MCQ 511 Mark
Give force $=\frac{\alpha}{\text{Density}+\beta^3}$ What are the dimensions of $\alpha,\beta$
- A
$[\text{ML}^2\text{T}^{-2}][\text{ML}^{\frac{-1}{3}}]$
- B
$[\text{M}^2\text{L}^4\text{T}^{-2}],[\text{M}^{\frac{1}{3}}\text{L}^{-1}]$
- ✓
$[\text{M}^2\text{L}^{-2}\text{T}^{-2}][\text{M}^{\frac{1}{3}}\text{L}^{-1}]$
- D
$[\text{M}^2\text{L}^{-2}\text{T}^{-2}][\text{ML}^{-2}]$
AnswerCorrect option: C. $[\text{M}^2\text{L}^{-2}\text{T}^{-2}][\text{M}^{\frac{1}{3}}\text{L}^{-1}]$
Dimensions of $\beta^3=$ Dimensions of density $=[\text{ML}^{-3}]$
$\beta=[\text{M}^{\frac{1}{3}}\text{L}^{-1}]$
Also, $\alpha=\text{Force}\times\text{Density}$
$=[\text{MLT}^{-2}][\text{ML}^{-3}]$
$=[\text{M}^2\text{L}^{-2}\text{T}^{-2}]$
View full question & answer→MCQ 521 Mark
In the gas equation $\Big(\text{p}+\frac{\text{a}}{\text{V}^2}\Big)(\text{V}-\text{b})=\text{RT}$ the dimensions of a are:
- A
$[\text{ML}^3\text{T}^{-2}]$
- B
$[\text{M}^{-1}\text{L}^3\text{T}^{-1}]$
- ✓
$[\text{ML}^5\text{T}^{-2}]$
- D
$[\text{M}^{-1}\text{L}^{-5}\text{T}^2]$
AnswerCorrect option: C. $[\text{ML}^5\text{T}^{-2}]$
View full question & answer→MCQ 531 Mark
In the standard equation $S_n=\text{u}+\frac{\text{a}}{2}[2\text{n}-1],$ what dimensions do you view for $S_n?$
- A
$\ce{M^{\circ} L^1 T^{\circ}}$
- B
$\ce{M^{\circ} L^{-1} T^1}$
- ✓
$\ce{M^{\circ} L^1 T^{-1}}$
- D
$\ce{M^{\circ} L^{\circ} T^1}$
AnswerCorrect option: C. $\ce{M^{\circ} L^1 T^{-1}}$
View full question & answer→MCQ 541 Mark
The units of electrical permittivity are:
- ✓
$\ce{N^{-2} m^{-2} C^2}$
- B
$\ce{{Nm}^{-2} {C}^2}$
- C
$\ce{{C}^2 /{Nm}^2}$
- D
$\ce{{n} / {Cm}^2}$
AnswerCorrect option: A. $\ce{N^{-2} m^{-2} C^2}$
$\text{F}=\frac{1}{4\pi\in}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$\in\text{ = }\frac{\text{q}_1\text{q}_2}{4\pi\text{Fr}^2}$
$=\frac{\text{C}^2}{\text{Nm}^2}$
$=\text{N}^{-2}\text{m}^{-2}\text{C}^2$
View full question & answer→MCQ 551 Mark
On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct:
- A
$\text{y}=\frac{\text{a}\sin2\pi\text{t}}{\text{T}}.$
- B
$\text{y}=\text{a}\sin\text{vt}.$
- C
$\text{y}=\frac{\text{a}}{\text{T}}\sin\Big(\frac{\text{t}}{\text{a}}\Big).$
- ✓
$B$ and $C$
AnswerCorrect option: D. $B$ and $C$
The argument of trigonometric functions $(\sin, \cos$ etc.$)$ should be di mensionless.
$y$ is displacement and according to the principle of homegeneity of dimensions $\ce{LHS}$ and $\ce{RHS}.$
$[\text{Y}]=[\text{L}],[\text{a}]=[\text{L}]$
$\Big[\frac{2\pi\text{t}}{\text{T}}\Big]=\frac{[\text{T}]}{[\text{T}]}=[\text{T}^0]$
$[\text{vt}]=[\text{v}][\text{t}]=[\text{LT}^{-1}][\text{T}]=[\text{L}]$
$\Big[\frac{\text{a}}{\text{T}}\Big]=\frac{[\text{a}]}{[\text{T}]}=\frac{[\text{L}]}{[\text{T}]}=[\text{LT}^{-1}]$
$\Big[\frac{\text{t}}{\text{a}}\Big]=[\text{L}^{-1}\text{T}]$
$[\text{LHS}]\neq[\text{RHS}]$
Hence, $(c)$ is not the correct option.
