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JEE Advanced 2013Paper 2
Exam
JEE Advanced
Year
2013
Paper
Paper 2
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2013 · Paper 2 · Physics · Laws of Motion

Paragraph: A small block of mass 1 ~kg is released from rest at the...

Paragraph: A small block of mass 1 kg1 \mathrm{~kg} is released from rest at the top of a rough track. The track is a circular arc of radius 40 m40 \mathrm{~m}. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point QQ, as shown in the figure below, is 150 J150 \mathrm{~J}. (Take the acceleration due to gravity, g=10 m s−2g=10 \mathrm{~m} \mathrm{~s}^{-2} ). Paragraph: A small block of mass 1 ~kg is released from rest at the top of a rough track. The track is a circular arc

Question: The magnitude of the normal reaction that acts on the block at the point QQ is

Options

  1. A7.5N7 . 5N
  2. B8.6N8 . 6N
  3. C11.5N11 . 5N
  4. D225N225N

Correct answer

A. 7.5N7 . 5N

Step-by-step solution

From FBD
N−mgsin30°=mv2R⇒N=1×(10)240+1×102⇒N=7.5NN - mgsin30 ° = \dfrac{m{v}^{2}}{R} ⇒ N = \dfrac{1 \times {\left(10\right)}^{2}}{40} + \dfrac{1 \times 10}{2} ⇒ N = 7 .5 N
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