Verified Solution • 2025–26 Session moderate

QAIPMT-2011-Ph-1900: NEET 2011 Physics - Class 12 AIPMT-2011-Ph-1900 Solution: The instantaneous angular position of a point on a rotating…

The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t)=2t3−6t2\theta(t) = 2t^3 - 6t^2 The torque on the wheel becomes zero at t = 1 s t = 0.5 s t = 0.25 s t = 2 s

Multiple Choice Options

  • A 14πε02qL(1+5){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 + \sqrt 5 } \right) 4 � _ 0 _ 1 _ L 2 q _ ( 1 + 5 _ )
  • B 14πε02qL(1+15){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 + {1 \over {\sqrt 5 }}} \right) 4 � _ 0 _ 1 _ L 2 q _ ( 1 + 5 _ 1 _ )
  • C 14πε02qL(1−15){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 - {1 \over {\sqrt 5 }}} \right) 4 � _ 0 _ 1 _ L 2 q _ ( 1 _ 5 _ 1 _ )
  • D zero

Step-by-Step Verified Solution

Distance of point A from the two +q charges = L. Distance of point A from the two �q charges = L 2 + ( 2 L ) 2 = 5 L _ V A = ( K q L _ 2 ) _ ( K q 5 L _ 2 ) = 2 K q L [ 1 _ 1 5 ] = 1 4 � _ 0 . 2 q L ( 1 _ 1 5 )

Related Learning Resources & Exam Preparation

NEET PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring