JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 5)
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]
3.62 mm
2.19 mm
5.34 mm
4.97 mm
Explanation
Image
We have, PA = PB. [Points A & B at same horizontal level]
$$\therefore$$ $${P_{atm}} - {{2T} \over {{r_1}}} + \rho g(x + \Delta h) = {P_{atm}} - {{2T} \over {{r_2}}} + \rho gx$$
$$\therefore$$ $$\rho g\Delta h = 2T\left[ {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right]$$
$$ = 2 \times 7.3 \times {10^{ - 2}}\left[ {{1 \over {2.5 \times {{10}^{ - 3}}}} - {1 \over {4 \times {{10}^{ - 3}}}}} \right]$$
$$\therefore$$ $$\Delta h = {{2 \times 7.3 \times {{10}^{ - 2}} \times {{10}^3}} \over {{{10}^3} \times 10}}\left[ {{1 \over {2.5}} - {1 \over 4}} \right]$$
= 2.19 $$\times$$ 10$$-$$3 m = 2.19 mm
Hence, option (b).
We have, PA = PB. [Points A & B at same horizontal level]
$$\therefore$$ $${P_{atm}} - {{2T} \over {{r_1}}} + \rho g(x + \Delta h) = {P_{atm}} - {{2T} \over {{r_2}}} + \rho gx$$
$$\therefore$$ $$\rho g\Delta h = 2T\left[ {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right]$$
$$ = 2 \times 7.3 \times {10^{ - 2}}\left[ {{1 \over {2.5 \times {{10}^{ - 3}}}} - {1 \over {4 \times {{10}^{ - 3}}}}} \right]$$
$$\therefore$$ $$\Delta h = {{2 \times 7.3 \times {{10}^{ - 2}} \times {{10}^3}} \over {{{10}^3} \times 10}}\left[ {{1 \over {2.5}} - {1 \over 4}} \right]$$
= 2.19 $$\times$$ 10$$-$$3 m = 2.19 mm
Hence, option (b).
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