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One physical quantity represented by an equation as $ { \pi \over 2 } ( p -q ) c $ where p, q and c are length then quantity is ..
$ if p =q = c =l $ $ then ( p -q) c = L^2 =Area $
The dimensional formula of magnetic flux is ...............
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 $[M^1 L^2 T^{-2} A^{-1}]$.
Which physical quantity has unit of pascal - second ?
$ if F = nA { dv \over dx } $ $ \therefore n = { F \over A { dv \over dx } } = pascal sound $
Dimensional formula of CV ? where C - capacitance and V - potential different
The equation of a wave is given by $ Y = A sin \omega \left[ { x \over v } - k \right] $ where $ \omega $ is the angular velocity and $ \nu $ is the linear velocity. Write the dimensional formula of K
$ y = A sin \omega ( { x \over v } - k ) $ $ \therefore { x \over v } = k $ $ k = { x \over v } = M^0 L ^0 T^ 1 $
If P and q are different physical quantities then which one of following is only possible dimensionaly ?
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.
From $ \left( p + { a \over v^2 } \right) ( v-b) $   constant equation is dimensionally correct find the dimensional formula for b ? where P = preasure V = volume
$ \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 $
Pressure P = A cosBx + c sinDt where xin meter and t in time then find dimensional formula of D/B
$ 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} $
Find the dimensional formula for energy per unit surface area per unit time
Energy per unit surface area per unit time is also known as energy flux. The dimensional formula for energy is $ML^2T^{-2}$. Surface area has the dimension of $L^2$ and time has the dimension of $T$. Therefore, the dimensional formula for energy flux is obtained by dividing the dimensional formula of energy by the dimensional formula of surface area and time: $$ rac{ML^2T^{-2}}{L^2 imes T} = M^1 L^0 T^{-3}.$$
Pressure $ P = { at ^2 \over bx } $ where x = distance, t= time find the dimensional formula for a/b
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: $$ rac{a}{b} = ML^{-1}T^{-4}.$$
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