Mohit M.

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Chemistry

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Question:

Which of the atom in each pair has a larger size? $$a) K^+$$ $$or$$ $$Cl^-$$ $$b) Na$$ $$or$$ $$Na^+$$ $$c) S$$ $$or$$ $$S^{2-} $$

Mohit M.

Answer:

a) $$K^+$$ and $$Cl^–$$ are isoelectronic (18 electrons each). For $$K^+$$, 19 protons at the centre are pulling 18 electrons. In $$Cl^–$$ there are 17 protons at the centre holding 18 electrons. So $$K^+$$ has its electrons pulled closer to the nucleus resulting in lesser radius than $$Cl^-$$ b) A cation is an ion with a positive charge(fewer electrons than protons). This will cause a decrease in atomic size because there are now fewer electrons for the protons to pull towards the nucleus and will result in a stronger pull of the electrons towards the nucleus. Hence $$Na^+$$ has a lesser radius as compared to its neutral atom. c) An anion is made because of the gain of electrons. The gain of an electron means more electrons in the outermost shell. There are more electrons to pull towards the nucleus so the pull becomes slightly weaker than of the neutral atom and causes an increase in atomic radius. $$S^{2-}$$ is our winner.

Electrical Engineering

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Question:

We have a 40W, 200V bulb connected in series with an 80W, 200V bulb. This combination is across a 200 V power supply(AC). Which of them will glow brighter?

Mohit M.

Answer:

We will brush up some concepts first: Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. $$V = I \times R$$ We also know that current passing through the resistors in series is same(Equal). So we can safely assume that voltage drop across each bulb depends upon its resistance value(Current being constant). Now we also know that Power across a resistor, $$P = V^2/R $$, Power is inversely proportional to resistance. A 40W bulb should have a higher resistance than an 80W bulb for same ratings of voltage. Now by Ohm's law, 40W will have a larger voltage drop meaning more brightness.

Algebra

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Question:

$$f(x) = x/(1-2^x)-x/2$$ Determine whether the function is even or odd.

Mohit M.

Answer:

A function is even when f(x) = f(−x) for all x. Geometrically it can be realized as symmetry about y axis. A function is odd when −f(x) = f(−x) for all x. Geometrically we get an origin symmetry. Now a function does not need to be even or odd. In fact most of the functions are neither even nor odd. To check the above conditions, we simply have to replace x with -x in the function, that is to determine f(-x). If it comes out to be f(x), this function is even. -f(x) will mean an odd function. If we don't get any of them, this function is Mr. Neither Even Nor Odd :) Lets apply this concept to our problem now. We will determine f(-x) for our case: $$f(-x) = -x/(1-2^{-x})-(-x)/2$$ $$= -x\times2^x/(2^x-1) + x/2$$ $$= -x\times (2^x-1+1)/(2^x-1) + x/2$$ Simplifying, we get: $$f(-x)= -x-x/(2^x-1) + x/2$$ $$= -x/(2^x-1)-x/2$$ $$= x/(1-2^x)-x/2$$ $$=f(x)$$ This is clearly an even function! Algebra is fun! :D

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