$\Rightarrow\text{LHS}\neq\text{RHS}.$
So, option $(b)$ is also not correct.
View full question & answer→MCQ 561 Mark
The physical quantities not having same dimensions are:
AnswerCorrect option: A. Momentum and Planck's constant.
Momentum $\text{= mv}=[\text{MLT}^{-1}]$
Planck's constant $=\text{h}=\frac{\text{E}}{\text{V}}=\frac{\text{ML}^2\text{L}^{-2}}{\text{T}^{-1}}$
View full question & answer→MCQ 571 Mark
To reduce the least count error, instruments need higher:
MCQ 581 Mark
The dimensional formula of Avagadro's number is:
- A
$\ce{[M^1L^1T^1]}$
- ✓
$\ce{[mole^{-1}]}$
- C
$\ce{[mole]}$
- D
$\ce{[M^0L^1T^0]}$
AnswerCorrect option: B. $\ce{[mole^{-1}]}$
View full question & answer→MCQ 591 Mark
If momentum $(P),$ area $(A)$ and time $(T)$ are taken to be fundamental quantities, then energy has the dimensional formula:
- A
$(\text{P}^1\text{A}^{-1}\text{T}^1).$
- B
$(\text{P}^2\text{A}^{1}\text{T}^1).$
- C
$(\text{P}^1\text{A}^\frac{-1}{2}\text{T}^1).$
- ✓
$(\text{P}^1\text{A}^\frac{1}{2}\text{T}^{-1}).$
AnswerCorrect option: D. $(\text{P}^1\text{A}^\frac{1}{2}\text{T}^{-1}).$
According to the problem, fundamental quantities are momentum $(p),$ area $(A)$ and time $(T)$ and we have to express energy in these fundamental quantities.
Let energy $E$,
$\text{E}\propto\text{p}^\text{a}\text{A}^\text{A}\text{T}^\text{c}\Rightarrow\text{E}=\text{kp}^\text{a}\text{A}^\text{A}\text{T}^\text{c}$
where, $k$ is dimensionless constant of proportionality.
Dimensional formula of energy, $[\text{E}]=[\text{ML}^2\text{T}^{-2}]$ and $[\text{p}]=[\text{MLT}^{-1}]$
$[\text{A}]=[\text{L}^2],\ [\text{T}]=[\text{T}]$ and $[\text{E}]=[\text{K}][\text{p}]^\text{a}[\text{A}]^\text{b}[\text{T}]^\text{c}$
Putting all the dimensions, we get
$\text{ML}^2\text{T}^2=[\text{MLT}^{-1}]^\text{a}[\text{L}^2]^\text{b}[\text{T}]^\text{c}$
$\text{M}^\text{a}\text{L}^{\text{a}+2\text{b}}\text{T}^{\text{-a}+\text{c}}$
According to the principle of homogeneity of dimensions, we get,
$\text{a}=1\ \ \ ...(\text{i)}$
$\text{a}+2\text{b}=2\ \ \ ...(\text{ii)}$
$-\text{a}+\text{c}=-2\ \ \ ...(\text{iii)}$
By solving these equations $(i), (ii)$ and $(iii),$ we get
$\text{a}=1,\ \text{b}=\frac{1}{2},\ \text{c}=-1$
Dimensional formula for $E$ is $[\text{p}^1\text{A}^\frac{1}{2}\text{t}^{-1}].$
View full question & answer→MCQ 601 Mark
Which of the following pairs of physical quantities does not have same dimensional formula?
- A
- B
Angular momentum and Planck’s constant.
- ✓
Tension and surface tension.
- D
Impulse and linear momentum.
AnswerCorrect option: C. Tension and surface tension.
