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Question

# A table with a smooth horizonal surface is placed in a cabin which moves in a horizontal circle of a large radius R. A smooth pulley of small radius is fixed to the table. Two masses m and 2m placed on the table are connected through a light string going over the pulley. System is released from rest with respect to the cabin. The initial tension in the string will be:

A
13mω2R
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B
43mω2R
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C
23mω2R
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D
mω2R
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Solution

## The correct option is B 43mω2RLet mass of block A=m, mass of block B=2m. Direction of centrifugal force (F & F′) will be radially outward on both the blocks, since acceleration of rotating frame (a=ω2R) is radially inwards towards centre of circle. ⇒F=mω2R on block A and F′=2mω2R on block B R= radius of horizontal circle in which cabin is rotating. Blocks will be moving with same magnitude of linear acceleration (a) relative to rotating frame, due to string constraint. FBD of blocks: ∵ Radius of circle is very large, hence length of string and radius of pulley can be neglected. Radius of each mass can be taken to be equal to R, from centre of horizontal circle. FBD of block A: T−mω2R=ma ...(i) FBD of block B: 2mω2R−T=2ma ...(ii) On adding Eq. (i) & (ii) mω2R=3ma ⇒a=ω2R3 ....(iii) Substituting in Eq (i), T−mω2R=mω2R3 ⇒T=mω2R3+mω2R ∴T=43mω2R So, initial tension in the string is 43mω2R

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