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Question: If \(\eta \) represents the coefficient of viscosity and \(T\) the surface tension, then the dimensi...

If η\eta represents the coefficient of viscosity and TT the surface tension, then the dimension of Tη\dfrac{T}{\eta } is the same as that of:
A). Length
B). Mass
C). Velocity
D). Speed

Explanation

Solution

Find out the dimensional formula for given expression. Write down the dimensional formula for all the options given in the question. Compare the dimensional formula of the given expression with that of the dimensional formula for every given option to get the answer.

Complete step-by-step solution:
T is the surface tension. Surface tension is nothing but the force per unit length. Hence, we can write the dimensional formula for the surface tension as:
T=forcelengthT=\dfrac{force}{length} ----- ---(i)
Also, force=mass×accelerationforce=mass\times acceleration
We know, [mass]=[M][mass]=[M] and [acceleration]=[LT−2][acceleration]=[L{{T}^{-2}}]
Hence,
[force]=[MLT−2][force]=[ML{{T}^{-2}}]
Using equation (i)
[T]=[MLT−2][L][T]=\dfrac{[ML{{T}^{-2}}]}{[L]}
⇒[T]=[MT−2]\Rightarrow [T]=[M{{T}^{-2}}] ---------(ii)
Now, for η\eta :
η\eta is the coefficient of viscosity. We know that the relation between the viscous force and the coefficient of viscosity is given by:
F=ηAΔvΔlF=\eta A\dfrac{\Delta v}{\Delta l} ---------(iii)
Here,
The dimension of change in velocity is given by: [Δv]=[LT−1][\Delta v]=[L{{T}^{-1}}]
Also, the dimension of force is given by: [F]=[MLT−2][F]=[ML{{T}^{-2}}]
Using these dimensional formula in equation (iii), we get:
[MLT−2]=[η][L2][LT−1][L][ML{{T}^{-2}}]=[\eta ][{{L}^{2}}]\dfrac{[L{{T}^{-1}}]}{[L]}
⇒[η]=[MLT−2][L][LT−1][L2]\Rightarrow [\eta ]=\dfrac{[ML{{T}^{-2}}][L]}{[L{{T}^{-1}}][{{L}^{2}}]}
⇒[η]=[ML−1T−1]\Rightarrow [\eta ]=[M{{L}^{-1}}{{T}^{-1}}] --------(iv)
Now, the dimensional formula for the expression Tη\dfrac{T}{\eta } is given by:
[T][η]=[MT−2][ML−1T−1]\dfrac{[T]}{[\eta ]}=\dfrac{[M{{T}^{-2}}]}{[M{{L}^{-1}}{{T}^{-1}}]}
⇒Tη=[LT−1]\Rightarrow \dfrac{T}{\eta }=[L{{T}^{-1}}] ---------(v)
Now, the dimensional formula of the parameters mentioned in the option of the question are as follows:
[length]=[L][length]=[L]
[mass]=[M][mass]=[M]
[velocity]=[LT−1][velocity]=[L{{T}^{-1}}]
[speed]=[LT−1][speed]=[L{{T}^{-1}}]
On comparing equation (v) with the dimensional formula of all the options, we can conclude that the dimensional formula of the expression Tη\dfrac{T}{\eta } is similar to that of the dimensional formula for velocity as well as speed.
Hence, the correct options are (c) and (d).

Note: Do not skip the writing of the dimensional formula of the known parameters such as force and velocity. Writing down these formulas on paper reduces the chances of minor errors, which can result in wrong answers. Always cross check the dimensional formula obtained by comparing it with general definition as done in equation (i).