Physics · Friction

JEE Advanced 2020 — Paper 1 — Question 12

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 xRx_{R} 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(incm)x_{R}(\mathrm{in} \mathrm{cm}) is \qquad —.

Answer: 25.6

Numerical answer — enter this value.

Step-by-step solution

Initially

40 N1=50 N240 \mathrm{~N}_{1}=50 \mathrm{~N}_{2}

For first move

μk N3=μs N4..(i)xN3=40 N4..(ii)x=32 cm\begin{aligned} & \mu_{\mathrm{k}} \mathrm{~N}_{3}=\mu_{\mathrm{s}} \mathrm{~N}_{4} ..(i) \\& \mathrm{xN}_{3}=40 \mathrm{~N}_{4} ..(ii) \\& \mathrm{x}=32 \mathrm{~cm} \end{aligned}

For second move

μs N5=μk N6..(i)32( N5)=XR N6..(ii)XR=25.6 cm\begin{aligned} & \mu_{\mathrm{s}} \mathrm{~N}_{5}=\mu_{\mathrm{k}} \mathrm{~N}_{6} ..(i) \\& 32\left(\mathrm{~N}_{5}\right)=\mathrm{X}_{\mathrm{R}} \mathrm{~N}_{6} ..(ii) \\& \mathrm{X}_{\mathrm{R}}=25.6 \mathrm{~cm} \end{aligned}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Physics
Chapter
Friction
Topic
Problems with Critical Understanding of Frictional Force