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Class 12th Electromagnetic Induction

A constant magnetic field of 1T is applied in the x > 0 region. A metallic circular ring of radius 1m is moving with a constant velocity of 1 m/s along the x-axis. At t = 0, the centre of O of the ring is at x = -1m. What will be the value of the induced emf in the ring at t = 1s? (Assume the velocity of the ring does not change.)

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alok kumar singh

Contributor-Level 10

$ε = B v (2 R) = 1 * 1 * 2 * 1 = 2 V$

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8 months ago

Class 12th Physics

Figure shows a rod AB, which is bent in a 120° circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric field $\vec{E}$ at the centre of curvature O?

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alok kumar singh

Contributor-Level 10

$λ = \frac{- Q}{R * 2 π / 3}$

$\vec{E} = \frac{2 k λ}{R} s i n (\frac{θ}{2}) (- \hat{i})$

$\Rightarrow \vec{E} = \frac{2 k}{R} * \frac{- 3 Q}{2 π R} s i n 60 ° (- \hat{i})$

$\Rightarrow \vec{E} = \frac{3 \sqrt[]{3} Q}{8 π^{2} ε_{0} R^{2}} (\hat{i})$

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8 months ago

Class 12th Dual Nature of Radiation and Matter

A monochromatic neon lamp with wavelength of 670.5nm illuminates a photo-sensitive material which has a stopping voltage of 0.48V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6nm?

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alok kumar singh

Contributor-Level 10

$e V_{S 2} = \frac{h c}{λ_{1}} - ?$

$e V_{S 2} = \frac{h c}{λ_{2}} - ?$

$e V_{S 2} - e V_{S 1} = \frac{h c}{λ_{2}} - \frac{h c}{λ_{1}}$

$\Rightarrow V_{S 2} = 0.48 + 12.43 * 1 0^{- 7} * (\frac{1}{474.6} - \frac{1}{670.5}) * 1 0^{- 9}$

= 1.25 V

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8 months ago

Class 12th Physics Ray Optics and Optical Instruments

Find the distance of the image from object O, formed by the combination of lenses in the figure:

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Vishal Baghel

Contributor-Level 10

$\frac{1}{V_{1}} + \frac{1}{30} = \frac{1}{10}$

V₁ = 15 cm

$\frac{1}{V_{2}} - \frac{1}{10} = - \frac{1}{10}$

$\Rightarrow V_{2} = \infty$

$\Rightarrow V_{3} = 30 c m$

$\Rightarrow O V_{3} = 75 c m$

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8 months ago

Class 12th Physics

For a transistor a and b are given as $α = \frac{I_{C}}{I_{E}} a n d β = \frac{I_{C}}{I_{B}} .$ Then the correct relation between a and b will be:

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alok kumar singh

Contributor-Level 10

β $= \frac{I_{C}}{I_{E}} = \frac{I_{C}}{I_{E} - I_{C}} = \frac{I_{C} / I_{E}}{1 - l_{C} / I_{E}} = \frac{α}{1 - α}$

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Class 12th Physics

For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be:

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alok kumar singh

Contributor-Level 10

Current at full deflection,

$l_{m a x} = \frac{50}{2} = 25 m A$

$R = \frac{V}{l_{m a x}} = \frac{50}{25} = 2 K Ω$

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8 months ago

Class 12th Electromagnetic Waves

Electric field in a plane electromagnetic wave is given by

E = 50 sin(500x – 10 * 10¹⁰t) V/m

The velocity of electromagnetic wave in this medium is:

(Given C = speed of light in vacuum)

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Vishal Baghel

Contributor-Level 10

$V = \frac{c o e f f i c i e n t o f t i m e}{c o e f f i c i e n t o f x} = \frac{ω}{k}$

$\Rightarrow V = \frac{10 * 1 0^{10}}{500} = 2 * 1 0^{8} = \frac{2 * c}{3}$

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8 months ago

Class 12th Physics

For a transistor in CE mode to be used as an amplifier, it must be operated in:

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Vishal Baghel

Contributor-Level 10

Active region of the CE transistor is linear region and is best suited for its use as an amplifier.

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Class 12th Physics

A coaxial cable consists of an inner wire of radius 'a' surrounded by an outer shell of inner and outer radii 'b' and 'c' respectively. The inner wire carries an electric current i₀, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x < a and(ii) a < x < b?

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alok kumar singh

Contributor-Level 10

For x < a,

B₁ = $\frac{μ_{0} i_{0} x}{2 π a^{2}}$

For a < x < b,

$B_{2} = \frac{μ_{0} i_{0}}{2 π x}$

$\frac{B_{1}}{B_{2}} = \frac{x^{2}}{a^{2}}$

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Class 12th Physics

The colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is:

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alok kumar singh

Contributor-Level 10

$R = 75 * 1 0^{2} \pm 5 %$

$\Rightarrow R = 7500 \pm 375 Ω$

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