If frequency F- velocity V- and density D are considered fundamental units- the dimensional formula momentum will beDVF-2DV-2F-1-D-2V-2F-2DV-4F-3-

Question

Toppr Admin · Accepted Answer

Momentum -F-a-V-b-D-c-x27F9-MLT-x2212-1-T-x2212-1-a-LT-x2212-1-b-ML-x2212-3-cComparing both sides-c-1b-x2212-3c-1-x2234-b-4Also-a-b-1-x2234-a-x2212-3Therefore-Momentum -F-x2212-3-V-4-D-1-DV4F-x2212-3

If frequency $F$ , velocity $V$ , and density $D$ are considered fundamental units, the dimensional formula for momentum will be
$D V F^{2}$
$D V^{2} F^{- 1}$
$D^{2} V^{2} F^{2}$
$D V^{4} F^{- 3}$

Momentum $= [F]^{a} [V]^{b} [D]^{c}$

$⟹ [M L T^{- 1}] = [T^{- 1}]^{a} [L T^{- 1}]^{b} [M L^{- 3}]^{c}$

Comparing both sides,

$c = 1$

$b - 3 c = 1$

$∴ b = 4$

Also,

$a + b = 1$

$∴ a = - 3$

Therefore,

Momentum $= [F]^{- 3} [V]^{4} [D]^{1} = D V^{4} F^{- 3}$

If frequency F, velocity V, and density D are considered fundamental units, the dimensional formula for momentum will beDVF2DV2F−1D2V2F2DV4F−3

Momentum =[F]a[V]b[D]c⟹[MLT−1]=[T−1]a[LT−1]b[ML−3]cComparing both sides,c=1b−3c=1∴b=4Also,a+b=1∴a=−3Therefore,Momentum =[F]−3[V]4[D]1=DV4F−3

If frequency $F$ , velocity $V$ , and density $D$ are considered fundamental units, the dimensional formula for momentum will be
$D V F^{2}$
$D V^{2} F^{- 1}$
$D^{2} V^{2} F^{2}$
$D V^{4} F^{- 3}$

Momentum $= [F]^{a} [V]^{b} [D]^{c}$

$⟹ [M L T^{- 1}] = [T^{- 1}]^{a} [L T^{- 1}]^{b} [M L^{- 3}]^{c}$

Comparing both sides,

$c = 1$

$b - 3 c = 1$

$∴ b = 4$

Also,

$a + b = 1$

$∴ a = - 3$

Therefore,

Momentum $= [F]^{- 3} [V]^{4} [D]^{1} = D V^{4} F^{- 3}$