IIT-JEE 2010 · Paper 2 · Physics Q52

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PhysicsSingle correct+3 / -1Question 52 of 57

Passage

Paragraph for Questions 50 to 52

When liquid medicine of density $\rho$ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension $T$ when the radius of the drop is $R$ . When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

After the drop detaches, its surface energy is

1.4\times10^{-6}

2.7\times10^{-6}

5.4\times10^{-6}

8.1\times10^{-6}

Answer and worked solution

Official answer: B

Surface energy of the free drop (one surface):
$E=T\cdot4\pi R^2=0.11\times4\pi\times(1.4\times10^{-3})^2\approx2.7\times10^{-6}\text{ J}.$

(B)

Topics:Surface tension

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IIT-JEE 2010 · Paper 2, Physics question 52