In a screw gauge, $$5$$ complete rotations of the screw cause it to move a linear distance of $$0.25$$ $$cm.$$ There are $$100$$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $$4$$ main scale divisions and $$30$$ circular scale divisions. Assuming negligible zero error, the thickness of the wire is :

The relative error in the determination of the surface area of sphere is $$\alpha $$. Then the relative error in the determination of its volume is :

A physical quantity P is described by the relation

P = a$$^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}$$ b² c³ d^$$-$$4

If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be :

Time (T), velocity (C) and angular momentum (h) are chosen as fundamentalquantities instead of mass, length and time. In terms of these, the dimensions of mass would be :