Question

Two intervals of time are measured as $\Delta t_1=(2.00\pm 0.02)s$ and $\Delta t_2=(4.00\pm 0.02)s$ . The value of $\sqrt{(\Delta t_1)(\Delta t_2)}$ with collect significant figures and error is

• A. $(2.828\pm 0.01)s$
• B. $(2.83\pm 0.01)s$
• C. $(2.828\pm 0.0075)s$
• D. $(2.83\pm 0.0075)s$

Answer 1

Answer: (B)

Here, $\Delta t_1=(2.00\pm 0.02)s$ and
$\Delta t_2=(4.00\pm 0.02)s$
Given, $T=\sqrt{\left(\Delta t_1\right)\left(\Delta t_2\right)}$
$=\sqrt{(2.00)(4.00)}=2.828427s$
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,
$\therefore T=2.83s$
Now,
$\Delta T=\pm \frac{1}{2}\left(\frac{0.02}{2.00}+\frac{0.02}{4.00}\right)\times 2.8284$
$\Delta T=0.02121s$
So, after rounding off
$\Delta T=0.02$
Hence, $T=(2.83\pm 0.02)s$