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In hydrogen atom, the area enclosed by nth orbit is An The graph between l o g invisible function application open parentheses fraction numerator A subscript n end subscript over denominator A subscript 1 end subscript end fraction close parentheses and logn will be

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The correct answer is:

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In hydrogen atom, the radius of nth Bohr orbit is Vn The graph between l o g invisible function application open parentheses fraction numerator r subscript n end subscript over denominator r subscript 1 end subscript end fraction close parentheses and log space n space will be

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In a Bohr atom the electron is replaced by a particle of mass 150 times the mass of the electron and the same charge If a0 is the radius of the first Bohr orbit of the orbital atom, then that of the new atom will be

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The figure shows energy levels of a certain atom, when the system moves from level 2E is emitted The to E, a photon of wavelength wavelength of photon produced during its transition from level 4/3 E to E level is:

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The ionisation energy of Li to the power of 2 plus end exponent atom in ground state is :

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A cylindrical solid of length L and radius a is having varying resistivity given by r = r0 x where r0 is a positive constant and x is measured from left end of solid. The cell shown in the figure is having emf V and negligible internal resistance. The electric field as a function of x is best described by :

A cylindrical solid of length L and radius a is having varying resistivity given by r = r0 x where r0 is a positive constant and x is measured from left end of solid. The cell shown in the figure is having emf V and negligible internal resistance. The electric field as a function of x is best described by :

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A charged oil drop falls with terminal velocity V0 in the absence of electric field An electric field E keeps it stationary The drop acquires additional charge q and starts moving upwards with velocity V0 The initial charge on the drop was

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if ell parallel to mfind the value of x in given figure.

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The number of positive integral solutions of the inequation fraction numerator x squared left parenthesis 3 x minus 4 right parenthesis cubed left parenthesis x minus 2 right parenthesis to the power of 4 over denominator left parenthesis x minus 5 right parenthesis to the power of 5 left parenthesis 2 x minus 7 right parenthesis to the power of 6 end fraction less or equal than 0 is –

The number of positive integral solutions of the inequation fraction numerator x squared left parenthesis 3 x minus 4 right parenthesis cubed left parenthesis x minus 2 right parenthesis to the power of 4 over denominator left parenthesis x minus 5 right parenthesis to the power of 5 left parenthesis 2 x minus 7 right parenthesis to the power of 6 end fraction less or equal than 0 is –

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Set of values of x satisfying the inequality fraction numerator left parenthesis x minus 3 right parenthesis squared left parenthesis 2 x plus 5 right parenthesis squared left parenthesis x minus 7 right parenthesis over denominator open parentheses x squared plus x plus 1 close parentheses left parenthesis 3 x plus 6 right parenthesis squared end fraction less or equal than 0 is left square bracket a comma b right parenthesis union left parenthesis b comma c right square bracket then 2a + b + c is equal to

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Number of integral values of x for which the inequality log10 open parentheses fraction numerator 2 x minus 2007 over denominator x plus 1 end fraction close parenthesesless or equal than 0 holds true, is

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log4 (2x2 + x + 1) – log2 (2x – 1) less or equal than – tan fraction numerator 7 pi over denominator 4 end fraction

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The solution set of the inequation log1/3 (x2 + x + 1) + 1 > 0 is

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If open parentheses a to the power of log subscript b end subscript invisible function application x end exponent close parentheses to the power of 2 end exponent–5x to the power of log subscript b end subscript invisible function application a end exponent + 6 = 0 where a > 0, b > 0 & ab not equal to 1. Then the value of x is equal to

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No. of ordered pair satisfying simultaneously the system of equation 2 to the power of square root of x end exponent. 2 to the power of square root of y end exponent= 256 & log10square root of x y end root – log10 1.5 = 1 is.

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