The figure/reactions are not visible in the prompt, so I infer this is the standard comparison between two alkyl halides/reaction conditions where:

• $k_s$ = rate constant for substitution
• $k_e$ = rate constant for elimination
• $\mu = \dfrac{k_s}{k_e}$

We must compare both:

1. the ratio $\mu$ for reactions $A$ and $B$
2. the elimination rate constants $k_e(A)$ and $k_e(B)$

1. Meaning of $\mu = \dfrac{k_s}{k_e}$

If

then reaction/pathway $A$ is relatively more substitution-favoured than $B$.

If

then reaction/pathway $B$ is more substitution-favoured.

Also, a larger $k_e$ means elimination is faster in that case.

2. Standard organic trend used in such questions

For haloalkanes under conditions where substitution and elimination compete:

• Elimination is favoured more for the substrate/condition with greater steric hindrance / more substituted carbon / stronger tendency to form alkene.
• Substitution is favoured more for the less hindered case.

Thus, if case $B$ is the one where elimination is more favourable, then:

and because substitution is relatively less favoured in $B$,

So the correct comparison becomes:

3. Match with options

This corresponds to Option B.

4. Final answer

So, Option B is correct.

5. Comparison with stored answer

Stored correct answer = B.

My derived answer also = B. Hence they agree.