Question

The period of a body under shm i, e presented by The=Pke power a D ke power b s ke power c and p is pressure, D is density, s is surface tension find the value of a, b, c

Answer 1

Answer:

$\large a=-\dfrac{3}{2}$

$\large b=\dfrac{1}{2}$

$\large c=1$

Explanation:

Given :

$\large T=P^a.D^b.S^c$

Where P is pressure, D is density and S is surface tension.

We have to find value of a , b and  c.

We know Dimensional formula for

$\large Time = \left[M^0L^0T^1\right]$

$\large Pressure= \left[M^{1}L^{-1}T^{-2}\right]$

$\large Density = \left[ML^{-3}\right]$

$\large Surface \ tension= \left[MT^{-2}\right]$

Putting these value in given

$\large T=P^a.D^b.S^c\\\\\\\large \left[M^0L^0T^{1}\right]=\left[M^{1}L^{-1}T^{-2}\right]^a.\left[ML^{-3}\right]^b\left[MT^{-2}\right]^c\\\\\\\left[M^0L^0T^{1}\right]=\left[M]^{a+b+c}.[L]^{-a-3b}.[T]^{-2a-2c}\right]$

Comparing both side we get

a + b + c = 0

a + c = - b .... ( i )

- a - 3 b = 0

- 3 b = a .....( ii )

- 2 a - 2 c = 1

a + c = - 1 / 2  .....( iii )

From ( i ) and ( iii ) we get

b = 1 / 2

putting b = 1 / 2 in ( ii )

a = - 3 b

a = - 3 / 2

putting a = - 3 / 2 in ( iii )

a + c = -1 / 2

$\large c=-\dfrac{1}{2}+\dfrac{3}{2} \\\\\\\large c=\dfrac{-1+3}{2}\\\\\\\large c=\dfrac{2}{2}\\\\\\\large c=1$

Thus we get answer.

Answer 2

Explanation:

refer to the attachment: