Question

The relation between power P ,distance x and time t is given as P=b+x²/at.The dimensions of b/a are

Answer 1

Answer:

Given equation;

$\longrightarrow\tt P = \dfrac{b + x^2}{at}$

Rearranging the given equation;

$\longrightarrow\tt Pat=b + x^2$

Now,

$\longrightarrow\tt [Pat]=[b] =[ x^2]$

Or,

$\longrightarrow\tt [b] = [ x^2]$

$\longrightarrow\bf [b] =[M^0L^2T^0]$

And,

$\longrightarrow\tt [a] =\dfrac{ [x^2] }{[Pt]}$

$\longrightarrow\tt [a] =\dfrac{ [M^0L^2T^0] }{[M^1L^2T^{-3}] [M^0L^0T^1] }$

$\longrightarrow\tt [a] =\dfrac{ [M^0L^2T^0] }{[M^1L^2T^{-2}] }$

$\longrightarrow\tt [a] = [M^0L^2T^0] [M^{ - 1}L^{ - 2}T^{2}]$

$\longrightarrow\bf [a] = [M^{ - 1}L^{0}T^{2}]$

Now,

$\longrightarrow\tt \dfrac{[b]}{[a] } = \dfrac{[M^0L^2T^0] }{[M^{ - 1}L^{0}T^{2}] }$

$\longrightarrow\tt \dfrac{[b]}{[a] } = [M^0L^2T^0] [M^{1}L^{0}T^{ - 2}]$

$\longrightarrow\underline{\boxed{\bf \dfrac{[b]}{[a] } = [M^{1}L^{2}T^{ - 2}]}}$

Answer 2

Given : The relation between power P ,distance x and time t is given as

P=(b+x²)/at

To Find : The dimensions of b/a

Solution:

P=(b+x²)/at

b and x² are added hence their dimensions are same

=> P  = b / at     will satisfy dimensions

=> Pt = b/a

=> b/a  = Pt

dimensions of b/a   are same as dimension of  Pt

power * time = Work  

work  = Force * displacement

= mass  * acceleration * displacement

mass = M

acceleration = m/s²  = LT⁻²

displacement = L

work  =  M LT⁻² L  = M L²T⁻²

=> dimensions of b/a =  M L²T⁻²

dimensions of b/a =  M L²T⁻²

Learn More

which physical identity has the unit ms-2

brainly.in/question/23799769

The physical quantity having the same unit in all the systems of unit ...

brainly.in/question/4617397

The reference standard used for the measurement of a physical ...

brainly.in/question/4616981