1. Identify molecular shapes

All three molecules, $CH_4$, $CCl_4$, and $CHCl_3$, have a central carbon atom with four sigma bonds, so each has approximately tetrahedral geometry.

2. Recall how dipole moment depends on symmetry

The molecular dipole moment is the vector sum of all bond dipole moments.

• If a molecule is perfectly symmetrical, bond dipoles cancel.
• If the molecule is unsymmetrical, complete cancellation does not occur.

3. Analyze each molecule

(i) $CH_4$

• Tetrahedral and perfectly symmetrical.
• All four $C-H$ bonds are identical.
• Their dipole moments cancel completely.

Hence, $\mu(CH_4)=0$

(ii) $CCl_4$

• Also tetrahedral and perfectly symmetrical.
• All four $C-Cl$ bonds are identical.
• Their bond dipoles also cancel completely.

Hence, $\mu(CCl_4)=0$

(iii) $CHCl_3$

• Tetrahedral, but not symmetrical like the above two.
• It has three $C-Cl$ bonds and one $C-H$ bond.
• Since the bonds are not all identical, the dipoles do not cancel completely.

Therefore, $\mu(CHCl_3) > 0$

4. Compare the three dipole moments

Thus, $CH_4 = CCl_4 < CHCl_3$

5. Match with the given options

This corresponds to:

Option B: $CH_4 = CCl_4 < CHCl_3$