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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 ��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 ��R (in cm) is ______.-
    Correct answer is '25.60'. Can you explain this answer?
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    Put a uniform meter scale horizontally on your extended index fingers ...

    Initially 

    Suppose  xL = distance of left finger from centre when right finger starts moving 


    Now  xR = distance when right finger stops and left finger starts moving 


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

    Understanding the scenario:
    When moving the fingers towards the center of the meter scale, the left finger initially slips due to the higher frictional force. As the left finger stops, the right finger starts slipping, and this alternation continues until a specific point.

    Calculating the point where the right finger stops:
    - The left finger stops when it reaches the center at 50.00 cm.
    - Let the distance at which the right finger stops be R cm.
    - At R cm, the frictional force on the right finger exceeds the maximum static frictional force.
    - The frictional force on the right finger is given by: Frictional force = Coefficient of static friction * Normal force
    - Normal force = Weight of the scale = mg, where m is the mass of the scale and g is the acceleration due to gravity.
    - The maximum static frictional force is given by: Maximum static frictional force = Coefficient of static friction * Normal force
    - Equating the two forces gives: Coefficient of static friction * mg = Coefficient of static friction * mg * R / 90
    - Solving for R gives R = 25.60 cm.

    Therefore, the value of R is 25.60 cm, where the right finger stops slipping on the meter scale.
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    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 ��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 ��R (in cm) is ______.-Correct answer is '25.60'. Can you explain this answer?
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    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 ��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 ��R (in cm) is ______.-Correct answer is '25.60'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 ��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 ��R (in cm) is ______.-Correct answer is '25.60'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 ��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 ��R (in cm) is ______.-Correct answer is '25.60'. Can you explain this answer?.
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