An inclined plane makes an angle of 30^{circ} | vedclass

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

Question

(D) For a body rolling down an inclined plane without slipping,the linear acceleration $a$ is given by the formula: $a = \frac{g \sin 	heta}{1 + \fra

Vedclass · Accepted Answer

$\frac{5 g}{14}$

Vedclass · Answer

(D) For a body rolling down an inclined plane without slipping,the linear acceleration $a$ is given by the formula: $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a solid sphere,the moment of inertia about its center of mass is $I = \frac{2}{5} MR^2$.Substituting this into the formula: $a = \frac{g \sin 	heta}{1 + \frac{2/5 MR^2}{MR^2}} = \frac{g \sin 	heta}{1 + 2/5} = \frac{g \sin 	heta}{7/5} = \frac{5}{7} g \sin 	heta$.Given $	heta = 30^{\circ}$,we have $\sin 30^{\circ} = 0.5 = \frac{1}{2}$.Therefore,$a = \frac{5}{7} g 	imes \frac{1}{2} = \frac{5g}{14}$.

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 ........

Similar Questions

During the change of $O_2$ to $O_2^-$ ion,the electron adds to which one of the following orbitals?

$A$ magnetized wire of magnetic moment $M$ and length $L$ is bent into an arc of a circle subtending an angle of $60^\circ$ at the center. The new magnetic moment is:

An integrating factor of the equation $(1+y+x^2 y) dx+(x+x^3) dy=0$ is

Find out the value of $K_C$ for the following reaction from the value of $K_P$
$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$
[Given : $K_P = 8 \times 10^{12} \ atm$ at $500 \ K$ use $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$ ]

The energy required to accelerate a car from $10 \, m/s$ to $20 \, m/s$ is how many times the energy required to accelerate the car from rest to $10 \, m/s$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module