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JEE Advanced 2020Paper 1
Exam
JEE Advanced
Year
2020
Paper
Paper 1
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2020 · Paper 1 · Physics · Laws of Motion

Put a uniform meter scale horizontally on your extended index...

Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00cm0 . 00cm and the right one at 90.00cm90 . 00cm. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance xR{x}_{R} from the center (50.00cm)( 50 . 00cm ) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.400 . 40 and 0.320 . 32, respectively, the value of xR{x}_{R} (in cm) is___________.Put a uniform meter scale horizontally on your extended index fingers with the left one at 0 . 00cm and the right onePut a uniform meter scale horizontally on your extended index fingers with the left one at 0 . 00cm and the right onePut a uniform meter scale horizontally on your extended index fingers with the left one at 0 . 00cm and the right one

Correct answer

0

Step-by-step solution

Put a uniform meter scale horizontally on your extended index fingers with the left one at 0 . 00cm and the right one Put a uniform meter scale horizontally on your extended index fingers with the left one at 0 . 00cm and the right one 5x′=405{x}^{'} = 40 x′=405{x}^{'} = \dfrac{40}{5} ∴4x′=32∴ 4{x}^{'} = 32 Put a uniform meter scale horizontally on your extended index fingers with the left one at 0 . 00cm and the right one 5{x}^{' '} = 32 {x}^{' '} = \dfrac{32}{5} ∴ 4{x}^{' '} = \dfrac{4}{5} \times 32 = {x}_{R} xR=25.6{x}_{R} = 25 . 6
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