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Paragraph for Questions numbers 14 to 16 :

Mass of a string varies with its length as m = 0.1L where m is in kg and L is in cm. A block of mass 5kg is hanging to its lower end. A force F = 12 N acts on the block in vertically downward direction as shown in the figure. The entire system is kept in an elevator which moves up with an acceleration of 2 m/s^2. Length of the string is 10 cm. Height of the elevator is 4m. Assume mass of (string + mass) system to be particle like at the point of interest. Take g = 10 m/s^2.

14. Tension at a distance 10 cm above the block is
(a) 60 N (b) 70 N (c) 72 N (d) 84 N

15. Force on the string from the ceiling of the elevator is
(a) 60 N (b) 70 N (c) 72 N (d) 84 N

16. Suppose the string connecting mass to ceiling of the elevator breaks at the point P (at ceiling; see figure) after some time but the force F continues to act. The acceleration of the (string + block) system just after the string breaks assuming (string + block) system to be particle is
(a) 2 m/s2 upward
(b) zero
(c) 10 m/s2 downward
(d) 12 m/s2 downward

4 Answers

1
skygirl ·

hmm.. if the length is measured from top..... dun u think it will break ??

3
iitimcomin ·

luks more like .... the linear density of string is 0.1 ..... therez no variable density acc. to ur wordings ...........

14 .. shud be 84 ............frm wat i interpreted ...

3
iitimcomin ·

FOR SECOND PART U CAN CONSIDER A POINT MASS OF 6kg ... hangingin frm a massless thread ...

clearly tension = 6*12 + 12 = 84N ....

now at point of contact wit elevator..........R-T = 0 .......R=T=84N...

21
tapanmast Vora ·

14 me 2 gettin 84.... but shayad koi trap ho sakta hai ......

iiti.... wata bt da rest of ur ans?

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