On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10   cm .…

1. $910 \mathrm{\Omega }$
2. $990 \mathrm{\Omega }$
3. $505 \mathrm{\Omega }$
4. $550 \mathrm{\Omega }$

Solution

$\frac{a}{b}=\frac{l}{100-l}$

$a+b=1000$

$\frac{b}{a}=\frac{l-10}{110-l}$

$\frac{100-l}{l}=\frac{l-10}{110-l}$

$l^2-10l=11000+l^2$

$-100l=110l$

$200l=11000$

$l=\frac{110}{2}=55$

$\frac{55}{45}=\frac{a}{b}$

$\Rightarrow a=550\mathrm{\Omega }b=450\mathrm{\Omega }$

Asked in: JEE Main 2018 (08 Apr)

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