If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be:
[2008]
Volume of a sphere \(V = \frac{4}{3}pi r^3\). The percentage error in volume is (3) times the percentage error in radius. Given \(\delta r/r \times 100 % = 2 % \). So, Percentage error in \(V = 3 \times 2 % = 6% \).
The area of a rectangular field (in \(\text{m}^2\)) of length \(55.3 \text{m}\) and breadth \(25 \text{m}\) after rounding off the value for correct significant digits is :
[NEET 2022]
Length \(L = 55.3 \text{m}\) (3 significant figures). Breadth \(B = 25 \text{m}\) (2 significant figures). Area \(A = L \times B = 55.3 \times 25 = 1382.5 \text{m}^2\). For multiplication, the result must be rounded to the least number of significant figures, which is 2. Rounding \(1382.5 \text{m}^2\) to 2 significant figures gives \(1400 \text{m}^2\) or \(14 \times 10^2 \text{m}^2\).
Taking into account of the significant figures, what is the value of \(9.99 \text{m} – 0.0099 \text{m}\)?
[NEET 2020]
The numbers are \(9.99 \text{m}\) (2 decimal places) and \(0.0099 \text{m}\) (4 decimal places). For subtraction, the result should have the same number of decimal places as the number with the fewest decimal places (2 in this case). \(9.9900 - 0.0099 = 9.9801\). Rounding \(9.9801\) to 2 decimal places gives \(9.98 \text{m}\).
A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading: \(0 \text{mm}\)
Circular scale reading: \(52 \text{divisions}\)
Given that \(1 \text{mm}\) on main scale corresponds to \(100 \text{divisions}\) on the circular scale. The diameter of the wire from the above data is:
[NEET 2020]
Pitch = \(1 \text{mm}\), Number of divisions = \(100\). Least Count (LC) = Pitch / Number of divisions = \(1 \text{mm} / 100 = 0.01 \text{mm}\). Total reading = MSR + (CSR \(\times\) LC) = \(0 \text{mm} + (52 \times 0.01 \text{mm}) = 0.52 \text{mm}\). Convert to cm: \(0.52 \text{mm} = 0.052 \text{cm}\).
A screw gauge has least count of \(0.01 \text{mm}\) and there are \(50 \text{divisions}\) in its circular scale. The pitch of the screw gauge is:
[NEET 2020]
Least Count (LC) = \(0.01 \text{mm}\), Number of divisions on circular scale = \(50\). The formula for LC is: LC = Pitch / Number of divisions. Therefore, Pitch = LC \(\times\) Number of divisions = \(0.01 \text{mm} \times 50 = 0.5 \text{mm}\).
A student measured the diameter of a small steel ball using a screw gauge of least count \(0.001 \text{cm}\). The main scale reading is \(5 \text{mm}\) and zero of circular scale division coincides with \(25 \text{divisions}\) above the reference level. If screw gauge has a zero error of \(-0.004 \text{cm}\), the correct diameter of the ball is :
[NEET 2018]
Least Count (LC) = \(0.001 \text{cm}\). Main Scale Reading (MSR) = \(5 \text{mm} = 0.5 \text{cm}\). Circular Scale Reading (CSR) = \(25 \text{divisions}\). Observed Reading = MSR + (CSR \(\times\) LC) = \(0.5 \text{cm} + (25 \times 0.001 \text{cm}) = 0.525 \text{cm}\). Correct Reading = Observed Reading - Zero Error = \(0.525 \text{cm} - (-0.004 \text{cm}) = 0.525 \text{cm} + 0.004 \text{cm} = 0.529 \text{cm}\).