A cylinder of mass M$_{c}$ and sphere of mass M$_{s}$ are placed at points A and B of two inclines, respectively. (See figure). If they roll on the incline without slipping such that their accelerations are the same, then the ratio $\frac{ sin {\theta }_{\text{c}} }{ sin {\theta }_{\text{s}} }$ is : ![JEE Main 2014 Physics, Rotational Motion — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Rotational_Motion/648b5a64417cc3fb48d65d85/question_1__q_648b5a64417cc3fb48d65d85__cdn-question-pool.getmarks.app__6d18db36-4801-4c30-9e19-1804e345c5c1-image__5d475f38de_final_ppt_sync.png)

A cylinder of mass M_ and… - JEE Main 2014 Physics PYQ

JEE Main 2014 — Physics Mechanics

2014mcqhard

A cylinder of mass M $_{c}$ and sphere of mass M $_{s}$ are placed at points A and B of two inclines, respectively. (See figure). If they roll on the incline without slipping such that their accelerations are the same, then the ratio $\frac{ sin {\theta }_{\text{c}} }{ sin {\theta }_{\text{s}} }$ is :

JEE Main 2014 Physics, Rotational Motion — question figure

Official previous-year question

Held on 9 Apr 2014 · Verified 6 Jul 2026.

Options

A
$\frac{ 8 }{ 7 }$
B
$\sqrt{ \frac{ 8 }{ 7 } }$
C
$\sqrt{ \frac{ 15 }{ 14 } }$
D
$\frac{ 1 5 }{ 1 4 }$

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Solution

$\text{a} = \frac{ \text{g sin } \theta }{ ( 1 + \text{I/} {\text{mR}}^{2} ) }$

${\text{a}}_{\text{S}} = \frac{ \text{g sin } {\theta }_{\text{s}} }{ ( 1 + 2 / 5 ) }$

${\text{a}}_{\text{C}} = \frac{ g sin {\theta }_{\text{C}} }{ ( 1 + 1 / 2 ) }$

Now, a $_{S}$ = a $_{C}$

$\frac{ g sin {\theta }_{\text{S}} }{ ( \frac{ 7 }{ 5 } ) } = \frac{ g sin {\theta }_{\text{C}} }{ ( \frac{ 3 }{ 2 } ) }$

$\frac{ 5 }{ 7 } \text{ sin } {\theta }_{\text{S}} = \frac{ 2 }{ 3 } \text{ sin } {\theta }_{\text{C}}$

$\therefore \frac{ sin {\theta }_{\text{C}} }{ sin {\theta }_{\text{S}} } = \frac{ 5 \times 3 }{ 7 \times 2 } = \frac{ 1 5 }{ 1 4 }$

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Solution

$\text{a} = \frac{ \text{g sin } \theta }{ ( 1 + \text{I/} {\text{mR}}^{2} ) }$

${\text{a}}_{\text{S}} = \frac{ \text{g sin } {\theta }_{\text{s}} }{ ( 1 + 2 / 5 ) }$

${\text{a}}_{\text{C}} = \frac{ g sin {\theta }_{\text{C}} }{ ( 1 + 1 / 2 ) }$

Now, a $_{S}$ = a $_{C}$

$\frac{ g sin {\theta }_{\text{S}} }{ ( \frac{ 7 }{ 5 } ) } = \frac{ g sin {\theta }_{\text{C}} }{ ( \frac{ 3 }{ 2 } ) }$

$\frac{ 5 }{ 7 } \text{ sin } {\theta }_{\text{S}} = \frac{ 2 }{ 3 } \text{ sin } {\theta }_{\text{C}}$

$\therefore \frac{ sin {\theta }_{\text{C}} }{ sin {\theta }_{\text{S}} } = \frac{ 5 \times 3 }{ 7 \times 2 } = \frac{ 1 5 }{ 1 4 }$

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Solution

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Other Physics topics for JEE Main

Electromagnetism

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Thermodynamics

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