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9 Mar 2008 19:07:34 IST

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PROB ON WHEATSTONES'S METER BRIDGE

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1.Two equal resistances are connected in the gaps of a meter bridge .if the resistance in the left gap is increasedf by 10%,the balancing point shift----

1. 10% to right

2. 10% to left

3. 9.6 to right

4. 4.8% to right

2.the emf of a cell is 2V and its internal resistance is 2 ohm.A resistance of 8 ohm is joined to battery in parallel.this is contacted in secondary circuit of potentiometer .if 1V standard cell balances for 100cm of potentiometer wire ,the balance point of above cell is--------------

1. 120cm

2. 240cm

3. 160cm

4.116cm

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Comments (6)

kaushik krishna

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9 Mar 2008 19:17:30 IST

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1) consider resis. 2 be 10 ohms each..balance pt. at 50 cm

when increased by 10% the left becomes 11 and right resistor remains 10 ohms..

now balanced pt is 11/21 * 100 = 52.4 approx

increase = (2.4 * 100)/50 = 4.8% to the right

kaushik krishna

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9 Mar 2008 19:34:47 IST

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sum 1 confirm my answer

Gaurav |spideyunlimited| Ragtah

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10 Mar 2008 15:45:29 IST

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1) since Metre Bridge has wire of 100 cm so if the 2 resistances are equal, the balance point is at 50 cm.

Now if left resistance is increased by 10% it becomes 1.1 times its original value.

So balance point L :

1.1 R/ R = L / 100 - L

110 - 1.1 L = L

110 = 2.1 L

L = 52.4 approx

So the balance point has shifted right by 2.4 cm

So in terms of %, it has shifted right by (2.4 / 50 x 100) %

= 4.8 % to the right

Chirag Sangani

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Joined: 12 Dec 2007

Posts: 611

10 Mar 2008 20:37:12 IST

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q2 - 3) 160cm.
the circuit is such that the potentiometer is balanced across the 8 ohm resistor. Total resistance = 10 ohm, hence current = 0.2 A. PD across 8 ohm resistor = 1.6 V. This is balanced by potentiometer. 1V of pd is balanced by 100 cm. Hence 1.6V corresponds to 160 cm

Sai Ganesh Bandiatmakur

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Joined: 17 Feb 2007

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11 Mar 2008 17:52:58 IST

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R₁/R₂=l/100-l where l is balancing length..first length L1=50 cm, next resistances are 1.1R and R,we get L2-=52.381cm %age inc in length=100*( 2.381/50)=approx 4.8% to the right

2. Pd. across 2nd circuit=1.6 v ,
For potentiometer, PD

Balancing length, so we get 160cm

abhishek gupta

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Joined: 30 Sep 2007

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22 Mar 2008 01:27:26 IST

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160cm.....

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