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Standard XII

Physics

Order of Magnitude

The velocity ...

Question

The velocity of central maxima at $t = 5 s$ is

50 m

s^{- 1}

along Y-axis

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50 m

s^{- 1}

along negative Y-axis

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3 \times 10^{8} m

s^{- 1}

along Y-axis

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3 \times 10^{8} m

s^{- 1}

along negative Y-axis

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Solution

The correct option is A $50 m$ $s^{- 1}$ along Y-axis
AS the screen will move downward with an acceleration $g = 10 m s^{- 2}$
Now velocity of screen will be the velocity of central maxima.
Thus at $t = 5 s$ , $\Rightarrow v = 0 + g (5) = 50 m s^{- 1}^j$

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The velocity of central maxima at t=5s is

The correct option is A 50m s−1 along Y-axisAS the screen will move downward with an acceleration g=10ms−2Now velocity of screen will be the velocity of central maxima.Thus at t=5s, ⇒v=0+g(5)=50ms−1^j

The velocity of central maxima at $t = 5 s$ is

The correct option is A $50 m$ $s^{- 1}$ along Y-axis
AS the screen will move downward with an acceleration $g = 10 m s^{- 2}$
Now velocity of screen will be the velocity of central maxima.
Thus at $t = 5 s$ , $\Rightarrow v = 0 + g (5) = 50 m s^{- 1}^j$