A symmetrical square wave of frequency $25 k H z$ and a peak-to-peak amplitude of $10 V$ is fed to $Y - i n p u t$ of a single channel oscilloscope. The screen appears as shown in figure. Then the $X a n d Y$ sensitivities and the trigger settings respectively are

5 μ s / c m, 2 V / c m a n d + v e s l o p e

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5 μ s / c m, 2 V / c m a n d - v e s l o p e

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10 μ s / c m, 1 V / c m a n d + v e s l o p e

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20 μ s / c m, 2 V / c m a n d - v e s l o p e

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Solution

The correct option is B $5 μ s / c m, 2 V / c m a n d - v e s l o p e$
Here,
$I n Y - d i r e c t i o n \to a m p l i t u d e i s t h e r e$
$I n X - d i r e c t i o n \to t i m e a x i s$

$\Rightarrow 2.5 c m o n Y a x i s c o r r e s p o n d t o 5 V .$
$S o, s e n s i t i v i t y = \frac{5 V}{2.5 c m} = 2 V / c m$

$\Rightarrow 4 c m o n x a x i s c o r r e s p o n d t o T / 2.$

$\frac{T}{2} = \frac{1}{2 f} = \frac{1}{2 \times 25 \times 10^{3}}$

$= 0.02 m s e c = 20 μ s e c$

$S o, s e n s i t i v i t y i n X - d i r e c t i o n .$

$= \frac{T / 2}{4 c m} = \frac{20 μ s e c}{4 c m} = \frac{5 μ s e c}{c m}$

Trigger circuit triggers the UJT oscillator which is a $- v e$ resistance device. So slope will be $- v e .$

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