Each of the given beakers contains two metal strips of same size fastened together and immersed in hydrochloric acid.
After 5 minutes, amount of Q ions formed is greater than P ions in beaker 1, S ions is greater than R ions in beaker 2, R ions is greater than Q ions in beaker 3, and T ions is greater than S ions in beaker 4.
Correct order of reactivity of these metals is

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$P > T > R > S > Q$

$R < P < T < S < Q$

$T > R > Q > P > S$

$P < Q < R < S < T$

answer is D.

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Detailed Solution

Based on reactivity series of metals, $P < Q < R < S < T$ .

Formation of T ions is more than P, Q, R and S.

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