21. A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length l. If the wire is replaced by another wire of same material but with double the length and half the thickness , the balancing point become 2l, then the value of l is 
(a) 40 cm   
(b) 20 cm 
(c) 300/7 cm 
(d) 240/7 cm

Let resistance of first wire be R1 and that of second be R2R1=ρlA=R1=ρlA=ρlπr2Acording to questionR2=ρ2lπr22=8ρlπr2=8R1R1R2=18We know that, in a mater bridgeR=l100-l X, where X is the known resitance connected to the other arm of meter bridgeLet R1=l100-l Xthen R2=2l100-2l XR1R2=l100-l X2l100-2l X18=1100-l 2100-2l =12×100-2l100-l14=100-2l100-l100-l=400-8l7l=300l=3007cm

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