MCQ 1011 Mark
The dimensional formula $M^0 L^2 T^{-2}$ stands for
- A
- B
- ✓
- D
Coefficient of thermal conductivity
Answer(c) Latent Heat $L=\frac{Q}{m}=\frac{\text { Energy }}{\text { mass }}=\frac{\left[M L^2 T^{-2}\right]}{[M]}=\left[L^2 T^{-2}\right]$
View full question & answer→MCQ 1021 Mark
Identify the pair which has different dimensions
- A
Planck's constant and angular momentum
- B
Impulse and linear momentum
- ✓
Angular momentum and frequency
- D
Pressure and Young's modulus
AnswerCorrect option: C. Angular momentum and frequency
(c) Angular momentum $=\left[M L^2 T^{-1}\right]$, Frequency $=\left[T^{-1}\right]$
View full question & answer→MCQ 1031 Mark
If the length of $\operatorname{rod} A$ is $3.25 \pm 0.01 cm$ and that of $B$ is $4.19 \pm 0.01$ $cm$ then the $\operatorname{rod} B$ is longer than $\operatorname{rod} A$ by
- A
$0.94 \pm 0.00 cm$
- B
$0.94 \pm 0.01 cm$
- ✓
$0.94 \pm 0.02 cm$
- D
$0.94 \pm 0.005 cm$
AnswerCorrect option: C. $0.94 \pm 0.02 cm$
View full question & answer→MCQ 1041 Mark
The unit of the coefficient of viscosity in S.l. system is
- A
$m / k g-s$
- B
$m-s / kg ^2$
- C
$k g / m-s^2$
- ✓
$kg / m - s$
AnswerCorrect option: D. $kg / m - s$
(d) $[\eta]=M L^{-1} T^{-1}$ so its unit will be $\mathrm{kg} / \mathrm{m}$-sec.
View full question & answer→MCQ 1051 Mark
According to Newton, the viscous force acting between liquid layers of area $A$ and velocity gradient $\Delta v / \Delta z$ is given by $F=-\eta A \frac{\Delta v}{\Delta z}$ where $\eta$ is constant called coefficient of viscosity. The dimension of $\eta$ are
- A
$\left[M L^2 T^{-2}\right]$
- ✓
$\left[M L^{-1} T^{-1}\right]$
- C
$\left[M L^{-2} T^{-2}\right]$
- D
$\left[M^0 L^0 T^0\right]$
AnswerCorrect option: B. $\left[M L^{-1} T^{-1}\right]$
(b) $F=-\eta A \frac{\Delta v}{\Delta z}$
$ \Rightarrow[\eta]=M L^{-1} T^{-1}$
As $F=\left[M L T^{-2}\right],$
$A=L^2, \frac{\Delta v}{\Delta z}=T^{-1}$
View full question & answer→MCQ 1061 Mark
The dimensional formula of angular velocity is
- ✓
$M^0 L^0 T^{-1}$
- B
$M L T^{-1}$
- C
$M^0 L^0 T^1$
- D
$M L^0 T^{-2}$
AnswerCorrect option: A. $M^0 L^0 T^{-1}$
(a)
Angular velocity $=\frac{\theta}{t}$,
${\omega}=[M^0 L^0 T^0]$
$[T]=\left[T^{-1}\right]$
View full question & answer→MCQ 1071 Mark
The expression $\left[M L^2 T^{-2}\right]$ represents
View full question & answer→MCQ 1081 Mark
One million electron volt $(1 MeV )$ is equal to
- A
$10^5 eV$
- ✓
$10^6 eV$
- C
$10^4 eV$
- D
$10^7 eV$
AnswerCorrect option: B. $10^6 eV$
(b) $1 \mathrm{MeV}=10^6 \mathrm{eV}$
View full question & answer→MCQ 1091 Mark
A wire has a mass $0.3 \pm 0.003 g$, radius $0.5 \pm 0.005 mm$ and length $6 \pm 0.06 cm$. The maximum percentage error in the measurement of its density is
Answer(d)$\because \text { Density, } \rho=\frac{M}{V}=\frac{M}{\pi r^2 L}$
$\Rightarrow \frac{\Delta \rho}{\rho}=\frac{\Delta M}{M}+2 \frac{\Delta r}{r}+\frac{\Delta L}{L} $
$=\frac{0.003}{0.3}+2 \times \frac{0.005}{0.5}+\frac{0.06}{6} =0.01+0.02+0.01=0.04$
$\therefore \text { Percentage error }=\frac{\Delta \rho}{\rho\times 100}=0.04 \times 100=4 \%$
View full question & answer→MCQ 1101 Mark
In the relation $P=\frac{\alpha}{\beta} e^{-\frac{\alpha Z}{k \theta}} \quad P$ is pressure, $Z$ is the distance, $k$ is Boltzmann constant and $\theta$ is the temperature. The dimensional formula of $\beta$ will be
- ✓
$\left[M^0 L^2 T^0\right]$
- B
$\left[M^1 L^2 T^1\right]$
- C
