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JEE physics

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The moment of inertia of a uniform cylinder of length

1

$fraction numerator 3 over denominator square root of 2 end fraction$
$square root of 3 over 2 end root$
$fraction numerator square root of 3 over denominator 2 end fraction$

Solution

C.

square root of 3 over 2 end root

$straight I space equals space fraction numerator straight m calligraphic l squared over denominator straight I 2 end fraction space plus mR squared over 4 or space straight I space equals space straight m over 4 open parentheses calligraphic l squared over 3 plus straight R squared close parentheses space space.... space left parenthesis 1 right parenthesis Also comma space straight m space equals space πR squared calligraphic l rho rightwards double arrow space R squared space equals space fraction numerator m over denominator pi </p> </div> </div> <div class=$

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A body of mass m = 10^–2 kg is moving in a medium and experiences a frictional force F = –kv². Its initial speed is v₀ = 10 ms^–1. If, after 10 s, its energy is 1/8 mv₀²,the value of k will be

10^-4 kg m^-1

10^–1 kg m^–1 s^–1

10^-3 kg m^-1

10^-3 kg s^-1

Solution

A.

10^-4 kg m^-1

1 half mv subscript straight f superscript 2 space equals space 1 over 8 mv subscript 0 superscript 2 straight v subscript straight f space equals space straight v subscript 0 over 2 space equals space 5 space straight m divided by straight s left parenthesis 10 to the power of negative 2 end exponent right parenthesis space dv over dt space equals space minus space kv squared integral subscript 10 superscript 5 space dv over straight v squared space equals negative 100 space straight k integral subscript 0 superscript 10 space dt 1 fifth space minus space 1 over 10 space equals space 100 space straight k space left parenthesis 10 right parenthesis straight k space equals space 10 to the power of negative 4 end exponent

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A slender uniform rod of mass M and length
$fraction numerator 3 straight g over denominator 2 calligraphic l end fraction space cos space straight theta$
$fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space cos space straight theta$
$fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta$
$fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space sin space straight theta$

Solution

C.

fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta

Taking torque about pivot τ = Iα

mg space sin space straight theta space calligraphic l over 2 space equals space fraction numerator straight m calligraphic l squared over denominator 3 end fraction space straight alpha straight alpha space equals space fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sinθ

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The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)

Solution

B.

straight g space equals space GMx over straight R cubed space inside space the space Earth space straight g equals space GM over straight r squared space outside space the space Earth

Where M is mass of earth

Question

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A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λ_A to λ_B after the collision is

$straight lambda subscript straight A over straight lambda subscript straight B space equals space 2 over 3$
$straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 half$
$straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 third$
$straight lambda subscript straight A over straight lambda subscript straight B space equals space 2$

Solution

D.

straight lambda subscript straight A over straight lambda subscript straight B space equals space 2

By conservation of linear momentum

mv space equals space mv subscript 1 space plus straight m over 2 straight v subscript 2 2 straight v space equals space 2 straight v subscript 1 space plus straight v subscript 2 space... space left parenthesis 1 right parenthesis By space law space of space collision straight e space equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction straight u space equals space straight v subscript 2 minus straight v subscript 1 space space... space left parenthesis 2 right parenthesis by space equ. space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis straight v subscript 1 space equals space straight v over 3 semicolon space straight v subscript 2 space equals space fraction numerator 4 straight v over denominator 3 end fraction straight lambda subscript 1 space equals space straight h over straight p subscript 1 semicolon space straight lambda subscript 2 space equals space straight h over straight o subscript 2 straight lambda subscript 1 over straight lambda subscript 2 space equals space 2 over 1

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