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Sahiba Bhayana 8 years ago
Posted by Shiva Pandey 8 years ago
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Varun Banda 8 years ago
λmax1=hcE=6.6×10−34×3×1082×1.6×10−19=6.18×10−7 mλmax2=hcE=6.6×10−34×3×1082.5×1.6×10−19=3.96×10−7 mλmax3=hcE=6.6×10−34×3×1083×1.6×10−19=4.125×10−7 m Given λ=6000A=6×10−7 m
For detection of optical signal the wavelength of incident energy radiation must be greater. It is true only for D2 .Only D2 can detect the radiation
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Shivang Gupta 8 years ago
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Varun Banda 8 years ago
Capacitance for drop => R/kwhere R is the radii.
From Volume conservation, Vnet = 64V
R^3 = 64r^3
R = 4r
For big drop, C = R/k.......(1)
For small dop, 5 = r/k........(2)
Divdng the eqns. we get,
C = 5*(R/r) => 5*4 = 20 mF.
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Amar Kumar 8 years ago
Gaussian surface :It is a closed surface in a three dimensional space through which the magnetic flux of any type of vector field can be found, this includes the electric field, the gravitational field, or magnetic field.
Basically Gaussian surface is an arbitrary closed surface used in accordance with Gauss's law for the particular field (for electricity, gravity, or Gauss' law for magnetism) by applying a surface integral, to calculate the total amount of the quantity of the source enclosed, i.e. the amount of electric charge (q) for of the electrostatic field or the amount of gravitational mass for the source of the gravitational field, or vice versa and then calculate the net fields for the source distribution in physics.
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Vineit Kumar Gupta 8 years ago
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Vineit Kumar Gupta 8 years ago
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