Units and Measurement

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New answer posted•a year ago
S
a year ago

Dimension of B = [L? ¹]
Dimension of D = [T? ¹]
Dimension of A = [MLT? ²]
The dimensional formula of AD/B
= [MLT? ²] [T? ¹]/ [L? ¹] = [ML²T? ³]

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New answer posted•a year ago
S
a year ago

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            

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New answer posted•a year ago
S
a year ago

W=α2β    e−−βx2kT

[βx2]= [kT]= [ML2T−2]

β=MT−2

[W]= [α2β]⇒ [ML2T−2]= [α2] [MT−2]⇒ [α2]= [L2]⇒ [α]= [L]

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New answer posted•a year ago
S
a year ago

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 %

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

14πε0|e|2hc=14πε0|e|2r2×r2hcλ×λ=Fr×rE×λ

Dimension of ‘Fr’ and ‘E’ are same dimensionless

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New answer posted•a year ago
S
a year ago

From principle of homogeneity

[ F ] = [ α t 2 ] = [ β t ]

[α]=[F][t2] and [β]=[F][t]

∴ [ α ] [ t ] = [ β ]

∴αtβ= dimensionless 

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New answer posted•a year ago
S
a year ago

V . C = M S D − V S D     … (i)

given :  (N+1)VSD=NMSD

VSD= (NN+1)MSD     … (ii)

From (1) and (2)

V . C = ( MSD ) − N N + 1 ( MSD )

=MSD (1−NN+1)=MSDN+1

= 0 . 0 1 N + 1 = 1 1 0 0 ( N + 1 )

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New answer posted•a year ago
New answer posted•a year ago