Question

Time period of body is t=p^a d^b s^c. where p is pressure d is densith and s is surgace tension. find a,b,and c

Answer 1

Answer:

a = -3/2, b = 1/2 and c = 1

Explanation:

Dimesions of time period t $=[T]$

Dimensions of Pressure p $=[ML^{-1}T^{-2}]$

Dimensions of density d $=[ML^{-3}]$

Dimensions of surface tension s $=[MT^{-2}]$

∵ $t=p^ad^bs^c$

∴ equating the dimensions on both sides we get

$[T]=[ML^{-1}T^{-2}]^a[ML^{-3}]^b[MT^{-2}]^c$

$\implies [T]=[M^{a+b+c}L^{-a-3b}T^{-2a-2c}]$

equating dimensions on both sides we get

$a+b+c=0$              ............ (1)

$-a-3b=0 \implies a+3b=0$           ............... (2)

$-2a-2c=1 \implies a+c=-1/2$         ............... (3)      

Putting the value of a+c in eq (1)  

$-1/2+b=0 \implies b=1/2$

Putting the value of b in eq (2)

$a+3\times (1/2)=0 \implies a=-3/2$

Putting the value of a in eq (3)

$-3/2+c=-1/2$

$\implies c=-1/2+3/2 = 1$

Thus,

a = -3/2

b = 1/2

c = 1