Units and Measurement

A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5 mm is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively:
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The dimension of stopping potential $V_0$ ${}_{}$ in photoelectric effect in units of Planck's constant ' $h$ $$ ', speed of light ' $c$ $$ ' and Gravitational constant ' $G$ $$ ' and ampere $A$ is:

The dimension of $\frac{{\mathrm{B}}^2}{2{\mu }_0}$ , where $\mathrm{B}$ is magnetic field and ${\mu }_0$  is the magnetic permeability of vacuum, is
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The resistance R = V/I where V =(50 ± 2V) and I = (20 ± 0.2)A. The percentage error in R is ‘x’ %. The value of ‘x’ to the nearest integer is ----------.

One main scale division of a vernier callipers is ‘a’ cm and nth division of the vernier scale coincide with(n -1)th division of the main scale. The least count of the callipers in mm is :

In order to determine the young's modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm.) What will be the fractional error in the value of young's modulus determined by this experiment?
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A physical 'y' is represented by the formula y = m²r⁻⁴gˣl⁻³/² if the percentage errors found in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p.

Match List I with List II.
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Student A and Student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is _______. [Figure shows position of reference 'O' when jaws of screw gauge are closed]. Given pitch = 0.1 cm.
(Figure A shows a positive zero error with the 5th division of the circular scale coinciding with the reference line. Figure B shows a negative zero error with the 92nd division of the circular scale coinciding with the reference line.)

Assertion A: If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.
Reason R: Least Count = Pitch / (Total divisions on circular scale)

In the light of the above statement, choose the most appropriate answer from the options given below:
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