If 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 \([E^1 V^{-1} T^0]\).
The angle \(1’\) (minute of arc) in radian is nearly equal to,
[2020- Covid]
Convert \(1'\) to degrees, then to radians. \(1' = (1/60)^\circ = (1/60) \times (\pi/180)\text{ rad} \approx 2.91 \times 10^{-4}\text{ rad}\)
The unit of thermal conductivity is
[2019]
Thermal conductivity \(k\) relates heat flux to temperature gradient. So, \(k\) has units of \(W/(m \cdot K) = W m^{-1} K^{-1}\).
The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are:
[2012 pre]
Damping force \(F_d = -bv\). The constant \(b\) has dimensions \([F_d]/[v] = (MLT^{-2})/(LT^{-1}) = MT^{-1}\), which is \(kg s^{-1}\).
The density of a material in CGS system of units is \(4\text{ g/cm}^3\). In a system of units in which unit of length is \(10\text{ cm}\) and unit of mass is \(100\text{ g}\), the value of density of material will be:
[2011 Mains]
Given \(\rho = 4\text{ g/cm}^3\). New mass unit \(M' = 100\text{ g}\) and new length unit \(L' = 10\text{ cm}\). \(\rho = 4 \frac{1/100 M'}{(1/10 L')^3} = 4 \frac{1/100}{1/1000} \frac{M'}{L'^3} = 40 \frac{M'}{L'^3}\).
The dimension \([ML^{-2}T^{-2}A^2]\) belong to the:
[2022]
The dimension of magnetic permeability is \(MLT^{-2}A^{-2}\). The given dimension \(ML^{-2}T^{-2}A^2\) does not match a standard physical quantity listed. Assuming a typo and it refers to magnetic permeability.
If \(E\) and \(G\) respectively denote energy and gravitational constant, then \(\frac{E}{G}\) has the dimensions of:
[2021]
Energy \([E] = ML^2T^{-2}\) and Gravitational Constant \([G] = M^{-1}L^3T^{-2}\). So \([E/G] = M^2L^{-1}T^0\).
Which two of the following five physical parameters have the same dimensions?
A. Energy density
B. Refractive index
C. Dielectric constant
D. Young’s modulus
E. Magnetic field
[2008]
Energy density (A) is \(ML^{-1}T^{-2}\). Young's modulus (D) is Stress/Strain, so also \(ML^{-1}T^{-2}\). Therefore, (A) and (D) have the same dimensions.
Dimensions of resistance in an electrical circuit, in terms of dimension of mass \(M\), of length \(L\), of time \(T\) and of current \(I\), would be:
[2007]
Resistance \(R = V/I = (W/q)/I = W/(I^2T)\). Work \([W] = ML^2T^{-2}\). So, \([R] = (ML^2T^{-2})/(I^2T) = ML^2T^{-3}I^{-2}\).
If Force \((F)\), Velocity \((V)\), and Time \((T)\), are taken as fundamental units, then the dimensions of mass are:
[2014]
Given fundamental units: Force \(F = [MLT^{-2}]\), Velocity \(V = [LT^{-1}]\), Time \(T = [T]\). We want to find dimensions of Mass \(M = F^x V^y T^z\). Equating dimensions: \([M] = [MLT^{-2}]^x [LT^{-1}]^y [T]^z = [M^x L^{x+y} T^{-2x-y+z}]\). Comparing powers: \(x=1\), \(x+y=0 ⇒ 1+y=0 ⇒ y=-1\), \(-2x-y+z=0 ⇒ -2(1)-(-1)+z=0⇒ -2+1+z=0 ⇒ z=1\). Thus, mass dimensions are \(FV^{-1}T\).