The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as $R_1$ has the colour code (Orange, Red, Brown). The resistors $R_2$ and $R_4$ are $80\, \Omega $ and $40\,\Omega $, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as $R_3$ would be
JEE MAIN 2019, Diffcult
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$R_{1}=32 \times 10=320 \,\Omega$
$\mathrm{R}_{3}=\frac{\mathrm{R}_{\mathrm{y}}}{\mathrm{R}_{2}} \times \mathrm{R}_{1}=\frac{40 \times 320}{80}=160 \,\Omega$
$\therefore $ Colour code of $\mathrm{R}_{3}$ be Brown, Blue, Brown.
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