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Match the items in Column-I with those in Column-II and select the correct option.
Column-I Column-II
a) A point on the parabola $y^{2} = 18 x$ at which the ordinate increases at twice the rate of the abscissa is p) $\frac{l ν}{O P^{2}}$
b) From the top of 64 m high tower, a stone is thrown upwards vertically with the velocity of 48 m/s.
The greatest height (in metre) attained by the stone,assuming the value of gravitational acceleration g = 32 m/s² is
q) (0,2)
c) If the rate of change of x is more than the rate of change of y on the curve $x^{3} = 12 y, (x > 0)$ then x lies
in the interval
r) 100
d) A point P is moving with uniform velocity $ν$ along a straight line AB..O is a point on a perpendicular to AB
at A and a distance $l$ from it. The angular velocity of P about O is
s) $(\frac{9}{8}, \frac{9}{2})$
t) (2,-4)

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$a \to q, b \to r, c \to p, d \to s$

$a \to r, b \to q, c \to s, d \to p$

$a \to s, b \to r, c \to q, d \to p$

$a \to p, b \to s, c \to q, d \to r, s$

answer is C.

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Detailed Solution

a) $y^{2} = 18 x \Rightarrow 2 y \frac{d y}{d t} = 18 \frac{d x}{d t} \Rightarrow 2 y (2. \frac{d x}{d t}) = 18. \frac{d x}{d t} \Rightarrow y = \frac{9}{2}, x = \frac{9}{8}$

B) maximum height, v = 0, using

$\begin{array}{l} v^{2} = u^{2} - 2 g h \Rightarrow 0 = 48^{2} - 2 (32) (h) \\ h = 36 m \end{array}$

maximum height = 36 + 64 = 100m

c) $x^{3} = 12 y, 3 x^{2} \frac{d x}{d t} = 12. \frac{d y}{d t} \Rightarrow \frac{d x / d t}{d y / d t} = \frac{4}{x^{2}}$

$\frac{d x}{d t} > \frac{d y}{d t} \Rightarrow \frac{d x / d t}{d y / d t} > 1 \Rightarrow \frac{4}{x^{2}} > 1 \Rightarrow 4 > x^{2} \Rightarrow x^{2} - 4 < 0 \Rightarrow (x + 2) (x - 2) < 0$

$\Rightarrow - 2 < x < 2 \Rightarrow 0 < x < 2 \Rightarrow x \in (0, 2)$

d) Question Image

$\cos θ = \frac{l}{o p}, \tan θ = \frac{x}{l}$

$\sec^{2} θ \frac{d θ}{d t} = \frac{1}{l} \frac{d x}{d t} = \frac{υ}{l} \Rightarrow \frac{d θ}{d t} = \frac{υ}{l} \cos^{2} θ = \frac{υ}{l} {(\frac{l}{o p})}^{2} = \frac{l υ}{{(o p)}^{2}}$ .

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