Measurement MCQs for NEET — Physics Questions with Answers

Given the equation for pressure $P = rac{at^2}{bx}$, we need to find the dimensional formula for $rac{a}{b}$. The dimensional formula for pressure $P$ is $ML^{-1}T^{-2}$. The dimensional formula for time $t$ is $T$ and for distance $x$ is $L$. Rewriting the given equation in terms of dimensional formulas, we have: $$ML^{-1}T^{-2} = rac{a imes T^2}{b imes L}$$ Solving for $rac{a}{b}$, we get: $$rac{a}{b} = ML^{-1}T^{-4}.$$

$ F = A ( 1 - e^{-Bxt^2 } )$ $ Bxt^2 = dimensional less $ $ B = { M^0 L^0 T^0 \over xt^2 } = M^0 L^{-1} T^{-2} $

$ Density = 3.125 \times { gm \over cm^3 } $ $ = { 3.125 \times 10 ^ {-3} kg \over 10 ^ {-6} m^3 } $ $ = 3125 kg / m^3 $

$$ volume of cube V = a^ 3$$ $$ total surface area of cube A = 6a^2$$ $ \therefore V = A $ $ a^3 = 6a^ 2$ $a = 6$ $ \therefore V = ( 6) ^3 = 216 unit $

$ E = G^p h^q c^r $ $ E = M^1 L^2 T^{-2} $ $ G = M^ {-1} L^3 T^{-2} $ $ h = M^1 L^2 T^{-1} $ $ c = M^0 L^1 T^{-1} $ take it $ ( M^1 L^2 T^{-2} ) = ( M^{-1} L^3 T^{-2} )^p = ( M^1 L^2 T^{-1} )^q (M^0 L^1 T^{-1} )^c $ $ = M ^ { -p +q } L^{3p + 2q+r} T^{-2p -q-r } $ $ \therefore P = {1 \over 2 } , q = {1 \over 2 } , r = { 5 \over 2 } $

$ F \alpha m^a \nu^b r^c $ $ F = M^1 L^1 T^{-2} $ $ \nu = M^0 L^1 T^{-1} $ $ r = M^0 L^1 T^0 $ $ m = M^1 L^0 T^0 $ take it $ ( M^1 L^1 T^{-2} ) = (M^1)^a (L^1 T^{-1} )^b (L^1)^c $ $ = M^a L^{b+c } T^{-b} $ $ \therefore a =1 , b =2 , c = -1 $