Question

Some parameters are given in Column - I which are to be matched with their correct answers given in Column - II.
Choose the correct option.
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{a)}& \text{Velocity at}t=0\text{of a particle following}& \text{p)}& \alpha \\ & \text{equation}x=u(t-8)+\alpha (t-2{)}^2& & \\ \text{b)}& \text{Acceleration of a particle moving according to equation}& \text{q)}& 2\alpha \\ & x=4(t-5)+\alpha (t-2{)}^2\text{at}t=0& & \\ \text{c)}& \text{Velocity of the particle moving according to equation}& \text{r)}& u\\ & x=ut+5\alpha t^{2}ln(1+\frac{t}{2})+\alpha t^2\text{at}t=0& & \\ \text{d)}& \text{Acceleration of the particle moving according to the equation}& \text{s)}& u-4\alpha \\ & x=ut+5\alpha t^{2}ln(1+\frac{t}{2})+\alpha t^2\text{at}t=0& & \end{array}$

• A. $a\to sb\to qc\to rd\to q$
• B. $a\to pb\to rc\to qd\to q$
• C. $a\to pb\to rc\to sd\to q$
• D. $a\to pb\to qc\to rd\to s$

Answer 1

The correct option is A $a\to sb\to qc\to rd\to q$

$a\to sb\to qc\to rd\to q$
a) $x=u(t-8)+\alpha (t-2{)}^2$
$v=u+2\alpha (t-2)\times 1$
$\Rightarrow v=u+2\alpha (t-2)$
$\Rightarrow \text{at}t=0$,
$v=u-4\alpha$

b) $x=4(t-5)+\alpha (t-2{)}^2$
$v=4+2\alpha (t-2)$
$\Rightarrow a=2\alpha$
at $t=0,a=2\alpha$

c) $x=ut+5\alpha t^{2}ln(1+\frac{t}{2})+\alpha t^2$
$\Rightarrow v=u+10\alpha t\left(ln(1+\frac{t}{2})\right)+\frac{5\alpha t^2.\frac{1}{2}}{1+\frac{t}{2}}+2\alpha t$
at $t=0$,
$v=u+0+0+0=u$

d) $x=ut+5\alpha t^{2}ln(1+\frac{t}{2})+\alpha t^2$
$\Rightarrow v=u+10\alpha tln(1+\frac{t}{2})+\frac{5\alpha t^2}{2+t}+2\alpha t$
$a=10\alpha ln(1+\frac{t}{2})+10\alpha t.\frac{\frac{1}{2}}{1+\frac{t}{2}}+\left[\frac{(2+t)\left(10\alpha t\right)-5\alpha t^2}{(2+t{)}^2}\right]+2\alpha$
at $t=0$,
$a=0+0+0+2\alpha =2\alpha$