This is physics one.: No. 1111----Determine the least count of the vernier callipers and measure the length and diameter of a small cylinder (Average of these seth may be as metal rod and length 2 to 3 cm and diameter 1 to 2 cm)...
No.2222-----Determine the pitch and least count of the given gange and measure the mean radius of the given......
Wise, taking three sets of reading in perpendicular direction.
Ok this all
And maths
Planning delivery route for post man and milkman
THIS is maths project
And first one was of physics
Answers
Answer:
SOLUTION 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.
SOLUTION 2:
SORRY I DON'T KNOW THE 2nd ANSWER
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