Physics Ncert Solutions Class 12th

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is……… times that of a smaller drop.

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With the help of conservation of volume, we can write

$27 * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 3 r . . . . . . . (1)$

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U1 = $\frac{q^{2}}{8 π ε_{0} r}$ $\frac{^{}}{_{}}$

Potential energy of bigger drop = $U_{2} = \frac{Q^{2}}{8 π ε_{0} R} = \frac{27 * 27 * q^{2}}{8 π ε_{0} (3 r)} = 243 (\frac{q^{2}}{8 π ε_{0} r}) = 243 U_{1}$ $_{} \frac{^{}}{_{}} \frac{^{}}{_{}} \frac{^{}}{_{}}_{}$

$\Rightarrow \frac{U_{2}}{U_{1}} = 243$

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The zener diode has a V_z = 30V. The current passing through the diode for the following circuit is…………mA.

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Physics Ncert Solutions Class 12th Communication Systems

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$l = \frac{90 - 30}{4000} = 15 m A$

$I_{1} = \frac{30}{5000} = 6 m A & l_{2} = 9 m A$

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If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are……….

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Band Width = 2 * n * the highest modulation frequency

$\Rightarrow n = \frac{90 k H_{z}}{2 * 5 k H z} = 9$

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Draw the output signal Y in the given combination of gates.

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Physics Ncert Solutions Class 12th Electric Flux

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$y = \bar{\bar{A} + B} = A . \bar{B}$

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Given below are two statements:

Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.

Statement II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r (

In the light of the above statements, choose the correct answer from the option given below:

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All the charge given to a conducting sphere resides on outer surface.

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The maximum and minimum amplitude of an amplitude modulated wave is 16V and 8V respectively. The modulation index for this amplitude wave x * 10^-2. The value of x is----------.

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Modulation Index $(μ) = \frac{A_{m a x} - A_{m i n}}{A_{m a x} + A_{m i n}} = \frac{16 - 8}{16 + 8} = \frac{1}{3} = 0.33 = 33 * 1 0^{- 2}$

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The circuit contains two diodes each with a forward resistance of 50 $Ω$ and with infinite reverse resistance. If the battery voltage is 6 V, the current through the 120 $Ω$ resistance is---------mA.

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D₁ is forward biased & D₂is reverse biased.

$l = \frac{6}{120 + 130 + 50} = \frac{6}{300} = 2 * 1 0^{- 2} A = 20 * 1 0^{- 3} A = 20 m A$

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The temperature q at the junction of two insulating sheets,having thermal resistances R₁ and R₂ as well top and bottom temperatures q₁ and q₂ (as shown in figure) is given by

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taking θ₁ > θ₂

Heat current = $\frac{θ_{1} - θ}{R_{1}} = \frac{θ - θ_{2}}{R_{2}} \Rightarrow \frac{θ_{1}}{R_{1}} + \frac{θ_{2}}{R_{2}} = θ (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\Rightarrow θ = \frac{θ_{1} R_{2} + θ_{2} R_{1}}{R_{1} + R_{2}}$

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LED is constructed from Ga – As – P semiconducting material. The energy gap of this LED is 1.9 eV. Calculate the wavelength of light emitted and its colour.

[h = 6.63 * 10^-34 Js and c = 3 * 10⁸ ms^-1]

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Physics Ncert Solutions Class 12th Communication Systems

Contributor-Level 10

$λ = \frac{h c}{E} = \frac{6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{1.9 * 1.6 * 1 0^{- 19}}$ = 654 nm (red)

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A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5dB per km and cable length is 20km. The power received at receiver is 10^-^xW. The value of x is…………………

[Given in dB = 10 log_{10 ^{$(\frac{P_{0}}{P_{i}})]$}}

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