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Q.

A block of mass m = 15 kg is suspended in an elevator with the help of three identical light elastic cords (spring constant k = 100 N/m each) attached vertically. One of them, cord 1 is tied to the ceiling of the elevator and the other two cords 2 and 3 are tied to the elevator floor as shown in the figure. When the elevator is stationary the tension force in each of the lower cords is T = 7.5 N. Take g = 10 m/s2. Now the elevator starts moving with given four accelerations shown in column I. Column II given the displacement of the block with respect to elevator when it is accelerating. Column III gives tension in the cords when elevator is accelerating.
                                           Question Image

 Column 1
(acceleration of the elevator in different cases)
 
 Column 2
(displacement of the block with respect to the elevator upto new equilibrium position)
 
 Column 3
(Tension in the cord)
 
I)1 m/s2 upwardi)7.5 cm downwardP)2.5 N
II)1.5 m/s2 upwardii)2.5 cm downwardQ)5 N
III)2 m/s2 upwardiii)15 cm downwardR)172.5 N
IV)33 m/s2 downwardiv)5 cm downwardS)0 N

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a

II  ii  S, iv  P, III  iii  Q 

b

II  i  R, I  iv  P, III  iii  S

c

II  iii  Q, I  ii  Q, III  ii  P 

d

II  ii  R, I  iv  Q, III  iii  S 

answer is A.

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Detailed Solution

Question Image

 

 

 

 

 

 

 

T1 = 15 + 150                              T1 = 165N

ΔT1=165100=1.65m   Δx2=Δx3=7.5100=7.5cm

Question Image

 

 

 

 

 

 

 

K(1.65 + x) = 15(11) + 2K(7.5 x 10-2 – x)
165 + 100x = 165 + (15) – 200x
300x = 15
x = 5100  = 5cm

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