Question

h=2 T cos(theta)/rpg
Check the correctness dimensionally

Answer 1

Explanation:

Given that,

h= \dfrac{r\rho g}{2s cos\theta}h=

2scosθ

rρg

The units of s,r,g and density

Surface tension S=[M][T]^{-2}S=[M][T]

−2

Radius R= [L]R=[L]

Density \rho=[M][L]^{-3}ρ=[M][L]

−3

Gravity due to acceleration g=[L][T]^{-2}g=[L][T]

−2

h= \dfrac{r\rho g}{S}h=

S

rρg

h= \dfrac{ [L]\times [M][L]^{-3}\times[L][T]^{-2}}{[M][T]^{-2}}h=

[M][T]

−2

[L]×[M][L]

−3

×[L][T]

−2

h=[L^{-1}]h=[L

−1

]

This relation is incorrect.

The correct relation is define as:

h= \dfrac{2scos\theta}{r\rho g}h=

rρg

2scosθ

.....(I)

Now put the all value in equation (I)

h= \dfrac{S}{r\rho g}h=

rρg

S

h= \dfrac{[M][T]^{-2}}{ [L]\times [M][L]^{-3}\times[L][T]^{-2}}h=

[L]×[M][L]

−3

×[L][T]

−2

[M][T]

−2

h=[L]h=[L]

This relation is correct.

Hence, Given relation is incorrect