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Posted by Shubham Pandey 6 years, 6 months ago
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Gaurav Seth 6 years, 6 months ago
Assume that one cup holds 250 ml of water.
250 ml = 250 gram of water since its density is 1 in cgs.
number of moles of water in 250 gram = 250 / molar weight of water
n = 250/18 = 13.88 moles
1 mol has 6.022 x 1023 number of water molecules
So 250 g water will have 13.88 x 6.022 x 1023 number of H2O molecules
= 8.36 x 1024 number of molecules
One molecule of water has 2 hydrogen atoms (2 protons) and one oxygen atom (8 protons).
Total protons = 10 = total electrons (since it is neutral)
So 250 g water will have 10 x 8.36 x 1024 = 8.36 x 1025 protons.
charge of proton = 1.6 x 10-19 C
So total positive charge of 250 g water = 1.6 x 10-19 x 8.36 x 1025 = 13.37 x 106 C
which is equal to 13.3 MC (mega coulomb)
So one cup of water has 13.3 mega coulomb positive and negative charge, so that the overall charge is zero.
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Average life or mean life (τ) of a radioactive element is the ratio of total life time of all the atoms and total number of atoms present initially in the sample.
Relation between average life and decay constant τ = 1 / λ
Relation between half-life and average life τ = 1.44 T
The number of atoms left undecayed after n half-lifes is given by
N = No (1 / 2)n = No (1 / 2) t/T
where, n = t / T, here t = total time.
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Lenz's law states that the direction of induced emf is such, that it produces a current which opposes the change in magnetic flux that produced it.
Mathematically,
e = -N{tex}\frac{d\varphi}{dt}{/tex}
Suppose a magnet is moved towards a closed coil. The direction of induced emf in it by Lenz’s law opposes its inward motion. As a result, the magnet will be moved towards the coil against this opposing force. In such a case work will have to be done. This work is converted into electrical energy in the form of induced emf. Thus, Lenz’s law is in accordance with the law of conservation of energy.
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Two co-axial cylindrical shells A and B of radii a and b are possessed by a cylindrical capacitor. Assume l be length of the cylindrical shell. Due to the introduction of +q charge on the inner cylindrical shell A, equal but opposite charge -q is induced on the inner surface of the outer cylindrical shell B. The induced charge +q on its outer surface will flow to earth if the shell B is earthed.
The capacitance of the cylindrical capacitor is given as follows if V is potential difference between the cylindrical shell A and B.
{tex}C = \frac{q}{V}{/tex}
By applying the Gaussian a theorem, we first need to find electric field E in the space between two shells to find out potential difference between the cylindrical shells A and B . Let a cylinder of radius r (such that b > r > a) and length l as the Gauss surface. Charge enclosed by the Gaussian surface is {tex}\lambda{/tex}1 and if {tex}\lambda{/tex} is charge per unit length on the shell A.
The electric flux will cross through only curved surface of the cylinder (Gauss surface). As the area of curved surface of cylinder is {tex}2\pi rl{/tex}, we have by Gaussian theorem.
{tex}\oint E \cdot ds = \frac{q}{{{\varepsilon _0}}}{/tex}
{tex}E \cdot 2\pi rl = \frac{{\lambda l}}{{{\varepsilon _0}}}{/tex}
{tex}E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}{/tex}
The field lines are radial and normal to the axis of charged cylinder.
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Kiran??? Gautam 6 years, 6 months ago
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