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Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its centre and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around its periphery as shown in the figure. Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad/s. The value of n, to the nearest integer, is ---------. [Given : In one complete revolution, the disk rotates by 6.28 rad]

Solution

We have, $\omega ^2 = 2\alpha \theta$

Or, $\omega ^2 = 2\alpha .(n.2\pi)$

$\Rightarrow n = \frac {\omega ^2}{4\pi \alpha }$

Torque $\tau = rF = I\alpha$

$\Rightarrow \alpha = \frac {rF}{I}$

So, $n = \frac {I \omega ^2}{4\pi rF}=\frac {mr^2 \omega ^2}{8\pi rF}=\frac {mr \omega ^2}{8\pi F}=\frac {20\times 0.2 \times 50^2}{4\times 6.28 \times 20}$

$\Rightarrow n=19.9 \approx 20$

Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$