If force (\(F\)), velocity (\(\mathrm{v}\)), and time (\(T\)) are taken as fundamental units, the dimensions of mass will be:
1. | \([FvT^{-1}]\) | 2. | \([FvT^{-2}]\) |
3. | \([Fv^{-1}T^{-1}]\) | 4. | \([Fv^{-1}T]\) |
The energy required to break one bond in DNA is \(10^{-20}~\text{J}\). This value in eV is nearly:
1. \(0.6\)
2. \(0.06\)
3. \(0.006\)
4. \(6\)
1. | W m–1 K–1 | 2. | J m K–1 |
3. | J m–1 K–1 | 4. | W m K–1 |
If dimensions of critical velocity \({v_c}\) of a liquid flowing through a tube are expressed as \(\eta^{x}\rho^yr^{z}\), where \(\eta, \rho~\text{and}~r\) are the coefficient of viscosity of the liquid, the density of the liquid, and the radius of the tube respectively, then the values of \({x},\) \({y},\) and \({z},\) respectively, will be:
1. | \(1,-1,-1\) | 2. | \(-1,-1,1\) |
3. | \(-1,-1,-1\) | 4. | \(1,1,1\) |
Dimensions of stress are:
1. | \( {\left[{ML}^2 {T}^{-2}\right]} \) | 2. | \( {\left[{ML}^0 {T}^{-2}\right]} \) |
3. | \( {\left[{ML}^{-1} {T}^{-2}\right]} \) | 4. | \( {\left[{MLT}^{-2}\right]}\) |
1. | both units and dimensions |
2. | units but no dimensions |
3. | dimensions but no units |
4. | no units and no dimensions |
A physical quantity of the dimensions of length that can be formed out of \(c, G,~\text{and}~\dfrac{e^2}{4\pi\varepsilon_0}\)is [\(c\) is the velocity of light, \(G\) is the universal constant of gravitation and \(e\) is charge]:
1. \(c^2\left[G \dfrac{e^2}{4 \pi \varepsilon_0}\right]^{\dfrac{1}{2}}\)
2. \(\dfrac{1}{c^2}\left[\dfrac{e^2}{4 G \pi \varepsilon_0}\right]^{\dfrac{1}{2}}\)
3. \(\dfrac{1}{c} G \dfrac{e^2}{4 \pi \varepsilon_0}\)
4. \(\dfrac{1}{c^2}\left[G \dfrac{e^2}{4 \pi \varepsilon_0}\right]^{\dfrac{1}{2}}\)