Question

The distance x moved by a particle in time t is given by, x = a (1 – e-bt).
The dimensions of ab are the same as those of
0
(A) velocity
(B) momentum
(C) force
(D) impulse


pls help me with this question​

Answer 1

Answer:

answer is c

Explanation:

because force is the answer

Answer 2

Concept

Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.

Given

We have given the distance of a particle $x=a(1-ebt)$.

Find

We are asked to find the dimension of ab and match the dimension with the given options.

Solution

We will simply the given equation.

$x=a-aebt$

Now, by principle of homogeneity

$x=a=aebt$

  $x=a\\\ [L] =a$

and,

 $x=aebt\\\[L]=[L]eb[T]$

As e is dimensionless quantity so this become,

$[L]=[L]b[T]\\\frac{[L]}{[L][T]} =b\\b=\frac{1}{[T]} \\b=[T]^{-1}$

Therefore, the dimensions of ab is  $[L][T]^{-1}$.

Velocity=displacment / time

$v=\frac{[L]}{[T]} \\v=[L][T]^{-1}$

Hence, the dimensions of ab match with the dimensions of velocity which is option (A).

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