When two resistance R _ 1 and R _ 2 connected in series and...

A meter bridge consists of a wire of length $1m$ and of uniform cross-sectional area stretched taut and clamped between two thick metallic strips bent at right angles with two gaps across which resistors are to be connected. It works on the principle of the Wheatstone bridge. Therefore, we can write

$\dfrac{{R}_{1} + {R}_{2}}{10} = \dfrac{60}{40} ⇒ {R}_{1} + {R}_{2} = 15Ω$

Then for the second case, we can write

$\dfrac{{R}_{1}{R}_{2}}{\left({R}_{1} + {R}_{2}\right) \times 3} = \dfrac{40}{60} ⇒ {R}_{1}{R}_{2} = 30{Ω}^{2}$