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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2010 · Paper 1 · Physics · Laws of Motion

A block of mass m is on an inclined plane of angle θ . The...

A block of mass mm is on an inclined plane of angle θ\theta. The coefficient of friction between the block and the plane is μ\mu and tan⁡θ>μ\tan \theta>\mu. The block is held stationary by applying a force PP parallel to the plane. The direction of force pointing up the plane is taken to be positive. As PP is varied from P1=mg(sin⁡θ−μcos⁡θ)P_1=m g(\sin \theta-\mu \cos \theta) \quad to P2=mg(sin⁡θ+μcos⁡θ)P_2=m g(\sin \theta+\mu \cos \theta), the frictional force ff versus PP graph will look like A block of mass m is on an inclined plane of angle θ . The coefficient of friction between the block and the plane is μ

Options

  1. AElectrode potential data are given below: Fe^ +3 (aq)+ ^ - Fe^ +2 (aq) ; E^ ^ =+0.77 V Al^ 3+ (aq)+3 ^ - Al_ (s) ; E^ ^
  2. BElectrode potential data are given below: Fe^ +3 (aq)+ ^ - Fe^ +2 (aq) ; E^ ^ =+0.77 V Al^ 3+ (aq)+3 ^ - Al_ (s) ; E^ ^
  3. CElectrode potential data are given below: Fe^ +3 (aq)+ ^ - Fe^ +2 (aq) ; E^ ^ =+0.77 V Al^ 3+ (aq)+3 ^ - Al_ (s) ; E^ ^
  4. DElectrode potential data are given below: Fe^ +3 (aq)+ ^ - Fe^ +2 (aq) ; E^ ^ =+0.77 V Al^ 3+ (aq)+3 ^ - Al_ (s) ; E^ ^

Correct answer

A. Electrode potential data are given below: Fe^ +3 (aq)+ ^ - Fe^ +2 (aq) ; E^ ^ =+0.77 V Al^ 3+ (aq)+3 ^ - Al_ (s) ; E^ ^

Step-by-step solution

When P=mg(sin⁡θ−μcos⁡θ)f=μmgcos⁡θ (upwards)  when \quadP=mgsin⁡θf=0\begin{aligned}P & =m g(\sin \theta-\mu \cos \theta)\\ f & =\mu m g \cos \theta \quad \text{ (upwards) }\\ \text{ when } \quadP & =m g \sin \theta\\ f & =0 \end{aligned}and whenP=m g(\sin \theta+\mu \cos \theta)f=μmgcos⁡f=\mu m g \cos\theta \text{ (downwards) }Hencefrictionisfirstpositive,thenzeroandthennegative.Hence friction is first positive, then zero and then negative.\therefore$ correct option is (a).
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