Physics MCQs for NEET — Practice Questions with Answers

Linear momentum has the dimensional formula $[p] = MLT^{-1}$, and angular momentum has the dimensional formula $[L] = ML^2T^{-1}$. The ratio of linear momentum to angular momentum is $rac{[p]}{[L]} = rac{MLT^{-1}}{ML^2T^{-1}} = L^{-1}$. Therefore, the dimensional formula of the ratio of linear momentum to angular momentum is $M^0L^{-1}T^0$.

$ U rms = \sqrt { \langle u^2 \rangle } = root mean square speed $

$ if p =q = c =l $ $ then ( p -q) c = L^2 =Area $

The dimensional formula of magnetic flux (Φ) is derived from its definition, which is the product of magnetic field (B) and area (A). The dimensional formula for magnetic field (B) is $[M^1 L^0 T^{-2} A^{-1}]$, and area (A) has the dimensional formula $[L^2]$. Combining these, the dimensional formula for magnetic flux becomes $[M^1 L^2 T^{-2} A^{-1}]$.

$ if F = nA { dv \over dx } $ $ \therefore n = { F \over A { dv \over dx } } = pascal sound $

$ y = A sin \omega ( { x \over v } - k ) $ $ \therefore { x \over v } = k $ $ k = { x \over v } = M^0 L ^0 T^ 1 $

When dealing with different physical quantities, only division (p/q) is dimensionally possible. This is because for addition (p+q) or subtraction (p-q) to be valid, both quantities must have the same dimensions. Similarly, for equality (p=q), both sides must have the same dimensions. Therefore, only the division (p/q) is dimensionally possible if P and q are different.

$ \left( P + { a \over v^2 } \right) ( v - b ) = constant $ $ PV - Pb + { a \over \nu } - { ab \over \nu^2 } = constant $ $ \therefore PV - Pb $ $ \therefore V = b = M^0 L^3 T^0 $

$ cos Bx = dimensional less $ $ Bx = M^0 L^0 T^0 $ $ B = { M^0 L^0 T^0 \over X} = M^0 L^{-1} T^0$ same as $ D = M^0 L^0 T^ {-1} $ $\therefore { D \over B } = M^0 L^1 T^{-1} $