KVPY 2020-21 SB SX Exam Previous Papers

Question : 95 of 120

Marks: +1, -0

A bottle has a thin nozzle on top. It is filled with water, held horizontally at a height of

1 {m}

and squeezed slowly by hands so that the water jet coming out of the nozzle hits the ground at a distance of

2 {m}

. If the area over which the hands squeeze it is

10 {cm}^{2}

, the force applied by hands is close to (take

{g}=10 {m} \/ {s}^{2}

and density of water

=1000 {\kg} \/ {m}^{3})

Solution:

Apply Bernoulli between point-1 and point-2.

{P}_{1}+{1}/{2} \ρ{V}_{1}^{2}={P}_{2}+{1}/{2} \ρ{V}_{2}^{2}

{P}_{1}={P}_{{atm}}+{\text "Force by hand "}/{\text " Area "}

{V}_{1}

tends to zero becomes area of point- 2 is very small.

{P}_{2}={P}_{\text "atm "}

{P}_{{\atm}}+{F} \/ {A}={P}_{{\atm}}+(1 \/ 2) \ρ{V}_{2}^{2} \

⇒{V}_{2}^{2}={2 {F}}/{{PA}}

......(i)
From Kinematics.

2=\√{{2({h})}/{{g}}} \×{V}_{2}

∴ {V}_{2}^{2}=20

....(ii)
Using (i) & (ii) we get

20={2 {F}}/{{pA}} \∴{F}=10 {N}

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