Question 6

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.

Answer

Let ratio be x

R_{p} = x (Resistance of bulb P)

R_{q }= 2x (Resistance of bulb Q)

as, P = VI = V. V / R

P_{p} / P_{q} = V2 / R_{p} x R_{q} / V2 [In series potential will be same]

P_{p} / P_{q} = 2x / x = 2 : 1

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">The work function of caesium metal is 2.14 eV. When light of frequency 6 ×10

^{14}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?

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