If $A$ and $B$ are events of a random experiment such that $P(A \cup B) = \frac{4}{5}$,$P(\bar{A} \cup \bar{B}) = \frac{7}{10}$ and $P(B) = \frac{2}{5}$,then $P(A)$ equals

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

$A$ telescope has an objective lens of focal length $150 \ cm$ and an eyepiece of focal length $5 \ cm$. If a $50 \ m$ tall tower at a distance of $1 \ km$ is observed through this telescope in normal setting,the angle formed by the image of the tower is $\theta$,then $\theta$ is close to..... (in $^{\circ}$)

The equivalent resistance between $A$ and $B$ is

If the harmonic mean of the roots of the equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ is $4$,then the value of $b$ is:

If $arg(z) < 0$,then $arg(-z) - arg(z)$ is equal to