Verified Solution • 2025–26 Session moderate

QJEEM11Ph2ef9: JEE Main 2011 Physics - Class 11 JEEM11Ph2ef9 Solution: Water is flowing continuously from a tap having an internal…

Water is flowing continuously from a tap having an internal diameter 8×10−3  m.8 \times {10^{ - 3}}\,\,m. The water velocity as it leaves the tap is 0.4  ms−10.4\,\,m{s^{ - 1}} . The diameter of the water stream at a distance 2×10−1  m2 \times {10^{ - 1}}\,\,m below the tap is close to :

Multiple Choice Options

  • A 7.5×10−3m7.5 \times {10^{ - 3}}m
  • B 9.6×10−3m9.6 \times {10^{ - 3}}m
  • C 3.6×10−3m3.6 \times {10^{ - 3}}m
  • D 5.0×10−3m5.0 \times {10^{ - 3}}m

Step-by-Step Verified Solution

From Bernoulli's theorem,
P0+12ρv12ρgh=P0+12ρv22+0{P_0} + {1 \over 2}\rho v_1^2\rho gh = {P_0} + {1 \over 2}\rho v_2^2 + 0
v2=v12+2gh{v_2} = \sqrt {v_1^2 + 2gh}
=0.16+2×10×0.2= \sqrt {0.16 + 2 \times 10 \times 0.2}
=2.03 m/s= 2.03\,m/s
From equation of continuity
A2v2=A1v1{A_2}{v_2} = {A_1}{v_1}
πD224×v2=πD124v1\pi {{D_2^2} \over 4} \times {v_2} = \pi {{D_1^2} \over 4}{v_1}
⇒  D1=D2v1v2=3.55×10−3m\Rightarrow \,\,{D_1} = {D_2}\sqrt {{{{v_1}} \over {{v_2}}}} = 3.55 \times {10^{ - 3}}m

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