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

If z=$\frac{A\times B}{A+B}$, and errors in the determination of A and B are respectively $△\mathrm{A} \mathrm{and} △\mathrm{B}$, then the fractional error in the calculation of z is :

$1. \frac{\mathrm{\Delta }A}{A}+\frac{\mathrm{\Delta }B}{B}+\frac{\mathrm{\Delta }A+\mathrm{\Delta }B}{A+B}$
$$
$2. \frac{\mathrm{\Delta }A}{A}+\frac{\mathrm{\Delta }B}{B}-\frac{(\mathrm{\Delta }A+\mathrm{\Delta }B)}{A+B}$
$$
$3. \frac{\mathrm{\Delta }A}{A}+\frac{\mathrm{\Delta }B}{B}$
$$
$4. Zero$

Two resistors \(R_1 = (3.0\pm0.3)~\Omega\) and \(R_2 = (5.0 \pm0.1)~\Omega\) are connected in parallel. The equivalent resistance, \(R_{eq}\), will be:
Hint: \({1 \over R_{eq}} = {1 \over R_{1}} + {1 \over R_{2}} \)

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. \(-0.14~\text{mm}\)

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

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