Match List – I with List – II :

List – I                                                                               List – II

(a) h (Planck’s constant)                                               (i) $\left[ML{T}^{-1}\right]$  

(b) E (kinetic energy)                                                     (ii) $\left[M{L}^2{T}^{-1}\right]$        

(c) V (electric potential)                                                (iii) $\left[M{L}^2{T}^{-2}\right]$  

(d) P (linear momentum)                                             (iv) $\left[M{L}^2{I}^{-1}{T}^{-3}\right]$  

Choose the correct answer from the options given below:

Please find the solution below:

According to question, we can write

 10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$                        

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$  

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$  

Average speed $=\left(\frac{4{\mathrm{v}}^2}{3\mathrm{v}}\right)$ $\left(\frac{{}^{}}{}\right)$

$=\frac{4\mathrm{v}}{3}$

(d) Initial velocity  $=-v\hat{j}$ $\stackrel{}{}$

Final velocity $=v\hat{i}$ $\stackrel{}{}$

Change in velocity  $=v\hat{i}-(-v\hat{j})$

$\stackrel{}{}\stackrel{}{}$$\stackrel{}{}\stackrel{}{}$ $=v(\hat{i}+\hat{j})$ Momentum gain is along $\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}$ $\stackrel{}{}\stackrel{}{}$

$$ $\Rightarrow$  Force experienced is along $\hat{i}+\hat{j}$ $\stackrel{}{}\stackrel{}{}$

$$  $\Rightarrow$ Force experienced is in North-East direction.

Air resistance resists the motion of an object. In this case, the net acceleration is lesser than 'g' and it shrinks as the speed increases. This makes the object to speed up more slowly. Ultimately, it reaches a constant terminal velocity which is lower for large-area ones and higher for heavy and streamlined ones.