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NEET-XII-Physics

01: Introduction to Physics

by HC Verma - page 2
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  • #1-a
    linear momentum, (b) frequency and (c) pressure.
    Ans : Linear momentum = mv
    Here, [m] = [M] and [v] = [``LT^{-1}``]
    ∴ Dimension of linear momentum, [mv] = [``MLT^{-1}``] (b) Frequency = `` \frac{1}{\, \mathrm{Time}}``
    ∴ Dimension of frequence = `` \left[\frac{1}{T}\right]=\left[{\, \mathrm{M}}^{0}{\, \mathrm{L}}^{0}{\, \mathrm{T}}^{-1}\right]`` (c) Pressure = `` \frac{\, \mathrm{Force}}{\, \mathrm{Area}}``
    `` \, \mathrm{Dimesion}\, \mathrm{of}\, \mathrm{force}=\left[{\, \mathrm{MLT}}^{-2}\right]\phantom{\rule{0ex}{0ex}}\, \mathrm{Dimesion}\, \mathrm{of}\, \mathrm{area}=\left[{\, \mathrm{L}}^{2}\right]\phantom{\rule{0ex}{0ex}}\therefore \, \mathrm{Dimesion}\, \mathrm{of}\, \mathrm{pressure}=\frac{\left[{\, \mathrm{MLT}}^{-2}\right]}{\left[{\, \mathrm{L}}^{2}\right]}=\left[{\, \mathrm{ML}}^{-1}{\, \mathrm{T}}^{-2}\right]``
  • #1-b
    frequency and
    Ans : Frequency = `` \frac{1}{\, \mathrm{Time}}``
    ∴ Dimension of frequence = `` \left[\frac{1}{T}\right]=\left[{\, \mathrm{M}}^{0}{\, \mathrm{L}}^{0}{\, \mathrm{T}}^{-1}\right]``
  • #1-c
    pressure.
    Ans : Pressure = `` \frac{\, \mathrm{Force}}{\, \mathrm{Area}}``
    `` \, \mathrm{Dimesion}\, \mathrm{of}\, \mathrm{force}=\left[{\, \mathrm{MLT}}^{-2}\right]\phantom{\rule{0ex}{0ex}}\, \mathrm{Dimesion}\, \mathrm{of}\, \mathrm{area}=\left[{\, \mathrm{L}}^{2}\right]\phantom{\rule{0ex}{0ex}}\therefore \, \mathrm{Dimesion}\, \mathrm{of}\, \mathrm{pressure}=\frac{\left[{\, \mathrm{MLT}}^{-2}\right]}{\left[{\, \mathrm{L}}^{2}\right]}=\left[{\, \mathrm{ML}}^{-1}{\, \mathrm{T}}^{-2}\right]``