MCQ 11 Mark
Which of the following is not a fundamental unit?
MCQ 21 Mark
Dimensions of kinetic energy are the same as that of
MCQ 31 Mark
View full question & answer→MCQ 41 Mark
The error in the measurement of the sides of a rectangle is 1%. The error in the measurement of its area is
MCQ 51 Mark
$\left[L^1 M^1 T^{-2}\right]$ is the dimensional formula for
View full question & answer→MCQ 61 Mark
A calorie is a unit of heat energy and its value is $18 J$, where $1 J =1 kg m ^2 s ^{-2}$. Suppose we use a new system of units in which unit of mass equals $x kg$, the unit of length equals $y m$ and the unit of the time is $z sec$. Then the value of a calorie in the new system of units is
- ✓
$4.18 \frac{z^2}{x y^2}$
- B
$4.18 \frac{x y^2}{z^2}$
- C
$4.18 \frac{z^2}{x}$
- D
$4.18 \frac{y^2}{x z^2}$
AnswerCorrect option: A. $4.18 \frac{z^2}{x y^2}$
$1 J =(1 \ kg )(1 m )^2(1 \sec )^{-2}$
$1 \alpha=(x \ kg )(y m )^2(z \sec )^{-2}$
$\therefore \frac{1 J }{1 \alpha}=\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)^2(z)^2=\frac{z^2}{x y^2}$
$\therefore 1 J =\frac{z^2}{x y^2} \alpha \text { or } 1 cal =4.18 \frac{z^2}{x y^2}$
View full question & answer→MCQ 71 Mark
The ratio of dimensions of Planck's constant to that of moment of inertia is the dimensions of
Answer(c) : Ratio of the dimensions of Planck's constant to that of moment of inertia is,
$\frac{[h]}{[I]}=\frac{\left[ ML ^2 T ^{-1}\right]}{\left[ ML ^2\right]}=\left[ T ^{-1}\right]=$ Dimension of frequency.
View full question & answer→MCQ 81 Mark
The dimensions of self or mutual inductance are given as
- A
$\left[ L ^{-2} M ^1 T ^{-2} I ^{-2}\right]$
- B
$\left[ L ^2 M ^2 T ^{-2} I ^{-2}\right]$
- ✓
$\left[ L ^2 M ^1 T ^{-2} I ^{-2}\right]$
- D
$\left[ L ^2 M ^2 T ^{-2} I ^{-1}\right]$
AnswerCorrect option: C. $\left[ L ^2 M ^1 T ^{-2} I ^{-2}\right]$
(c) : We know that,
$
\varepsilon=-\frac{d \phi}{d t}=-M \frac{d i}{d t} \Rightarrow M=(-) \frac{\varepsilon d t}{d i}
$
$\therefore$ Dimension of mutual inductance is
$M=\frac{\left[ ML ^2 T ^{-3} I ^{-1}\right] \times[ T ]}{[ I ]}=\left[ M ^1 L ^2 T ^{-2} I ^{-2}\right]$
View full question & answer→MCQ 91 Mark
The force $F$ acting on a body of density $d$ are related by the relation $F=\frac{y}{\sqrt{d}}$. The dimensions of $y$ are
- ✓
$\left[ L ^{-\frac{1}{2}} M ^{\frac{3}{2}} T ^{-2}\right]$
- B
$\left[ L ^{-1} M ^{\frac{1}{2}} T ^{-2}\right]$
- C
$\left[ L ^{-1} M ^{\frac{3}{2}} T ^{-2}\right]$
- D
$\left[ L ^{-\frac{1}{2}} M ^{\frac{1}{2}} T ^{-2}\right]$
AnswerCorrect option: A. $\left[ L ^{-\frac{1}{2}} M ^{\frac{3}{2}} T ^{-2}\right]$
(a) : Given, force, $F=\frac{y}{\sqrt{d}}$ where $F=$ force and $d=$ density
$\therefore \quad[y]=[F \sqrt{d}]=\left[ MLT ^{-2}\right]\left[ ML ^{-3}\right]^{1 / 2}=\left[ M ^{3 / 2} L ^{-1 / 2} T ^{-2}\right]$
View full question & answer→MCQ 101 Mark
Dimensions of gyromagnetic ratio are
- A
$\left[ L ^1 M ^0 T ^1 I ^{ I }\right]$
- ✓
$\left[ L ^0 M ^{-1} T ^1 I ^1\right]$
- C
$\left[ L ^1 M ^0 I ^0 T ^{-1}\right]$
- D
$\left[ L ^{-1} M ^0 T ^1 I ^1\right]$
AnswerCorrect option: B. $\left[ L ^0 M ^{-1} T ^1 I ^1\right]$
(b) : Gyromagnetic ratio of particle or a system is defined as the ratio of its magnetic moment to its angular momentum.
