For which value of \(k\), linear pair \(x+y-4=0,2x+ky-3=0\) has no solution?

The condition for no solution is \(\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\neq \frac{{c}_{1}}{{c}_{2}}\)

Here, \({a}_{1}=1,{b}_{1}=1,{c}_{1}=-4\) and \({a}_{2}=2,{b}_{2}=k,{c}_{2}=-3\)

Applying the condition:
\(\frac{1}{2}=\frac{1}{k}\)

Cross-multiplying gives:
\(k=2\)

Check the second part:

\(\frac{1}{k}\neq \frac{-4}{-3}\)

For k=2, this is \(\frac{1}{2}\neq \frac{4}{3}\)

which is true.

Therefore, the correct answer is (B).