# Tutor profile: Kuldeep P.

## Questions

### Subject: SAT II Mathematics Level 2

A right circular cylinder has radius 3 and height 8. If A and B are two points on its surface, what is the maximum possible straight-line distance between A and B?

For the two points to be as far apart as possible, they should be present on the opposite lids (circular faces of the given cylinder). More over, they should be diametrically opposite ends of the circular faces. These two positions are related by the following horizontal and vertical distance: Horizontal distance between the two points: diameter of the circular regions: = 2*radius= 2*3=6 Vertical distance= height of the cylinder= 8. Since this forms a right angled triangle, the total distance between the two points is the hypotenuse of this triangle. The length of hypotenuse is given by Pythagoras theorem: h=sqrt(base^2+perpendicular^2) =sqrt(6^2+8^2) =sqrt(36+64)=sqrt(100)=10. So, the maximum possible straight line distance between any two points A and B on this surface is 10 units.

### Subject: Electrical Engineering

A 2H inductor is connected in series with a 4Ω resistor to a 100V constant D.C. supply. (a). What is the time constant of the circuit? (b). Determine the initial and final current flowing in the circuit.

The circuit is a single loop containing a resistor, an inductor and a DC voltage source. V=100. R=4 L=2 Writing Kirchhoff's Voltage Law equation for this loop: V= i*R+L*di/dt where i is the current flowing in the circuit. Since the circuit contains the inductive element, the current will vary with time (i is a function of time). Substituting the known values: 100=4i+2di/dt. This differential equation is solved in the following manner: We simplify the expression to obtain di/dt on one side: 50-2i= di/dt Rearranging further, taking i terms to one side and t terms to another: di/(50-2i) = dt Integrating both sides, we get { log(50-2i) / (-2) } = { t } Since this was a definite integral, we specify the limits: lower limit: initial time: t=0; (since the inductor acts as a short circuit in the beginning (at t=0), the circuit acts as if there is just a resistor) Hence the initial current can be determined by ohms law: lower limit: initial current:= V/R = 100/4= 25. upper limit: time: t upper limit: current: i Taking exponential on both sides, 50-2i=50exp(-2t) This leads us to the expression for the current flowing in the circuit. i = 25 - 25 exp(-2t) Ans(a): The formula for Time constant for a R-L circuit if T.C= L/R = 2/4 = 0.5 here. Ans(b): The initial and final value of the current flowing in the circuit can be obtained by substituting t=0 and t=infinity in the expression for i. For t=0, the equation becomes i=25-0=25 Amperes. For t=infinity, the equation becomes i=25-25=0.

### Subject: Physics

A pendulum of length with a bob of mass m is oscillating with small amplitude. Determine the effect of the following changes in the pendulum on its Time period: (A) Doubling the mass m of the bob (B) Doubling the initial force used to set the pendulum in motion (C) Doubling the amplitude of the pendulum’s swing (D) Quadrupling the mass m of the bob (E) Quadrupling the length of the pendulum

The time it takes a pendulum to swing back to its original position is called the period of the pendulum. The period of a pendulum is directly proportional to the square root of its length. Thus, if the length of the pendulum is increased by a factor of 4,its period will be increased by a factor of 2. Th e period is independent of the mass of the bob, and for small oscillations, it is independent of the amplitude and hence also of the force required to set it in motion. Hence, only in option (E) will there be a change in the pendulum's time peroid.

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