In order to make the parabolas formed by sing | vedclass

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(C) In Thomson's mass spectrograph,the equation of the parabola is given by $y = \frac{k q}{m} x^2$,where $k = \frac{B^2 L D}{E}$.For the parabolas to

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$9 : 4$

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(C) In Thomson's mass spectrograph,the equation of the parabola is given by $y = \frac{k q}{m} x^2$,where $k = \frac{B^2 L D}{E}$.For the parabolas to coincide,the value of $\frac{k q}{m}$ must be the same for both cases.Let the singly ionized ion have charge $q_1 = e$ and mass $m_1$,and the doubly ionized ion have charge $q_2 = 2e$ and mass $m_2$.Given the ratios of electric fields $\frac{E_1}{E_2} = \frac{1}{2}$ and magnetic fields $\frac{B_1}{B_2} = \frac{3}{2}$.Equating the constants: $\frac{B_1^2}{E_1 m_1} q_1 = \frac{B_2^2}{E_2 m_2} q_2$.Substituting the values: $\frac{(3/2)^2}{1/2 \cdot m_1} \cdot e = \frac{1^2}{1 \cdot m_2} \cdot 2e$.$\frac{9/4}{1/2 \cdot m_1} = \frac{2}{m_2} \implies \frac{9}{2 m_1} = \frac{2}{m_2}$.Therefore,$\frac{m_1}{m_2} = \frac{9}{4}$.

List-$I$	List-$II$
$a$. Fleming's left-hand rule	$e$. Direction of induced current
$b$. Fleming's right-hand rule	$f$. South pole
$c$. Clockwise current	$g$. North pole
$d$. Anticlockwise current	$h$. Direction of force

In order to make the parabolas formed by singly ionized ions in one spectrograph and doubly ionized ions in another Thomson's mass spectrograph coincide,the electric fields and magnetic fields are kept in the ratios $1 : 2$ and $3 : 2$ respectively. Then the ratio of the masses of the ions is

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Match the following:
List-$I$ List-$II$
$a$. Fleming's left-hand rule $e$. Direction of induced current
$b$. Fleming's right-hand rule $f$. South pole
$c$. Clockwise current $g$. North pole
$d$. Anticlockwise current $h$. Direction of force

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