Question

The velocity of a body moving in a viscous medium is given by V=A/B × [1-e^-t/B] where t is the time, A and B are constants. Then the dimensions of A are:
A) M°L°T° B) M° L^1 T°
C) M° L^1 T^-2 D) M^1 L^1 T^-1

Answer 1

Answer:

t/B is diamensionless

so diamension of B =[T]

diamension of A/B = diamension of V

A/[T] = LT^-1

A = LT^-1 ×T^1

A = M^0 L^1 T^0

so correct option is B

Answer 2

Answer:

The dimensions of A are an option (b) $\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$

Explanation:

The equation acquired by equating a physical quantity with its dimensional formula exists named a dimensional equation.

It $e^{\frac{\bar{-t}}{\bar{B}}}$ is the exponential term. So, the $\frac{\mathrm{t}}{\mathrm{B}}$ is constant.

So, B has the dimensions as:

B=[T]

On the other hand, the dimension  $\frac{A}{B}$will be the same as that of the velocity.

$\frac{\mathrm{A}}{\mathrm{B}}$has dimensions as velocity

$\left[\mathrm{LT}^{-1}\right]=\frac{\mathrm{a}}{[\mathrm{T}]}$

dimensional fo0rmula of $\mathrm{A}$ is $\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$

Hence, the dimensions of A are an option (b) $\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}$

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