Principle involved

The acidity of carboxylic acids depends on the stability of the conjugate base, $RCOO^-$.
Electron-withdrawing groups increase acidity by stabilizing the negative charge through the $-I$ effect.

Compare the given acids

Option A: $HCOOH$

Formic acid has no alkyl group attached. It is more acidic than simple alkyl carboxylic acids like acetic acid because there is no electron-donating alkyl group.

Option D: $CH_3COOH$

Acetic acid has a methyl group, which shows a $+I$ effect and decreases acidity compared to formic acid.

Therefore, $HCOOH > CH_3COOH$

Effect of chlorine substituent

Chlorine is strongly electron-withdrawing by the $-I$ effect, so chloro-substituted carboxylic acids are more acidic than unsubstituted ones.

The strength of the $-I$ effect decreases with distance from the $COOH$ group.

Option B: $CH_3CH_2CH(Cl)CO_2H$

Here, Cl is on the $\alpha$-carbon (adjacent to $COOH$). So its $-I$ effect is very strong.

Option C: $ClCH_2CH_2CH_2COOH$

Here, Cl is on the $\gamma$-carbon (farther from $COOH$). So its $-I$ effect is much weaker.

Hence, $CH_3CH_2CH(Cl)CO_2H > ClCH_2CH_2CH_2COOH$

Overall acidity order

Combining all effects: $CH_3CH_2CH(Cl)CO_2H > ClCH_2CH_2CH_2COOH > HCOOH > CH_3COOH$

Conclusion

The strongest acid is: $\boxed{CH_3CH_2CH(Cl)CO_2H}$ So, the correct option is B.