Visit the website manishverma.site for latest posts, courses, admission & more.

Consider a particle in circular motion as shown in the figure. The position vector sweeps equal area in equal time. Select correct option(s).

(A) the speed is constant
(B) acceleration is constant
(C) $\left| {\overrightarrow {{a_r}} } \right|$ = constant
(D) $a_t = 0$

Let $d\theta$ be the angle swept by the radius in time dt.

Area swept $dA = \frac{1}{2}{r^2}d\theta$

$\frac{{dA}}{{dt}} = \frac{1}{2}{r^2}\frac{{d\theta }}{{dt}} = \frac{1}{2}{r^2}\omega$ where $\omega$ is the instantaneous angular speed.

$\frac{1}{2}{r^2}\omega$ = constant

$\therefore \omega$ = constant or speed v = constant

If speed is constant, $a_t = 0$

${a_r} = \frac{{{v^2}}}{r}$ = constant (magnitude)

The direction of acceleration vector is changing and hence it is not a constant.

(A), (C), (D).

### 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 = ${}^{12}{C_6} = 924$