$\left[M^1 L^0 T^{-1}\right]$
- D
$\left[M^0 L^2 T^{-1}\right]$
AnswerCorrect option: A. $\left[M^0 L^2 T^0\right]$
In given equation, $\frac{\alpha z}{k \theta}$ should be dimensionless
$\therefore \alpha=\frac{k \theta}{z} \Rightarrow[\alpha]=\frac{\left[M L^2 T^{-2} K^{-1} \times K\right]}{[L]}$
$=\left[M L T^{-2}\right] $
$\text { and }P=\frac{\alpha}{\beta} \Rightarrow[\beta]$
$=\left[\frac{\alpha}{p}\right]$$=\frac{\left[M L T^{-2}\right]}{\left[M L^{-1} T^{-2}\right]}$
$=\left[M^0 L^2 T^0\right] .$
View full question & answer→MCQ 1111 Mark
Which of the following group have different dimension
- A
Potential difference, EMF, voltage
- B
Pressure, stress, young's modulus
- C
- ✓
Dipole moment, electric flux, electric field
AnswerCorrect option: D. Dipole moment, electric flux, electric field
View full question & answer→MCQ 1121 Mark
Let $\left[\varepsilon_0\right]$ denotes the dimensional formula of the permittivity of the vacuum and $\left[\mu_0\right]$ that of the permeability of the vacuum. If $M=$ mass,$L=$ length, $T=$ Time and $I=$ electriccurrent, then
- A
$\left[\varepsilon_0\right]=M^{-1} L^{-3} T^2 I$
- ✓
$\left[\varepsilon_0\right]=M^{-1} L^{-3} T^4 I^2$
- C
$\left[\mu_0\right]=M L T^{-3} I^{-2}$
- D
$\left[\mu_0\right]=M L^2 T^{-1} I$
AnswerCorrect option: B. $\left[\varepsilon_0\right]=M^{-1} L^{-3} T^4 I^2$
View full question & answer→MCQ 1131 Mark
The pair(s) of physical quantities that have the same dimensions, is (are)
- A
Reynolds number and coefficient of friction
- B
Latent heat and gravitational potential
- C
Curie and frequency of a light wave
- ✓
Answer(D) Reynolds number and coefficient of friction are dimensionless.Latent heat and gravitational potential both have dimension $\left[L^2 T^{-2}\right]$.Curie and frequency of a light wave both have dimension $\left[T^{-1}\right]$. But dimensions of Planck's constant is $\left[M L^2 T^{-1}\right]$ and torque is $\left\lfloor M L^2 T\right\rfloor$
View full question & answer→MCQ 1141 Mark
Column 1 Column II(i) Curie $M L T^{-2}$
(ii) Light year $M$
(iii) Dielectric strength Dimensionless
(iv) Atomic weight $T$
(v) Decibel (E) $M L^2 T^{-2}$
(F) $M T^{-3}$
(G) $T^{-1}$
(H) $L$
(l) $M L T^{-3} \Gamma^{-1}$
(J) $L T^{-1}$
Choose the correct match
- ✓
(i) $G$, (ii) $H$, (iii) $C$, (iv) $B$, (v) $C$
- B
(i) $D$, (ii) $H$, (iii) $l$, (iv) $B$, (v) $G$
- C
(i) $G$, (ii) $H$, (iii) $l$, (iv) $B$, (v) $G$
- D
AnswerCorrect option: A. (i) $G$, (ii) $H$, (iii) $C$, (iv) $B$, (v) $C$
View full question & answer→MCQ 1151 Mark
A highly rigid cubical block $A$ of small mass $M$ and side $L$ is fixed rigidly onto another cubical block $B$ of the same dimensions and of low modulus of rigidity $\eta$ such that the lower face of $A$ completely covers the upper face of $B$. The lower face of $B$ is rigidly held on a horizontal surface. A small force $F$ is applied perpendicular to one of the side faces of $A$. After the force is withdrawn block $A$ executes small oscillations. The time period of which is given by
- A
$2 \pi \sqrt{\frac{M \eta}{L}}$
- B
$2 \pi \sqrt{\frac{L}{M \eta}}$
- C
$2 \pi \sqrt{\frac{M L}{\eta}}$
- ✓
$2 \pi \sqrt{\frac{M}{\eta L}}$
AnswerCorrect option: D. $2 \pi \sqrt{\frac{M}{\eta L}}$
(d) By substituting the dimensions of mass $[M]$, length $[L]$ and coefficientofrigidity$\left\lfloor M L^{-1} T^{-2}\right\rfloor$ we get $T=2 \pi \sqrt{\frac{M}{\eta L}}$ is the right formula for time period of oscillations