$
\therefore \quad \gamma=\frac{\text { Magnetic moment }}{\text { Angular momentum }}
$
or $[\gamma]=\frac{\left[ L ^2 I \right]}{\left[ L ^2 MT ^{-1}\right]}=\left[ L ^0 M ^{-1} T ^1 I ^1\right]$
View full question & answer→MCQ 111 Mark
The dimensions of torque are same as that of
Answer(a) : Moment of force is known as torque.
Thus, dimensions of force and torque are same. Dimension of torque $=$ Dimension of moment of force
$\begin{aligned} & =\left[ ML ^2 T ^{-2}\right] \\ & \text { Dimension of moment of force }=\left[ ML ^2 T ^{-2}\right]\end{aligned}$
View full question & answer→MCQ 121 Mark
Let $x=\left[\frac{a^2 b^2}{c}\right]$ be the physical quantity. If the percentage error in the measurement of physical quantities $a, b$ and $c$ is 2,3 and 4 percent respectively. Then percentage error in the measurement of $x$ is
- A
$7 \%$
- ✓
$14 \%$
- C
$21 \%$
- D
$28 \%$
AnswerCorrect option: B. $14 \%$
(b) : Percentage error in $x$ is given by
$
\begin{aligned}
& \frac{\Delta x}{x}=2 \frac{\Delta a}{a}+2 \frac{\Delta b}{b}+\frac{\Delta c}{c} \\
& \frac{\Delta x}{x} \times 100=2 \times 2 \%+2 \times 3 \%+4 \%=14 \%
\end{aligned}
$
View full question & answer→MCQ 131 Mark
The following observations were taken for determining surface tension $T$ of water by capillary method:
diameter of capillary, $D = 1.25 \times 10^{-2} m$
rise of water, $h = 1.45 \times 10^{-2} m$
Using $g = 9.80 \ m/s^2$ and the simplified relation
$T =\frac{ rhg }{2} \times 10^3 \ N/m,$ the possible error in surface tension is closest to:
- A
$0.15\%$
- ✓
$1.5\%$
- C
$2.4\%$
- D
$10\%$
AnswerCorrect option: B. $1.5\%$
$D = 1.25 \times 10^{-2} m;$
$h = 1.45 \times 10^{-2} m$
The maximum permissible error in $D$
$= \triangle D = 0.01 \times 10^{-2} m$
The maximum permissible error in $h$
$= \triangle h = 0.01 \times 10^{-2} m$
$g$ is given as a constant and is errorless.
$T =\frac{ rhg }{2} \times 10^3 N / m =\frac{ dhg }{4} \times 10^3 N / m$
$\therefore \%$ error $\frac{\Delta T }{ T }=\frac{\Delta d }{ d }+\frac{\Delta h }{ h }$
$ \therefore \frac{\Delta T }{ T } \times 100 =\frac{\Delta d }{ d } \times 100+\frac{\Delta h }{ h } \times 100$
$ =\left(\frac{0.01 \times 10^{-2}}{1.25 \times 10^{-2}}+\frac{0.01 \times 10^{-2}}{1.45 \times 10^{-2}}\right) \times 100$
$ =\frac{100}{125}+\frac{100}{145}$
$\therefore \frac{\Delta T }{ T }= 0.8 \%+0.7 \%$
$=1.5 \%$
View full question & answer→MCQ 141 Mark
A physical quantity of the dimensions of length that can be formed out of $c, G$ and $\frac{ e ^2}{4 \pi \varepsilon_0}$ is $[c$ is velocity of light, $G$ is universal constant of gravitation and $e$ is charge$]:$
- ✓
$\frac{1}{ c ^2}\left[ G \frac{ e ^2}{4 \pi \varepsilon_0}\right]^{1 / 2}$
- B
$c ^2\left[ G \frac{ e ^2}{4 \pi \varepsilon_0}\right]^{1 / 2}$
- C
$\frac{1}{c^2}\left[\frac{ e ^2}{ G 4 \pi \varepsilon_0}\right]^{1 / 2}$
- D
$\frac{1}{ c } G \frac{ e ^2}{4 \pi \varepsilon_0}$
AnswerCorrect option: A. $\frac{1}{ c ^2}\left[ G \frac{ e ^2}{4 \pi \varepsilon_0}\right]^{1 / 2}$
Let the physical quantity formed of the dimensions of length be given as.
