JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 14)
Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1
. The cross-sectional area of
the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be : (Take g = 10 ms–2)
5 × 10–4 m2
2 × 10–5 m2
5 × 10–5 m2
1 × 10–5 m2
Explanation
Using Bernoullie’s equation $${v_2} = \sqrt {v_1^2 + 2gh} $$
Equation of continuity
A1V1 = A2V2
(1 cm3)(1m/s) = $$\left( {{A_2}} \right)\left( {\sqrt {{{\left( 1 \right)}^2} + 2 \times 10 \times {{15} \over {100}}} } \right)$$
$$ \Rightarrow {A_2}\left( {\ln c{m^2}} \right) = {1 \over 2}$$
$$ \Rightarrow {A_2} = 5 \times {10^{ - 5}}{m^2}$$
Equation of continuity
A1V1 = A2V2
(1 cm3)(1m/s) = $$\left( {{A_2}} \right)\left( {\sqrt {{{\left( 1 \right)}^2} + 2 \times 10 \times {{15} \over {100}}} } \right)$$
$$ \Rightarrow {A_2}\left( {\ln c{m^2}} \right) = {1 \over 2}$$
$$ \Rightarrow {A_2} = 5 \times {10^{ - 5}}{m^2}$$
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