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Class 12th Physics Ray Optics and Optical Instruments

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a alight of wavelength 2440 $\overset{o}{A}$ is used to see the object will be:

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Raj Pandey

Contributor-Level 9

Resolving power, = $\frac{d}{1.22 λ} = \frac{24.4 * 1 0^{- 2}}{1.22 * 2440 * 1 0^{- 10}}$

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Class 12th Physics Ray Optics and Optical Instruments

The refracting angle of a prims is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be-

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Raj Pandey

Contributor-Level 9

At minimum deviation –

$δ = 2 i - A \Rightarrow i = \frac{δ + A}{2}, r = \frac{A}{2}$

Using snell's low : - $s i n i = μ = s i n r$

$\Rightarrow \frac{δ + A}{2} = \frac{π - A}{2} \Rightarrow δ = π - 2 A$

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Class 12th Physics Electromagnetic Waves

A radar sends an electromagnetic single of electric (E₀) = 2.25 V/m and magnetic field (B₀) = 1.5 * 10^-⁸ T which strikes a target on line of sight at a distance of 3km in a medium. After that, a part of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time t = 0 from radar, then after how much time echo will reach to the radar?

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Raj Pandey

Contributor-Level 9

$E_{0} = 2.25 \frac{v}{m}$ Speed of declromagnetic signal, $v = \frac{E_{0}}{B_{0}} = \frac{2.25 v / m}{1.5 * 1 0^{- 8} T}$

$B_{0} = 1.5 * 1 0^{- 8} T \Rightarrow v = \frac{225}{150} * 1 0^{8} = 1.5 * 1 0^{8} m / s$

Time to reach each to the same

radar, $t = \frac{2 * 3 * 1 0^{3} m}{1.5 * 1 0^{8} m / s} = 4 * 1 0^{- 5} S$

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Class 12th Physics

An infinitely long hollow conducting cylinder with radius R carries a uniform current along its surface.

Choose the correct representation of magnetic field (B) as a function of radial distance (r) from the axis of cylinder.

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Raj Pandey

Contributor-Level 9

Using Ampere's law for long hollow cylinder carrying current on its surface

B_in = 0

$B_{o u t} = \frac{μ_{0} i}{2 π r}$

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a year ago

Class 12th Physics Electric Charge and Field

The three charges q/2, q and q/2 are placed at the corners A, B and C of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the corner D of the square, is:

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Raj Pandey

Contributor-Level 9

$E = \frac{k q}{2 a^{2}}$

$E' = \frac{k q}{{(\sqrt[]{2} a)}^{2}} = \frac{k q}{2 a^{2}}$
$E_{N e t} = 2 E c o s 45 ° + E'$

$= \frac{k q}{a^{2}} (\frac{1}{\sqrt[]{2}} + \frac{1}{2})$

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New answer posted

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Class 12th Physics Electric Charge and Field

Given below are two statements:

Statement – I : A point change is brought in an electric field. The value of electric field at a point near to the change may increase if the charge is positive.

Statement – I : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.

Choose the correct answer from the options given below

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Raj Pandey

Contributor-Level 9

Kindly consider the following Image

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a year ago

Class 12th Biomolecules

What is the zwitter ion class 12?

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Swayam Gupta

Contributor-Level 9

A zwitterion is an ion that contains both negative and positive charges. This is a dipolar ion and is considered neutral.

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Class 12th Biomolecules

What is a peptide bond?

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Swayam Gupta

Contributor-Level 9

It is a covalent chemical bond that connects amino acids to form polypeptide chains.

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Class 12th Biomolecules

What is the function of biomolecules?

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Swayam Gupta

Contributor-Level 9

Different functions of biomolecules include storing genetic information (by nucleic acids), providing energy (by carbohydrates), and building and repairing tissue (by proteins).

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Class 12th Physics

An inductor of 0.5 mH, a capacitor of 200 $μ F$ and a resistor of 2 $Ω$ are connected in series with a 220V ac source. If the current is in phase with the emf, the frequency of ac source will be__________ * 10² Hz.

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alok kumar singh

Contributor-Level 10

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2 π \sqrt[]{L C}} = \frac{1}{2 π \sqrt[]{(0.5 * 1 0^{- 3}) * (200 * 1 0^{- 6})}}$

$\Rightarrow f = \frac{1 0^{4}}{2 π \sqrt[]{10}} \approx 5 * 1 0^{2} H z$ $[T a k i n g π = \sqrt[]{10}]$

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