$(a)$ Screw moves $0.5\,mm$ on main scale in one complete rotation
$(b)$ Total divisions on circular scale $=50$
$(c)$ Main scale reading is $2.5\,mm$
$(d)$ $45^{\text {th }}$ division of circular scale is in the pitch line
$(e)$ Instrument has $0.03 \;mm$ negative error
Then the diameter of wire is $...........\,mm$
| List $I$ | List $II$ |
| $A$ Torque | $I$ ${\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}$ |
| $B$ Magnetic fileld | $II$ $\left[\mathrm{L}^2 \mathrm{~A}^1\right]$ |
| $C$ Magneti moment | $III$ ${\left[\mathrm{M}^1 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]}$ |
| $D$ permeability of free space | $IV$ $\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$ |
Choose the correct answer from the options given below :
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.
where $X = \frac{{{A^2}{B^{\frac{1}{2}}}}}{{{C^{\frac{1}{3}}}{D^3}}}$, will be