Two masses M_1 and M_2 carry positive charges | vedclass

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Question

(d)

Time of fall $=\sqrt{\frac{2 h}{a_{	ext {net }}}}$

Net acceleration of charged masses is

$a_{	ext {net }}=g-\frac{q E}{m}$

As, $M_1$

Vedclass · Accepted Answer

$M_1 Q_2 > M_2 Q_1$

Vedclass · Answer

(d) Time of fall $=\sqrt{\frac{2 h}{a_{ ext {net }}}}$ Net acceleration of charged masses is $a_{ ext {net }}=g-\frac{q E}{m}$ As, $M_1$ hits the floor before $M_2$. $\Rightarrow \sqrt{\frac{2 h}{a_1}}>\sqrt{\frac{2 h}{a_2}}$ $\Rightarrow \frac{1}{a_1} > \frac{1}{a_2}$ $\Rightarrow a_2 > a_1$ When reciprocal is taken in equality sign is reversed, then. $g-\frac{Q_1 E}{M_1} > g-\frac{Q_2 E}{M_2}$ $\Rightarrow \quad -\frac{Q_1 E}{M_1} > -\frac{Q_2 E}{M_2}$ $\Rightarrow \quad \frac{Q_1 E}{M_1} < \frac{Q_2 E}{M_2}$ Here, multiplication with $-1$ reverse sign of inequality. $ ext { So, } \frac{Q_1}{Q_2} < \frac{Q_2}{M_2}$ $\Rightarrow M_2 Q_1 < M_1 Q_2$ $\Rightarrow M_1 Q_2 > M_2 Q_1$

Two masses $M_1$ and $M_2$ carry positive charges $Q_1$ and $Q_2$, respectively. They are dropped to the floor in a laboratory set up from the same height, where there is a constant electric field vertically upwards. $M_1$ hits the floor before $M_2$. Then,

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