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STD 11 - 6. system of particles and rotational motion — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 6. system of particles and rotational motion4 Marks
MCQ
A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, $\mathrm{A}$ is the point of contact, $\mathrm{B}$ is the centre of the sphere and $\mathrm{C}$ is its topmost point. Then,

$(A)$ $\vec{V}_C-\vec{V}_A=2\left(\vec{V}_B-\vec{V}_C\right)$

$(B)$ $\vec{V}_C-\vec{V}_B=\vec{V}_B-\vec{V}_A$

$(C)$ $\left|\vec{V}_C-\vec{V}_A\right|=2\left|\vec{V}_B-\vec{V}_C\right|$

$(D)$ $\left|\vec{V}_C-\vec{V}_A\right|=4\left|\vec{V}_B\right|$

  • ✓
    $(B,C)$
  • B
    $(B,D)$
  • C
    $(A,C)$
  • D
    $(A,D)$
✓

Answer

Correct option: A.
$(B,C)$
a
If $\vec{V}_0$ is the velocity of centre of the sphere, then

$\vec{V}_C=2 \vec{V}_0, \vec{\rightarrow} B=\vec{V}_0 \text { and } \rightarrow A=0$

$\therefore \vec{V}_C-\vec{V}_B=2 \vec{V}_0-\vec{V}_0=\vec{V}_0$

$\vec{V}_B-\vec{V}_A=\vec{V}_0-\overrightarrow{0}=\vec{V}_0$

$\therefore \vec{V}_C-\vec{V}_B=\vec{V}_B-\vec{V}_A$

$(b)$ is the correct opton.

Now, $\left|\vec{V}_C-\vec{V}_A\right|=\left|2 \vec{V}_0-0\right|=\left|2 \vec{V}_0\right|=2\left|\vec{V}_0\right|$

and $\left|\vec{V}_C-\vec{V}_A\right|=2\left|\vec{V}_B-\vec{V}_C\right|$

$(c)$ is the correct option.

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