The heat dissipated in a resistance can be determined from the relation: H= I²RT/42 cal. If the maximum errors in the measurement of current, resistance and time are 2%, 1% and 1% respectively, what would be the maximum error in the dissipated heat ?

The correct option is 6%.

To find the maximum percentage error in the dissipated heat, we start from the given physical relation for heat:

$H=\frac{I^{2}RT}{42}$

Where:
- $H$ is the dissipated heat,
- $I$ is the current,
- $R$ is the resistance,
- $T$ is the time, and
- 42 is a constant (which has no error associated with it).

Taking the natural logarithm (ln) on both sides of the equation, we get:

$\ln \left(H\right)=2\ln \left(I\right)+\ln \left(R\right)+\ln \left(T\right)-\ln \left(42\right)$

Differentiating both sides to find the relative error relation:

$\frac{\Delta H}{H}=2\frac{\Delta I}{I}+\frac{\Delta R}{R}+\frac{\Delta T}{T}$

To find the maximum percentage error, we multiply each term by 100:

$\left(\frac{\Delta H}{H}\times 100\right)=2\left(\frac{\Delta I}{I}\times 100\right)+\left(\frac{\Delta R}{R}\times 100\right)+\left(\frac{\Delta T}{T}\times 100\right)$

We are given the maximum percentage errors in the measurement of current, resistance, and time as follows:
- Percentage error in current $\left(\frac{\Delta I}{I}\times 100\right)=2\%$
- Percentage error in resistance $\left(\frac{\Delta R}{R}\times 100\right)=1\%$
- Percentage error in time $\left(\frac{\Delta T}{T}\times 100\right)=1\%$

Substituting these values into our maximum error formula:

$\text{Maximum \% error in}H=2(2\%)+1\%+1\%$

$\text{Maximum \% error in}H=4\%+1\%+1\%=6\%$

Thus, the maximum error in the dissipated heat is 6%.