Question

The correct match is:

List - I	List - II
a) A body covers first half of distance with a speed V1​ and second half of distance with a speed V2​	e) Average velocity is  $\sqrt {\dfrac{{gh}}{2}} $
b) A body covers first half of the time with a speed V1​ and second half of the time with a speed V2​	f) Average speed is​ $\dfrac{{{v_1} + {v_2}}}{2}$
c) A body is projected vertically up from ground with certain velocity. Considering its total motion,	g) Average speed is $\dfrac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}$
d) A body freely is released from a height h	h) Average velocity is zero

$A.\,{\text{ }}a \to f{\text{ }},{\text{ }}b \to g{\text{ }},{\text{ }}c \to e{\text{ }},{\text{ }}d \to h$
$B.\,{\text{ }}a \to g{\text{ }},{\text{ }}b \to f{\text{ }},{\text{ }}c \to h{\text{ }},{\text{ }}d \to e$
$C.\,{\text{ }}a \to h{\text{ }},{\text{ }}b \to g{\text{ }},{\text{ }}c \to h{\text{ }},{\text{ }}d \to e$
$D.\,{\text{ }}a \to e{\text{ }},{\text{ }}b \to f{\text{ }},{\text{ }}c \to h{\text{ }},{\text{ }}d \to g$

Answer 1

Hint: In this question we are given the two lists named list 1 and list 2 and in list 1 we are given different scenario which includes body covering distance in different velocity and free fall while list two consist of average velocity which would come after solving each scenario in the list 1 and then we have to match them.

Formula used:
$\text{average speed} = \dfrac{\text{total distance}}{\text{total time taken}}$
This formula is used to find average speed
$\text{average velocity} = \dfrac{\text{total displacement}}{\text{total time taken}}$
This formula is used to find average velocity

Complete step by step answer:
Case a:A body covers the first half of distance with a speed $v_1$​​ and second half of distance with a speed $v_2$​. Let the total distance be $s$.So now body will cover $\dfrac{s}{2}$ each with the velocity of ${v_1}$ and ${v_2}$.Therefore,
$\text{average speed} = \dfrac{\text{total distance}}{\text{total time taken}}$
${v_{avg}} = \dfrac{s}{{\dfrac{{\dfrac{s}{2}}}{{{v_1}}} + \dfrac{{\dfrac{s}{2}}}{{{v_2}}}}}$
$ \Rightarrow \dfrac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}$
Hence (a) matches with (g).

Case b: A body covers first half of the time with a speed $v_1$​ and second half of the time with a speed $v_2$.
​\[\text{total distance covered} = {v_1}\dfrac{t}{2} + {v_2}\dfrac{t}{2}\]
Here total time is equal to t
\[{v_{avg}} = \dfrac{{ {v_1}\dfrac{t}{2} + {v_2}\dfrac{t}{2}}}{t}\]
\[ \Rightarrow \dfrac{{ {v_1} + {v_2}}}{2}\]
Hence (b) matches with (f).

Case c: A body is projected vertically up from ground with certain velocity. Considering its total motion
$\text{average velocity}= \dfrac{\text{total displacement}}{\text{total time taken}}$
As the ball comes back from where it is projected so the total displacement is zero therefore average velocity is equal to zero.
$\text{average velocity} = 0$
Hence (c) matches with (h).

Case d: If there is free fall, then acceleration due to gravity is $g$. Also the acceleration is rate of change of velocity
$g = \dfrac{{dv}}{{dt}}$
$ \Rightarrow \dfrac{d}{{dt}}(\dfrac{{dh}}{{dt}})$
$ \Rightarrow \dfrac{{{d^2}h}}{{d{t^2}}}$
On integrating we have
$ \Rightarrow \dfrac{{{d^2}h}}{{d{t^2}}} = h$
It is for zero initial condition
${v_{avg}} = \dfrac{h}{{\sqrt {\dfrac{{2h}}{g}} }}$
$ \Rightarrow \sqrt {\dfrac{{gh}}{2}} $
Hence (d) matches with (e).

Hence, the correct answer is option A.

Note: Many students make the mistake that gravitational constant should be taken negative or positive when taken downward but it only depends which you decide should be negative or positive. Another common mistake is between average velocity and average speed. Average velocity is total displacement upon time while average speed is total distance upon time.