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Physics-

General

Easy

Question

The dumbbell is placed on a frictionless horizontal table. Sphere A is attached to a frictionless pivot so that B can be made to rotate about A with constant angular velocity. If B makes one revolution in period P, the tension in the rod is

$\frac{4 π^{2} M d}{P^{2}}$
$\frac{8 π^{2} M d}{P^{2}}$
$\frac{4 π^{2} M d}{P}$
$\frac{2 M d}{P}$

The correct answer is: $fraction numerator 8 pi to the power of 2 end exponent M d over denominator P to the power of 2 end exponent end fraction$

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If $2 f left parenthesis x right parenthesis plus x f open parentheses fraction numerator 1 over denominator x end fraction close parentheses minus 2 f open parentheses open vertical bar square root of 2 sin invisible function application pi open parentheses x plus fraction numerator 1 over denominator 4 end fraction close parentheses close vertical bar close parentheses equals 4 c o s to the power of 2 end exponent invisible function application open parentheses fraction numerator pi x over denominator 2 end fraction close parentheses plus x c o s invisible function application fraction numerator pi over denominator x end fraction comma for all x element of R minus left curly bracket 0 right curly bracket$ then which of the following statement(s) is/are true?

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parallel

The domain of the function $square root of 1 minus f left parenthesis x right parenthesis end root text where end text f to the power of 3 end exponent left parenthesis x right parenthesis equals 1 minus x to the power of 3 end exponent minus 3 x f left parenthesis x right parenthesis. x element of R text is end text left parenthesis f left parenthesis 1 right parenthesis not equal to negative 1 right parenthesis$

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If $f left parenthesis x right parenthesis equals s i n to the power of negative 1 end exponent invisible function application x text and end text g left parenthesis x right parenthesis equals left square bracket s i n invisible function application x left parenthesis c o s invisible function application x right parenthesis right square bracket plus left square bracket c o s invisible function application left parenthesis s i n invisible function application x right parenthesis right square bracket comma text end text$ then range of f(g(x)) is

If $f left parenthesis x right parenthesis equals s i n to the power of negative 1 end exponent invisible function application x text and end text g left parenthesis x right parenthesis equals left square bracket s i n invisible function application x left parenthesis c o s invisible function application x right parenthesis right square bracket plus left square bracket c o s invisible function application left parenthesis s i n invisible function application x right parenthesis right square bracket comma text end text$ then range of f(g(x)) is

Domain of the function $f left parenthesis x right parenthesis equals fraction numerator 1 over denominator square root of l n invisible function application c o t to the power of negative 1 end exponent invisible function application x end root end fraction$ is

Domain of the function $f left parenthesis x right parenthesis equals fraction numerator 1 over denominator square root of l n invisible function application c o t to the power of negative 1 end exponent invisible function application x end root end fraction$ is

parallel

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