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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 6. system of particles and rotational motion4 Marks
MCQ
A solid sphere rolls without slipping, first horizontally and then up to a point $X$ at height $h$ on an inclined plane before rolling down, as shown in the figure below. The initial horizontal speed of the sphere is
  • ✓
    $\sqrt{10 g h / 7}$
  • B
    $\sqrt{7 g h / 5}$
  • C
    $\sqrt{5 g h / 7}$
  • D
    $\sqrt{2 g h}$
✓

Answer

Correct option: A.
$\sqrt{10 g h / 7}$
a
(a)

Let initial horizontal speed of sphere is $v$.

Then, total kinetic energy of sphere on horizontal part

$=( KE )_{\text {translation }}+( KE )_{\text {rotation }}$

$=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2$

$=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2}=\frac{7}{10} m v^2$

If sphere rises upto height $h$ then, by conservation of energy, we have

$m g h=\frac{7}{10} m v^2$

$\text { or } \quad v=\sqrt{\frac{10}{7} g h}$

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