$[ L ]=[ C ]^{ X }[ G ]^{ y }\left[\frac{ e ^2}{4 \pi \varepsilon_0}\right]^z .......(i)$
Now,
Dimensions of velocity of light $[c]^x = [LT^{-1}]^x$
Dimensions of universal gravitational constant
$[G]^y = [L^3T^2M^{-1}]^y$
Dimensions of $\left[\frac{ e ^2}{4 \pi \varepsilon_0}\right]^z=\left[ ML ^3 T^{-2}\right]^z$
Substitrning these in equation $(i)$
$[L] [LT^{-1}]^x [M^{-1}L^3T^{-2}]^y [ML^3T^{-2}]^z$
$= L^{x+3y+3z} M^{-y+z} T^{-x-2y-2z}$
Solving for $x, y, z$
$x + 3y + 3z = 1$
$-y + z = 0$
$x + 2y + 2z = 0$
Solving the above equation,
$x=-2, y=\frac{1}{2}, z=\frac{1}{2}$
$\therefore L=\frac{1}{c^2}\left[G \frac{e^2}{4 \pi \varepsilon_0}\right]^{1 / 2}$
View full question & answer→MCQ 151 Mark
Let $x=\left[\frac{a^2 b^2}{c}\right]$ be the physical quantity. If the percentage error in the measurement of physical quantities $a, b$ and $c$ is $2, 3$ and $4$ percent respectively then percentage en$-$or in the measurement of $x$ is
- A
$7\%$
- ✓
$14\%$
- C
$21\%$
- D
$28\%$
AnswerCorrect option: B. $14\%$
Given: $x=\frac{a^2 b^2}{c}$
Percentage error is given by.
$\frac{\Delta x}{x}=\frac{2 \Delta a}{a}+\frac{2 \Delta b}{b}+\frac{\Delta c}{c}$
$=(2 \times 2)+(2 \times 3)+4$
$=4+6+4=14$
$\therefore \frac{\Delta x }{ x } \%=14 \%$
View full question & answer→MCQ 161 Mark
The density of the material in the shape of a cube is determined by measuring three sides of the cube and its mass. 1f the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:
Answer$4.5 \%$
Hint: Density $(\rho)=\frac{\text { mass }}{\text { Volume }}=\frac{ m }{l^3} \quad \ldots$ (for cube $V =l^3$ )
Percentage relative error in density will be,
$
\begin{aligned}
\frac{\Delta \rho}{\rho} \times 100 & =\frac{\Delta m }{ m } \times 100+3 \frac{\Delta l}{l} \times 100 \\
& =1.5+(3 \times 1) \\
& =1.5+3=4.5 \%
\end{aligned}
$
View full question & answer→MCQ 171 Mark
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 screw gauge has a zero error of— 0.004 cm, the correct diameter of the ball is
Answer0.29 cm
Hint:
Least count of screw gauge = 0.001 cm = 0.01mm
Main scale reading = 5 mm.
Zero error = – 0.004 cm = -0.04 mm
Zero correction = +0.04 mm
Observed reading = Mainscale reading + (Division × least count)
Observed reading = 5 + (25 × 0.01) = 5.25 mm
Corrected reading = Observed reading + Zero correction
Corrected reading = 5.25 + 0.04
= 5.29 mm = 0.529 cm
View full question & answer→MCQ 181 Mark
A student measures time for $20$ oscillations of a simple pendulum as $30 s. 32 s, 35 s$ and $31 s. 1f$ the minimum division in the measuring clock is I s, then correct mean time in second is
- A
$32 ± 3$
- B
$32 ± 1$
- ✓
$32 ± 2$
- D
$32 ± 5$
AnswerCorrect option: C. $32 ± 2$
$32 ± 2$
Hint :
$\text { Mean }(t)=\frac{30+32+35+31}{4}=32$
$ \text { Mean error }(\Delta t)=\frac{\left|\Delta t_1\right|+\left|\Delta t_2\right|+\left|\Delta t_3\right|+\left|\Delta t_4\right|}{4}$
$ =\frac{2+0+3+1}{4}=\frac{6}{4}=1.5$
Hence rounding off,
$\triangle t = ± 2 s$
$\therefore t ± \triangle t = 32 ± 2 s$
View full question & answer→MCQ 191 Mark
The main scale of a vernier callipers has $n$ divisions $cm. n$ divisions of the vernier scale coincide with $(n – 1)$ divisions of main scale. The least count of the vernier callipers is,
AnswerCorrect option: D. $\frac{1}{n^2} \ cm$
Hint:
$1 \text { V.S.D. }=\frac{(n-1)}{n} \text { M.S.D. }$
$\text { LC. }=1 \text { M.S.D. }-1 \text { V.S.D. }$
