Question

Determine the least count of the Vernier Callipers and measure the length and diameter of a small cylinder (average of three sets)may be a metal rod of length 2 to 3 cm and diameter 1 to 2 cm.

Answer 1

Vernier callipers.A spherical body ( it can be a pendulum bob)A cylinder.A small rectangular metallic block of known massA beaker or a calorimeter. The Procedure

 

1 M.S.D. = 1 mm
10 V.S.D.= 9 M.S.D.
1 V.S.D.= 9/10 M.S.D. = 0.9 mm.
Vernier Constant, V.C.= 1 M.S.D.-1 V.S.D. = (1-0.9) mm = 0.1 mm = 0.01cm.
 Zero error
(i).........cm,  
(ii).........cm,  
(iii)...........cm.
Mean zero error (e)=..........cm.
Mean zero correction (c) = -e=.........cm.Dimension to be measuredSl NoMain Scale ReadingMSR cmVernier Scale ReadingVSR cmVSR x L.CcmToatl ReadingMSR + (V S R x L.C)cmMeancm Diameter of the bob                  Diameter of the cylinder                  Length of thye cylinder                  Length of the block            Breadth of the block            Thickness of the block            Internal diameter of the beaker            Internal depth of the beaker            Calculations 

Mean corrected diameter------------cm

Volume of sphere,=---------cm3= ------m3.

Mean corrected length of the block, l=............cm

Mean corrected breadth of the block,  b= .......cm

Mean corrected thickness of the block,  h= .........cm

Volume of block ,  =........................cm3  =..........m3

Density of the block material,=..................cm

Mean corrected internal diameter,D=................cm

Mean correcteddepth,d=........cm 

Volume of beaker / calorimeter ,= ..........cm3=............m3. 

The Result

 The volume of the beaker / calorimeter is ...........m3.

Volume of Sphere=.......................... m3

Volume of block is ................................m3

The volume of the  beaker / calorimeter is ...........cm3.  

1 M.S.D. = 1 mm
10 V.S.D.= 9 M.S.D.
1 V.S.D.= 9/10 M.S.D. = 0.9 mm.
Vernier Constant, V.C.= 1 M.S.D.-1 V.S.D. = (1-0.9) mm = 0.1 mm = 0.01cm.
 Zero error
(i).........cm,  
(ii).........cm,  
(iii)...........cm.
Mean zero error (e)=..........cm.
Mean zero correction (c) = -e=.........cm.Dimension to be measuredSl NoMain Scale ReadingMSR cmVernier Scale ReadingVSR cmVSR x L.CcmToatl ReadingMSR + (V S R x L.C)cmMeancm Diameter of the bob                  Diameter of the cylinder                  Length of the cylinder                  Length of the block            Breadth of the block            Thickness of the block            Internal diameter of the beaker            Internal depth of the beaker            Calculations 

Mean corrected diameter------------cm

Volume of sphere,=---------cm3= ------m3.

Mean corrected length of the block, l=............cm

Mean corrected breadth of the block,  b= .......cm

Mean corrected thickness of the block,  h= .........cm

Volume of block ,  =........................cm3  =..........m3

Density of the block material,=..................cm

Mean corrected internal diameter,D=................cm

Mean correcteddepth,d=........cm 

Volume of beaker / calorimeter ,= ..........cm3=............m3. 

The Result

 The volume of the beaker / calorimeter is ...........m3.

Volume of Sphere=.......................... m3

Volume of block is ................................m3

The volume of the  beaker / calorimeter is ...........cm3.  

Answer 2

Explanation:

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