Question

*Answer the question which is given in the attachment*

Answer 1

The equation given is:

$\gamma = P \log (Qx+Rt)$


Here,
$\gamma = Displacement \, \, \implies [\, \gamma \, ] = [ \, L \, ] \\ \\ x = Position \, \, \implies [ \, x \, ] = [ \, L \, ] \\ \\ t = Time \, \, \implies [ \, t \, ] = [ \, T \, ]$


Now, we know that $\log$ is a dimensionless quantity. Anything that is inside $\log$ must be dimensionless. 

So, $(Qx+Rt)$ is a dimensionless quantity. There are two terms added to give a dimensionless quantity. This means that both $Qx$ and $Rt$ must be individually dimensionless too.

We have:

$[ \, Qx \, ] = [ \, M^0 \, L^0 \, T^0 \, ] \\ \\ \\ \implies [ \, Q \, ] \, \, [ \, L \, ] = [ \, M^0 \, L^0 \, T^0 \, ] \\ \\ \\ \implies [ \, Q \, ] = [ \, L^{-1} \, ]$

Similarly


$[ \, Rt\, ] = [ \, M^0 \, L^0 \, T^0 \, ] \\ \\ \\ \implies [ \, R \, ] \, \, [ \, T \, ] = [ \, M^0 \, L^0 \, T^0 \, ] \\ \\ \\ \implies [ \, R \, ] = [ \, T^{-1} \, ]$

So, finally we have:


$[ \, \gamma \, ] = [ \, L \, ]$

$[ \, P \, ] = [ \, L \, ]$

$[ \, Q \, ] = [ \, L^{-1} \, ]$

$[\, R \, ] = [ \, T^{-1} \, ]$




Now, let us check the options.



1)


$\gamma R \\ \\ \\ \implies [ \, \gamma R \, ] = [ \, L \, ] \, \, [ \, T^{-1} \, ] \\ \\ \\ \implies \boxed{[ \, \gamma R \, ] = [ \, L \, T^{-1} \, ]}$


2)


$PR \\ \\ \\ \implies [ \, P R \, ] = [ \, L \, ] \, \, [ \, T^{-1} \, ] \\ \\ \\ \implies \boxed{[ \, P R \, ] = [ \, L \, T^{-1} \, ]}$


3) 

$\frac{R}{Q} \\ \\ \\ \implies \left[ \, \frac{R}{Q} \, \right] = \frac{[ \, T^{-1} \, ]}{[ \, L^{-1} \, ]} \\ \\ \\ \implies\boxed{\left[ \, \frac{R}{Q} \, \right] = [ \, L \, T^{-1} \, ]}$


4)


$Q R \\ \\ \\ \implies [ \, Q R \, ] = [ \, L^{-1} \, ] \, \, [ \, T^{-1} \, ] \\ \\ \\ \implies \boxed{[ \, Q R \, ] = [ \, L^{-1} \, T^{-1} \, ]}$


Thus, clearly, Option (4) is the odd one out. Its dimensions are different from others. So, our answer is Option (4).