Question 1:
easyThe mechanical quantity, which has dimensions of reciprocal of mass \( \text{M}^{-1} \) is
The dimension of the universal gravitational constant \(G\) is \([\text{M}^{-1}\text{L}^3\text{T}^{-2}]\). Therefore, its dimensions contain the reciprocal of mass.
Question 2:
easyWhich of the following statement is not true?
Pressure is defined as thrust per unit area. Since thrust is a normal force component, it is always perpendicular to the surface. Pressure does not have a unique direction associated with it in space, making it a scalar quantity.
Question 3:
easyDimensional formula for torque is
Torque is defined as force multiplied by perpendicular distance: \(\tau = F \times r\). Its dimensions are \([MLT^{-2}] \times [L] = [ML^2T^{-2}]\).
Question 4:
easyFind dimensions of constants \(a\) & \(b\) in given equation: \(v = at^2\cos t + \frac{b}{t\sin\theta}\), where \(v \rightarrow\) velocity, \(t \rightarrow\) time.
By the principle of homogeneity, each term on the right must have the dimensions of velocity \(v\). Thus, \([at^2] = [v] ⇒ [a] = [LT^{-3}]\), and \([\frac{b}{t}] = [v] ⇒ [b] = [L]\).
Question 5:
easyAssertion: Mass, length and time may be taken as fundamental quantities.
Reason: Mass, length and time are independent of one another.
Mass, length, and time are chosen as fundamental quantities because they cannot be defined in terms of each other and are completely independent.
Question 6:
easyMatch the column:\n(a) Pressure -> (i) \([ML^2T^{-1}]\)\n(b) Angular Momentum -> (ii) \([M^{-1}L^{-3}T^4A^2]\)\n(c) Magnetic Field -> (iii) \([MT^{-2}A^{-1}]\)\n(d) Permittivity of free space -> (iv) \([ML^{-1}T^{-2}]\)\n(v) \([MT^{-2}A^{-1}]\)
Pressure = \(F/A = [ML^{-1}T^{-2}]\); Angular Momentum = \(mvr = [ML^2T^{-1}]\); Magnetic field = \(F/qv = [MT^{-2}A^{-1}]\); Permittivity = \([M^{-1}L^{-3}T^4A^2]\).
Question 7:
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 8:
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 9:
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 10:
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, \([\frac{E}{G}] = \frac{[M L^2 T^{-2}]}{[M^{-1} L^3 T^{-2}]} = [M^2 L^{-1} T^0]\).