View full question & answer→MCQ 1161 Mark
If the dimensions of length are expressed as $G^x c^y h^z$; where $G, c$ and $h$ are the universal gravitational constant, speed of light and Planck's constant respectively, then
- A
$x=\frac{1}{3}, y=\frac{1}{2}$
- B
$x=\frac{1}{2}, z=\frac{1}{2}$
- C
$y=-\frac{3}{2}, z=\frac{1}{2}$
- ✓
Answer$(d)$ Length $\propto G c h$
$L=\left [M^{-1} L^3 T^{-2}\right]^x \quad \left[L T^{-1}\right]^y \left[M L^2 T^{-1}\right]^z$
By comparing the power of $M, L$ and $T$ inboth sides we get
$-x+z=0,3 x+y+2 z=1$ and $-2 x-y-z=0$
By solving above three equationswe get
$x=\frac{1}{2}, y=-\frac{3}{2}, z=\frac{1}{2}$
View full question & answer→MCQ 1171 Mark
If $L, C$ and $R$ represent inductance, capacitance and resistance respectively, then which of the following does not represent dimensions of frequency
- A
$\frac{1}{R C}$
- B
$\frac{R}{L}$
- C
$\frac{1}{\sqrt{L C}}$
- ✓
$\frac{C}{L}$
AnswerCorrect option: D. $\frac{C}{L}$
(d) $f=\frac{1}{2 \pi \sqrt{L C}} \quad \therefore\left(\frac{C}{L}\right)$ does not represent the dimension of frequency
View full question & answer→MCQ 1181 Mark
$M L^{-1} T^{-2}$ represents
- A
- B
- C
- ✓
All the above three quantities
AnswerCorrect option: D. All the above three quantities
View full question & answer→MCQ 1191 Mark
Dimensions of $C R$ are those of
Answer(c)$\text { Capacity } \times \text { Resistance }=\frac{\text { Charge }}{\text { Potential }} \times \frac{\text { Volt }}{\mathrm{amp}} $
$=\frac{\text { amp }\times \text { second } \times \text { Volt }}{\text { Volt } \times \text { amp }}=\text { Second }$
View full question & answer→MCQ 1201 Mark
The physical quantity that has no dimensions
- A
- B
- C
- ✓
Strain has no dimensions.
AnswerCorrect option: D. Strain has no dimensions.
(d) Strain has no dimensions.
View full question & answer→MCQ 1211 Mark
The physical quantity which has dimensional formula as that of Energy $\overline{\text { Mass } \times \text { Length }}$
Answer(d) $\frac{\text { Energy }}{\text { mass } \times \text { length }}=\frac{\left[M L^2 T^{-2}\right]}{[M][L]}=\left[L T^{-2}\right]$
View full question & answer→MCQ 1221 Mark
Dimensions of time in power are
- A
$T^{-1}$
- B
$T^{-2}$
- ✓
$T^{-3}$
- D
$T^0$
AnswerCorrect option: C. $T^{-3}$
(c) Dimensions of power is $\left[M L^2 T^{-3}\right]$
View full question & answer→MCQ 1231 Mark
Dimensional formula $M L^2 T^{-3}$ represents
Answer(b) Power $=\frac{\text { Work }}{\text { Time }}=\frac{M L^2 T^{-2}}{T}=M L^2 T^{-3}$
View full question & answer→MCQ 1241 Mark
- ✓
- B
Dynes $/ cm ^2$
- C
$cm$ of $Hg$
- D
View full question & answer→MCQ 1251 Mark
- A
$M^0 L^0 T^{-1} A^{-1}$
- B
$M L T A^{-1}$
- C
$T^{-1} A$
- ✓
$T A$
Answer(d) Charge $=$ Current $\times$ Time $=[A T]$
View full question & answer→MCQ 1261 Mark
One femtometer is equivalent to
- A
$10^{15} m$
- ✓
$10^{-15} m$
- C
$10^{-12} m$
- D
$10^{12} m$
AnswerCorrect option: B. $10^{-15} m$
View full question & answer→MCQ 1271 Mark
If $L=2.331 cm , B=2.1 cm$, then $L+B=$
- A
$4.431 cm$
- B
$4.43 cm$
- C
$4.4 cm$
- ✓
$4 cm$
AnswerCorrect option: D. $4 cm$
(d)$\text { Kinetic energy } E=\frac{1}{2} m v^2$
$\therefore \frac{\Delta E}{E} \times 100=\frac{v^{\prime 2}-v^2}{v^2} \times 100 $
$=\left[(1.5)^2-1\right] \times100$
$\therefore \frac{\Delta E}{E} \times 100=125 \%$
(c) Quantity C has maximum power. So it brings maximum error in $P$.
