# Tutor profile: Krishna P.

## Questions

### Subject: GRE

12 accountants checked records in books at the rate of 60 books in 9 minutes. How many minutes will it take 27 accountants to check 180 books, if all the accountants worked at the same constant rate?

A general rate and work problem with multiple workers can be worked out using the simple formula, $$Work done (W) = Individual rate (R) * Number of workers (N) * Time taken (T)$$ Here, Work done, W = 60 books Number of workers, N = 12 Time taken, T = 9 minutes Let $$R$$ be the rate at which each worker works. So, $$W=RNT$$ gives $$60=R*12*9$$ $$R=5/9$$ books per minute per accountant Now we need to figure out the time taken by 27 accountants to check 180 books at the same rate. So $$180 = \frac{5}{9} * 27 *T$$ Solving, we get the time taken as 12 minutes

### Subject: Calculus

The intensity of illumination at a point $$x$$ units of length from a light source of candle power c is $$\frac{c}{x^{2}}$$. The two light sources, whose candle powers are $$k$$ and $$8k$$ are 60 units of length apart. Find the point between them where the intensity of illumination is a minimum. (Assume that intensity at any point is the sum of intensities due to the two sources).

Let $$I$$ denote the intensity of illumination due to the two sources. Let the point of minimum intensity be at a distance $$x$$ from the light source of candle power $$k$$ and let it be at a distance $$60-x$$ from the light source of candle power $$8k$$ (since it is given that the two light sources are $$60$$ units apart). Given that the intensity of illumination at a point $$x$$ units of length from a light source of candle power c is $$\frac{c}{x^{2}}$$, so $$I = \frac{k}{x^{2}} + \frac{8k}{{(60-x)}^{2}}$$ Differentiating both sides once with respect to $$x$$, $$\frac{dI}{dx} = \frac{-2k}{x^{3}} + \frac{16k}{{(60-x)}^{3}}$$..............(1) Differentiating both sides once more with respect to $$x$$, $$\frac{d^{2}I}{dx^{2}} = \frac{6k}{x^{4}} + \frac{48k}{{(60-x)}^{4}}$$............(2) When intensity is of light is minimum, $$\frac{dI}{dx} =0$$. Solving (1) with this condition, we get that $$x=20$$. When $$x=20$$, from (2) we get that $$\frac{d^{2}I}{dx^{2}} >0$$, thus proving that this is the point of minimum intensity of illumination.

### Subject: Physics

A cylinder of mass $$M$$ is suspended horizontally by means of two light, inextensible strings wrapped around it. Find the tension in the string.

Let M = the mass of the cylinder T = the tension in each string g = the acceleration due to gravity a = actual acceleration of the cylinder r = radius of the cylinder I = moment of inertia of the cylinder = $$ \frac{1}{2} Mr^{2}$$ (for all rigid cyinders) $$\alpha$$ = angular accelaration of the cylinder = $$a/r$$ (for all rigid bodies) Considering the free body diagram of the cylinder suspended by two strings, Forces pulling the cylinder u : $$2*T$$ (as there are two strings) Force pulling the cylinder down:$$ Mg$$ (due to its weight being pulled by gravity) So the equation of motion of any body is written by considering ALL the forces acting on the body and it is equal to the mass of the body (M) times its acceleration (a). Hence, $$Mg - 2T = Ma$$.........(1) Now considering rotational dynamics to write the equation of the body (again, using the free body diagram), The rotational force acting on the cylinder due to the tension in the strings:$$ 2*T*r$$ So, $$ 2Tr = I \alpha = \frac{1}{2} Mr^{2} * a/r $$ $$2Tr = \frac{1}{2} Mra$$ $$2T = \frac{1}{2} Ma$$ .............(2) Solving (1) and (2), $$ a =\frac{2}{3}g$$ $$T = \frac{mg}{6}$$

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