- $\text{Work}=\text{F}\times\Delta\text{x}=[\text{MLT}^2][\text{L}]=[\text{ML}^2\text{T}^2]$
Torque$=$ force$\times$ distance $=[\text{ML}^2\text{T}^2]$
- Angular momentum $=\text{mvr}=[\text{M}][\text{LT}^1][\text{L}]=[\text{ML}^2\text{T}^1]$
Planck's constant $=\frac{\text{E}}{\text{V}}=\frac{[\text{ML}^2\text{T}^{-2}]}{[\text{T}^{-1}]}=[\text{ML}^2\text{T}^{-1}]$
- Tension $($force$)= [\text{MLT}^{-2}]$
Surface tension $=\frac{\text{force}}{\text{length}}=\frac{[\text{MLT}^{-2}]}{[\text{L}]}=[\text{ML}^0\text{T}^{-2}]$
- Impulse $=\text{F}\times\Delta\text{t}=[\text{MLT}^{-2}][\text{T}]=[\text{MLT}^{-1}]$
Momentum $=$ mass$\times$ velocity $=[\text{M}][\text{LT}^{-1}]=[\text{MLT}^{-1}]$
So, among the above pairs only tension and surface tension does not have same dimensional formula. They both sound similar but they both have different meaning and different applications. View full question & answer→MCQ 611 Mark
The dimensions of energy per unit volume are the same as those of:
MCQ 621 Mark
Photon is quantum of radiation with energy $E = hν$ where $ν$ is frequency and $h$ is Planck’s constant. The dimensions of $h$ are the same as that of:
AnswerCorrect option: D. $B$ and $C$
We know that energy of radiation, $E = hv.$
So, we have to compare $h$ with dimensional formula of each option.
$[\text{h}]=\frac{[\text{E}]}{[\text{v}]}=\frac{\text{force}\times\text{displacement}}{\text{frequency}}$
$=\frac{[\text{ML}^2\text{T}^{-2}]}{[\text{T}^{-1}]}=[\text{ML}^2\text{T}^{-1}]$
- Dimension of linear impulse,
$[\text{I}]=[\text{Ft}]=[\text{MLT}^{-2}][\text{T}]=[\text{MLT}^{-1}]$
where, t is the time interval.
- Dinmension of angular impulse,
$[\text{J}]=[\text{I}\omega]=[\text{ML}^2][\text{T}^{-1}]=[\text{ML}^2\text{T}^{-1}]$
- Dimension of angular momentum,
$[\text{P}]=[\text{mv}]=[\text{M}][\text{LT}^{-1}]=[\text{ML}\text{T}^{-1}]$
- Dimension of angular momentum,
$[\text{L}]=[\text{mvr}]=[\text{M}][\text{LT}^{-1}][\text{L}]=[\text{ML}^2\text{T}^{-1}]$
Hence, dimension of angular impulse and angular momentum is same as Planck's constant $(h).$ View full question & answer→MCQ 631 Mark
In $4700m,$ significant digits are:
View full question & answer→MCQ 641 Mark
Which of the following has same dimension as that of Planck constant?
Answer$\text{As, }\text{E}=\text{hv}$ or $\text{h}=\frac{\text{E}}{\text{V}}=\Big[\frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-1}}\Big]=[\text{ML}^2\text{T}^{-1}]$
Angular momentum $=\text{mvr}=[\text{M}][\text{LT}^{-1}][\text{L}]=[\text{ML}^2\text{T}^{-1}]$
View full question & answer→MCQ 651 Mark
If the value of force is $100N$ and value of acceleration is $0.001\ ms^{-2}$, what is the value of mass in this system of units?
- A
$10^3\ kg$
- B
$10^4\ kg$
- ✓
$10^5\ kg$
- D
$10^6\ kg$
AnswerCorrect option: C. $10^5\ kg$
View full question & answer→MCQ 661 Mark
Which of the technique$(s)$ is/ are not used for measuring time intervals?
- A
- B
- ✓
- D
Decay of elementary particles.
AnswerSpring oscillator cannot be used to meausre time intervals. But in electrical oscillator electrical oscillations are produced in $L-C$ circuit and oscillators are maintained with the help of transistor. $1$ us can be measured with these oscillators.
Atomic clock are based on periodic vibrations taking place within the atoms.
Decay of elementary particles: The life span of particles varies from $10^{-16}$ to $10^{-24}$. By making use of their decay times, very small time intervals can be measured.
View full question & answer→MCQ 671 Mark
The number of significant figures in $3400$ is:
AnswerIn $x = 3400,$ zero are not significant. Therefore, number of significant figure $= 2.$
View full question & answer→MCQ 681 Mark
Equivalent resistance of two resistors $\text{R}_1=100\pm312$ and $\text{R}_2=200\pm4\Omega$ in series is:
- A
$(250\pm7)\Omega$
- B
$(320\pm6)\Omega$
- ✓
$(300\pm7)\Omega$
- D
$(300\pm1)-\Omega$
AnswerCorrect option: C. $(300\pm7)\Omega$
View full question & answer→MCQ 691 Mark
The $SI$ units of magnetic field is:
View full question & answer→MCQ 701 Mark
The mass and volume of a body are $4.237g$ and $2.5\ cm^3$, respectively. The density of the material of the body in correct significant figures is:
- A
$1.6048g\ cm^{-3}$.
