A ring and a disc are initially at rest, side | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$.For a ring, $k^2 = R^2$, so

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(C) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$.For a ring, $k^2 = R^2$, so $a_R = \frac{g \sin 	heta}{1 + 1} = \frac{g \sin 	heta}{2}$.For a disc, $k^2 = \frac{R^2}{2}$, so $a_D = \frac{g \sin 	heta}{1 + 0.5} = \frac{2g \sin 	heta}{3}$.The distance traveled along the incline is $d = \frac{h}{\sin 	heta}$.Using $d = \frac{1}{2} a t^2$, we have $t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2h}{a \sin 	heta}}$.For the ring: $t_R = \sqrt{\frac{2h}{(g \sin 	heta / 2) \sin 	heta}} = \sqrt{\frac{4h}{g \sin^2 	heta}}$.For the disc: $t_D = \sqrt{\frac{2h}{(2g \sin 	heta / 3) \sin 	heta}} = \sqrt{\frac{3h}{g \sin^2 	heta}}$.Given $	heta = 60^{\circ}$, $\sin 	heta = \frac{\sqrt{3}}{2}$, so $\sin^2 	heta = \frac{3}{4}$.$t_R = \sqrt{\frac{4h}{g(3/4)}} = \sqrt{\frac{16h}{3g}}$ and $t_D = \sqrt{\frac{3h}{g(3/4)}} = \sqrt{\frac{4h}{g}}$.Given $t_R - t_D = \frac{2-\sqrt{3}}{\sqrt{10}}$, we have $\sqrt{\frac{h}{g}} \left( \frac{4}{\sqrt{3}} - 2 ight) = \frac{2-\sqrt{3}}{\sqrt{10}}$.$\sqrt{\frac{h}{10}} \left( \frac{4-2\sqrt{3}}{\sqrt{3}} ight) = \frac{2-\sqrt{3}}{\sqrt{10}}$.$\sqrt{h} \left( \frac{2(2-\sqrt{3})}{\sqrt{3}} ight) = 2-\sqrt{3} \Rightarrow \sqrt{h} = \frac{\sqrt{3}}{2} \Rightarrow h = \frac{3}{4} = 0.75 \,m$.

Column-$I$	Column-$II$
$(A)$ Ring	$(I)$ $\frac{Mg \sin \theta}{3.5}$
$(B)$ Solid sphere	$(II)$ $\frac{Mg \sin \theta}{2}$
$(C)$ Solid cylinder	$(III)$ $\frac{Mg \sin \theta}{3}$
$(D)$ Hollow cylinder	$(IV)$ $\frac{Mg \sin \theta}{2.5}$

Similar Questions

An object of uniform density rolls up a curved path with an initial velocity $v_0$ as shown in the figure. If the maximum height attained by the object is $\frac{7v_0^2}{10g}$ ($g =$ acceleration due to gravity), the object is a . . . . . . .

$A$ spherical ball of mass $20 \ kg$ is stationary at the top of a hill of height $100 \ m$. It rolls down a surface to the ground,then climbs up another hill of height $30 \ m$ and finally rolls down to a horizontal base at a height of $20 \ m$ above the ground. The velocity attained by the ball is

$A$ ball of radius $11 \ cm$ and mass $8 \ kg$ rolls from rest down a ramp of length $2 \ m$. The ramp is inclined at $35^{\circ}$ to the horizontal. When the ball reaches the bottom,its velocity is .......... $m/s$. $(\sin 35^{\circ} = 0.57)$

$A$ sphere,a disc,a ring,and a hollow sphere of the same radius are rolled down from the same height on an inclined plane simultaneously. The order in which these objects reach the bottom is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module