Q.

W1, W2 and W3   are  the  different  vertical widths of  windows  1,  2  and  3 respectively  placed  in  a vertical  plane.  A  particle is  thrown  up  in  the vertical plane. Let t1, t2 and t3  are the time taken to cross the window W1, W2 and W3 respectively and ΔV1, ΔV2 and ΔV3 are the magnitude of  change in speed after respective window cross.  
Mark the CORRECT option(s).
 

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a

Average speed of the particle passing the windows may be equal if W1 < W2 < W3

b

If  W1 = W2 = W3, the time taken by particle to cross the windows will satisfy t1 < t2 < t3 .

c

Average speed of the particle passing the windows may be equal if W1 > W2 > W3

d

If  W1 = W2 = W3, the change in speed of the particle while crossing the windows  will satisfy ΔV1 < ΔV2 < ΔV3.

answer is C, D.

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Detailed Solution

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As going up, speed of the particle is decreasing and hence the time taken in crossing the windows will be  t1<t2<t3 (if  W1=W2=W3)
Simultaneously  ΔVt
   ΔV1<ΔV2<ΔV3 
As the body moves up its speed is continuously decreasing, hence at lower part of building its average speed will always be higher compared to upper part of building.
 

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