1. Use Joule’s law of heating

   The thermal energy developed in a resistor is $Q = I^2 R t$ where:

   • $Q$ = heat produced,
   • $I$ = current,
   • $R$ = resistance,
   • $t$ = time.

2. First case

   When current $I_1 = 4\,\text{A}$ flows for $t=10\,\text{s}$, the heat produced is given as $H$.

   So, $H = (4)^2 R (10)$

3. Second case

   Now current is increased to $I_2 = 16\,\text{A}$, with the same resistor and same time $t=10\,\text{s}$.

   Let the new heat produced be $Q_2$.

   $Q_2 = (16)^2 R (10)$

4. Take ratio

   $\frac{Q_2}{H} = \frac{(16)^2 R (10)}{(4)^2 R (10)}$

   Cancelling common factors: $\frac{Q_2}{H} = \frac{16^2}{4^2} = \left(\frac{16}{4}\right)^2 = 4^2 = 16$

   Therefore, $Q_2 = 16H$

5. Check options

   • A: $\frac{H}{4}$ ❌
   • B: $16H$ ✅
   • C: $H$ ❌
   • D: $4H$ ❌

Hence, the correct answer is Option B.