- B
$1.69g\ cm^{-3}$.
- ✓
$1.7g\ cm^{-3}$.
- D
$1.695g\ cm^{-3}$.
AnswerCorrect option: C. $1.7g\ cm^{-3}$.
The significant figures in given numbers $4.237g$ and $2.5\ cm^3$ are four and two respectively so result must have only two significant figures.
$\text{Density}=\frac{\text{mass}}{\text{volume}}=\frac{4.237}{2.5},$ Density $= 1.6948 = 1.7g\ cm^{-3}$ rounding off upto $2$ significant figures.
View full question & answer→MCQ 711 Mark
Which of the following statement is incorrect regarding mass?
AnswerCorrect option: B. The $SI$ unit of mass is candela.
View full question & answer→MCQ 721 Mark
Which of the following have same dimensions?
- A
Specific heat and latent heat.
- ✓
- C
Moment of inertia and moment of momentum.
- D
Tension and surface tension.
View full question & answer→MCQ 731 Mark
Which of the following is not a dimensional constant?
- A
- ✓
$\pi$
- C
- D
Gas constant $(R).$
View full question & answer→MCQ 741 Mark
On checking the dimensional consistency of equation, it is based on the principle of:
- A
Homogeneity of equations.
- ✓
Homogeneity of dimensions.
- C
Homogeneity of expressions.
- D
AnswerCorrect option: B. Homogeneity of dimensions.
View full question & answer→MCQ 751 Mark
A is the fundamental quantity. Here, $A$ refers to:
AnswerMass is the fundamental quantity as it does not depend upon other physical quantities. However, other three quantities, i.e. velocity, acceleration and linear momentum are not fundamental quantities as these show their dependency on fundamental quantities.
View full question & answer→MCQ 761 Mark
How many wavelength of $Kr^{86}$ are there in $1m?$
- A
$1553164.13.$
- ✓
$1650763.73.$
- C
$2348123.73.$
- D
$652189.63.$
AnswerCorrect option: B. $1650763.73.$
The number of wavelengths of $Kr^{86}$ in $1 m$ is $1650763.73.$
View full question & answer→MCQ 771 Mark
The SI units of the universal gravitational constant G are:
- A
$\text{kg}\text{ m}^2\text{ s}^{-2}$
- B
$\text{kg}^{-1}\text{m}^3\text{s}^{-2}$
- ✓
$\text{Nm}^2\text{kg}^{-2}$
- D
$\text{N }\text{kg}^2\text{m}^{-2}$
AnswerCorrect option: C. $\text{Nm}^2\text{kg}^{-2}$
(C) $\text{Nm}^2\text{kg}^{-2}$
View full question & answer→MCQ 781 Mark
Measure of two quantities along with the precision of respective measuring instrument is: $\text{A} = 2.5\text{ms}^{-1} \pm 0.5\text{ms}^{-1}$ $\text{B} = 0.10\text{s} \pm 0.01\text{s}$ The value of $A B$ will be,
- ✓
$(0.25 \pm 0.08)\text{m}.$
- B
$(0.25 \pm 0.5)\text{m}.$
- C
$(0.25 \pm 0.05)\text{m}.$
- D
$(0.25 \pm 0.135)\text{m}$
AnswerCorrect option: A. $(0.25 \pm 0.08)\text{m}.$
$\text{A}=(2.5\pm0.5)\text{ms}^{-1}$
$\text{B}=(0.10\pm0.01)\text{s}$
$\text{X}=\text{AB}=2.5\times0.10=0.25$
$\frac{\Delta\text{x}}{\text{x}}=\frac{\Delta\text{A}}{\text{A}}+\frac{\Delta\text{B}}{\text{B}}$
$\frac{\Delta\text{x}}{\text{x}}=\frac{0.5}{2.5}+\frac{0.01}{0.10}$
$\frac{\Delta\text{x}}{\text{x}}=\frac{0.075}{0.25},\Delta\text{x}=0.007\cong0.008$
$($Rounding off upto $2$ significant figures$)$
$\therefore\text{AB}=(2.5\pm0.08)\text{m}.$
Hence, verifies the option $(a).$
View full question & answer→MCQ 791 Mark
Which one of the following quantities has not been expressed in proper units?
