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If the dimensions of a physical quantity quantity are given by MaLbTc, then the physical quantity will be:

  • Option 1)

    Velocity if a=1, b=0,c=-1

  • Option 2)

    Acceleration if a=1,b=1,c=-2

  • Option 3)

    Force if a=0,b=-1,c=-2

  • Option 4)

    Pressure if a=1,b=-1,c=-2

 

Answers (1)

best_answer

As we learnt

Time period of a simple pendulum -

T=Km^{a}l^{b}g^{0}

Equating exponents of similar quantities

a=0     b=1/2    c=-1/2

	herefore T= 2pi sqrt{l/g}

- wherein

T= time : period

l= length

g=: acceleration : due: to: gravity

 

 

Pressure, P= \frac{Force}{Area}=\frac{mass \times acceleration}{Area}

P= \frac{MLT^{-2}}{L^{2}}={M}^{1}{L}^{-1}T^{-2}= M^{a}L^{b}T^{c}

\therefore a=1, b=1, c=-2


Option 1)

Velocity if a=1, b=0,c=-1

This option is incorrect

Option 2)

Acceleration if a=1,b=1,c=-2

This option is incorrect

Option 3)

Force if a=0,b=-1,c=-2

This option is incorrect

Option 4)

Pressure if a=1,b=-1,c=-2

This option is correct

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