IIT-JEE 2010 Paper 2 Offline | Properties of Matter Question 35 | Physics

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

-0.75

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 the force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

If the radius of the opening of the dropper is $$r$$, the vertical force due to the surface tension on the drop of radius R (assuming $$r$$ << R) is

$$2\pi rT$$

$$2\pi RT$$

$${{2\pi {r^2}T} \over R}$$

$${{2\pi {R^2}T} \over r}$$

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

-0.75

If r = 5 $$ \times $$ 10⁻⁴ m, $$\rho $$ = 10³ kg m⁻³ , g = 10 m/s² , T = 0.11 Nm⁻¹ , the radius of the drop when it detaches from the dropper is approximately

1.4 $$ \times $$ 10⁻³ m

3.3 $$ \times $$ 10⁻³ m

2.0 $$ \times $$ 10⁻³ m

4.1 $$ \times $$ 10⁻³ m

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

-0.75

After the drop detaches, its surface energy is

1.4 $$ \times $$ 10⁻⁶ J

2.7 $$ \times $$ 10⁻⁶ J

5.4 $$ \times $$ 10⁻⁶ J

8.1 $$ \times $$ 10⁻⁶ J

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

A glass tube of uniform internal radius (r) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,

IIT-JEE 2008 Paper 2 Offline Physics - Properties of Matter Question 4 English