Verified Solution • 2025–26 Session moderate

QJEEAdv20Ph3a97: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph3a97 Solution: 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 0.00 cm and the right one at 90.00 cm. 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 from the center (50.00 cm) 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 0.40 and 0.32, respectively, the value of xR (in cm) is ______.

Step-by-Step Verified Solution

JEE Advanced 2020 Paper 1 Offline Physics - Laws of Motion Question 13 English Explanation 1

Here, μ\mu s = 0.40, μ\mu k = 0.32

∵\because μ\mu kN3 = μ\mu sN4

and x1N3 = 40 ×\times N4

So, μkx1=μs40⇒0.32x1=0.440{{{\mu _k}} \over {{x_1}}} = {{{\mu _s}} \over {40}} \Rightarrow {{0.32} \over {{x_1}}} = {{0.4} \over {40}}

⇒0.32x1=1100⇒x1=32\Rightarrow {{0.32} \over {{x_1}}} = {1 \over {100}} \Rightarrow {x_1} = 32 cm

JEE Advanced 2020 Paper 1 Offline Physics - Laws of Motion Question 13 English Explanation 2

μsN5x1N5=μsN6xgN6⇒0.4010×32×0.32xg{{{\mu _s}{N_5}} \over {{x_1}{N_5}}} = {{{\mu _s}{N_6}} \over {{x_g}{N_6}}} \Rightarrow {{0.40} \over {10 \times 32}} \times {{0.32} \over {{x_g}}}

⇒xg=25.6\Rightarrow {x_g} = 25.6

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