General Physics Ques 9
- Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II.
(2007, 6M)
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Answer:
Correct Answer: 9.( $A-p, q, \quad B-r, s, C-r, s, D-r, s$ )
Solution:
$ F=\frac{G M_e M_s}{r^2} $
$ =\text { Gravitational force between sun and earth } $
$ \Rightarrow \quad G M_e M_s=F r^2 $
$ \therefore \quad [\mathrm{GM} _{\mathrm{e}} \mathrm{M} _{\mathrm{s}} ]= [\mathrm{Fr}^2 ]= [\mathrm{MLT}^{-2} ] [\mathrm{L}^2 ]= [\mathrm{ML}^3 \mathrm{T}^{-2} ] $
(B) $ \begin{aligned} & \text { B) } v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}=\text { rms speed of gas molec } \\ & \therefore \quad \frac{3 R T}{M}=v_{\mathrm{rms}}^2 \\ & \text { or }\left[\frac{3 \mathrm{RT}}{\mathrm{M}}\right]=\left[\mathrm{v}_{\mathrm{rms}}^2\right]=\left[\mathrm{LT}^{-1}\right]^2=\left[\mathrm{L}^2 \mathrm{I}^{-2}\right] \end{aligned} $
(C) $F=B q v=$ magnetic force on a charged particle
$ \therefore \quad \frac{F}{B q}=v \text { or }\left[\frac{\mathrm{F}^2}{\mathrm{B}^2 \mathrm{q}^2}\right]=[\mathrm{v}]^2=\left[\mathrm{L}^2 \mathrm{T}^{-2}\right] $
(D) $v_o=\sqrt{\frac{G M_e}{R_c}}=$ orbital velocity of earth’s satellite
$ \therefore \quad \frac{G M_e}{R_{\varepsilon}}=v_o^2 \quad \text { or }\left[\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R} _{\mathrm{e}}}\right]=\left[\mathrm{v} _{\mathrm{o}}^2\right]=\left[\mathrm{L}^2 \mathrm{T}^{-2}\right] $
(p) $W=q V \Rightarrow$ (Coulomb) $($ Volt $)=$ Joule or $[($ Volt ) (Coulomb) (Metre) $]=[$ (Joule) (Metre) $]$ $ =\left[\mathrm{ML}^2 \mathrm{T}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^3 \mathrm{T}^{-2}\right] $
(q) $\left.\left[\text { (kilogram) }(\text { metre })^3 \text { (second }\right)^{-2}\right]=\left[\mathrm{ML}^3 \mathrm{T}^{-2}\right]$
(r) $\left.\left[(\text { metre })^2 \text { (second }\right)^{-2}\right]=\left[\mathrm{L}^2 \mathrm{T}^{-2}\right]$ (s) $U=\frac{1}{2} C V^2 \Rightarrow$ (farad) $(\text { volt })^2=$ Joule or $\left[(\right.$ farad $\left.)(\text { volt })^2(\mathrm{kg})^{-1}\right]=\left[(\right.$ Joule $\left.)(\mathrm{kg})^{-1}\right]$
$ =\left[\mathrm{ML}^2 \mathrm{T}^{-2}\right]\left[\mathrm{M}^{-1}\right]=\left[\mathrm{L}^2 \mathrm{T}^{-2}\right] $
BETA
