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

$A$ $2 \ m$ long rod of radius $1 \ cm$ which is fixed from one end is given a twist of $0.8 \ radians$. The shear strain developed will be

Let $P = \{\theta : \sin \theta - \cos \theta = \sqrt{2} \cos \theta\}$ and $Q = \{\theta : \sin \theta + \cos \theta = \sqrt{2} \sin \theta\}$ be two sets,then:

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If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin to a plane, then the equation of that plane is

If $y=A \cos n x+B \sin n x$, then $y_2+n^2 y$ is equal to