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  • A point object is located at a distance of 100 m from a screen. A lens of focal length 23 cm mounted on a movable frictionless stand is kept between the source and the screen. The stand is attached

A point object is located at a distance of 100 m from a screen. A lens of focal length 23 cm mounted on a movable frictionless stand is kept between the source and the screen. The stand is attached to a spring of natural length 50cm and spring constant 800 N/m as shown. Mass of the stand with lens is 2 kg. How much impulse P should be imparted to the stand so that a real image of the object is formed on the screen after a fixed time gap? (Neglect width of the stand).

Option: 1

5 \mathrm{kgm} / \mathrm{s}


Option: 2

6\, \mathrm{kgm} / \mathrm{s}


Option: 3

7\, \mathrm{kgm} / \mathrm{s}


Option: 4

8\, \mathrm{kgm} / \mathrm{s}


Answers (1)

best_answer

Let the distance of the lens from the object be \ell when a real image is formed on the screen. Then \mathrm{\frac{1}{100-\ell}=\frac{1}{-\ell}=\frac{1}{23}}

solving, we get \mathrm{\ell=(50 \pm 10 \sqrt{2}) \mathrm{cm}}.

Now, if the lens performs SHM and real image is formed after a fixed time gap then this time gap must be one fourth of the time period.

\mathrm{\therefore} Phase difference between the two positions of real image must be \mathrm{\pi / 2}. As the two positions are symmetrically located about the origin, phase difference of any of these positions from origin must be \mathrm{\pi / 4}.

  \mathrm{\Rightarrow \quad 10 \sqrt{2} \mathrm{~cm}=A \sin \frac{\pi}{4} \Rightarrow A=20 \mathrm{~cm}}
To achieve this velocity at the mean position
\mathrm{\quad v_{0}=A \omega=A \sqrt{\frac{K}{m}}}

\mathrm{\therefore}  Required impulse
\mathrm{p=m v_{0}=A \sqrt{K m}=8 \mathrm{kgm} / \mathrm{s} }.

Posted by

rishi.raj

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