The most accurate reading of the length of  \(6.28\) cm long fiber is: 
1. \(6\) cm
2. \(6.5\) cm 
3. \(5.99\) cm 
4. \(60\) cm

Which of the following is the most precise measurement?

(1) $3\times <mspace\; width="1em"></mspace>{10}^{-3}m$
$$

(2) $0.0030<mspace\; width="1em"></mspace>m<mspace\; width="1em"></mspace>$
$$

(3) $30\times <mspace\; width="1em"></mspace>{10}^{-4}<mspace\; width="1em"></mspace>m<mspace\; width="1em"></mspace>$
$$

(4) $300\times <mspace\; width="1em"></mspace>{10}^{-5}<mspace\; width="1em"></mspace>m$

The values of a number of quantities are used in a mathematical formula. The quantity that should be most precise and accurate in measurement is the one 

(1) Having smallest magnitude 

(2) Having largest magnitude 

(3) Used in the numerator 

(4) Used in the denominator 

The thickness of a pencil measured by using a screw gauge (least count \(0.001\) cm) comes out to be \(0.802\) cm. The percentage error in the measurement is:
1. \(0.125 \%\)
2. \(2.43\%\)
3. \(4.12\%\)
4. \(2.14\%\)

The percentage error in the measurement of the voltage V is 3% and in the measurement of the current is 2%. The percentage error in the measurement of the resistance is 

(1) 3%

(2) 2%

(3) 1%

(4) 5%

The relative error in the measurement of the side of cube is 0.027. The relative error in the measurement of its volume is 

(1) 0.027

(2) 0.054

(3) 0.081

(4) 0.246

Zero error in an instrument introduces 

(1) Systematic error 

(2) Random error 

(3) Least count error 

(4) Personal error 

A packet contains silver powder of mass 20.23 g $\pm$ 0.01 g. Some of the powder of mass 5.75 g $\pm$ 0.01 g is taken out from it. The mass of the powder left-back is 

(1) 14.48 g $\pm$ 0.00 g 

(2) 14.48 $\pm$ 0.02 g

(3) 14.5 g $\pm$ 0.1 g

(4) 14.5 g $\pm$ 0.1 g

The radius of a sphere is (2.6 $\pm$ 0.1) cm . The percentage error in its volume is 

(1) $\frac{0.1}{2.6}\times 100\%<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

(2) $3\times \frac{0.1}{2.6}\times 100\%<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

(3) $<mspace\; width="1em"></mspace>\frac{0.1}{3\times 2.6}\times 100\%<mspace\; width="1em"></mspace>$

(4) $\frac{0.1}{2.6}\%$

We can reduce random errors by 

(1) Taking large number of observations  

(2) Corrected zero error 

(3) By following proper technique of experiment 

(4)  Both (1) & (3)