AnswerCorrect option: C. Energy: $\ce{kgm/ s}$
Energy $= \ce{[M^1L^2T^{-2}] = Kgm^2g^{-2}}$
View full question & answer→MCQ 801 Mark
Choose the correct option.
- A
$3.00 - 25 = -5.0$
- ✓
$3.00 - 2.5 = 0.50$
- C
$3.00 + 25 = 5.50$
- D
$3.00 + 2.5 = 5.500$
AnswerCorrect option: B. $3.00 - 2.5 = 0.50$
View full question & answer→MCQ 811 Mark
Young’s modulus of steel is $1.9 \times 10^{11} \ce{N/m^2}$. When expressed in $\ce{CGS}$ units of $\ce{dynes/cm^2}$, it will be equal to $\ce{(1N = 10^5 dyne, 1m^2 = 10^4cm^2)}$.
- A
$1.9 × 10^{10}$.
- B
$1.9 × 10^{11}$.
- ✓
$1.9 × 10^{12}$.
- D
$1.9 × 10^{13}$.
AnswerCorrect option: C. $1.9 × 10^{12}$.
According to the problem,
Young's modulus, $\ce{Y = 1.9 \times 10^{11} N/m^2}$
$1 N$ in $SI$ system of units $= 10^5$ dyne in $\ce{C.G.S}$ system.
Hence, $\ce{Y = 1.9 \times 10^{11} \times 10^5 dyne/m^2}$
In $\ce{C.G.S}.$ length is measured in unit $'cm',$ so we should also convert m into $cm.$
$\therefore\text{Y}=1.9\times10^{11}\Big(\frac{10^5\text{dyne}}{10^4\text{cm}^2}\Big)$
$[\because1\text{m}=100\text{cm}]$
$=1.9\times10^{12}\ \text{dyne/cm}^2$
View full question & answer→MCQ 821 Mark
Which of the following has unit but no dimension?
AnswerAngle has unit of radian but has no dimensions because,
$\theta=\frac{\text{l}}{\text{r}}$
i.e., it is the ratio of two quantities of same dimensions.
View full question & answer→MCQ 831 Mark
The sum of the numbers $436.32, 227.2$ and $0.301$ in appropriate significant figures is:
- A
$663.821.$
- ✓
$664.$
- C
$663.8.$
- D
$663.82.$
AnswerCorrect option: B. $664.$
The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as present in the number having the least number of decimal places.
The final result should, therefore, be rounded off to one decimal place, i.e. $664.$ View full question & answer→MCQ 841 Mark
If the unit of force is $100N,$ unit of length is $10m$ and unit of time is $100s$. What is the unit of mass in this system of units?
- ✓
$10^5\ kg$
- B
$10^7\ kg$
- C
$10^2\ kg$
- D
$10^9\ kg$
AnswerCorrect option: A. $10^5\ kg$
View full question & answer→MCQ 851 Mark
Which of the following combinations have the dimensions of time? L, C, R represent inductance, capacitance and resistance respectively?
AnswerCorrect option: B. $\sqrt{\text{LC}}$
Explanation: (B) $\sqrt{\text{LC}}$
We know that:
$\text{R}=[\text{M}^{1}\text{L}^{2}\text{T}^{-3}\text{A}^{-2}]$
$\text{L}=[\text{M}^{1}\text{L}^{2}\text{T}^{-2}\text{A}^{-2}]$
$\text{C}=[\text{M}^{1}\text{L}^{-2}\text{T}^{4}\text{A}^{2}]$
$\therefore\text{RC}=\text{T}\text{ and }\sqrt{\text{LC}}=\text{T}$
View full question & answer→MCQ 861 Mark
The surface area of a solid cylinder of radius $2.0\ cm$ and height $A \ cm$ is equal to $1.5 \times 10^4(mm)^2$. Here, $A$ refers to:
- A
$0.9\ cm$
- ✓
$10\ cm$
- C
$30\ cm$
- D
$15\ cm$
AnswerCorrect option: B. $10\ cm$
View full question & answer→MCQ 871 Mark
