Answer:
It takes <u>951 seconds</u> to boil a cup of water.
Explanation:
Given:
EMF of the battery (E) = 10.0 V
Internal resistance of the battery (r) = 0.04 Ω
Resistance of the circuit = 'R'
Current measured in the circuit (I) = 11.0 A
Energy required to boil water (U) = 100 kJ = 100 × 10³ J [1 kJ = 10³ J]
Time taken for boiling (t) = ?
We know that, the emf of the battery is given as:

Plug in the given values and solve for 'R'. This gives,

Now, energy required to boil the water is given as:

Plug in the given values and solve for 't'. This gives,

So, it takes 951 seconds to boil a cup of water.