Unit and Measurement MCQ Questions & Answers in Basic Physics | Physics
Learn Unit and Measurement MCQ questions & answers in Basic Physics are available for students preparing for IIT-JEE, NEET, Engineering and Medical Entrance exam.
31.
If $$Z = {A^3},$$ then $$\frac{{\Delta Z}}{Z} = \_\_\_\_\_$$
A
$$\frac{{\Delta {A^3}}}{A}$$
B
$${\left( {\frac{{\Delta A}}{A}} \right)^3}$$
C
$$3\left( {\frac{{\Delta A}}{A}} \right)$$
D
$${\left( {\frac{{\Delta A}}{A}} \right)^{\frac{1}{3}}}$$
32.
If $$P,Q,R$$ are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?
A
$$\frac{{\left( {P - Q} \right)}}{R}$$
B
$$PQ - R$$
C
$$\frac{{PQ}}{R}$$
D
$$\frac{{\left( {PR - {Q^2}} \right)}}{R}$$
Answer :
$$\frac{{\left( {P - Q} \right)}}{R}$$
$$\left[ P \right] \ne \left[ Q \right]$$ so, $${P - Q}$$ is not meaningful
33.
A certain body weighs $$22.42\,g$$ and has a measured volume of $$4.7\,cc.$$ The possible error in the measurement of mass and volume are $$0.01\,g$$ and $$0.1\,cc.$$ Then, maximum error in the density will be
34.
If $$x$$ and $$R$$ stands for distance. Then which of the following is dimensionally same as $$\int {\frac{{Rdx}}{{{x^2}}}} $$ ?
A
$$R{x^2}$$
B
$$2xR$$
C
$$\frac{R}{x}$$
D
$$ - \frac{{{R^2}}}{x}$$
Answer :
$$\frac{R}{x}$$
$$\left[ {\int {\frac{{Rdx}}{{{x^2}}}} } \right] = \frac{{\left[ R \right]\left[ x \right]}}{{\left[ {{x^2}} \right]}} = \frac{{\left[ R \right]}}{{\left[ x \right]}}$$
35.
A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $$0 \,mm$$
Circular scale reading : $$52$$ divisions
Given that 1 $$mm$$ on main scale corresponds to $$100$$ divisions of the circular scale. The diameter of wire from the above data is:
36.
The time dependence of physical quantity $$p$$ is given by $$p = {p_0}\exp \left( { - \alpha {t^2}} \right),$$ where $$\alpha $$ is a constant and $$t$$ is the time. The constant $$\alpha $$
A
is dimensionless
B
has dimensions $$\left[ {{T^{ - 2}}} \right]$$
C
has dimensions $$\left[ {{T^2}} \right]$$
D
has dimensions of $$p$$
Answer :
has dimensions $$\left[ {{T^{ - 2}}} \right]$$
$$p = {p_0}\exp \left( { - \alpha {t^2}} \right)$$
As powers of exponential quantity is dimensionless, so $${ \alpha {t^2}}$$ is dimensionless.
$$\eqalign{
& {\text{or}}\,\alpha {t^2} = {\text{dimensionless}} = \left[ {{M^0}{L^0}{T^0}} \right] \cr
& \therefore \alpha = \frac{1}{{{t^2}}} = \frac{1}{{\left[ {{T^2}} \right]}} = \left[ {{T^{ - 2}}} \right] \cr} $$
37.
What are the dimensions of $$\frac{A}{B}$$ in the relation $$F = A\sqrt x + B{t^2},$$ where $$F$$ is the force, $$x$$ is the distance and $$t$$ is time?
A
$$M{L^2}{T^{ - 2}}$$
B
$${L^{ - \frac{1}{2}}}{T^2}$$
C
$${L^{ - \frac{1}{2}}}{T^{ - 1}}$$
D
$$L{T^{ - 2}}$$
Answer :
$${L^{ - \frac{1}{2}}}{T^2}$$
$$\eqalign{
& \left[ F \right] = \left[ A \right] \times {\left[ x \right]^{\frac{1}{2}}} = \left[ B \right]\left[ {{t^2}} \right] \cr
& \Rightarrow \frac{{\left[ A \right]}}{{\left[ B \right]}} = \frac{{\left[ {{t^2}} \right]}}{{{{\left[ x \right]}^{\frac{1}{2}}}}} = \left[ {{L^{ - \frac{1}{2}}}{T^2}} \right] \cr} $$
38.
The dimensional formula for angular momentum is
A
$$\left[ {{M^0}{L^2}{T^{ - 2}}} \right]$$
B
$$\left[ {M{L^2}{T^{ - 1}}} \right]$$
C
$$\left[ {ML{T^{ - 1}}} \right]$$
D
$$\left[ {M{L^2}{T^{ - 2}}} \right]$$
Answer :
$$\left[ {M{L^2}{T^{ - 1}}} \right]$$
Angular momentum $$L = r \times p = r \times mv$$
$$\therefore $$ Dimensional formula for angular momentum $$ = \left[ L \right]\left[ M \right]\left[ {L{T^{ - 1}}} \right] = \left[ {M{L^2}{T^{ - 1}}} \right]$$
39.
If velocity $$\left( V \right),$$ force $$\left( F \right)$$ and energy $$\left( E \right)$$ are taken as fundamental units, then dimensional formula for mass will be
A
$${V^{ - 2}}{F^0}{E^3}$$
B
$${V^0}F{E^2}$$
C
$$V{F^{ - 2}}{E^0}$$
D
$${V^{ - 2}}{F^0}E$$
Answer :
$${V^{ - 2}}{F^0}E$$
Let $$\left( M \right) = {V^a}{F^b}{E^c}$$
Putting the dimensions of $$V,F$$ and $$E,$$ we have
$$\eqalign{
& \left( M \right) = {\left( {L{T^{ - 1}}} \right)^a} \times {\left( {ML{T^{ - 2}}} \right)^b} \times {\left( {M{L^2}{T^{ - 2}}} \right)^c} \cr
& {\text{or}}\,\,{M^1} = {M^{b + c}}\,{L^{a + b + 2c}}\,{T^{ - a - 2b - 2c}} \cr} $$
Equating the powers of dimensions, we have $$b + c = 1$$
$$a + b + 2c = 0; - a - 2b - 2c = 0$$
which give $$a = - 2,b = 0\,{\text{and}}\,c = 1.$$
Therefore $$\left( M \right) = \left( {{V^{ - 2}}{F^0}E} \right).$$
40.
The resistance of a metal is given by $$R = \frac{V}{I},$$ where $$V$$ is potential difference and $$I$$ is the current. In a circuit the potential difference across resistance is $$V = \left( {8 \pm 0.5} \right)V$$ and current in resistance, $$I = \left( {2 \pm 0.2} \right)A.$$ What is the value of resistance with its percentage error ?