Answer: P=VI= V

^{2}/R For the same V, R is inversely proportional to P. Therefore, the bulb 60 W, 220 V has a greater resistance.

(ii) explain the reason for the difference in resistance.

V = I R ---- Ohm's law R = V/I = 2/.2 =10 Ω

(ii) Resistance of the filament of the bulb increases with increase in temperature. Hence when it glows its resistances is greater than when it is cold.

Question: Calculate the resistance of 1 km long copper wire of radius 1 mm. (Resistivity of copper = 1.72 x 1 0

^{-8}

^{}

Answer: L = 1 km = 1000 m

R = 1 mm = 1 x 1 0

^{-3}

^{2}

Question: When a potential difference of 2 V is applied across the ends of a wire of 5 m length, a current of 1A is found to flow through it. Calculate: (i) the resistance per unit length of the wire (ii) the resistance of 2 m length of this wire (iii) the resistance across the ends of the wire if it is doubled on itself.

Answer: (i) V = I R ----- Ohm's law

(ii) Resistance of 2 m length of the wire = 0.4 x 2

(iii) When the wire is doubled on itself:

(1) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.

(2) The length becomes half i.e.

## 1 comment:

it was very usesfull

I just need someone to help me with how i should approach this question. I have NO idea how i should solve it please help!

Two identical particles, each having charge +q, are fixed in space and separated by a distance d. A third point charge -Q is free to move and lies initially at rest on the perpendicular bisector of the two charges a distance x from the midpoint between the two fixed charges.

the question is b) How fast will the charge -Q be moving when it is at the midpoint between the two fixed charges, if initially it is released at a distance a<<d from the midpoint?

pls try to send reply soon to -ishwaryasun@gmail.com

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