The velocity of a body moving in viscous medium is given by v - P - Q 1 - e -t-Q where -t- is the time- P and Q are constants- Then the dimensions of P are

The relation between the time t and position x for a particle moving on the x-axis is given by $t = p x^{2} + q x$ , Where $p$ and $q$ are constants. The relation between velocity $b$ and acceleration $a$ is as

Q. A physical quantity

P

is given by the relation,

P = P_{0} e^{(- α t^{2})}

. If

t

denotes the time, the dimensions of constant

α

are

The coordinates of a particle moving in a plane are given by x(t) = a cos(pt) and y(t) = a sin (pt)

where a, and p are positive contants of appropriate dimensions. Then,

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The velocity of a body moving in viscous medium is given by v=PQ(1−e−t/Q) where ′t′ is the time, P and Q are constants. Then the dimensions of P are

The correct option is C M0LT0Here,Dimensions of [PQ]=dimensions of [v]And dim[tQ]=M0L0T0Hence, [Q]=[T]And, [P]=[Q][LT−1]=[L]Option C is the correct answer

The velocity of a body moving in viscous medium is given by $v = \frac{P}{Q} (1 - e^{- t / Q})$ where $^{'} t^{'}$ is the time, $P$ and $Q$ are constants. Then the dimensions of $P$ are

The correct option is C $M^{0} L T^{0}$
Here,

Dimensions of $[\frac{P}{Q}] =$ dimensions of $[v]$

And $d i m [\frac{t}{Q}] = M^{0} L^{0} T^{0}$

Hence, $[Q] = [T]$

And, $[P] = [Q] [L T^{- 1}] = [L]$

Option $C$ is the correct answer