Class 12th

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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Class 12th Dual Nature of Radiation and Matter

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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New answer posted

a year ago

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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Class 12th Physics Ray Optics and Optical Instruments

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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New answer posted

a year ago

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

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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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New answer posted

a year ago

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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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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New answer posted

a year ago

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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Class 12th Electromagnetic Waves

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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New answer posted

a year ago

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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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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New answer posted

a year ago

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

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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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New answer posted

a year ago

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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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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New answer posted

a year ago

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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Contributor-Level 10

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

$\Rightarrow R = 7500 \pm 375 Ω$

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New answer posted

a year ago

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 * 10^-⁵m and density 1.2 * 10³ kgm^-³? Take viscosity of liquid = 1.8 * 10^-⁵ Nsm^-². (Neglect buoyancy due to air).

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