Akshat A.

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Physics (Electricity and Magnetism)

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Question:

The coercivity of a small magnet where the ferromagnet gets demagnatised is 3 x 10^3 Am-1. the current required to be passed in a solenoid of length 10 cm and number of turns 100 , so that the magnet gets demagnetised when inside the solenoid, is

Akshat A.

Answer:

For solenoid , the magnatic field needed to be magnetised the magnet. B = u. * N/L *I where , N=100 L=10 cm = 0.1 m 3 * 10^3 = 100/0.1 * I => I = 3A.

Trigonometry

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Question:

he number of points in (−∞,∞), for which x^2−xsinx−cosx= 0 is (A) 6 (B) 4 (C) 2 (D) 0

Akshat A.

Answer:

Fortunately, the problem asks only the number of solutions and not what they are. And there are theorems which guarantee the existence(and sometimes the non-existence too) of zeros of functions which satisfy certain smoothness conditions such as continuity and differentiability. Two foremost such theorems the Intermediate Value Property (IVP) of continuous functions and Rolle’s theorem (applicable to a function which is the derivative of some other function which takes equal values at the two end points of an interval) .Let us first try the IVP for the function f(x). For large x, the dominating term isx2and so qualitatively f(x) behaves likex2. (To see this more precisely, note that f(x) =x^2(1−sinx/x−cosx/x^2) and since sinx and cosx are both bounded, the second factor tends to 1 as x→ ∞ In particular we see that f(x) is positive for all sufficiently large x, say for all x≥R for some R >0. Since f(0) =−1<0 while f(R)>0, the IVP implies that f(x) vanishes at least once in (0, R). To see if it vanishes more often, suppose a, b with 0< a < b are two zeros of f(x). Then by Rolle’s theorem,f′(c) = 0 for some c∈(a, b). But a direct calculation gives f′(x) = 2x−sinx − xcosx + sinx = x(2−cosx) for all x. The second factor is always positive and so f'(x)>0 for all x >0. That would contradict that f′(c) = 0. Thus we see that f(x) has precisely one zero in (0,∞) and hence precisely two zeros in (−∞,∞).

Algebra

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Question:

Three business entities X Ltd, Y Ltd. and Z Ltd, with 4, 3 and 5 employees respectively, merged into XYZ Ltd in order to jointly raise the capital for setting up a new modern production plant in Jaipur. After two years, on the question of management decisions on the new venture at Jaipur, the employees started adopting differing viewpoints and began to quarrel among themselves. Given the fact that there is no quarrel among the employees of the erstwhile (i.e. former) X Ltd, Y Ltd and Z Ltd, what could be the maximum number of quarrels that can take place within XYZ Ltd?

Akshat A.

Answer:

Option(C) is correct Before the merger: X Ltd. had no. of employees =4 Y Ltd. had no. of employees =3 Z Ltd. had no. of employees =5 After the merger, there are quarrels amongst the employees of the erstwhile X ltd, Y ltd, and Z ltd. An employee does not quarrel with any of the employees from his parent company. Company :- X,Y,Z Employees: 4, 3 ,5 Any employees in X can have a maximum possible no. of quarrels with the employees of Y ltd. and Z ltd. as given by: 1×3+1×5 As there are 4 employees in the merged company, who came from X ; So, the total number of quarrels involving employees of X company is: 4×[1×3+1×5]=4×8=32 quarrels -------(1) Further, employees of Y & Z ltd. will be involved with quarrels within each other. The total number of such quarrels: =3×[1×5]=5×[1×3]=15 quarrels --------(2) From (1) and (2), the total number of quarrels between the employees of three erstwhile companies is: =32+15=47 quarrels. Hence (C) option is correct.

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