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Question

An inclined plane makes an angle of $$30^{\circ}$$ with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to :


A
g3
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B
2g3
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C
5g7
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D
5g14
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Solution

The correct option is D $$\dfrac {5g}{14}$$
$$a=\cfrac { mg\sin\theta -F }{ m } \\ \alpha =\cfrac { FR }{ I } =\cfrac { 5FR }{ 2M{ R }^{ 2 } } =\cfrac { 5F }{ 2MR } $$
For no slipping , $$R\alpha =a$$
$$\cfrac { 5F }{ 2M } =g\sin\theta -\cfrac { F }{ M } \Rightarrow \cfrac { F }{ M } =\cfrac { 2 }{ 7 } g\sin\theta \\ a=g\sin\theta .\cfrac { F }{ M } =\cfrac { 5 }{ 7 } g\sin\theta =\cfrac { 5 }{ 7 } g\sin30\\ =\cfrac { 5 }{ 14 } g$$

835276_782249_ans_c410be3c68bd44ec8ef458db4c0b9128.PNG

Physics

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