MCQ
A thin circular ring first slips down a smooth incline then rolls down a rough incline of identical geometry from same height. Ratio of time taken in the two motion is ........
- A$\frac{1}{2}$
- B$1$
- ✓$\frac{1}{\sqrt{2}}$
- D$\frac{1}{4}$
✓
Answer
Correct option: C.
$\frac{1}{\sqrt{2}}$
c
(c)
(c)
case $(i)$ Transitional motion
acceleration $=g \sin \theta$
case $(ii)$ acceleration $=\frac{ g \sin \theta}{1+\frac{1}{ MR ^2}}$
$I = MR ^2$
$a =\frac{ g \sin \theta}{2}$
Now using $\rho=u t+\frac{1}{2} at ^2$
$a _1 t _1^2= a _2 t _2^2 \quad \theta$ and $\rho$ same for both are and $v =0$
$\frac{ t _1{ }^2}{ t _2^2}=\frac{ a _2}{ a _1}=\frac{ g \sin \theta}{2 g \sin \theta}$
$\frac{t_1}{t_2}=\frac{1}{\sqrt{2}}$
$\therefore \frac{t_1}{t_2}=\frac{1}{\sqrt{2}}$
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