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Physics 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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Physics 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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Physics 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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Physics Class 12th

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

Physics 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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Physics 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

Physics Physics Thermodynamics

Given below are two statements:

Statement – I : When $μ$ moment of an ideal gas undergoes adiabatic change from state $(P_{1}, V_{1}, T_{1})$ to state $(P_{2}, V_{2}, T_{2})$ then work done is w = $\frac{μ R (T_{2} - T_{1})}{1 - γ},$ where $γ = \frac{C_{p}}{C_{v}} a n d$ R = universal gas constant.

Statement – II : In the above case, when work done on the gas, the temperature of the gas would rise.

Choose the correct answer from the options given below:

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Raj Pandey

Contributor-Level 9

In case of adiabatic process -

$W o r k d o n e, W = \frac{η R (T_{2} - T_{1})}{1 - Y}$

$A s, Δ Q = Δ U + W$

for adiabatic process :- $Δ Q = 0$

$\Rightarrow Δ U = - W$

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

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New answer posted

a year ago

Physics Class 11th

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be:

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Raj Pandey

Contributor-Level 9

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 * \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 * 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 * 12 * 3.14 * 1 0^{- 4} = 2.8 * 1 0^{- 4} J$

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Physics Class 11th

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

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Raj Pandey

Contributor-Level 9

According to kepler's third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4 r_{B}^{3}$

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a year ago

Physics Class 11th

Match List – I with List – II

Choose the correct answer from the options given below:

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Raj Pandey

Contributor-Level 9

Use formula for M.I.

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