Question

The following bodies 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 ________.

• A. a ring
• B. a disc
• C. a solid cylinder
• D. a solid sphere

Answer 1

The correct option is D

a solid sphere

The explanation for the correct option

Step 1: Given data

Bodies of the same mass $m$ and radius $R$ are allowed to roll down without slipping simultaneously from the top of the inclined plane.

Step 2: Formula used

Acceleration of body rolling down an inclined plane is: $\text{a =}\frac{\mathrm{gsin\theta }}{(1+{\displaystyle \frac{1}{{\mathrm{mR}}^2}})}$

Step 3: Calculating time

Now, calculate the acceleration for each given object in the options,

$\Rightarrow$$I_R=m{R}^2,aR=$$\frac{\mathrm{gsin\theta }}{2}$

$\Rightarrow$$I_D=$ $\frac{{\mathrm{mR}}^2}{2}$, $a_D=$$\frac{2}{3}\mathrm{gsin\theta }$

$\Rightarrow$$I_{SC}=$ $\frac{{\mathrm{mR}}^2}{2}$, $a_{SC}=$ $\frac{2}{3}\mathrm{gsin\theta }$

$\Rightarrow$$I_{SS}=$ $\frac{{\mathrm{mR}}^2}{2}$, $a_{SS}=$ $\frac{5}{7}\mathrm{gsin\theta }$

We know that, $s=ut+$ $\left(\frac{1}{2}\right)$$a{t}^2$

Where, t = $\frac{\sqrt{2}\mathrm{S}}{\mathrm{a}}$

Therefore, t $\propto$ $\frac{1}{\sqrt{\mathrm{a}}}$ For least time, so, $a$ should be the highest.

Thus, solid spheres will take minimum time.

Hence, the body which will reach first at the bottom of the inclined plane is a solid sphere.

Therefore, option (D) is the correct answer.