(B) The dimensional formula for pressure is $[M^1 L^{-1} T^{-2}]$.Given: $M_1 = 50 \ g = 0.05 \ kg$,$L_1 = 10 \ cm = 0.1 \ m$,$T_1 = 5 \ s$.In $SI$ un

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(B) The dimensional formula for pressure is $[M^1 L^{-1} T^{-2}]$.Given: $M_1 = 50 \ g = 0.05 \ kg$,$L_1 = 10 \ cm = 0.1 \ m$,$T_1 = 5 \ s$.In $SI$ units: $M_2 = 1 \ kg$,$L_2 = 1 \ m$,$T_2 = 1 \ s$.The conversion formula is $n_2 = n_1 [M_1/M_2]^1 [L_1/L_2]^{-1} [T_1/T_2]^{-2}$.Substituting the values: $n_2 = 1 \times [0.05/1]^1 \times [0.1/1]^{-1} \times [5/1]^{-2}$.$n_2 = 0.05 \times (1/0.1) \times (1/25)$.$n_2 = 0.05 \times 10 \times 0.04 = 0.5 / 25 = 1/50 \ \text{Pascal}$.

Class 11 Physics Units, Dimensions and Measurement Dimensional Analysis, Uses and Limitations

Medium

In a certain system of units,one unit of mass is $50 \ g$,one unit of length is $10 \ cm$,and one unit of time is $5 \ s$. Then,in this system,$1$ unit of pressure equals to :-

A
$50 \ \text{Pascal}$
B
$1/50 \ \text{Pascal}$
C
$100 \ \text{Pascal}$
D
$1/100 \ \text{Pascal}$

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Dimensional Analysis, Uses and Limitations

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In a certain system of units,one unit of mass is $50 \ g$,one unit of length is $10 \ cm$,and one unit of time is $5 \ s$. Then,in this system,$1$ unit of pressure equals to :-

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