If mass and specific heat capacity of calormeter is negligible what would be maximum permissible error is `S` Use the data mentioned below `m_(1) rarr 0, S_(c) rarr 0, m_(2) = 1.00kg V = 10.0 V I = 10.0A`
`t = 1.00 xx 10^(2) sec theta_(0) = 15^9@)C` Corrected `theta_(t) = 65^(@)C`
(A) `4%`
(B) `5%` (C) `8%` (D) `12%` .

Explanation

To solve the problem of determining the maximum permissible error in the specific heat capacity \( S \) of the liquid in a calorimeter, we will follow these steps:

### Step 1: Understand the given data
We have the following data:
- Mass of the calorimeter \( m_1 = 0 \) (negligible)
- Specific heat capacity of the calorimeter \( S_c = 0 \) (negligible)
- Mass of the liquid \( m_2 = 1.00 \, \text{kg} \)
- Voltage \( V = 10.0 \, \text{V} \)
...