The damping force on an oscillator is directly proportional to the velocity. The unit of the constant of proportionality is:
- A
$\ce{kg ms^{-1}}$
- B
$\ce{kg ms^{-2}}$
- ✓
$\ce{kg s^{-1}}$
- D
$\ce{kg s}$
AnswerCorrect option: C. $\ce{kg s^{-1}}$
Given, damping force $\times$ velocity
$\text{F}\propto\text{v}\Rightarrow\text{F}=\text{kv}$
$\Rightarrow\text{k}=\frac{\text{F}}{\text{v}}$
Unit of $\text{k}=\frac{\text{Unit of F}}{\text{Unit of v}}=\frac{\text{kg }\text{-ms}^{-2}}{\text{ms}^{-1}}=\text{kg s}^{-1}$
View full question & answer→MCQ 881 Mark
You measure two quantities as $\text{A} = 1.0 \text{m}\pm 0.2 \text{m},\ \text{B} = 2.0 \text{m}\pm 0.2 \text{m}.$ We should report correct value for $\sqrt{\text{AB}}$ as:
- A
$1.4 \text{m}\pm 0.4\text{m.}$
- B
$1.41\text{m} \pm 0.15 \text{m}.$
- C
$1.4\text{m} \pm 0.3 \text{m}.$
- ✓
$1.4\text{m} \pm 0.2 \text{m}.$
AnswerCorrect option: D. $1.4\text{m} \pm 0.2 \text{m}.$
According to the problem, $\text{A}=1.0\text{m}\pm0.2\text{m},\ \text{B}=2.0\text{m}\pm0.2\text{m}$
Let, $\text{Z}=\sqrt{\text{AB}}=\sqrt{(1.0)(2.0)}=1.414\text{m}$
Rounding off to two significant digits $Z = 1.4m$
$\text{AS}\ \frac{\Delta\text{Z}}{\text{Z}}=\frac{1}{2}\frac{\Delta\text{A}}{\text{A}}+\frac{1}{2}\frac{\Delta\text{B}}{\text{B}}$
$=\frac{1}{2}\Big(\frac{0.2\text{m}}{1\text{m}}\Big)+\frac{1}{2}\Big(\frac{0.2\text{m}}{2\text{m}}\Big)=0.15$
$\Rightarrow\Delta\text{Z}=\text{Z}(0.15)=1.4\text{m}(0.15)=0.212$
Rounding off to one significant digit, $\Delta\text{Z}=0.2\text{m}$
The correct value for $\sqrt{\text{AB}}=1.4\pm0.2\text{m}.$
View full question & answer→MCQ 891 Mark
Which of the following ratios express pressure?
- ✓
$\frac{\text{Force}}{\text{Area}}.$
- B
$\frac{\text{Energy}}{\text{Volume}}.$
- C
$A$ and $B$
- D
$\frac{\text{Force}}{\text{Volume}}.$
AnswerCorrect option: A. $\frac{\text{Force}}{\text{Area}}.$
Let us first express the relation of pressure with other physical quantities one by one with the help of dimensional analysis.
We know that pressure,
- $\frac{\text{Force}}{\text{Area}}=\frac{[\text{MLT}^{-2}]}{[\text{L}^2]}=[\text{ML}^{-1}\text{T}^{-2}]$
So, this ratio express pressure (In fact this ratio actually represents pressure).
- $\frac{\text{Energy}}{\text{Area}}=\frac{[\text{ML}^{2}\text{T}^{-2}]}{[\text{L}^2]}=[\text{MT}^{-2}]$
Dimensions of this ratio are not same as pressure, so this ratio does not express pressure.
- $\frac{\text{Energy}}{\text{Volume}}=\frac{[\text{ML}^{2}\text{T}^{-2}]}{[\text{L}^3]}=[\text{ML}^{-1}\text{T}^{-2}]$
Dimensions of this ratio is the same as pressure, so this ratio also express pressure.
- $\frac{\text{Force}}{\text{Volume}}=\frac{[\text{ML}\text{T}^{-2}]}{[\text{L}^3]}=[\text{ML}^{-2}\text{T}^{-2}]$
Dimensions of this ratio are not same as pressure, so this ratio does not express pressure. View full question & answer→MCQ 901 Mark
If Planck’s constant $(h)$ and speed of light in vacuum $(c)$ are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?
