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Two resistances are connected in the two gaps of a meter bridge. The balance point is `20 cm` from the zero end. When a resistance `15 Omega` is connected in series with the smaller of two resistance, the null point+ shifts to `40 cm`. The smaller of the two resistance has the value.A. 12B. 24C. 36D. 48 |
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Answer» Correct Answer - B `l_(x)=40cm, I_(R)=60cm` `x/R=(l_(x))/(l_(R))=(40)/(60)=2/3 .....(i)` `therefore (x+30)/(R)=(60)/(40)=3/2` `therefore (x+30)/(R)=3/2` `therefore 2(X+30)rArr 3R` `R=(2(x+30))/(3).........(ii)` From (i) and (iii) `(X)/(2((x+30))/(3)))=2/3` `(3x)/(2(x+30))=2/3` `9x=4x+120` `5x=120 rArr x=24Omega` |
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