Here in this question , we need to compare the dimension of the Left hand side and right hand side
LHS
$v= [LT^{-1}]$
RHS
${{v}_{0}}+ at = [LT^{-1}] + [LT^{-2}][T]=[LT^{-1}] + [LT^{-1}] =[LT^{-1}] $
So, the dimensions on both the sides are same
So , we can say that expression $v={{v}_{0}}+ at$ is dimensionally correct
a. The expression is $\frac {1}{2}mv^2 = \frac {1}{2}mv_0^2 + \sqrt {mgh}$
LHS
$\frac {1}{2}mv^2 = M(LT^{-1})^2= ML^2T^{-2}$
RHS
$\frac {1}{2}mv_0^2 + \sqrt {mgh}=M(LT^{-1})^2 + [ML^2T^{-2}]^{1/2} = ML^2T^{-2} + M^{1/2}LT^{-1}$
Now this cannnot be correct as we cannot add quantities of two different dimension.
So this equation is incorrect
b.The expression is $v= v_0 + at^2$
LHS
$v= [LT^{-1}]$
RHS
$v_0 + at^2 = [LT^{-1}] + [LT^{-2}]T^2= [LT^{-1}] + [L]$
Now this cannnot be correct as we cannot add quantities of two different dimension.
So this equation is incorrect
c.The expression is $ma= v^2$
LHS
$ma= [M][LT^{-2}]=[MLT^{-2}]$
RHS
$v^2 = [L^2T^{-2}]$
Now dimension of LHS is not equal to dimension of RHS
So this equation is incorrect
Here $x=B{{t}^{2}}$
or
$B= \frac {x}{t^2}$
Therefore dimension of B will be
=$\frac {[L]}{[T^{2}]}= [LT^{-2}]$
SI unit of pressure is pascal
a. This is possible . Example Work and Torque have different unit Joule and Newton-meter but the dimension $[ML^2T^{-1}]$ is same
b. No it is not possible.
We can write
$a \; \alpha \; v^m r^n$
$a= Kv^m r^n$
Where K is the dimensionless constant
Now writing dimension of various quantities in the equation
$[LT^{-2}]=[LT^{-1}]^m [L]^n$
$[LT^{-2}]=[L^{m+n} T^{-m}]$
Comparing the dimensions of similar quantities on both the sides, we get
$m+n=1$
$-m=-2$ or $m=2$
Therefore 2+n=1 or n=-1
Hence the values of m and n are 2 and -1 respectively
Now $\rho = \frac {M}{V}$
Here M= 856 g = .856 kg
$V= (5.35)^3 cm^3= (5.35)^3 (1/100)^3 m^3$
$\rho = \frac {.856}{ (5.35)^3 (1/100)^3} kg/m^3 = 5590 kg/m^3$
\[T=2\pi \sqrt{\frac{l}{g}}\]
LHS
$T = [T]$
RHS
$2\pi \sqrt{\frac{l}{g}} = (\frac {[L]}{[LT^{-2}]})^{1/2} = ([T^{2}])^{1/2} =[T]$
So, this equation is dimensionally consistent
a. $\frac{{{v}^{2}}}{xa}$
$= \frac {(m/s)^2}{m m/s^2}= \frac {(m/s)^2}{(m/s)^2}$ =dimensionless
b.
$\sqrt {\frac {m}{m/s^2}}= \sqrt {s^2} =s$
c.Here constant 1/2 does not have any units
$(m/s^2) s^2 = m$