Verified Solution • 2025–26 Session moderate

QAIPM-2011-Ph-1885: NEET 2011 Physics - Class 11 AIPM-2011-Ph-1885 Solution: The dimensions of \(( _0 _0)^{-1/2}\) are \([L^{1/2}T^{-1/2}]\) …

The dimensions of

(\mu_0\epsilon_0)^{-1/2}

are

[L^{1/2}T^{-1/2}]

[L^{-1}T]

[LT^{-1}]

[L^{1/2}T^{1/2}]

Step-by-Step Verified Solution

Work done = Area under (F-d) graph = Area of rectangle ABCD + Area of triangle DCE = 2 _ ( 7 _ 3 ) + 1 2 _ 2 _ ( 12 _ 7 ) = 8 + 5 = 13 J