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Q.

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

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a

${ML}^{2} T$

b

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

c

${ML}^{2} T^{- 2}$

d

${MLT}^{- 2}$

answer is B.

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Detailed Solution

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

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

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