Convert \(25\text{ dyne}\) to a system where fundamental units of mass, length & time are \(100\text{ g}\), \(100\text{ cm}\) and \(1\text{ hour}\):
Using \(n_2 = n_1 [M_1/M_2]^1 [L_1/L_2]^1 [T_1/T_2]^{-2}\), we get \(n_2 = 25 \left(\frac{1}{100}\right) \left(\frac{1}{100}\right) \left(\frac{1}{3600}\right)^{-2} = 32400\).
A screw gauge gives the following reading while measuring diameter of a wire. Main Scale Reading = \(7\text{ mm}\), Circular Scale Reading = \(67\). Given that \(1\text{ mm}\) on main scale corresponds to \(100\text{ divisions}\) on circular scale. The diameter of the wire is:
Least count is \(LC = \frac{1\text{ mm}}{100} = 0.01\text{ mm}\). Total Reading is \(MSR + CSR \times LC = 7\text{ mm} + 67 \times 0.01\text{ mm} = 7.67\text{ mm}\).
The dimensional formula of \(\frac{1}{2}\epsilon_0 E^2\) is (All symbols have their usual meaning)
The term \(\frac{1}{2}\epsilon_0 E^2\) represents the electrostatic energy density (energy per unit volume). Therefore, its dimensional formula is \(\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]\).
The correct dimensional formula for Planck’s constant \(h\) will be
From Planck's equation, \(E = h\nu\), we have \(h = \frac{E}{nu}\) where \(E\) is energy and \(nu\) is frequency. Thus, \([h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]\).
If \(E\) and \(G\) respectively denote Energy and Universal gravitational constant, then \(\frac{E}{G}\) has the dimensions of
Dimensions of energy \([E] = [M L^2 T^{-2}]\) and universal gravitational constant \([G] = [M^{-1} L^3 T^{-2}]\). Therefore, \([\frac{E}{G}] = \frac{[M L^2 T^{-2}]}{[M^{-1} L^3 T^{-2}]} = [M^2 L^{-1} T^0]\).
A screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale.
The diameter of the wire from the above data is
Least Count \(\frac{1\text{ mm}}{100} = 0.01\text{ mm} = 0.001\text{ cm}\. Diameter = MSR + (CSR times LC) = 0\text{ mm} + 52 \times 0.001\text{ cm} = 0.052\text{ cm}\.
If force \([F]\), acceleration \([A]\) and time \([T]\) are chosen as the fundamental physical quantities. Find the dimensions of energy.
Energy has the dimensions of work, which is \(\text{Force} \times \text{displacement}\). Since displacement has the dimensions of \(text{acceleration} \times \text{time}^2 = [A][T^2]\), the dimensional formula of energy is \([F][A][T^2]\).
In the given expression of force \(F = a log \left(\frac{b}{x}\right)\) where \(x\) is displacement, the dimensions of \(a\) and \(b\) respectively are
The argument of the logarithm must be dimensionless, so \([b] = [x] = [L]\). Since the logarithm term is dimensionless, \([a] = [F] = [MLT^{-2}]\).
The side of a cube is \( 2.00 + 0.01\text{ cm}\). The volume and total surface area of cube respectively are
Volume \(V = a^3 = 8.00\text{ cm}^3\), \(\Delta V = 3 V \frac{\Delta a}{a} = 0.12\text{ cm}^3\). Surface Area \(S = 6a^2 = 24.0\text{ cm}^2\), \(\Delta S = 2 S \frac{\Delta a}{a} = 0.24\text{ cm}^2\).
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The absolute error has the same unit as the quantity itself.
Reason (R): Fractional error has no unit.
In the light of above statements, choose the correct answer from the options given below.
Absolute error \(\Delta x\) has the same unit as the physical quantity. Fractional error is the ratio \(\Delta x / x\) and is dimensionless (no unit). Both statements are true, but the lack of unit in fractional error does not explain why absolute error has a unit.