MCQ
The dimension $\ce{ML^{-1} T^{-2}}$ can correspond to$:$
- AMoment of a force.
- ✓Surface tension.
- CModulus of elasticity.
- DMoefficient of viscosity.
✓
Answer
Correct option: B.
Surface tension.
Dimension of modulus of elasticity$: \frac{\frac{\text{F}}{\text{A}}}{\frac{\triangle\text{l}}{\text{l}}}=\frac{\big[\text{MLT}^{-1}\big]}{\text{L}^2}=\big[\text{ML}^{-1}\text{T}^{-2}\big]$
Dimension of moment of force$: =\big[\text{MLT}^{-2}\big][\text{L}]=\big[\text{ML}^{2}\text{T}^{-2}\big]$
Dimension of surface tension$: \frac{\text{F}}{\text{L}}=\frac{\big[\text{MLT}^{-2}\big]}{\text{L}}=\big[\text{MT}^{-2}\big]$
Dimension of coefflcient of viscosity$: \frac{\text{FL}}{\text{A}\text{v}}=\frac{\big[\text{MLT}^{-2}[\text{L}]}{\big[\text{L}^2\big]\big[\text{LT}^{-1}\big]}=\big[\text{ML}^{-1}\text{T}^{-1}\big]$
Dimension of moment of force$: =\big[\text{MLT}^{-2}\big][\text{L}]=\big[\text{ML}^{2}\text{T}^{-2}\big]$
Dimension of surface tension$: \frac{\text{F}}{\text{L}}=\frac{\big[\text{MLT}^{-2}\big]}{\text{L}}=\big[\text{MT}^{-2}\big]$
Dimension of coefflcient of viscosity$: \frac{\text{FL}}{\text{A}\text{v}}=\frac{\big[\text{MLT}^{-2}[\text{L}]}{\big[\text{L}^2\big]\big[\text{LT}^{-1}\big]}=\big[\text{ML}^{-1}\text{T}^{-1}\big]$
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