12th Physics 2018 Set1 Delhi Board Paper Solution
Two Electric bulbs P and Q have their resistances in the ratio of 1:2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.
Let ratio be x
Rp = x (Resistance of bulb P)
Rq = 2x (Resistance of bulb Q)
as, P = VI = V. V / R
Pp / Pq = V2 / Rp x Rq / V2 [In series potential will be same]
Pp / Pq = 2x / x = 2 : 1
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(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
">The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
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(b) Stopping potential, and
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- Q:- ">
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(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
">A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
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(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
">The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?