MCQ
If $x \propto {t^{5/2}}$ , then
- A$v \propto {t^{3/2}}$
- B$a \propto \sqrt t $
- ✓both$ (A)$ and $(B)$
- D$v \propto \sqrt t $
$\therefore$ Velocity of the particle $v =\frac{ dx }{ dt }=2 Kt$
$\therefore$ Acceleration of the particle $a =\frac{ dv }{ dt }=2 K = constant$
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| Column $-I$ $R/H_{max}$ |
Column $-II$ Angle of projection $\theta $ |
| $A.$ $1$ | $1.$ ${60^o}$ |
| $B.$ $4$ | $2.$ ${30^o}$ |
| $C.$ $4\sqrt 3$ | $3.$ ${45^o}$ |
| $D.$ $\frac {4}{\sqrt 3}$ | $4.$ $tan^{-1}\,4\,=\,{76^o}$ |
