Write the dimensions of a - b in the relation E - b - x2-at- where E is the energy- x is the displacement and t is time-

Here [b] and [x]^2 = L^2 have same dimensions. Also [a] = [x]^2/[E]×[t] = L^2/(ML^2T^ 2)T = M^ 1T^1 a × b = [M^ 1L^2 T^1] hence correct answer is (b). ...

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Standard XII

Physics

Dimensional Analysis

Write the dim...

Question

Write the dimensions of a × b in the relation $E = \frac{b - x^{2}}{a t}$ , where E is the energy, x is the displacement and t is time.

$M L^{2} T$

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$M^{- 1} L^{2} T^{1}$

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$M L T^{- 2}$

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$M L^{2} T^{- 2}$

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Solution

The correct option is B
$M^{- 1} L^{2} T^{1}$

Here [b] and $[x]^{2} = L^{2}$ have same dimensions.

Also $[a] = \frac{[x]^{2}}{[E] \times [t]} = \frac{L^{2}}{(M L^{2} T^{- 2}) T} = M^{- 1} T^{1}$

$a \times b = [M^{- 1} L^{2} T^{1}]$

hence correct answer is (b).

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Write the dimensions of a × b in the relation E=b−x2at, where E is the energy, x is the displacement and t is time.

The correct option is B M−1L2T1 Here [b] and [x]2=L2 have same dimensions. Also [a]=[x]2[E]×[t] =L2(ML2T−2)T =M−1T1 a×b=[M−1L2T1] hence correct answer is (b).

Write the dimensions of a × b in the relation $E = \frac{b - x^{2}}{a t}$ , where E is the energy, x is the displacement and t is time.

The correct option is B
$M^{- 1} L^{2} T^{1}$

Here [b] and $[x]^{2} = L^{2}$ have same dimensions.

Also $[a] = \frac{[x]^{2}}{[E] \times [t]} = \frac{L^{2}}{(M L^{2} T^{- 2}) T} = M^{- 1} T^{1}$

$a \times b = [M^{- 1} L^{2} T^{1}]$

hence correct answer is (b).