The orbital angular momentum of a satellite revolving at a distance $r$ from the centre is $L$. If the distance is increased to $4r$,then the new angular momentum will be:

$A$ microscope consists of an objective of focal length $1.9 \ cm$ and an eyepiece of focal length $5 \ cm$. The two lenses are kept at a distance of $10.5 \ cm$. If the image is to be formed at the least distance of distinct vision $(25 \ cm)$,the distance at which the object is to be placed before the objective is: (in $cm$)

Two workers $A$ and $B$ are engaged to do a piece of work. Working alone,$A$ takes $8 \text{ h}$ more to complete the work than if both worked together. On the other hand,working alone,$B$ would need $4 \frac{1}{2} \text{ h}$ more to complete the work than if both worked together. How much time would they take to complete the job working together (in $text{ h}$)?

In the adjoining circuit diagram,the readings of the ammeter and voltmeter are $2 \ A$ and $120 \ V$,respectively. If the value of $R$ is $75 \ \Omega$,then the voltmeter resistance will be $\Omega$.

Two identical balls $A$ and $B$ are released from the positions shown in the figure. They collide elastically on the horizontal portion $MN$. The ratio of the heights attained by $A$ and $B$ after the collision will be (neglect friction):

For the gaseous reactions $(I)$ and $(II)$,the equilibrium constants are $X$ and $Y$ respectively.
$I$. $\frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$
$II$. $2 NO_{2(g)} \rightleftharpoons N_2 O_{4(g)}$
Using the above reactions,the equilibrium constant $Z$ for the reaction $(III)$ given below is:
$III$. $N_2 O_{4(g)} \rightleftharpoons N_{2(g)} + 2 O_{2(g)}$