The position of a particle at time t is given | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$,the exponent of the exponential function must be dimensionless. Therefore,the dimensions of $A

Vedclass · Accepted Answer

$[M^0 LT^{-1}]$ and $[M^0 L^0 T^{-1}]$

Vedclass · Answer

(D) In the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$,the exponent of the exponential function must be dimensionless. Therefore,the dimensions of $At$ must be $[M^0 L^0 T^0]$.Since $[t] = [T]$,we have $[A][T] = [1]$,which implies $[A] = [T^{-1}] = [M^0 L^0 T^{-1}]$.Next,the term $(1 - e^{-At})$ is dimensionless. Thus,the dimensions of $x$ must be equal to the dimensions of $\frac{v_0}{A}$.$[x] = \frac{[v_0]}{[A]} \implies [L] = \frac{[v_0]}{[T^{-1}]}$.Therefore,$[v_0] = [L][T^{-1}] = [M^0 LT^{-1}]$.Thus,the dimensions of $v_0$ and $A$ are $[M^0 LT^{-1}]$ and $[M^0 L^0 T^{-1}]$ respectively.

The position of a particle at time $t$ is given by the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$,where $v_0$ is a constant and $A > 0$. The dimensions of $v_0$ and $A$ respectively are:

Similar Questions

The dimensional formula for Boltzmann's constant is

Dimensional formula for the universal gravitational constant is

The focal power of a lens has the dimensions

Inductance $L$ can be dimensionally represented as

The dimensions of the universal gravitational constant are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module