## APPARATUS REQUIRED:

1. Vernier caliper

2. A hollow tube

3. Any balance for measuring mass

## THEORY:

### Principle of vernier

Vernier caliper consists of two scales and the vernier scale. These scales

are graduated in such a way that divisions of vernier scale coincide with (n-1) divisions of main scale in the principle of vernier caliper.

Then,

n usd consists with (n-1) msd.

1 usd consists with (n-11 / n ) msd.

Therefore, vernier constant (VC) = (1 msd – 1 usd) x value of 1 msd

={1 -(n-1) / n } x value of msd

= I / n x values of 1 msd

= value of 1 main scale division / total no. of division on vernier scale

Volume and density of hollow tube:-

Let l is the length, di is the internal and de is the external diameter of the tube. Then,

External volume of the tube, ve= πd²el / 4

ve = πd²el / 4

Internal volume of the tube, vi=πd² / 4 i

Hence, volume of the tube, v= π(d²/e -d²/i)2 / 4

= πl(de + di ) (de – di ) / 4

Density (D) of the material of the given tube is defined as mass (M) of the tube pen unit volume of the material of the tube.

i.e P= M / V

**OBSERVATIONS:**

### 1. For vernier constant

In the main scale, value of 10 main scale division = 1 cm

value of 1 main scale division = 0.1

In the vernier scale,

Total no. of vernier scale division (n)=10

therefore, V.S.= 1/ n x value of 1 msd = OVL

Observation for | No.of Obs. | Main scale reading (v) | Vernier scale reading | Value of vernier scale reading | Total z=(x+y)cm | Main value z cm | Connected mean value z +error cm |

length | 1. | 5.2 | 7 | 0.09 | 5.17 | 5.3 | 5.3±0=5.38 |

2. | 5.1 | 8 | 0.07 | 5.48 | |||

3. | 5.1 | 9 | 0.07 | 5.29 | |||

4. | 5.3 | 9 | 0.09 | 5.19 | |||

5. | 5.1 | 7 | 0.07 | 5.37 | |||

External diameter | 1. | 2.4 | 4 | 0.04 | 2.44 | 2.44 | 2.54±0=2.54 |

2. | 2.4 | 3 | 0.03 | 2.43 | |||

3. | 2.4 | 4 | 0.04 | 2.44 | |||

4. | 2.4 | 5 | 0.05 | 2.45 | |||

5. | 2.4 | 5 | 0.05 | 2.45 | |||

Internal diameter (di) | 1. | 2.1 | 8 | 0.08 | 2.18 | 2.174 | 2.174±0=2.174 |

2. | 2.1 | 7 | 0.07 | 2.17 | |||

3. | 2.1 | 9 | 0.09 | 2.19 | |||

4. | 2.1 | 6 | 0.06 | 2.16 | |||

5. | 2.1 | 7 | 0.07 | 2.17 |

**CALCULATIONS:**

FROM TABLE:

l = 5.3

de = 2.44

di = 2.174

Volume of the table, V= πl (de+di) (de-di) / 4

= π 5.3 (2.44 +2.174) (2.44-2.175) / 4

= 5.186cm³

Mass of the hollow tube = 8.65gm

Density of the material of the tube : P = m/v = 8.65/3.64 = 2.376

**PERCENTAGE ERROR:**

Standard value density of maternal of the tube, PS = 2.7gm/cm³

Observed value of density of maternal of the tube PO = 2.376gm km³

Percentage error = |Ps – Po /Ps| x 100%

= |2.7 – 2.376 / 2.7| x 100%

= 12%

## RESULT:

The volume and density of material of the given tube are found to be 3.64 cm³ and 2.37cm ³ and 2.37 gm/cm ³ experimentally

## CONCLUSIONS:

The length, internal and external diameters of a given hollow tube are measured by using vernier calipers and hence volume and density the material of the tube is determined.

**SOURCES OF ERRORS:**

1. Undve pressure applied to the jaus.

2. Non-uniformity in shape of the block.

3. Porallox may be three in taking observations.

**PRECAUTIONS:**

1. The zero error should be determined and applied carefully.

2. Two jaus should be pressed gently.

3. Observations should be taken at different points of the table.