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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,
$Q_1 > Q_2$
$Q_1 < Q_2$
$M_1 Q_1 > M_2 Q_2$
$M_1 Q_2 > M_2 Q_1$
Solution
(d)
Time of fall $=\sqrt{\frac{2 h}{a_{\text {net }}}}$
Net acceleration of charged masses is
$a_{\text {net }}=g-\frac{q E}{m}$
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.
$\text { So, } \frac{Q_1}{Q_2} < \frac{Q_2}{M_2}$
$\Rightarrow M_2 Q_1 < M_1 Q_2$
$\Rightarrow M_1 Q_2 > M_2 Q_1$
