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 $\%$)

While measuring the diameter of a wire using a screw gauge, the following readings were noted. The main scale reading is $1 \,mm$ and the circular scale reading is equal to $42$ divisions. The pitch of the screw gauge is $1 \,mm$ and it has $100$ divisions on the circular scale. The diameter of the wire is $\frac{x}{50} \,mm$. The value of $x$ is:

Figure $1$ shows the configuration of the main scale and Vernier scale before measurement. Figure $2$ shows the configuration corresponding to the measurement of the diameter $D$ of a tube. The measured value of $D$ is (in $cm$)

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a Vernier calliper, if a positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading.
Reason $(R)$: The zero error in a Vernier calliper might have happened due to a manufacturing defect or due to rough handling.
In the light of the above statements, choose the correct answer from the options given below:

In a screw gauge,the fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are $50$ divisions on the circular scale,and the main scale moves by $0.5 \, mm$ on a complete rotation. For a particular observation,the reading on the main scale is $5 \, mm$ and the $20^{th}$ division of the circular scale coincides with the reference line. Calculate the true reading in $mm$.

$A$ travelling microscope is used to determine the refractive index of a glass slab. If $40$ divisions are there in $1 \; cm$ on the main scale and $50$ Vernier scale divisions are equal to $49$ main scale divisions,then the least count of the travelling microscope is $\dots \times 10^{-6} \; m$.