Q121 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Physics
Two intervals of time are measured as $\Delta t_1=(2.00 \pm 0.02) \mathrm{s}$ and $\Delta t_2=(4.00 \pm 0.02) \mathrm{s}$. The value of $\sqrt{\left(\Delta t_1\right)\left(\Delta t_2\right)}$ with correct significant figures and error is
- A$(2.828 \pm 0.01) \mathrm{s}$
- Correct answerB$(2.83 \pm 0.01) \mathrm{s}$
- C$(2.828 \pm 0.0075) \mathrm{s}$
- D$(2.83 \pm 0.0075) \mathrm{s}$
Solution
Here, $\Delta t_1=(2.00 \pm 0.02) \mathrm{s}$ and $\Delta t_2=(4.00 \pm 0.02) \mathrm{s}$
Given, $T=\sqrt{\left(\Delta t_1\right)\left(\Delta t_2\right)}=\sqrt{(2.00)(4.00)}=2.828427 \mathrm{~s}$
Now, according to the relative error of product
$\pm \frac{\Delta T}{T}= \pm\left(\frac{1}{2} \frac{\Delta t_1}{t_1}+\frac{1}{2} \frac{\Delta t_2}{t_2}\right)$
Since, $\Delta t_1$ and $\Delta t_2$ have 3 significant figures, then $T$ also has 3 significant figures. Hence, after rounding off,
$\begin{array}{rlrl}
& \therefore & T & =2.83 \mathrm{~s} \\
& \text {Now, } & \Delta T & = \pm \frac{1}{2}\left(\frac{0.02}{2.00}+\frac{0.02}{4.00}\right) \times 2.8284 \\
\Delta T & =0.02121 \mathrm{~s}
\end{array}$
So, after rounding off,
$\begin{aligned}
\Delta T & =0.02 \\
T & =(2.83 \pm 0.02) \mathrm{s}
\end{aligned}$
Hence,
So, most matched answer is (b).
Hence, the correct answer is option: 'B'.