$A.$ Mass of electron $(m_e).$
$B.$ Universal gravitational constant $(G).$
$C.$ Charge of electron $(e).$
$D.$ Mass of proton $(m_p).$
- A
$A$ and $B$
- B
$A$ and $C$
- ✓
$B$ and $C$
- D
$A , B$ and $D$
AnswerCorrect option: C. $B$ and $C$
Dimension of
$\text{h}=\frac{\text{E}}{\text{v}}=\frac{[\text{ML}^2\text{T}^{-1}]}{[\text{T}^{-1}]}=[\text{ML}^2\text{T}^{-1}]$
$\text{c}=\frac{\text{s}}{\text{t}}=[\text{LT}^{-1}]$
$\text{G}=\frac{\text{Fr}^2}{\text{M}_1\text{M}_2}=\frac{[\text{ML}^3\text{T}^{-2}]}{[\text{M}][\text{M}]}=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$
$\text{hc}=[\text{ML}^2\text{T}^{-1}]\times[\text{LT}^{-1}]=[\text{ML}^3\text{T}^{-2}]$
$\frac{\text{hc}}{\text{G}}=\frac{[\text{ML}^3\text{T}^{-2}]}{[\text{M}^{-1}\text{L}^3\text{T}^{-1}]}=[\text{M}^2]$
$\text{M}=\sqrt{\frac{\text{hc}}{\text{G}}}=\Big[\text{h}^\frac{1}{2}\text{c}^\frac{1}{2}\text{G}^\frac{-1}{2}\Big]$
$\frac{\text{h}}{\text{c}}=\frac{[\text{ML}^2\text{T}^{-1}]}{[\text{LT}^{-1}]}=[\text{ML}]=\sqrt{\frac{\text{hc}}{\text{G}}}\times\text{L}$
$\text{L}=\frac{\text{h}}{\text{c}}\times\sqrt{\frac{\text{G}}{\text{hc}}}=\frac{\sqrt{\text{Gh}}}{\text{C}^\frac{3}{2}}=\Big[\text{G}^\frac{1}{2}\text{h}^\frac{1}{2}\text{c}^\frac{-3}{2}\Big]$
$\text{c}=[\text{LT}^{-1}]=\Big[\text{G}^\frac{1}{2}\text{h}^\frac{1}{2}\text{c}^\frac{-3}{2}\text{T}^{-1}\Big]$
$\text{T}=\Big[\text{G}^\frac{1}{2}\text{h}^\frac{1}{2}\text{c}^{-\frac{3}{2}-1}\Big]=\Big[\text{G}^\frac{1}{2}\text{h}^\frac{1}{2}\text{c}^{\frac{-5}{2}}\Big]$
Hence, physical quantities $(a, b$ and $d)$ can be used to represent $\ce{L, M, T}$ in terms of the choosen fundamental quantities.
View full question & answer→MCQ 911 Mark
The length and breadth of a rectangular sheet are $16.2\ cm$ and $10.1\ cm,$ respectively. The area of the sheet in appropriate significant figures and error is:
- ✓
$164 \pm 3 \text{cm}^2$.
- B
$163.62 \pm 2.6 \text{cm}^2.$
- C
$163.6 \pm 2.6 \text{cm}^2.$
- D
$163.62 \pm 3 \text{cm}^2.$
AnswerCorrect option: A. $164 \pm 3 \text{cm}^2$.
$1=16.2\text{cm }\Delta^1=0.1$
$\text{b}=10.1\text{cm}\ \Delta\text{b}=0.1$
$1=16.2\pm0.1$
$\text{b}=10.1\pm0.1$
$\text{A}=\text{Area}=1\times\text{b}=16.2\times10.1=163.62\text{cm}^2$
$=164\text{cm}^2\ ($in significant figures$)$
$\frac{\Delta\text{A}}{\text{A}}=\frac{\Delta\text{l}}{\text{l}}+\frac{\Delta\text{b}}{\text{b}}=\frac{0.1}{16.2}+\frac{0.1}{10.1}$
$\frac{\Delta\text{A}}{164}=\frac{10.1\times0.1+16.2\times0.1}{16.2\times10.1}$
$\Delta\text{A}=164\Big(\frac{1.01+1.62}{163.62}\Big)$
$\Delta\text{A}=2.63\text{cm}^2$
Now rounding off upto significant figures in $\Delta^1$ and $\Delta^\text{b}$ i.e., one
$\Delta\text{A}=3\text{cm}^2$
$\text{A}=(164\pm3)\text{cm}^2$. Hence, verifies the option $(a).$
View full question & answer→MCQ 921 Mark
Among the given following system of unit which is not based on unit of mass, length and time?
- A
$\text{CGS}$
- B
$\text{FPS}$
- C
$\text{MKS}$
- ✓
$\text{SI}$
AnswerCorrect option: D. $\text{SI}$
View full question & answer→MCQ 931 Mark
Number of fermi in one metre is:
- A
$10^6F$
- B
$10^{17}F$
- ✓
$10^{15}F$
- D
$10^{14}F$
AnswerCorrect option: C. $10^{15}F$
$1$ fermi $(F) = 10^{-15}$m
OR
$1\text{m}=\frac{1}{10^{-15}}=10^{15}\text{F}$
View full question & answer→MCQ 941 Mark
$N$ divisions on the main scale of a vernier calipers coincide with $(N + 1)$ divisions on the vernier scale. If each division on the main scale is of a unit, determine the least count of instrument.
