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Physics - NLM Question with Solution | TestHub

PhysicsNLMMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A block of mass $m$ is placed over rough inclined plane having inclination $30 °$ . The co-efficient of friction between the block and the inclined plane is $0.75$ . The contact force on the block is

Options:

Answer:

Solution:

$mgsin θ = \frac{\sqrt{3}}{2} mg = f$

$f_{l} = μmgcos θ$

$= \frac{3}{4} mg \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{8} mg$

$f < f_{r}$

$f = mgsin θ$

$N = mgcos θ$

$f_{net} = mg \sqrt{\sin^{2} θ + \cos^{2} θ}$

$= mg$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:NLMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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