Current Electricity

JEE Main level, numerical answer, medium

297 of 380

Question 297 of 380

In a meter bridge experiment SS is a standard resistance. RR is a resistance wire. It is found that balancing length is l=25cm.l=25cm. If RR is replaced by a wire of hall length and half diameter that of RR of same material, then the balancing distance l′l' (in cmcm ) will now be ______________.

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Answer: 40

XR=7525=3\dfrac{X}{R}=\dfrac{75}{25}=3
R=plA=4plπd2R=\dfrac{pl}{A}=\dfrac{4pl}{\pi {d}^{2}}
R′=4p(l2)π(d2)2=2R{R}^{'}=\dfrac{4p\left(\dfrac{l}{2}\right)}{\pi {\left(\dfrac{d}{2}\right)}^{2}}=2R
π(d2)2\pi {\left(\dfrac{d}{2}\right)}^{2}
then XR′=(100−ll)\dfrac{X}{R'}=\left(\dfrac{100-l}{l}\right)
100−ll=X2R=32\dfrac{100-l}{l}=\dfrac{X}{2R}=\dfrac{3}{2}
l=40.00cml=40.00cm

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