The determination of the value of acceleration due to gravity \((g)\) by simple pendulum method employs the formula,
 \(g=4\pi^2\frac{L}{T^2}\)
The expression for the relative error in the value of \(g\) is:

Time intervals measured by a clock give the following readings: 
\(1.25\) s, \(1.24\) s, \(1.27\) s, \(1.21\) s and \(1.28\) s. 
What is the percentage relative error of the observations? 
1. \(2\)%
2. \(4\)%
3. \(16\)%
4. \(1.6\)%

In an experiment, the percentage errors that occurred in the measurement of physical quantities \(A,\) \(B,\) \(C,\) and \(D\) are \(1\%\), \(2\%\), \(3\%\), and \(4\%\) respectively. Then, the maximum percentage of error in the measurement of \(X,\) where \(X=\frac{A^2 B^{\frac{1}{2}}}{C^{\frac{1}{3}} D^3}\), will be:
1. \(10\%\)
2. \(\frac{3}{13}\%\)
3. \(16\%\)
4. \(-10\%\)

A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate \(g,\) the acceleration due to gravity. If the maximum percentage errors in the measurement of the distance and the time are \(e_1\) and \(e_2\) respectively, the percentage error in the estimation of \(g\) is:
1. $e_1+2e_2$
2. $e_1+e_2$
3. $e_1-2e_2$
4. $e_2-e_1$

If the error in the measurement of the radius of a sphere is 2%, then the error in the determination of the volume of the sphere will be:

1. 4%

2. 6%

3. 8%

4. 2%