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1.Units, Dimensions and Measurement
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The density of a material is $CGS$ system of units is $4\,g/cm^3$. In a system of units in which unit of length is $10\,cm$ and unit of mass is $100\,g$, the value of density of material will be
A$400$
B$0.04$
C$0.4$
D$40$
Solution
$4 \mathrm{g} \mathrm{cm}^{-3}=\frac{?}{\mathrm{n}_{2}} \mathrm{M}_{2} \mathrm{L}_{2}^{-3}$
$\mathrm{n}_{2}=4\left[\frac{\mathrm{g}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{cm}}{\mathrm{L}_{2}}\right]^{-3}$
$=4\left[\frac{\mathrm{g}}{100 \mathrm{g}}\right]\left[\frac{\mathrm{cm}}{10 \mathrm{cm}}\right]^{-3}$
$=40$
$\mathrm{n}_{2}=4\left[\frac{\mathrm{g}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{cm}}{\mathrm{L}_{2}}\right]^{-3}$
$=4\left[\frac{\mathrm{g}}{100 \mathrm{g}}\right]\left[\frac{\mathrm{cm}}{10 \mathrm{cm}}\right]^{-3}$
$=40$
Standard 11
Physics
Similar Questions
Match the following two coloumns
Column $-I$ | Column $-II$ |
$(A)$ Electrical resistance | $(p)$ $M{L^3}{T^{ – 3}}{A^{ – 2}}$ |
$(B)$ Electrical potential | $(q)$ $M{L^2}{T^{ – 3}}{A^{ – 2}}$ |
$(C)$ Specific resistance | $(r)$ $M{L^2}{T^{ – 3}}{A^{ – 1}}$ |
$(D)$ Specific conductance | $(s)$ None of these |
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