Question: Two bulbs have ratings 100 W, 220 V and 60 W, 220 V respectively. Which one has a greater resistance?

Answer: P=VI= V

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.Question: A torch bulb has a resistance of 1 Ω when cold. It draws a current of 0.2 A from a source of 2 V and glows. Calculate

(i) the resistance of the bulb when glowing and

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

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

Answer: (i) When the bulb glows:

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

Answer: L = 1 km = 1000 m

R = 1 mm = 1 x 1 0

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.

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}A = p r

^{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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