JEE Main 31-Aug-2021 Shift 2 Solved Paper

When the velocity (v), time (T) and force (F) are chosen as base quantities. Then, mass is given by
m∝vxTyFz ...(i)
Using dimensional formula of all quantities,
[ML0T0]=[LT−1]x[T]y[MLT−2]z
[M1L0T0]=[MZLx+zT−x+y−2z]
Comparing the powers of dimensions on both sides, we get
z = 1
x+z=0 and −x+y−2z=0
⇒x+1=0
⇒x=−1
and −(−1)+y−2(1)=0
⇒1+y−2=0
⇒y=1
Substituting these values in Eq. (i), we get
m∝v−1T1F1
⇒m=[FTv−1]