A person measures the depth of a well by meas | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The total time $T$ taken is the sum of the time taken by the stone to fall $(t_1)$ and the time taken by the sound to travel back $(t_2)$.$T = t_1

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The total time $T$ taken is the sum of the time taken by the stone to fall $(t_1)$ and the time taken by the sound to travel back $(t_2)$.$T = t_1 + t_2 = \sqrt{\frac{2L}{g}} + \frac{L}{v}$Given $L = 20 \ m$,$g = 10 \ ms^{-2}$,and $v = 300 \ ms^{-1}$.Differentiating $T$ with respect to $L$:$\frac{dT}{dL} = \frac{1}{2} \sqrt{\frac{2}{gL}} + \frac{1}{v} = \frac{1}{\sqrt{2gL}} + \frac{1}{v}$Substituting the values:$\frac{dT}{dL} = \frac{1}{\sqrt{2 	imes 10 	imes 20}} + \frac{1}{300} = \frac{1}{20} + \frac{1}{300} = \frac{15+1}{300} = \frac{16}{300} \ s/m$Since $\delta T = 0.01 \ s$,we have $\delta L = \delta T / (dT/dL) = 0.01 	imes (300/16) = 3/16 \ m = 0.1875 \ m$.The fractional error is $\frac{\delta L}{L} = \frac{0.1875}{20} = 0.009375$.Percentage error = $0.009375 	imes 100 \% \approx 0.9375 \% \approx 1 \%$.

Similar Questions

Can error be completely eliminated?

The energy of a system is given as $E(t) = \alpha^3 e^{-\beta t}$,where $t$ is the time and $\beta = 0.3 \ s^{-1}$. The errors in the measurement of $\alpha$ and $t$ are $1.2 \%$ and $1.6 \%$,respectively. At $t = 5 \ s$,the maximum percentage error in the energy is: (in $\%$)

In the density measurement of a cube,the mass and edge length are measured as $(10.00 \pm 0.10) \, kg$ and $(0.10 \pm 0.01) \, m$ respectively. The error in the measurement of density is

The initial and final temperatures of water as recorded by an observer are $(38.6 \pm 0.2)^{\circ}C$ and $(82.3 \pm 0.3)^{\circ}C$. The rise in temperature with proper error limits is

Vedclass Test Series

Exam Paper Generator

Online Exam Module