find the thickness of the given wire and sphere using a screw gauge and hence to find their volumes. Screw gauge, thin wire, small sphere and meter scale. Volume of the given wire= IIrh in m Apparatus: 3 Theory: (a) Where r=radius of the wire (in m) h=length of the wire. (in m) (b) Volume of the sphere-(4/3) IIr (m) Where r=radius of the sphere (in m) Observations and calculations It is the smallest measurement that which any measuring instrument can measure accurately (value of one division=L.C.) --div Least count: Zero error=-- Pitch of the screw =distance moved/no: of rotations made =4mm/4 =lmm Least count (LC) =Pitch of the screw/Total no of circular scale divisions =lmm/100 =0.01mm Length of the given wire,h= cm mm (1) To find the diameter (thickness) of the given wire HSR(div) Diameter,d=PSR+ (CHSRX LC) (mm) Slno: PSR(mm) CHSR(div) CHSRXLC (mm) 2 6. Mean d= mm Radius of the wire,r=d/2= mm ----- 3 Volume of the wire, V= Ilr h 3 mm 3 m (2) To find the diameter of the sphere CHSR(div) CHSRXLC Diameter,d= PSR+(CHSRXLC) (mm) slno PSR(mm) HSR(div) (mm) 1 2 3 4 Mean d= mm Radius of the sphere,r3Dd/2 mm Volume of the sphere, V=(4/3) Ir. mm mm m3 Procedure: 1. Measure the Zero correction and least count of screw gauge. 2. Measure the diameter of the given wire and sphere using screw gauge. 3. Find PSR,HSR and CHSR using screw gauge. 4. Calculate PSR+(CHSRXLC). 5. Measure the length of the given wire by using metre scale, 6. Find volume of wire and sphere using the formulae V= IIr h& V=(4/3) IIr 7. Repeat the experiment 4 or 5 times. 3 Precautions 1. At a time rotate the screw in one direction to avoid backlash error 2. Zero error should be observed carefully and taken into consideration Sources of error 1. The wire may not be of uniform cross section 2. Backlash error always exists because it cannot be removed completely Result: (a) Diameter of the given wire,d=------- m 3 m Volume of the wire, V (b) Diameter of the sphere,d m Volume of the sphere,V m3 ------