In a meter bridge, the balancing length from the left end (standard resistance of 1 ohm m the gap) is found to be 20 cm. The value of the unknown resistance will be.

Question

In a meter bridge, the balancing length from the left end (standard resistance of 1 ohm m the gap) is found to be 20 cm. The value of the unknown resistance will be. (A)  0.25 Ω  (B)  0.4 Ω  (C)  0.5 Ω  (D)  0.8 Ω

Tardigrade Admin · Accepted Answer

Here: Standard resistance of the wire =1 Ω Length of the wire from the left end l=20 text cm The value of unknown resistance is given by R=( (l/100-l) )× S =(20/100-20)× 1=(20/80) =0.25 Ω

Q. In a meter bridge, the balancing length from the left end (standard resistance of 1 ohm m the gap) is found to be 20 cm. The value of the unknown resistance will be.

$0.25 Ω$

$0.4 Ω$

$0.5 Ω$

$0.8 Ω$

Here: Standard resistance of the wire $= 1 Ω$ Length of the wire from the left end $l = 20 c m$ The value of unknown resistance is given by $R = (\frac{l}{100 - l}) \times S$ $= \frac{20}{100 - 20} \times 1 = \frac{20}{80}$ $= 0.25 Ω$

Q. In a meter bridge, the balancing length from the left end (standard resistance of 1 ohm m the gap) is found to be 20 cm. The value of the unknown resistance will be.

0.25Ω

0.4Ω

0.5Ω

0.8Ω

Here: Standard resistance of the wire =1Ω Length of the wire from the left end l=20cm The value of unknown resistance is given by R=(100−ll​)×S =100−2020​×1=8020​ =0.25Ω

$0.25 Ω$

$0.4 Ω$

$0.5 Ω$

$0.8 Ω$

Here: Standard resistance of the wire $= 1 Ω$ Length of the wire from the left end $l = 20 c m$ The value of unknown resistance is given by $R = (\frac{l}{100 - l}) \times S$ $= \frac{20}{100 - 20} \times 1 = \frac{20}{80}$ $= 0.25 Ω$