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 measurement of the distance and the time are ${\mathrm{e}}_1$ and ${\mathrm{e}}_2$ respectively, the percentage error in the estimation of g is:

1. ${\mathrm{e}}_2-{\mathrm{e}}_1$                                                2. ${\mathrm{e}}_1+2{\mathrm{e}}_2$

3. ${\mathrm{e}}_1+{\mathrm{e}}_2$                                                4. ${\mathrm{e}}_1-2{\mathrm{e}}_2$

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:

The pitch of a screw gauge is \(1~\)mm and there are \(100\) divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads \(1\) mm and \(47\)^th division on the circular scale coincides with the reference line. The length of the wire is \(5.6\) cm. Curved surface area (in cm²) of the wire in appropriate number of significant figures will be:

1. \(2.4\) cm²

2. \(2.56\) cm²

3. \(2.6\) cm²

4. \(2.8\) cm²

Consider a screw guage without any zero error. What will be the final reading corresponding to the final state as shown? It is given that the circular head translates P msd in N rotations. One msd is equal to 1mm.

1.  $\left(\mathrm{P}/\mathrm{N}\right)\left(2+45/100\right) \mathrm{mm}$

2.  $\left(\mathrm{N}/\mathrm{P}\right)\left(2+45/\mathrm{N}\right) \mathrm{mm}$

3.  $\mathrm{P} \left(2/\mathrm{N}+45/100\right) \mathrm{mm}$

4.  $\left(2+\frac{45}{100}\times \frac{\mathrm{P}}{\mathrm{N}}\right) \mathrm{mm}$

A screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw, the state of the instrument is shown by diagram (I). When both the rods are inserted together in series then the state is shown by the diagram (II). What is the zero error of the instrument? \(1~\text{msd}= 100~\text{csd}= 1~\text{mm}\)

                                 
1. \(-0.16~\text{mm}\)
2. \(+0.16~\text{mm}\)
3. \(+0.14~\text{mm}\)
4. \(-0.14~\text{mm}\)

One cm on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of the vernier scale coincide with 8 small divisions of the main scale, what will be the least count of the vernier callipers?

1.  0.06 cm

2.  0.6 cm

3.  0.5 cm

4.  0.7 cm

Find the zero correction in the given figure. 

1.  0.4 mm

2.  0.5 mm

3.  -0.5 mm

4.  -0.4 mm

Find the thickness of the wire. The least count is \(0.01\) mm. The main scale reads: (in mm)
            

1. \(7.62\)
2. \(7.63\)
3. \(7.64\)
4. \(7.65\)

In the given Vernier scale, 10 division of vernier scale are matching with 9 divisions of main scale as shown in first figure. Find the diameter of the object in second figure. Assume the edge of the vernier as '0' of vernier.

(1)  15.6 mm

(2)  156 mm

(3)  1.56 mm

(4)  0.156 mm