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A student measures that distance travell...

A student measures that distance travelled in free fall of a body, initially at rest in given time. He uses this data to estimated `g`, the acceleration due to gravity. If the maximum percentage error in measurement of the distance and the time are `e_(1)` and `e_(2)`, respectively, the percentage error in the estimation of `g` is

A

`e_(1)+2e_(2)`

B

`e_(1)+e_(2)`

C

`e_(1)-2e_(2)`

D

`e_(2)-e_(1)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of estimating the percentage error in the acceleration due to gravity \( g \), we start with the formula for the distance \( h \) traveled by a body in free fall, which is given by: \[ h = \frac{1}{2} g t^2 \] Since the body is initially at rest, we can rearrange this equation to express \( g \): \[ g = \frac{2h}{t^2} \] ### Step 1: Identify the errors in measurements Let: - \( e_1 \) be the maximum percentage error in the measurement of distance \( h \). - \( e_2 \) be the maximum percentage error in the measurement of time \( t \). ### Step 2: Determine the relationship between \( g \), \( h \), and \( t \) From the equation \( g = \frac{2h}{t^2} \), we can see that \( g \) depends on both \( h \) and \( t \). ### Step 3: Calculate the percentage error in \( g \) To find the percentage error in \( g \), we can use the method of propagation of errors. The formula for the percentage error in \( g \) can be derived as follows: 1. The percentage error in \( h \) contributes directly to the error in \( g \). 2. The percentage error in \( t^2 \) contributes to the error in \( g \) as well, but since \( t \) is squared in the denominator, the error is doubled. Thus, the formula for the percentage error in \( g \) is: \[ \frac{\Delta g}{g} \times 100 = \frac{\Delta h}{h} \times 100 + 2 \times \frac{\Delta t}{t} \times 100 \] ### Step 4: Substitute the known errors Substituting the known errors \( e_1 \) and \( e_2 \): \[ \text{Percentage error in } g = e_1 + 2e_2 \] ### Final Result Thus, the maximum percentage error in the estimation of \( g \) is: \[ \text{Percentage error in } g = e_1 + 2e_2 \]
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