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