JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 23)
The following bodies,
(1) a ring
(2) a disc
(3) a solid cylinder
(4) a solid sphere,
of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ___________. [Mark the body as per their respective numbering given in the question]
_17th_March_Morning_Shift_en_23_1.png)
(1) a ring
(2) a disc
(3) a solid cylinder
(4) a solid sphere,
of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ___________. [Mark the body as per their respective numbering given in the question]
Trả lời
4
Giải thích
$$a = {{g\sin \theta } \over {\left( {1 + {I \over {m{R^2}}}} \right)}}$$
IR = mR2, aR = g sin$$\theta$$/2
ID = $${{m{R^2}} \over 2}$$, aD = $${2 \over 3}$$ g sin$$\theta$$
ISC = $${{m{R^2}} \over 2}$$, aSC = $${2 \over 3}$$ g sin$$\theta$$
ISC = $${2 \over 5}$$mR2, aSS = $${5 \over 7}$$ g sin$$\theta$$
S = ut + $${1 \over 2}$$at2,
t = $$\sqrt {{{2S} \over a}} $$
$$ \therefore $$ t $$\propto$$ $${1 \over {\sqrt a }}$$
solid sphere will take minimum time.
IR = mR2, aR = g sin$$\theta$$/2
ID = $${{m{R^2}} \over 2}$$, aD = $${2 \over 3}$$ g sin$$\theta$$
ISC = $${{m{R^2}} \over 2}$$, aSC = $${2 \over 3}$$ g sin$$\theta$$
ISC = $${2 \over 5}$$mR2, aSS = $${5 \over 7}$$ g sin$$\theta$$
S = ut + $${1 \over 2}$$at2,
t = $$\sqrt {{{2S} \over a}} $$
$$ \therefore $$ t $$\propto$$ $${1 \over {\sqrt a }}$$
solid sphere will take minimum time.