A) 2A
B) 3A
C) A
D) 4A
Correct Answer: B
Solution :
\[{{(\Delta {{T}_{f}})}_{x}}={{(\Delta {{T}_{f}})}_{y}}\] \[{{K}_{f}}{{m}_{x}}={{K}_{f}}{{m}_{y}}\] \[\frac{4\times 1000}{A\times 96}=\frac{12\times 1000}{{{M}_{B}}\times 88}\] \[{{M}_{b}}=3.27A\simeq 3A\]
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