Verified Solution • 2025–26 Session moderate

QJEEM21Phee6c: JEE Main 2021 Physics - Class 11 JEEM21Phee6c Solution: Two narrow bores of diameter 5.0 mm and 8.0 mm are joined…

Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube. [Take surface tension of water T = 7.3 ×\times 10 −- 2 Nm −- 1, angle of contact = 0, g = 10 ms2 and density of water = 1.0 ×\times 103 kg m −- 3]

Multiple Choice Options

  • A 3.62 mm
  • B 2.19 mm
  • C 5.34 mm
  • D 4.97 mm

Step-by-Step Verified Solution

Image

We have, PA = PB. [Points A & B at same horizontal level]

∴\therefore Patm−2Tr1+ρg(x+Δh)=Patm−2Tr2+ρgx{P_{atm}} - {{2T} \over {{r_1}}} + \rho g(x + \Delta h) = {P_{atm}} - {{2T} \over {{r_2}}} + \rho gx

∴\therefore ρgΔh=2T[1r1−1r2]\rho g\Delta h = 2T\left[ {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right]

=2×7.3×10−2[12.5×10−3−14×10−3]= 2 \times 7.3 \times {10^{ - 2}}\left[ {{1 \over {2.5 \times {{10}^{ - 3}}}} - {1 \over {4 \times {{10}^{ - 3}}}}} \right]

∴\therefore Δh=2×7.3×10−2×103103×10[12.5−14]\Delta h = {{2 \times 7.3 \times {{10}^{ - 2}} \times {{10}^3}} \over {{{10}^3} \times 10}}\left[ {{1 \over {2.5}} - {1 \over 4}} \right]

= 2.19 ×\times 10 −- 3 m = 2.19 mm

Hence, option (b).

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring