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In a new system of units, the net force applied on a block of mass 10 kg moving with acceleration 10 m/s² is given as 100 unit of force. When the same block is moving at a speed of 20 m/s, its kinetic energy become 20 unit of energy. A liquid has a surface tension of 10 SI units. What is the magnitude of its surface tension in new system ? (Dimensional formula of surface tension is ML⁰T^-2.)

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Detailed Solution

Let in new system, dimensions be M', L' and T'

$\begin{matrix} ∴ Given , 100 M^{'} L^{'} T^{' - 2} = 100 {MLT}^{- 2} . . . . . . . . . . (i) \\ 20 M^{'} L^{' 2} T^{' - 2} = 20 \times 100 {ML}^{2} T^{- 2} . . . . . . . . . . (ii) \end{matrix}$

Solving (i) & (ii); $L^{'} = 100 L$

$∴ M^{'} T^{' - 2} = \frac{1}{100} {MT}^{- 2}$

$\Rightarrow 10 {MT}^{- 2} = 10^{3} M^{'} T^{' - 2}$

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