2 5 6 Aim: 1 2 Apparatus: Theory: 4 5 Least count: Slno: PSR(mm) Pitch of...

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2 5 6 Aim: 1 2 Apparatus: Theory: 4 5 Least count: Slno: PSR(mm) Pitch of the screw Least count (LC) To 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. 3 Volume of the given wire-Ir²h in m³ (a) E-learning Physics Lab World University of Bangladesh Experiment no:2 SCREW GAUGE Where r-radius of the wire (in m) h-length of the wire. (in m) (b) Volume of the sphere-(4/3) Ir (m³) Observations and calculations It is the smallest measurement that which any measuring instrument can measure accurately (value of one division=L.C.) Zero error=-- --div =distance moved/no: of rotations made =4mm/4 =1mm =Pitch of the screw/Total no of circular scale divisions =1mm/100 = 0.01mm Length of the given wire,h= Where r-radius of the sphere (in m) (1) To find the diameter (thickness) of the given wire Radius of the wire,r=d/2= Volume of the wire.V=Ir²h (2) To find the diameter of the sphere PSR(mm) slno HSR (div) HSR(div) CHSR(div) CHSRXLC (mm) Radius of the sphere,r=d/2 Volume of the sphere,V=(4/3) Ir³ cm mm 3 mm 3 m mm CHSR(div) mm 3 mm 3 mm 3 m³ Diameter,d=PSR+ (CHSRX LC) (mm) Mean d= mm CHSRXLC Diameter,d= (mm) PSR+(CHSRXLC) (mm) Mean d= mm 2 5 6 Aim: 1 2 Apparatus: Theory: 4 5 Least count: Slno: PSR(mm) Pitch of the screw Least count (LC) To 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. 3 Volume of the given wire-Ir²h in m³ (a) E-learning Physics Lab World University of Bangladesh Experiment no:2 SCREW GAUGE Where r-radius of the wire (in m) h-length of the wire. (in m) (b) Volume of the sphere-(4/3) Ir (m³) Observations and calculations It is the smallest measurement that which any measuring instrument can measure accurately (value of one division=L.C.) Zero error=-- --div =distance moved/no: of rotations made =4mm/4 =1mm =Pitch of the screw/Total no of circular scale divisions =1mm/100 = 0.01mm Length of the given wire,h= Where r-radius of the sphere (in m) (1) To find the diameter (thickness) of the given wire Radius of the wire,r=d/2= Volume of the wire.V=Ir²h (2) To find the diameter of the sphere PSR(mm) slno HSR (div) HSR(div) CHSR(div) CHSRXLC (mm) Radius of the sphere,r=d/2 Volume of the sphere,V=(4/3) Ir³ cm mm 3 mm 3 m mm CHSR(div) mm 3 mm 3 mm 3 m³ Diameter,d=PSR+ (CHSRX LC) (mm) Mean d= mm CHSRXLC Diameter,d= (mm) PSR+(CHSRXLC) (mm) Mean d= mm

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Solution The screw gauge has more precision than the vernier caliper The separation between each thread is the same By rotating the nut clockwise or anticlockwise the screw can be moved either forward ... View the full answer