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Standard XII

Physics

Acceleration in 2D

The distance ...

Question

The distance traveled by a body in time $^{'} t^{'}$ is given by $x = a + b t + {c t}^{2}$ where $x$ is distance, $t$ is time $a, b$ and $c$ are constants. The dimensional formula for $a, b$ and $c$ respectively are :

L^{1}, L^{1} T^{- 1}, L^{1} T^{- 2}

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L^{1} T^{- 1}, L^{1} T^{- 2}, L

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L^{1} T^{- 2}, L^{1} T^{- 1}, L

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L, L, L

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Solution

The correct option is A $L^{1}, L^{1} T^{- 1}, L^{1} T^{- 2}$
$x = a + b t + c t^{2}$
$L^{1} = a$
$L^{1} = b t$
$\Rightarrow b = L^{1} T^{- 1}$
$c T^{2} = L \Rightarrow c = L^{1} T^{- 2}$

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The distance traveled by a body in time ′t′ is given by x=a+bt+ct2 where x is distance, t is time a,b and c are constants. The dimensional formula for a,b and c respectively are :

The correct option is A L1,L1T−1,L1T−2x=a+bt+ct2L1=aL1=bt⇒b=L1T−1cT2=L⇒c=L1T−2

The distance traveled by a body in time $^{'} t^{'}$ is given by $x = a + b t + {c t}^{2}$ where $x$ is distance, $t$ is time $a, b$ and $c$ are constants. The dimensional formula for $a, b$ and $c$ respectively are :

The correct option is A $L^{1}, L^{1} T^{- 1}, L^{1} T^{- 2}$
$x = a + b t + c t^{2}$
$L^{1} = a$
$L^{1} = b t$
$\Rightarrow b = L^{1} T^{- 1}$
$c T^{2} = L \Rightarrow c = L^{1} T^{- 2}$