Question

Let us assume that the acceleration due to gravity be $10m/s^2$. If acceleration due to gravity ($g$), the velocity ($v$) acquired by a body after falling from rest for $5$ seconds and the momentum ($p$) acquired by a mass of $1$ kg in falling freely from rest for $10$ seconds are taken as the fundamental units, then the value of the unit of time in terms of second?

• A. 1 s
• B. 2 s
• C. 5 s
• D. 10 s

Answer 1

The correct option is C 5 s

$g=10\frac{m}{s^2}$
$v=u+gt=5g\frac{m}{s}$
$p=momentumfor1kg=m{v}_1=m(0+10g)=10mg$

$t=g^{\alpha }{v}^{\beta }{p}^{\gamma }$
$M^0{L}^0{T}^1={\left(L{T}^{-2}\right)}^{\alpha }{\left(L{T}^{-1}\right)}^{\beta }{\left(M^1{L}^1{T}^{-1}\right)}^{\gamma }$
$0=\gamma ,0=\alpha +\beta +\gamma \Rightarrow \alpha +\beta =0$
$1=-2\alpha -\beta -\gamma$

$-2\alpha -\beta =1$
$\alpha =-1$
$\beta =1$
$t=g^{-1}{v}^1$

$=\frac{5\times 10\frac{m}{s}}{10\frac{m}{s^2}}=5s$