The plank in the figure moves a distance 100, | vedclass

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(B) Let the displacement of the plank be $s_1 = 100\,mm$ to the right.Let the displacement of the centre of mass of the sphere be $s_2 = 75\,mm$ to th

Vedclass · Accepted Answer

$\frac{7}{6}$

Vedclass · Answer

(B) Let the displacement of the plank be $s_1 = 100\,mm$ to the right.Let the displacement of the centre of mass of the sphere be $s_2 = 75\,mm$ to the left.The radius of the sphere is $R = 150\,mm$.Since there is no slipping at the point of contact between the sphere and the plank,the velocity of the point of contact on the sphere must be equal to the velocity of the plank.In terms of displacement,the condition for no slipping is given by:$R \Delta 	heta = s_1 + s_2$where $\Delta 	heta$ is the angular displacement of the sphere.Substituting the given values:$150 \cdot \Delta 	heta = 100 + 75$$150 \cdot \Delta 	heta = 175$$\Delta 	heta = \frac{175}{150} = \frac{7}{6}\,rad$Thus,the angular displacement of the sphere is $\frac{7}{6}\,rad$.

Molecule type	$\frac{C_P}{C_V}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

The plank in the figure moves a distance $100\,mm$ to the right while the centre of mass of the sphere of radius $150\,mm$ moves a distance $75\,mm$ to the left. The angular displacement of the sphere (in radian) is (there is no slipping anywhere):-

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