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1.Units, Dimensions and Measurement
hard
A force is represented by $\mathrm{F}=a \mathrm{x}^2+\mathrm{bt}^{1 / 2}$. Where $\mathrm{x}=$ distance and $\mathrm{t}=$ time. The dimensions of $\mathrm{b}^2 / \mathrm{a}$ are :
A $\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]$
B$\left[\mathrm{MLT}^{-2}\right]$
C$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$
D $\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]$
(JEE MAIN-2024)
Solution
$\mathrm{F}=\mathrm{ax}^2+\mathrm{bt}^{1 / 2}$
${[\mathrm{a}]=\frac{[\mathrm{F}]}{\left[\mathrm{x}^2\right]}=\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}$
${[\mathrm{b}]=\frac{[\mathrm{F}]}{\left[\mathrm{t}^{1 / 2}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-5 / 2}\right]}$
${\left[\frac{\mathrm{b}^2}{\mathrm{a}}\right]=\frac{\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-5}\right]}{\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^3 \mathrm{~T}^{-3}\right]}$
${[\mathrm{a}]=\frac{[\mathrm{F}]}{\left[\mathrm{x}^2\right]}=\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}$
${[\mathrm{b}]=\frac{[\mathrm{F}]}{\left[\mathrm{t}^{1 / 2}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-5 / 2}\right]}$
${\left[\frac{\mathrm{b}^2}{\mathrm{a}}\right]=\frac{\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-5}\right]}{\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^3 \mathrm{~T}^{-3}\right]}$
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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