$=1 \text { M.S.D. }-\frac{(n-1)}{n} \text { M.S.D. }$
$=\frac{1}{n} \text { M.S.D. }$
$=\frac{1}{n} \times \frac{1}{n} \ cm$
$\therefore \text { L.C. }=\frac{1}{n^2} \ cm$
View full question & answer→MCQ 201 Mark
In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X,
where $X =\frac{A^2 \frac{1}{B^2}}{C^{\frac{1}{3}} D^3}$, will be:
Answer16 %
Hint :
Given: $X=\frac{A^2 \frac{1}{B^2}}{C^{\frac{1}{3}} D^3}$
Error contributed by $A =2 \times\left(\frac{\Delta A }{ A } \times 100\right)$
$=2 \times 1 \%=2 \%$
Error contributed by $B=\frac{1}{2} \times\left(\frac{\Delta B }{ B } \times 100\right)$
$=\frac{1}{2} \times 2 \%=1 \%$
Error contributed by $C =\frac{1}{3} \times\left(\frac{\Delta C }{ C } \times 100\right)$
$=\frac{1}{3} \times 3 \%=1 \%$
Error contributed by D $=3 \times\left(\frac{\Delta D }{ D } \times 100\right)$
$=3 \times 4=12 \%$
∴ Percentage error in x is given as,
$\frac{\Delta x}{x} \times 100$ – (error contributed by A) – (error contributed by B) + (error contributed by C) + (error contributed by D)
= 2% + 1% + 1% + 12%
= 16%
View full question & answer→MCQ 211 Mark
__________ is the smallest measurement that can be made using the given instrument
MCQ 221 Mark
The Earth’s radius is $6371 \ km.$ The order of magnitude of the Earth’s radius is
- A
$10^3 m$
- B
$10^9 m$
- ✓
$10^7 m$
- D
$10^2 m$
AnswerCorrect option: C. $10^7 m$
$10^7 m$
View full question & answer→MCQ 231 Mark
$3.310 \times 10^2$ has $.........$ significant figures.
View full question & answer→MCQ 241 Mark
0.00849 contains ___________ significant figures.
MCQ 251 Mark
The number of significant figures in $11.118 \times 10^{-6}$ is
View full question & answer→MCQ 261 Mark
The diameter of the paper pin is measured accurately by using ________.
MCQ 271 Mark
Zero error of an instrument introduces .
MCQ 281 Mark
Accuracy of measurement is determined by
MCQ 291 Mark
Dimensions of $\sin \theta$ is
- A
$[L^2]$
- B
$[M]$
- C
$[ML]$
- ✓
$[M^0L^0T^0]$
AnswerCorrect option: D. $[M^0L^0T^0]$
View full question & answer→MCQ 301 Mark
$[L^1M^1T^{-1}]$ is an expression for $..............$
View full question & answer→MCQ 311 Mark
$S.I.$ unit of energy is joule and it is equivalent to
- A
$10^6 \ erg$
- B
$10^{-7} \ erg$
- ✓
$10^7 \ erg$
- D
$10^5 \ erg$
AnswerCorrect option: C. $10^7 \ erg$
View full question & answer→MCQ 321 Mark
An atomic clock makes use of _________.
MCQ 331 Mark
Which of the following is NOT a unit of time?
MCQ 341 Mark
The two stars $S_1$ and $S_2$ are located at distances $d_1$ and $d_2$ respectively. Also if $d_1 > d2_2$ then following statement is true.
- A
The parallax of $S_1$ and $S_2$ are same.
- B
The parallax of $S_1$ is twice as that of $S_2$
- C
The parallax of $S_1$ is greater than parallax of $S_2$
- ✓
The parallax of $S_2$ is greater than parallax of $S_1$
AnswerCorrect option: D. The parallax of $S_2$ is greater than parallax of $S_1$
The parallax of $S_2$ is greater than parallax of $S_1$
View full question & answer→MCQ 351 Mark
The distance of the planet from the earth is measured by __________.
MCQ 361 Mark
Which of the following is NOT a fundamental quantity?
MCQ 371 Mark
Which of the following is the fundamental unit?
MCQ 381 Mark
A physical quantity may be defined as
- ✓
the one having dimension.
- B
that which is immeasurable.
- C
- D
AnswerCorrect option: A. the one having dimension.
View full question & answer→