A student measured the diameter of a small steel ball using a screw gauge of least count \(0.001\) cm. The main scale reading is \(5\) mm and zero of circular scale division coincides with \(25\) divisions above the reference level. If the screw gauge has a zero error of \(-0.004\) cm, the correct diameter of the ball is:
1. | \(0.521\) cm | 2. | \(0.525\) cm |
3. | \(0.053\) cm | 4. | \(0.529\) cm |
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. \(1,1,1\)
1. | \([Ev^{-2}T^{-1}]\) | 2. | \([Ev^{-1}T^{-2}]\) |
3. | \([Ev^{-2}T^{-2}]\) | 4. | \([E^{-2}v^{-1}T^{-3}]\) |
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 dimensions of where is the permittivity of free space and E is the electric field, are:
1. [ML2T-2]
2. [ML-1T-2]
3. [ML2T-1]
4. [MLT-1]
1. | \(a=1,\) \(b=-1,\) \(c=-2\) | pressure if
2. | \(a=1,\) \(b=0,\) \(c=-1\) | velocity if
3. | \(a=1,\) \(b=1,\) \(c=-2\) | acceleration if
4. | \(a=0,\) \(b=-1,\) \(c=-2\) | force if
Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, length L, time T, and current I, would be:
1.
2.
3.
4.
The velocity \(v\) of a particle at time \(t\) is given by \(v=at+\dfrac{b}{t+c}\), where \(a,\) \(b\) and \(c\) are constants. The dimensions of \(a,\) \(b\) and \(c\) are respectively:
1. \(\left[{LT}^{-2}\right],[{L}] \text { and }[{T}]\)
2. \( {\left[{L}^2\right],[{T}] \text { and }\left[{LT}^2\right]} \)
3. \( {\left[{LT}^2\right],[{LT}] \text { and }[{L}]} \)
4. \( {[{L}],[{LT}] \text { and }\left[{T}^2\right]}\)