Units and Measurement

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New answer posted

2 weeks ago

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A
alok kumar singh

Contributor-Level 10

Unit of temperature is Fahrenheit

Unit of mass is atomic mass of unit

Unit of length is light year

Unit of time is shake

New answer posted

2 weeks ago

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A
alok kumar singh

Contributor-Level 10

In a measurement, a choice of change of different units does not change the number of significant figures.

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A
alok kumar singh

Contributor-Level 10

In a number without decimal, trailing zeros are insignificant. If a number has a decimal, then preceding zeros are insignificant.

Hence, in 40.40 all zeros are significant

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2 weeks ago

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V
Vishal Baghel

Contributor-Level 10

F = m a = M L T 2

T 2 = M L F

[ T ] = M 1 / 2 L 1 / 2 F 1 / 2

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R
Raj Pandey

Contributor-Level 9

(b)  L1=  Length of rod =20.123cm

L2=  Length of rod = 18.1cm

Difference in length  

= 20.123-18.1

 =2.0 cm

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V
Vishal Baghel

Contributor-Level 10

L C = 1 1 0 0 m m = 0 . 0 1 m m

Zero error = +0.08 mm.

Diameter = 1 + 72 * 0.01 – 0.08 = 1.64 mm

Radius = 0.82 mm

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2 weeks ago

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V
Vishal Baghel

Contributor-Level 10

T = 2 π L g o r T 2 = 4 π 2 ( L g )

g = 4 π 2 ( L T 2 )

Δ g g % = ( Δ L L + 2 Δ T T ) % = [ 1 m m 1 m + 2 * 0 . 0 1 1 . 9 5 ] * 1 0 0 % = ( 0 . 0 0 1 + 0 . 0 1 0 2 ) * 1 0 0 = 1 . 1 3 %

New answer posted

2 weeks ago

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V
Vishal Baghel

Contributor-Level 10

Since, x 2 α k T  should be dimensionless.

So, dimension of α , [ α ] = L 2 M L 2 T 2 = M 1 T 2

Dimension of α β 2  should be that of W.

So, [ α β 2 ] = M L 2 T 2

[ β 2 ] = M L 2 T 2 M 1 T 2 = M 2 L 2 T 4 [ β ] = M L T 2

 

New answer posted

2 weeks ago

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R
Raj Pandey

Contributor-Level 9

Let 'x' he the value of one division of main scale

              x = 1 2 0 c m = 0 . 0 5 c m  

              Let y be value of one division on venire scale given

              10 y = 9 x

              y = 9 x 1 0  

              Least count = x 9 x 1 0 = x 1 0 = 0 . 0 5 1 0  

              = 0.005 cm

              = 5 * 10-2 mm

New answer posted

2 weeks ago

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V
Vishal Baghel

Contributor-Level 10

Least  count of Vernier = 0.1mm

Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10-2 cm.

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