Q. In an experiment, to measure the height of a bridge by dropping stone into water underneath, if the error in measurement of time is $0.1s$ at the end of $2s$, then the error in estimation of height of bridge will be
A) $0.49\ m$
B) $0.98\ m$
C) $1.96\ m$
D) $2.12\ m$
Explanation
From $h = ut + \frac{1}{2}gt^{2}$ $h = 0 + \frac{1}{2} \times 9.8 \times (2)^{2} = 19.6\ m$ $\frac{\mathrm{\Delta}h}{h} = \pm 2\frac{\mathrm{\Delta}t}{t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because a = g = constant\rbrack$ $= \pm 2\left( \frac{0.1}{2} \right) = \pm \frac{1}{10}$ $\therefore\ \mathrm{\Delta}h = \pm \frac{h}{10} = \pm \frac{19.6}{10} = \pm 1.96\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.