Question

A syringe containing water is held horizontally with its nozzle at a height h=1.25 m above the ground as shown in the figure. The diameter of the piston is 5 times as that of the nozzle. The piston is pushed at a constant speed of 20 cms−1. If g=10 ms−2, then -

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Solution

The correct option is **D** The magnitude of velocity with which the water hits the ground is 5√2 ms−1.

Area of piston,

A=πD24 ,D→Diameter of piston

Area of nozzle,

a=πd24 ,d→Diameter of nozzle

Using equation of continuity,

AV=au

⇒u=AVa=D2d2×V=52×0.2=5 m/s

Option (A) is correct.

Time of flight,

T=√2hg=√2×1.2510=0.5 s

Option (B) is correct.

Range,

R=uT=5×0.5=2.5 m

Option (C) is correct.

Horizontal velocity,

vx=5 ms−1 [Remain unchanged]

Vertical velocity at t=0.5 s,

vy=0+ayt

⇒vy=gT=10×0.5=5 ms−1

Further, net speed with which water hits the ground,

v=√52+52=5√2 m/s

Option (D) is correct.

Area of piston,

A=πD24 ,D→Diameter of piston

Area of nozzle,

a=πd24 ,d→Diameter of nozzle

Using equation of continuity,

AV=au

⇒u=AVa=D2d2×V=52×0.2=5 m/s

Option (A) is correct.

Time of flight,

T=√2hg=√2×1.2510=0.5 s

Option (B) is correct.

Range,

R=uT=5×0.5=2.5 m

Option (C) is correct.

Horizontal velocity,

vx=5 ms−1 [Remain unchanged]

Vertical velocity at t=0.5 s,

vy=0+ayt

⇒vy=gT=10×0.5=5 ms−1

Further, net speed with which water hits the ground,

v=√52+52=5√2 m/s

Option (D) is correct.

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