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A potentiometer wire of length 1 m has a resistance of 10 Ω. It is connected to a 6 V battery in series with a resistance of 5 Ω. Determine the emf of the

primary cell which gives a balance point at 40 cm.


Given that,l =1 m,
m =100cm,
R =10 ohm,
R'=5 ohm.
I =E0/(R+R') = 6/(10+5) =6/15 A.
=60/15 Volt.

VAB/I =4/100 =0.04,
E0 =,(VAB /I )X
(x=balance point =40cm)
=0.04 ×40 = 1.6 Volt.
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The maximum line-of-sight distance dMdM between two antennas having heights hThT and hRhR above the earth is given by dM=2RhT−−−−−√+2RhR−−−−−√dM=2RhT+2RhR where hThT and hRhR are the heights of the transmitting and receiving antennae. Substituing, we get: dM=2×64×105×32−−−−−−−−−−−−−−√+2 times64×105×50−−−−−−−−−−−−−−−−−√mdM=2×64×105×32+2 times64×105×50m dM=64×10210−−√+8×103×10−−√mdM=64×10210+8×103×10m =144×102×10−−√m=45.5km
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