(c) Given, $L=2.331 \mathrm{~cm}=2.33$ (correct upto two decimal places)and $B=2.1 \mathrm{~cm}=2.10 \mathrm{~cm}$$\therefore L+B=2.33+2.10=4.43 \mathrm{~cm} .=4.4 \mathrm{~cm}$Since minimum significantfigure is 2 .
View full question & answer→MCQ 1281 Mark
The Martians use force $(F)$, acceleration $$ and time $(T)$ as their fundamental physical quantities. The dimensions of length on Martians system are
- A
$F T^2$
- B
$F^{-1} T^2$
- C
$F^{-1} A^2 T^{-1}$
- ✓
$A T^2$
AnswerCorrect option: D. $A T^2$
(d) Acceleration $=\frac{\text { distance }}{\text { time }^2} \Rightarrow A=L T^{-2}\Rightarrow L=A T^2$
View full question & answer→MCQ 1291 Mark
The dimensions of $C V^2$ matches with the dimensions of
- A
$L^2 I$
- B
$L^2 I^2$
- ✓
$L I^2$
- D
$\frac{1}{L I}$
AnswerCorrect option: C. $L I^2$
(c) Both are the formula of energy . $\left(E=\frac{1}{2} C V^2=\frac{1}{2} L I^2\right)$
View full question & answer→MCQ 1301 Mark
The dimensions of physical quantity $X$ in the equation Force $=\frac{X}{\text { Density }}$ is given by
- A
$M^1 L^4 T^{-2}$
- B
$M^2 L^{-2} T^{-1}$
- ✓
$M^2 L^{-2} T^{-2}$
- D
$M^1 L^{-2} T^{-1}$
AnswerCorrect option: C. $M^2 L^{-2} T^{-2}$
(c) $[X]=[F] \times[\rho]=\left[M L T^{-2}\right] \times\left[\frac{M}{L^3}\right]=\left[M^2 L^{-2} T^{-2}\right]$
View full question & answer→MCQ 1311 Mark
Erg $-m^{-1}$ can be the unit of measure for
Answer(a) Energy $(E)=F \times d \Rightarrow F=\frac{E}{d}$ so Erg/metre can be the unit offorce.
