Question

The velocity v of the a particle depends upen the time t according to the equation v= a + bt + ( c) /(d+1) Write the dimension of a, b,c and d.

Answer 1

Answer:

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Answer 2

The dimensions of a,b,c and d are $[M^0 L^1 T^-^1]$ , $[M^0 L^1 T^-^2]$ , $[M^0 L^1 T^-^1]$ and $[M^0 L^0 T^0]$ respectively

Step by step explanation:

Given equation :

$\textbf{\Large v = a + bt + }\frac{\textbf{\Large c }}{\textbf{\Large (d+1) }}$

 

it is given that

v = Velocity of the particle

t = Time

To know the dimension of all constants a,b,c and d,

We know that

$\textbf{\Large Two quantities can be added , if they have same dimension}$So,

The dimension of velocity v = $[M^0 L^1 T^-^1]$

The dimension of time t =       $[M^0 L^0 T^1]$

So, The dimension of v = dimension of     $\textbf{ a + bt + }\frac{\textbf{ c }}{\textbf{ (d+1) }}$

so, The dimension of a = dimesion of v ( as both a and v equate with each other)

So, the dimension of a = $[M^0 L^1 T^-^1]$

Similarly ( a ) is added to ( bt ),

Then dimesion of bt = dimension of a

The dimension of bt = dimension of  $\frac{\textbf{\large a}}{\textbf{\large t}}$

So, The dimension of b = $\frac{[M^0 L^1 T^-^1]}{[M^0 L^0 T^1]}$ =    $[M^0 L^1 T^-^2]$

As we know that two same dimensional quantities add up,

so in case of (d + 1 )

d has the same dimension as the constant = 1 has.

So, the dimension of d = $[M^0 L^0 T^0]$

In case of c , the denominator is constant so,

the dimension of c = dimension of v

So, the dimension of c = $[M^0 L^1 T^-^1]$

$\textbf{\Large So, the dimensions are :}$

$\textbf{\Large a = }$ $[M^0 L^1 T^-^1]$

$\textbf{\Large b = }$ $[M^0 L^1 T^-^2]$

$\textbf{\Large c = }$ $[M^0 L^1 T^-^1]$

$\textbf{\Large d = }$ $[M^0 L^0 T^0]$