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physics

mechanics

Easy

Question

A small sphere of radius R rolls without slipping inside a large hemispherical bowl of radius R. The sphere starts from rest at the top point of the hemisphere. What fraction of the total energy is rotational when the small sphere is at the bottom of the hemisphere

7/5
2/7
5/7
7/10

The correct answer is: 7/5

$table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell beta subscript text कhere end text end subscript equals 2 over 5 end cell row cell T times E subscript R equals 1 half m V squared left parenthesis 1 plus beta right parenthesis equals 1 half m V squared open parentheses 1 plus 2 over 5 close parentheses equals 7 over 5 cross times 1 half m V squared end cell row cell equals fraction numerator T times E subscript R over denominator K times E end fraction equals fraction numerator 1 half m V squared open parentheses 7 over 5 close parentheses over denominator 1 half m V squared end fraction equals 7 over 5 end cell end table$

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$straight A space uniform space rod space of space mass space straight M equals 2 kg space and space length space straight L space is space suspended space by space two space smooth space hinges space 1 space and space 2 space as space shown space in space figure. space straight A space force space straight F equals 4 straight N space is space applied space downward space at space straight a space distance space straight L divided by 4 space from space hinge space 2. space Due space to space the space application space of space force space straight F comma space hinge space 2 space breaks. space At space this space instant space comma space applied space force space straight F space is space also space removed. space The space rod space starts space to space rotate space downward space about space hinge space 1. space open parentheses g equals 10 m divided by s squared close parentheses$

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$straight A space uniform space rod space of space mass space straight M equals 2 kg space and space length space straight L space is space suspended space by space two space smooth space hinges space 1 space and space 2 space as space shown space in space figure. space straight A space force space straight F equals 4 straight N space is space applied space downward space at space straight a space distance space straight L divided by 4 space from space hinge space 2. space Due space to space the space application space of space force space straight F comma space hinge space 2 space breaks. space At space this space instant space comma space applied space force space straight F space is space also space removed. space The space rod space starts space to space rotate space downward space about space hinge space 1. space open parentheses g equals 10 m divided by s squared close parentheses$

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ii. $I subscript x equals I subscript y plus I subscript z$
b. possible for hollow sphere or solid sphere

iii. $I subscript y equals I subscript z plus I subscript x$
c. possible for all of the bodies mentioned

iv. $I subscript z equals I subscript x plus I subscript y$
d. not possible for any of the body mentioned

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Column I
Column II

i. $I subscript x equals I subscript y equals I subscript z$
a. possible for ring or disc

ii. $I subscript x equals I subscript y plus I subscript z$
b. possible for hollow sphere or solid sphere

iii. $I subscript y equals I subscript z plus I subscript x$
c. possible for all of the bodies mentioned

iv. $I subscript z equals I subscript x plus I subscript y$
d. not possible for any of the body mentioned

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Two particles A and B of mass m each, are joined by a rigid massless rod of length l. A particle P of mass m, moving with speed u normal to AB, strikes A and stick to it. The centre of mass of the 'A + B + P' system after collision is C. If u = 4m/s, m = 3 kg, l = 3 m, then (All quantities in column are in SI units)

PhysicsMechanics

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