02 May 2011

Solved Electricity numerical for class 10

Question: Two bulbs have ratings 100 W, 220 V and 60 W, 220 V respectively. Which one has a greater resistance?
Answer: P=VI=  V2/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 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.
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 0-8
Answer: L = 1 km = 1000 m
R = 1 mm = 1 x 1 0-3
A = p r

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 R=V/I=2/1= 2 Ohm
Resistance per unit length: 2/5= 0.4Ohm/m
(ii) Resistance of 2 m length of the wire = 0.4 x 2=0.8 ohm
(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.L/2 
Resistance of this wire =R' = p (l/2)/(2A) = 1/4(p(L/A)
But p(L/A) = 2 ohm
R' = 1/4 x 2=0.5 Ohm

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sumit said...

where are the questions?

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