1 joule of energy is to be converted into new system of units in which length is measured

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1.Units, Dimensions and Measurement

medium

$1$ $joule$ of energy is to be converted into new system of units in which length is measured in $10\, m$, mass in $10\, kg$ and time in $1$ $minute$ then numerical value of $1\, J$ in the new system is

A$36 \times 10^{-4}$

B$36 \times 10^{-3}$

C$36 \times 10^{-2}$

D$36 \times 10^{-1}$

Solution

$1 \mathrm{J}=2 \quad$ New system
$\mathrm{n}_{2}=\mathrm{n}_{1}\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}\right]^{2}\left[\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}\right]^{-2}$
$\quad=1\left[\frac{\mathrm{kg}}{10 \mathrm{kg}}\right]\left[\frac{\mathrm{m}}{10 \mathrm{m}}\right]^{2}\left[\frac{10}{60}\right]^{-2}$
$\quad=\left(\frac{1}{10}\right)^{1} \times\left(\frac{1}{10}\right)^{2} \times(60)^{2}=36 \times 10^{-1}$

Standard 11

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Match the following two coloumns

Column $-I$ Column $-II$

$(A)$ Electrical resistance $(p)$ $M{L^3}{T^{ – 3}}{A^{ – 2}}$

$(B)$ Electrical potential $(q)$ $M{L^2}{T^{ – 3}}{A^{ – 2}}$

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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

$1$ $joule$ of energy is to be converted into new system of units in which length is measured in $10\, m$, mass in $10\, kg$ and time in $1$ $minute$ then numerical value of $1\, J$ in the new system is

Solution

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