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Chapter 7: Problem 994

A big drop of radius \(R\) is formed by 1000 small droplets of coater then the radius of small drop is (A) \((\mathrm{R} / 2)\) (B) \((\mathrm{R} / 5)\) (C) \((\mathrm{R} / 6)\) (D) \((\mathrm{R} / 10)\)

Short Answer

Expert verified
The radius of the small droplet is \(\frac{R}{10}\). So, the correct option is (D).

Step by step solution

01

Write the volume relationship for the big drop and small droplets

Since the big drop is formed by combining 1000 small droplets, the total volume of the small droplets equals the volume of the big drop. The volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\), where V is the volume and r is the radius. Let the radius of the small droplet be r. Then we have: Volume of big drop = 1000 * Volume of small droplet \(\frac{4}{3}\pi R^3 = 1000 * \frac{4}{3} \pi r^3\)
02

Solve for the radius of the small droplet

We can now solve for the radius of the small droplet by dividing both sides of the equation by \(\frac{4}{3}\pi\). \(R^3 = 1000 * r^3\) Now take the cube root of both sides. \(R = 10 * r\) Solve for r. \(r = \frac{R}{10}\) The radius of the small droplet is \(\frac{R}{10}\). So, the correct option is (D).

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Most popular questions from this chapter

A thin liquid film formed between a u-shaped wire and a light slider supports a weight of \(1.5 \times 10^{-2} \mathrm{~N}\) (see figure). The length of the slider is \(30 \mathrm{~cm}\) and its weight negligible. The surface tension of the liquid film is. (A) \(0.0125 \mathrm{Nm}^{-1}\) (B) \(0.1 \mathrm{Nm}^{-1}\) (C) \(0.05 \mathrm{Nm}^{-1}\) (D) \(0.025 \mathrm{Nm}^{-1}\)

The compressibility of water \(4 \times 10^{-5}\) per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be.......... (A) \(4 \times 10^{-5} \mathrm{CC}\) (B) \(4 \times 10^{-5} \mathrm{CC}\) (C) \(0.025 \mathrm{CC}\) (D) \(0.004 \mathrm{CC}\)

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