Unit And Dimensions - NEET Physics Chapterwise MCQs & PYQs
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NEET Unit And Dimensions MCQs & PYQs
Practice NEET Unit And Dimensions Questions
Question 71:
easy
Turpentine oil is flowing through a tube of length (l) and radius (r). The pressure difference between the two ends of the tube is (P). The viscosity of oil is given by \(\eta = \frac{P(r^2 – x^2)}{4vl}\) where (v) is the velocity of oil at a distance (x) from the axis of the tube. The dimensions of (eta) are:
The time dependence of a physical quantity (p) is given by \(p = p_0 \text{exp } (-\alpha t^2)\), where (alpha) is constant and (t) is the time. The constant \(\alpha\):
[1993]
For \(\text{exp }(-\alpha t^2)\) to be dimensionless, \(\alpha t^2\) must be dimensionless. \([\alpha][t^2] = [M^0L^0T^0]\). Since \([t] = [T]\), \([\alpha][T^2] = [1]\). Thus, \([\alpha] = [T^{-2}]\).
(P) represents radiation pressure, (c) represents speed of light and (S) represents radiation energy striking per unit area per sec. The non-zero integers (x, y, z) such that \(P^x S^y c^z\) is dimensionless are:
[1992]
Dimensions: \(P = [ML^{-1}T^{-2}]\), \(c = [LT^{-1}]\), \(S = [MT^{-3}]\). For \(P^x S^y c^z\) to be dimensionless, powers of M, L, T must be zero. \(M: x+y=0\). \(L: -x+z=0\). \(T: -2x-3y-z=0\). Solving gives \(y=-x\) and \(z=x\). Taking \(x=1\) yields \(y=-1\), \(z=1\).
The frequency of vibration (f) of a mass (m) suspended from a spring of spring constant (k) is given by a relation \(f = a.m^x k^y\), where (a) is a dimensionless constant. The values of (x) and (y) are:
[1990]
Frequency \(f = [T^{-1}]\). Mass (m = [M]). Spring constant \(k = [MT^{-2}]\). Comparing dimensions of \(f = m^x k^y\): \([T^{-1}] = [M]^x [MT^{-2}]^y = [M^{x+y} T^{-2y}]\). Solving (x+y=0) and (-2y=-1) gives (y = 1/2) and (x = -1/2).
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 :
Pressure is defined as force per unit area, where the force is perpendicular to the surface. Since it acts equally in all directions, pressure is a scalar quantity.