An open hemispherical storage tank has radius 13 m . Oil flows into the tank such that the depth ' $\boldsymbol{h}$ ' of oil in the tank changes at the rate of $3 \mathrm{~m} / \mathrm{hr}$. When the depth $\boldsymbol{h}=1 \mathrm{~m}$, the rate of change of the area of the top surface of the oil is
$72 \pi \mathrm{~m}^2 / \mathrm{hr}$
$75 \pi \mathrm{~m}^2 / \mathrm{hr}$
$24 \pi \mathrm{~m}^2 / \mathrm{hr}$
$26 \pi \mathrm{~m}^2 / \mathrm{hr}$
$$ \text { The second derivative of } \sin 3 \boldsymbol{x} \boldsymbol{\operatorname { c o s }} \mathbf{5 x} \text { is: } $$
$2 \sin 2 x+32 \sin 8 x$
$2 \sin 2 x+16 \sin 8 x$
$2 \sin 2 x-16 \sin 8 x$
$2 \sin 2 x-32 \sin 8 x$
If $\mathop {\lim }\limits_{x \to 0}\left(\frac{p \sin 2 x+1-\cos 2 x}{x+\tan x}\right)=1$ then the value of ' $p$ ' is
$2$
$-1$
$1$
$\frac{1}{2}$
Let the line $L_1$ be a line passing through the point $(\mathbf{0},-\mathbf{6})$ and making an angle of $\mathbf{1 5 0}^{\circ}$ with the positive $x$-axis. Then the equation of a line $L_2$ parallel to $L_1$ and crossing the $y$-axis 2 units below the origin is:
$$ x \sqrt{3}+y+6=0 $$
$$ x-\sqrt{3} y+6 \sqrt{3}=0 $$
$$ x-\sqrt{3} y-2 \sqrt{3}=0 $$
$$ x+\sqrt{3} y+2 \sqrt{3}=0 $$
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