If force is proportional to square of velocity, then the dimensions of proportionality constant is

Question

If force is proportional to square of velocity, then the dimensions of proportionality constant is (A) [ ML -1 T ] (B) [ ML -1 T0 ] (C)  [MLT0]  (D)  [M0LT-1]

Tardigrade Admin · Accepted Answer

As F propto v2 or F=k v2 ∴ k=(F/v2) [k]=([F]/[v2])=([ MLT -2]/[ L 2 T -2]) =[ ML -1 T 0]

Q. If force is proportional to square of velocity, then the dimensions of proportionality constant is

$[M L^{- 1} T]$

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

$[M L T^{0}]$

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

As $F \propto v^{2}$
or $F = k v^{2}$
$∴ k = \frac{F}{v ^{2}}$
$[k] = \frac{[ F ]}{[ v ^{2} ]} = \frac{[ M L T ^{- 2} ]}{[ L ^{2} T ^{- 2} ]}$
$= [M L^{- 1} T^{0}]$

Q. If force is proportional to square of velocity, then the dimensions of proportionality constant is

[ML−1T]

[ML−1T0]

[MLT0]

[M0LT−1]

As F∝v2 or F=kv2 ∴k=v2F​ [k]=[v2][F]​=[L2T−2][MLT−2]​ =[ML−1T0]

$[M L^{- 1} T]$

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

$[M L T^{0}]$

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

As $F \propto v^{2}$
or $F = k v^{2}$
$∴ k = \frac{F}{v ^{2}}$
$[k] = \frac{[ F ]}{[ v ^{2} ]} = \frac{[ M L T ^{- 2} ]}{[ L ^{2} T ^{- 2} ]}$
$= [M L^{- 1} T^{0}]$