A circular disc reaches from top to bottom of an inclined plane of length 'L' when it slips down the plane, it takes time 't1'. When it rolls down the plane, it takes time t2. The value of t 1 t 2 i s 3 x The value of x will be__________.

Case – I : When disk slides down t 1 = 2 L g s i n θ . . . . . . . . . . ( i ) Case – II : When disk rolls down t 2 = 2 L g s i n θ 1 + β 2 = 2 L g s i n θ 1 + ( k R ) 2 = 2 L g s i n θ 1 + ( R / 2 R ) 2 = 2 L g s i n θ 1 + 1 2 = 4 L 3 g s i n θ . . . . . . ( i i ) ⇒ t 1 t 2 = 2 L g s i n θ * 3 g s i n θ 4 L = 3 2

A circular disc reaches from top to bottom of an inclined plane of length 'L' when it slips down the plane, it takes time 't₁'. When it rolls down the plane, it takes time t₂. The value of $\frac{t_{1}}{t_{2}} i s \sqrt[]{\frac{3}{x}}$ The value of x will be__________.

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Alok Kumar Singh | Contributor-Level 10

6 months ago

Case – I : When disk slides down

$t_{1} = \sqrt[]{\frac{2 L}{g s i n θ}} . . . . . . . . . . (i)$

Case – II : When disk rolls down

$t_{2} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + β^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + {(\frac{k}{R})}^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + {(\frac{R / \sqrt[]{2}}{R})}^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + \frac{1}{2}}}} = \sqrt[]{\frac{4 L}{3 g s i n θ}} . . . . . . (i i)$

$\Rightarrow \frac{t_{1}}{t_{2}} = \sqrt[]{\frac{2 L}{g s i n θ}} * \sqrt[]{\frac{3 g s i n θ}{4 L}} = \sqrt[]{\frac{3}{2}}$

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