1. Current in the solenoid

Given $I(t)=I_0 t(1-t)=I_0(t-t^2), \,\, 0\le t\le 1$

2. Magnetic field inside a long solenoid

For a long solenoid, $B(t)=\mu_0 n I(t)$ so the magnetic field is proportional to $I(t)$.

3. Magnetic flux through the ring

The ring has radius $2R$, while the solenoid has radius $R$. For a long solenoid, magnetic field is essentially confined within the solenoid cross-section, so flux through the ring is only through area of the solenoid: $\Phi(t)=B(t)\cdot \pi R^2=\mu_0 n \pi R^2 I(t)$

Thus, $\Phi(t)\propto I(t)$

4. Induced EMF in the ring

By Faraday's law, $V_R=-\frac{d\Phi}{dt}$ Hence, $V_R\propto -\frac{dI}{dt}$

Now, $\frac{dI}{dt}=I_0(1-2t)$ So, $V_R\propto -(1-2t)=2t-1$

Therefore:

• At $t=0.5$, $\frac{dI}{dt}=0 \Rightarrow V_R=0$
• For $0<t<0.5$, $\frac{dI}{dt}>0$ so flux increases
• For $0.5<t<1$, $\frac{dI}{dt}<0$ so flux decreases

Thus the sign of induced EMF changes at $t=0.5$, meaning the direction of induced current reverses at $t=0.5$.

5. Check options

• A: Direction unchanged and $V_R=0$ at $t=0.25$
  False, because reversal occurs at $t=0.5$, not unchanged.

• B: Direction unchanged and $V_R$ maximum at $t=0.5$
  False, because $V_R=0$ at $t=0.5$.

• C: At $t=0.25$ direction reverses and $V_R$ is maximum
  False, reversal is not at $t=0.25$.

• D: At $t=0.5$ direction of $I_R$ reverses and $V_R$ is zero
  True.

6. Conclusion

The correct option is D.