A disc arranged in a vertical plane has two g | vedclass

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

Question

(B) Let the length of both grooves be $L$. The acceleration of a particle sliding down a smooth groove inclined at an angle $	heta$ with the vertical

Vedclass · Accepted Answer

$1 : \sqrt{2}$

Vedclass · Answer

(B) Let the length of both grooves be $L$. The acceleration of a particle sliding down a smooth groove inclined at an angle $	heta$ with the vertical is $a = g \cos 	heta$.For the vertical chord $AB$,the angle with the vertical is $	heta_1 = 0^\circ$. Thus,$a_{AB} = g \cos 0^\circ = g$.The time taken is $t_{AB} = \sqrt{\frac{2L}{a_{AB}}} = \sqrt{\frac{2L}{g}}$.For the chord $CD$,the angle with the vertical is $	heta_2 = 60^\circ$. Thus,$a_{CD} = g \cos 60^\circ = g/2$.The time taken is $t_{CD} = \sqrt{\frac{2L}{a_{CD}}} = \sqrt{\frac{2L}{g/2}} = \sqrt{\frac{4L}{g}} = 2\sqrt{\frac{L}{g}}$.Taking the ratio,$\frac{t_{AB}}{t_{CD}} = \frac{\sqrt{2L/g}}{2\sqrt{L/g}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

$A$ disc arranged in a vertical plane has two grooves of the same length directed along the vertical chord $AB$ and the chord $CD$ as shown in the figure. Particles slide down along $AB$ and $CD$ starting from rest. The ratio of the time $t_{AB}/t_{CD}$ is

Similar Questions

Find the work done by friction if a $1 \, kg$ block reaches the end of an inclined plane of length $10 \, m$ and inclination $30^{\circ}$ with constant velocity.

$A$ particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is $0.5$,the acceleration is ($g=$ Acceleration due to gravity).

$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \, N$,the mass of the block (in $kg$) is (take $g = 10 \, m/s^2$).

$A$ block of mass $m$ slides down an inclined plane at an angle of $30^{\circ}$ with an acceleration of $\frac{g}{4}$. The value of the coefficient of kinetic friction will be:

$A$ cylindrical vessel filled with water is released on an inclined surface of angle $\theta$ as shown in the figure. The friction coefficient of the surface with the vessel is $\mu (< \tan \theta)$. Then the contact angle made by the surface of water with the incline will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module