A 10 m long wire of uniform cross a potentiometer. The wire is connected in series with a battery of 5 V along with an external resistance of 480 Ω. If an unknown emf E is balanced at 6.0 m length of the wire

(i) 1.2 V

(ii) 1.02 V

(iii) 0.2 V

(iv) 0.12 V

Asked by Pragya Singh | 1 year ago |  144

Solution :-

Right answer is (iv) 0.12 V

Explanation:-

Let PQ is a potentiometer wore of length 10m,

I = $$\dfrac{E}{R+R'}=\dfrac{5}{480+20}$$

$$\dfrac{5}{500}$$

$$\dfrac{5}{100}=0.01A$$

$$V_{pq}=IR_{pq}=0.01\times 20$$

= 0.2V

If 10 m potentiometer wire balances = 0.2 V

Then 1 m potentiometer wire balances =$$\dfrac{0.2}{10}V$$

Then 6 m potentiometer wire balances = $$\dfrac{0.2}{10}\times 6V$$

$$\dfrac{1.2}{10}V= 0.12 V$$

Answered by Abhisek | 1 year ago

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