Thickness of a pencil measured by a screw gauge (least count $0.001 \ cm$) comes out to be $0.802 \ cm$. The percentage error in the measurement is (in $\%$)

The diagrams show readings of a screw gauge. Figure $(i)$ shows the zero error reading when the screw gauge is closed,and figure $(ii)$ shows the reading when the screw gauge is being used to measure the diameter of a ball-bearing. What is the diameter of the ball-bearing in $mm$? There are $50$ divisions on the circular scale.

$A$ specially designed Vernier calliper has the main scale least count of $1 \,mm$. On the Vernier scale,there are $10$ equal divisions and they match with $11$ main scale divisions. Then,the least count of the Vernier calliper is ........... $mm$.

Match the instruments in Column-$I$ with their respective least counts in Column-$II$.

Column-$I$	Column-$II$
$(1)$ Traveling Microscope	$(a)$ $0.01\,cm$
$(2)$ Screw Gauge	$(b)$ $0.001\,cm$
	$(c)$ $0.0001\,cm$

In a screw gauge, the zero of the main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions on the circular scale and the pitch of the screw gauge is $0.1 \text{ mm}$. When the diameter of a sphere is measured, the reading of the main scale is $5 \text{ mm}$ and the $50^{th}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the sphere is . . . . . . $\text{mm}$.

The pitch of a screw gauge is $0.5 \ mm$ and there are $100$ divisions on its circular scale. The instrument reads $+2$ divisions on the circular scale when nothing is placed between its jaws. In measuring the diameter of a wire,there are $8$ divisions on the main scale and the $83^{rd}$ division of the circular scale coincides with the reference line. Then the diameter of the wire is (in $mm$)