View full question & answer→MCQ 1321 Mark
The dimensional formula of relative density is
- A
$M L^{-3}$
- B
$L T^{-1}$
- C
$M L T^{-2}$
- ✓
Answer(d) Relative density $=\frac{\text { Densityof substance }}{\text { density of water }=\left[M^0 L^0 T^0\right]$
View full question & answer→MCQ 1331 Mark
The dimensions of emf in MKS is
- A
$M L^{-1} T^{-2} Q^{-2}$
- B
$M L^2 T^{-2} Q^{-2}$
- C
$M L T^{-2} Q^{-1}$
- ✓
$M L^2 T^{-2} Q^{-1}$
AnswerCorrect option: D. $M L^2 T^{-2} Q^{-1}$
(d)
$e=L \frac{d i}{d t} \Rightarrow[e]=\left[M L^2 T^{-2} A^{-2}\right]\left[\frac{A}{T}\right]$
${[e]=\left[\frac{M L^2 T^{-2}}{A T}\right]=\left[M L^2 T^{-2} Q^{-1}\right]}$
View full question & answer→MCQ 1341 Mark
The dimensions of Planck's constant and angular momentum are respectively
- A
$M L^2 T^{-1}$ and $M L T^{-1}$
- ✓
$M L^2 T^{-1}$ and $M L^2 T^{-1}$
- C
$M L T^{-1}$ and $M L^2 T^{-1}$
- D
$M L T^{-1}$ and $M L^2 T^{-2}$
AnswerCorrect option: B. $M L^2 T^{-1}$ and $M L^2 T^{-1}$
View full question & answer→MCQ 1351 Mark
Which physical quantities have the same dimension
- ✓
- B
- C
Latent heat and specific heat
- D
Answer(a)
$ { Couple of force }=|\vec{r} \times \vec{F}|=\left[M L^2 T^{-2}\right]$
$\text { Work }=[\vec{F} \cdot \vec{d}]=\left[M L^2 T^{-2}\right]$
View full question & answer→MCQ 1361 Mark
- A
- B
$kg \rightarrow m$
- ✓
$kg - m / s$
- D
AnswerCorrect option: C. $kg - m / s$
(c) Impulse $=$ Force $\times$ time $=\left(k g-m / s^2\right) \times s=k g-m / s$
View full question & answer→MCQ 1371 Mark
Which of the following is not the unit of time
- A
- B
- C
- ✓
Parallactic second,Solar day
AnswerCorrect option: D. Parallactic second,Solar day
View full question & answer→MCQ 1381 Mark
Joule-second is the unit of
Answer(d) $\tau=\frac{d L}{d t} \Rightarrow d L=\tau \times d t=r \times F \times d t$i.e. the unit of angular momentum is joule-second.
View full question & answer→MCQ 1391 Mark
Which of the following represents a volt
Answer(b) $\frac{\text { watt }}{\text { ampere }}=$ volt
View full question & answer→MCQ 1401 Mark
The velocity of a particle depends upon as $v=a+b t+c t^2$; if the velocity is in $m$ / sec , the unit of $a$ will be
- ✓
$m / sec$
- B
$m / sec ^2$
- C
$m^2 / sec$
- D
$m / \sec ^3$
AnswerCorrect option: A. $m / sec$
(a) Quantities of similar dimensions can be added or subtracted so unit of $a$ will be same as that of velocity.
View full question & answer→MCQ 1411 Mark
Whose dimensions is $M L^2 T^{-1}$
Answer(b) Angular momentum $=m v r=M L T^{-1} \times L=M L^2 T^{-1}$
View full question & answer→MCQ 1421 Mark
Dimensions of frequency are
- A
$M^0 L^{-1} T^0$
- ✓
$M^0 L^0 T^{-1}$
- C
$M^0 L^0 T$
- D
$M T^{-2}$
AnswerCorrect option: B. $M^0 L^0 T^{-1}$
(b) Frequency $=\frac{1}{\pi}=\left[M^0 L^0 T^{-1}\right]$
View full question & answer→MCQ 1431 Mark
The unit of reduction factor of tangent galvanometer is
MCQ 1441 Mark
Henry/ohm can be expressed in
Answer(a) $\frac{L}{R}$ is a time constant of $L-R$ circuit so Henry/ohm can be expressed as second.
View full question & answer→MCQ 1451 Mark
Answer(d) Unit of e.m.f. = volt $=$ joule $/$ coulomb
View full question & answer→MCQ 1461 Mark
Newton/metre $e^2$ is the unit of
View full question & answer→MCQ 1471 Mark
Which one of the following does not have the same dimensions
- A
- B
- C
Relative density and refractive index
- ✓
Planck constant and energy
AnswerCorrect option: D. Planck constant and energy
(d)
${[\text { Planck constant }]=\left[M L^2 T^{-1}\right] \text { and }} $
${[\text { Energy }]=\left[M L^2 T^{-2}\right]}$
View full question & answer→MCQ 1481 Mark
The dimensions of calorie are
- ✓
$M L^2 T^{-2}$
- B
$M L T^{-2}$
- C
$M L^2 T^{-1}$
- D
$M L^2 T^{-3}$
AnswerCorrect option: A. $M L^2 T^{-2}$
(a) Calorie is the unit of heat i.e., energy.So dimensions of energy $=M L^2 T^{-2}$
View full question & answer→MCQ 1491 Mark
- A
- B
- ✓
Joule per second and watt both
- D
AnswerCorrect option: C. Joule per second and watt both
(c) Watt $=$ Joule/sec.
View full question & answer→MCQ 1501 Mark
Newton-second is the unit of
Answer(c) Impulse = change in momentum $=F \times t$So the unit of momentum will be equal to Newton-sec
View full question & answer→