Numerical Problems Based on Combination of Errors for Class 11 Physics

Given $A = 12.0$, $\Delta A = 0.1$ and $B = 8.5$, $\Delta B = 0.5$.

(i) For Addition ($A + B$):

Value $= A + B = 12.0 + 8.5 = 20.5\text{ cm}$

In addition, absolute errors are added: $\Delta Z = \Delta A + \Delta B$

Result $= (20.5 \pm 0.6)\text{ cm}$

(ii) For Subtraction ($A – B$):

Value $= A – B = 12.0 – 8.5 = 3.5\text{ cm}$

In subtraction, absolute errors are also added (errors always maximize): $\Delta Z = \Delta A + \Delta B$

Result $= (3.5 \pm 0.6)\text{ cm}$

Temperature difference $t’ = t_2 – t_1$

The error in the difference is the sum of the absolute errors of the individual measurements:

Therefore, the temperature difference is $(30 \pm 1)^\circ\text{C}$.

Let $L_1 = 15.2\text{ cm}$, $\Delta L_1 = 0.2\text{ cm}$ and $L_2 = 10.7\text{ cm}$, $\Delta L_2 = 0.2\text{ cm}$.

Difference in length $L’ = L_1 – L_2$

Total absolute error $\Delta L’ = \Delta L_1 + \Delta L_2$

Result: $(4.5 \pm 0.4)\text{ cm}$

Given $L = 5.7\text{ cm}$, $\Delta L = 0.1\text{ cm}$ and $B = 3.4\text{ cm}$, $\Delta B = 0.2\text{ cm}$.

Area $A = L \times B$

In multiplication, relative errors are added:

Absolute error $\Delta A = 19.38 \times 0.0763 = 1.478 \approx 1.5\text{ cm}^2$

Result: $(19.4 \pm 1.5)\text{ cm}^2$

Velocity is defined as $v = \frac{s}{t}$.

In division, the maximum fractional (relative) error is the sum of the fractional errors of the individual quantities:

Multiplying by 100 to get the percentage error:

The time period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$.

Squaring both sides and rearranging for $g$, we get:

The constant $4\pi^2$ has no error. For quantities raised to a power, the percentage error is the exponent multiplied by the percentage error of the quantity. Thus, the total percentage error in $g$ is:

Given the percentage error in length ($\frac{\Delta l}{l} \times 100$) is $1\%$, and the percentage error in time period ($\frac{\Delta T}{T} \times 100$) is $2\%$: