$M L^{- 1} T^{- 2}$ is the dimensional formula of

force

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coefficient of friction

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modulus of elasticity

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energy

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Solution

The correct option is B modulus of elasticity
Modulus of elasticity is given by $Y = \frac{F}{A} . \frac{L}{Δ L}$
Dimension of force $F = [M L T^{- 2}]$
Dimension of Area $A = [L^{2}]$
Dimension of length $L = [L]$
Thus dimension of modulus of elasticity $Y = \frac{[M L T^{- 2}] [L]}{[L^{2}] [L]} = [M L^{- 1} T^{- 2}]$

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