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Question

Physics Question on Current electricity

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10cm10\, cm. The resistance of their series combination is 1kΩ1 \, k\Omega . How much was the resistance on the left slot before interchanging the resistances ?

A

990Ω990\,\Omega

B

505Ω505 \,\Omega

C

550Ω550 \,\Omega

D

910Ω910 \,\Omega

Answer

550Ω550 \,\Omega

Explanation

Solution

R1R2=l(100l)\frac{R_{1}}{R_{2}}=\frac{l}{(100-l)} R2R1=(l10)(110l)\frac{R_{2}}{R_{1}}=\frac{(l-10)}{(110-l)} (100l)(110l)=l(l10)(100-l)(110-l)=l(l-10) 11000+l2210l=l210l11000+l^2-2\,10 \,l=l^2-10\,l l=55cm\Rightarrow l=55 \,cm R1=R2(5545)R_{1}=R_{2}\left(\frac{55}{45}\right) R1+R2=1000ΩR_{1}+R_{2}=1000\, \Omega R1=550ΩR_{1}=550 \,\Omega