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The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring diameter of a sphere, the main scale reading is $1.7 \,cm$ and coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$
To determine the Young's modulus of a wire, the formula is $Y = \frac{FL}{A\Delta L};$ where $L$ = length, $A = $area of cross-section of the wire, $\Delta L = $change in length of the wire when stretched with a force $F$. The conversion factor to change it from $CGS$ to $MKS$ system is .............. $10^{-1}\mathrm{N/m}^{2}$
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: [position $]=\left[X^\alpha\right] ;[$ speed $]=\left[X^\beta\right]$; [acceleration $]=\left[X^{ p }\right]$; [linear momentum $]=\left[X^{ q }\right]$; [force $]=\left[X^{ I }\right]$. Then -
In a vernier callipers, $(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1 \mathrm{MSD}$ represents $0.1 \mathrm{~mm}$, the vernier constant (in $\mathrm{cm}$ ) is:
In the equation $y = pq$ $tan\,(qt)$, $y$ represents position, $p$ and $q$ are unknown physical quantities and $t$ is time. Dimensional formula of $p$ is