Unit and Measurement MCQ Questions & Answers in Basic Physics | Physics

From principle of homogeneity of dimensions. Dimensions of $$p =$$  dimensions of $$\frac{a}{{{V^2}}}$$
$$\eqalign{ & p = \frac{a}{{{V^2}}} \Rightarrow a = p{V^2} \cr & = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]{\left[ {{L^3}} \right]^2} \cr & = \left[ {M{L^5}{T^{ - 2}}} \right] \cr} $$

$${\text{Pressure = }}\frac{{{\text{Force}}}}{{{\text{Area}}}} = \frac{F}{A} = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$

$$\eqalign{ & {\text{Pressure}}\,\left( p \right) = \frac{{{\text{Force}}}}{{{\text{Area}}}} = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \cr & {\text{Velocity,}}\,v = \left[ {L{T^{ - 1}}} \right] \cr} $$
From principle of homogeneity, the dimensions of $${r^2}$$ and $${x^2}$$ are same.
So, the dimensions of viscosity,
$$\eqalign{ & \eta = \frac{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ {L{T^{ - 1}}} \right]\left[ L \right]}} \cr & = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right] \cr} $$

$$\eqalign{ & {\text{As}}\,\,\frac{a}{{{V^2}}} = P \cr & \therefore a = P{V^2} = \frac{{dyne}}{{c{m^2}}}{\left( {c{m^3}} \right)^2} = dyne \times c{m^4} \cr} $$

We can thus say that the viscous force $$\left( F \right)$$  is the function of radius $$\left( r \right),$$  velocity $$\left( v \right)$$  and viscosity $$\left( \eta \right).$$
$${\text{or}}\,\,F = f\left( {\eta ,r,v} \right)\,\,{\text{or}}\,\,F = k{\eta ^x}{r^y}{v^z}\,......\left( {\text{1}} \right)$$
Where $$k$$ is a constant.
Now, dimensions of the constituents are
$$\eqalign{ & \therefore \left[ {ML{T^{ - 2}}} \right] = {\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^x}{\left[ L \right]^y}{\left[ {L{T^{ - 1}}} \right]^z} \cr & = \left[ {{M^x}{L^{ - x + y + z}}{T^{ - x - z}}} \right] \cr} $$
Equating the exponents of similar quantities of both sides we get, $$x = 1; - x + y + z = 1$$     and $$ - x - z = - 2$$
Solving for $$x,y\,\& \,z,$$  we get $$x = y = z = 1$$
Equation (1) becomes $$F = k\eta rv$$
Experimentally, it was found that
$$k = 6\pi \,\,{\text{or}}\,\,F = 6\pi \eta rv,$$     which is the famous Stokes' law.

The significant number in the potential, $$V = iR;$$   should be the minimum of either $$i$$ or $$R.$$ So corresponding to $$i = 3.23\,A,$$   we have only three significant numbers in $$V = 35.02935\,V.$$   Thus the result is $$V = 35.0\,V.$$

$$\frac{h}{{{e^2}}} = \frac{{M{L^2}{T^{ - 1}}}}{{{{\left( {AT} \right)}^2}}} = M{L^2}{T^{ - 3}}{A^{ - 2}}$$
= Resistance (ohm)

$$\left[ a \right] = \frac{{\left[ {RT} \right]}}{{\left[ V \right]}} = \left[ P \right]$$

As we know that,
Dimension of $${\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$
Dimension of $$E = \left[ {ML{T^{ - 3}}{A^{ - 1}}} \right]$$
So, dimension of $$\frac{1}{2}{\varepsilon _0}{E^2} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right] \times {\left[ {ML{T^{ - 3}}{A^{ - 1}}} \right]^2}$$
$$ = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$

Measured length of rod = 3.50 $$cm$$
For Vernier scale with 1 Main Scale Division = 1 $$mm$$
9 Main Scale Division = 10 Vernier Scale Division ,
Least count = 1 MSD $$-$$ 1 VSD = 0.1 $$mm$$