\text{E}^2=\text{p}^2\text{c}^2+\text{m}_0^2\text{c}^4$
$\text{E}=\frac{1}{2}\text{mv}^2$

Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity then stopping potential will:

Increase.
Decrease.
Be zero.
Become exactly half.

When ultraviolet rays incident on metal plate then photoelectric effect does not occur, it occur by incident of:

Infrared rays
X-rays
Radio wave
Micro wave

If frequency $(\upsilon>\upsilon_0)$ of incident light becomes n times the initial frequency $(\upsilon)$, then K.E. of the emitted photoelectrons becomes ($\upsilon_0$ threshold frequency).

N times of the initial kinetic energy.
More than n times of the initial kinetic energy.
Less than n times of the initial kinetic energy.
Kinetic energy of the emitted photoelectrons remains unchanged.

A monochromatic light is used in a photoelectric experiment. The stopping potential.

Is related to the mean wavelength.
Is related to the shortest wavelength.
Is not related to the minimum kinetic energy of emitted photoelectrons.
Intensity of incident tight.

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Hydrogen spectrum consists of discrete bright lines in a dark background, and it is specifically known as hydrogen emission spectrum. There is one more type of hydrogen spectrum that exists where we get dark lines on the bright background, it is known as absorption spectrum. Balmer found an empirical formula by the observation of a small part of this spectrum, and it is represented by $\frac{1}{\lambda}=\text{R}\bigg(\frac{1}{2^2}-\frac{1}{\text{n}^2}\bigg)$ where n = 3, 4, 5 For Lyman series, the emission is from first state to $n^{th}$ state, for Paschen series, it is from third state to $n^{th}$ state, for Brackett series, it is from fourth state to $n^{th}$ state and for Pfund series, it is from fifth state to $n^{th}$ state.

Number of spectral lines in hydrogen atom is:

8
6
15
$\infty$

Which series of hydrogen spectrum corresponds to ultraviolet region?

Balmer series.
Brackett series.
Paschen series.
Lyman series.

Which of the following lines of the H-atom spectrum belongs to the Balmer series?

1025A
1218A
4861A
18751A

Rydberg constant is.

A universal constant.
A universal constants.
Different for different elements.
None of these.

Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be.

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Mr. Verma (50kg) and Mr. Mathur (60kg) are sitting at the two extremes of a 4m long boat (40kg) standing still in water. To discuss a mechanics problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the process?

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Refraction of light is the change in the path of light as it passes obliquely from one transparent medium to another medium. According to law of refraction $\frac{\sin\text{i}}{\sin\text{r}}=\ ^1\mu_2,$ where $^1\mu_2$ is called refractive index of second medium with respect to first medium. From refraction at a convex spherical surface, we have $\frac{\mu_2}{\text{v}}=\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}.$ Similarly from refraction at a concave spherical surface when object lies in the rarer medium, we have $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$ and when object lies in the denser medium, we have $\frac{\mu_1}{\text{v}}-\frac{\mu_2}{\text{u}}=\frac{\mu_1-\mu_2}{\text{R}}.$

Refractive index of a medium depends upon:

Nature of the medium.
Wavelength of the tight used.
Temperature.
All of these.

A ray of light of frequency $5 \times 10^{14}Hz$ is passed through a liquid. The wavelength of light measured inside the liquid is found to be $450 \times 10^{-9}m.$ The refractive index of the liquid is:

1.33
2.52
2.22
0.75

A ray of light is incident at an angle of 60º on one face of a rectangular glass slab of refractive index 1.5. The angle of refraction is:

$\sin^-1 (0.95)$
$\sin^-1 (0.58)$
$\sin^-1 (0.79)$
$\sin^-1 (0.86)$

A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of sphere is: