1. Given

   • Initial speed of bullet = $u$
   • After penetrating $s_1 = 3\text{ cm}$, speed becomes $v = \dfrac{u}{2}$
   • Retarding force is constant, so deceleration is constant.

2. Use kinematics for first part of motion

   For constant acceleration, $v^2=u^2+2as$

   Substituting for the first $3\text{ cm}$: $\left(\frac{u}{2}\right)^2=u^2+2a(3)$

   $\frac{u^2}{4}=u^2+6a$

   $6a=\frac{u^2}{4}-u^2=-\frac{3u^2}{4}$

   $a=-\frac{u^2}{8} \quad \text{(in cm-based units)}$

3. Now find the further distance before stopping

   Let the additional distance be $s_2$. Initial speed for this part = $\dfrac{u}{2}$ Final speed = $0$

   Again use $v^2=u^2+2as$

   $0=\left(\frac{u}{2}\right)^2+2a s_2$

   $0=\frac{u^2}{4}+2\left(-\frac{u^2}{8}\right)s_2$

   $0=\frac{u^2}{4}-\frac{u^2}{4}s_2$

   $1-s_2=0$

   $s_2=1\text{ cm}$

4. Check with energy method (quick verification)

   Since resistance is constant, loss of kinetic energy is proportional to distance.

   • Initial KE $\propto u^2$
   • KE after $3\text{ cm}$ is $\propto \left(\frac{u}{2}\right)^2=\frac{u^2}{4}$
   • So KE lost in first $3\text{ cm}$ is $1-\frac14=\frac34$ of initial KE.
   • Remaining KE is $\frac14$ of initial KE.

   Therefore, further distance $s_2=3\times \frac{\frac14}{\frac34}=1\text{ cm}$

5. Option-wise conclusion

   • A: $2.0\text{ cm}$ ❌
   • B: $3.0\text{ cm}$ ❌
   • C: $1.0\text{ cm}$ ✅
   • D: $1.5\text{ cm}$ ❌

Final Answer: $\boxed{1.0\text{ cm}}$