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Question

Water is filled upto height H units in a tank placed on the ground and whose side walls are vertical. A hole is made in one of the vertical wall such that the emerging stream of water strikes the ground at the maximum range. If the level in the tank is changing at the rate of R units per second, at that instant the rate at which range will be changing will be:

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Solution

The correct option is **D** Zero

According to to
Torricelli's theorem, velocity of efflux at point A ( shown in
the figure ) is

vA=√(2gh)

After emerging from the
orifice the water adopts parabolic path. if it takes time t secs in
falling a vertical height (H=h) and covers a horizontal range
R1, then

H−h=12gt2

⇒t=√2(H−h)g

R1=vAt=√(2gh)∗√2(H−h)g=2√h(H−h) ....(1) where
R1 is the range.

The above results show that Range is same
whether the hole is made at distance h from the top or at (H-h) from
the top.

∴h=H−h

⇒h=12H ....(2)

If h is changing at
the rate dhdt=R

differentiating eqn (1) wrt t, we
get

dR1dt=(h(H−h))−12(dhdt(H−h)+h(−dhdt))=(h(H−h))−12(R(H−h)−Rh)
....(3)

Substituting h=H2 from eqn(2) in eqn(3) we
get

dR1dt=(H2)−1R(H−2h)=(H2)−1R(H−H)=0

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