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Physics
Question 1 of 2Two physical quantities having the same dimensional formula $\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]$ are
Explanation:
Stress and pressure both have the dimensions of force per unit area. ⢠A force F has dimensions $$[F] = [M\,L\,T^{-2}].$$ ⢠Dividing by area (dimensions $L^2$) gives $$\frac{F}{L^2} = [M\,L\,T^{-2}]\;[L^{-2}] = [M\,L^{-1}\,T^{-2}].$$ Hence stress and pressure share the dimensional formula $$[M^1\,L^{-1}\,T^{-2}],$$ so the correct answer is Option C.
Physics
Question 2 of 2In the expression $\mathrm{P}=\mathrm{El}^2 \mathrm{~m}^{-5} \mathrm{G}^{-2}$ where $\mathrm{E}, 1, \mathrm{~m}$ and G represent Energy, Angular Momentum, Mass and Gravitational Constant, the dimensions of P are
Explanation:
LetâÂÂs write down the dimensions of each symbol and then combine them. Energy, $E$: $ [E] = \mathrm{M}^1 \,\mathrm{L}^2 \,\mathrm{T}^{-2} $ Angular momentum, $\ell$: $ [\ell] = \mathrm{M}^1 \,\mathrm{L}^2 \,\mathrm{T}^{-1} $ Mass, $m$: $ [m] = \mathrm{M}^1 $ Gravitational constant, $G$ (from $F = G\,m_1m_2/r^2$): $ [G] = \frac{[\!F\!]\,[r]^2}{[m]^2} = \frac{\mathrm{M}^1\mathrm{L}^1\mathrm{T}^{-2}\;\mathrm{L}^2}{\mathrm{M}^2} = \mathrm{M}^{-1}\,\mathrm{L}^3\,\mathrm{T}^{-2} $ Now for $ P = E\,\ell^2\,m^{-5}\,G^{-2}, $ we add exponents of $\mathrm{M}$, $\mathrm{L}$, $\mathrm{T}$: ⢠M-exponent: $1$ (from $E$) $+\;2\times1$ (from $\ell^2$) $-5\times1$ (from $m^{-5}$) $-2\times(-1)$ (from $G^{-2}$) = $1 + 2 - 5 + 2 = 0$ ⢠L-exponent: $2$ (from $E$) $+\;2\times2$ (from $\ell^2$) $-2\times3$ (from $G^{-2}$) = $2 + 4 - 6 = 0$ ⢠T-exponent: $-2$ (from $E$) $+\;2\times(-1)$ (from $\ell^2$) $-2\times(-2)$ (from $G^{-2}$) = $-2 - 2 + 4 = 0$ So $ [P] = \mathrm{M}^0\,\mathrm{L}^0\,\mathrm{T}^0, $ which is dimensionless. Answer: Option C.