Question

Calculate the percentage error in the determination of g= 4π 2 l/t 2 when l and t are measured with 2% and 3% respectively.

Answer 1

Answer:

The percentage error is ±8

Explanation:

According to the problem the error is in the determination of g= 4π^2l/t^2

where  both the values are measured with 2% and 3% respectively

now to calculate error the expression will be

Δg/g=Δl/l+2Δt/t

putting the value of I and t

Δg/g=±2/100+2 x ±3/100

therefore the percentage error=Δg/g×100

      =(±2)+(±6)=±8

Hence the answer will be ±8

Answer 2

Given :

The value of g is given by :

$g=4\pi^2\dfrac{l}{t^2}$

Here,

l and t are measured with 2% and 3% respectively.

To find :

It is required to find the percentage error in the determination of g.

Solution:

The percentage error in value of g is given by :

$\dfrac{\Delta g}{g}=1\dfrac{\Delta l}{l}+2\dfrac{\Delta t}{t}\\\\\dfrac{\Delta g}{g}=1\times 2+2\times 3\\\\\dfrac{\Delta g}{g}=8\%$

So, the percentage error in g is 8 percent.

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