- ✓
$\frac{\text{a}}{(\text{N}+1)}$
- B
$\frac{3\text{a}}{4(\text{N}+1)}$
- C
$\frac{\text{a}}{(\text{N}+1)^2}$
- D
$\Big(\frac{\text{a}}{\text{N}+1}\Big)^2$
AnswerCorrect option: A. $\frac{\text{a}}{(\text{N}+1)}$
View full question & answer→MCQ 951 Mark
The number of particles crossing per unit area perpendicular to $X-$axis in unit time is: $\text{N}=-\text{D}\frac{\text{n}_2-\text{n}_1}{\text{x}_2-\text{x}_1}$ Where $n_1$ and $n_2$ are number of particles per unit volume for the value of $x_1$ and $x_2$ respectively. The dimensions of diffusion constant $D$ are:
- A
$\text{M}^{0}\text{L}\text{T}^2$
- B
$\text{M}^{0}\text{L}^2\text{T}^{-4}$
- C
$\text{M}^{0}\text{L}\text{T}^{-3}$
- ✓
$\text{M}^{0}\text{L}^2\text{T}^{-1}$
AnswerCorrect option: D. $\text{M}^{0}\text{L}^2\text{T}^{-1}$
Since $N$ is no. of particles per unit area per unit time, thus its dimensions are $\text{M}^{0}\text{L}^2\text{T}^{-1}$.
$n_1$ and $n_2$ are no. f particles per unit volume. Thus they have the dimensions $\text{M}^{0}\text{lL}^{-3}$.
$x$ has dimension of length, i.e $L.$
Thus, $D$ has dimensions of $\frac{\text{Nx}}{\text{n}}=\frac{\text{M}^{0}\text{lL}^2\text{T}^{-1}\text{L}}{\text{M}^{0}\text{lL}^{-3}}=\text{L}^2\text{T}^{-1}$
View full question & answer→MCQ 961 Mark
A wire has a mass $0.3 \pm 0.003\text{g},$ radius $0.5 \pm 0.005\text{mm}$ and length $6 \pm 0.06\text{cm}.$ The maximum percentage error in the measurement of its density is
View full question & answer→MCQ 971 Mark
Which of the following sets have different dimensions?
- ✓
Dipole moment, Electric field and Electric flux.
- B
Pressure, Young's modulus, Stress.
- C
- D
Emf, Potential difference and potential.
AnswerCorrect option: A. Dipole moment, Electric field and Electric flux.
Heat, work and energy are same things, so they have same dimensions.
$\ce{Emf,}$ potential difference and potential have the same dimensions.
$\text{Pressure}=\frac{\text{force}}{\text{area}},\text{stress}=\frac{\text{force}}{\text{area}}$
$\text{Y}=\frac{\text{Stress}}{\text{Strain}}=\frac{\frac{\text{force}}{\text{area}}}{\text{dimensionless}}=\frac{\text{force}}{\text{ area}}$
So, they have same dimensions.
But dimension of Dipole moment $= \ce{[M^0L^1T^1A^1]}$
Dimenslion of electric field $= \ce{[M^1L^1T^{-3}-A^{-1}]}$
and dimension of electric flux $= \ce{[M^1L^3T^{-3}-A{-1}]}$
Hence they are different.
View full question & answer→MCQ 981 Mark
Which of the following has metre kelvin as the unit?
MCQ 991 Mark
Among the given following units which one is not unit of length?
MCQ 1001 Mark
The solid angle subtended by the periphery of an area $1\ cm^2$ at a point situated symmetrically at a distance of $5\ cm$ from the area is:
- A
$2 \times 10^{-2}$ steradian.
- ✓
$4 \times 10^{-2}$ steradian.
- C
$6 \times 10^{-2}$ steradian.
- D
$8 \times 10^{-2}$ steradian.
AnswerCorrect option: B. $4 \times 10^{-2}$ steradian.
Solid angle, $\text{d}\Omega=\frac{\text{dA}}{\text{r}^2}=1\text{cm}^2/(\text{ 5cm})^2$
$=0.04$ steradian
$=4\times10^{-2}$ steradian
View full question & answer→