Question 11:
easyConvert \(25\text{ dyne}\) 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 \(n_2 = 25 \left(\frac{1}{100}\right) \left(\frac{1}{100}\right) \left(\frac{1}{3600}\right)^{-2} = 32400\).
Question 12:
moderateIf energy \((E)\), velocity \((V)\) and time \((T)\) are chosen as fundamental quantities the dimensional formula for momentum \((P)\) is:
Since Energy \(E = F \cdot d = P \cdot V\), momentum \(P = E V^{-1} T^0\). Thus, the dimensional formula is \([E^1 V^{-1} T^0]\).
Question 13:
easyA 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 \(MSR + CSR \times LC = 7\text{ mm} + 67 \times 0.01\text{ mm} = 7.67\text{ mm}\).
Question 14:
easyThe 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 \(\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]\).
Question 15:
easyThe 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, \([h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]\).
Question 16:
easyIf \(E\) 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, \([\frac{E}{G}] = \frac{[M L^2 T^{-2}]}{[M^{-1} L^3 T^{-2}]} = [M^2 L^{-1} T^0]\).
Question 17:
easyA 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}\.
Question 18:
easyIf 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 \([F][A][T^2]\).
Question 19:
easyIn 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, \([a] = [F] = [MLT^{-2}]\).
Question 20:
easyThe 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 \(S = 6a^2 = 24.0\text{ cm}^2\), \(\Delta S = 2 S \frac{\Delta a}{a} = 0.24\text{ cm}^2\).