MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 9

When a wave traverses a medium the displacement of a particle located at x at a time is given by y = a sin (bt - cx), where a, band are constants of the wave, which of the following is a quantity with dimensions?

0%
$\displaystyle \frac {y}{a}$
0%
$bt$
0%
$cx$
0%
$\displaystyle \frac {b}{c}$

Explanation

Given, y = a sin (bt - cx)
Comparing the given equation with general wave equation
y = a sin (

$\displaystyle \frac {2 \pi t}{T}\, -\, \displaystyle \frac {2 \pi x}{\lambda}$ )
we get

$b\,=\, \displaystyle \frac {2 \pi}{T},c\,= \displaystyle \frac {2 \pi}{\lambda}$
(a) Dimensions of

$\displaystyle \frac {y}{a}\, =\, \displaystyle \frac {metre}{metre}\, =\, \displaystyle \frac {L}{L}$
(b) Dimensions of bt

$=\, \displaystyle \frac {2 \pi}{T}\, \cdot\, t\, =\, \displaystyle \frac {T}{T}$
(c) Dimensions of cx

$=\, \displaystyle \frac {2 \pi}{\lambda}\, \cdot\, x\, =\, \displaystyle \frac {L}{L}$
(d) Dimensions of

$\displaystyle \frac {b}{c}\, =\, \displaystyle \frac {2 \pi}{T}/\displaystyle \frac {2 \pi}{\lambda}\, =\, \lambda/T\, =\, LT^{-1}$
Thus, option (d) has dimensions.

If energy (E) , force (F) and linear momentum (P) are fundamental quantities, then match the following and give correct answer.

A	B
Physical quantity	Dimensional formula
a) Mass	d) $\displaystyle { E }^{ 0 }{ F }^{ -1 }{ p }^{ 1 }$
b) Length	e) $\displaystyle { E }^{ -1 }{ F }^{ 0 }{ p }^{ 2 }$
c) Time	f) $\displaystyle { E }^{ 1 }{ F }^{ -1 }{ p }^{ 0 }$

0%
a-d, b-e, c-f
0%
a-f, b-e, c-d
0%
a-e, b-f, c-d
0%
a-e, b-d, c-f

Explanation

Dimensions of

$E=[ML^2T^{-2}]$

Dimensions of

$F=[MLT^{-2}]$

Dimensions of

$P=[MLT^{-1}]$

Thus

$E^0F^{-1}P^1=[T]$

$E^{-1}F^0P^2=[M]$

$E^{-1}F^0P^2=[L]$

What is the dimensional formula of Inductance?

0%
$[ML^2T^{-2}A^{-2}]$
0%
$[ML^2TA^{-2}]$
0%
$[ML^2T^{-1}A^{-2}]$
0%
$[ML^2T^{-2}A^{-1}]$

Explanation

Hint:

Use the formula of inductance in terms of voltage and current and and use there dimensions.

Step 1: Formula used,

The formula of the inductance is written as,

$L=\dfrac { V }{ \dfrac { dI }{ dt } }$

step 2: Finding the dimensions,

Substituting the dimensions of voltage and the rate of change of current in the above relation,

$[L]=[\dfrac { M{ L }^{ 2 }{ T }^{ -3 }{ A }^{ -1 } }{ { A }{ T }^{ -1 } }] =[M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -2 }]$

$\textbf{Hence option A correct}$

$x=at+b{ t }^{ 2 }$ where

$x$ is measured in

$m$ and

$t$ in

$s$ , then the dimension of

$\left( { b }/{ a } \right)$ is:

0%
$[L{ T }^{ -2 }]$
0%
$[L{ T }^{ -1 }]$
0%
$[T]$
0%
$[{ T }^{ -1 }]$

Explanation

Given equation is:

$x=at+b{ t }^{ 2 }$
The standard equation is:

$S=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$

$\Rightarrow \left[ a \right] =\left[ u \right] =L{ T }^{ -1 }$

$\Rightarrow \left[ b \right] =\left[ a \right] =L{ T }^{ -2 }$

$\Rightarrow \left[ \dfrac { b }{ a } \right] =\dfrac { L{ T }^{ -2 } }{ L{ T }^{ -1 } } =[{ T }^{ -1 }]$

The dimensions of

$\displaystyle \frac { a }{ b }$ in the equation

$\displaystyle P=\frac { a-{ t }^{ 2 } }{ bx }$ where

$P$ is pressure,

$x$ is distance and

$t$ is time, are:

0%
$\displaystyle \left[ { M }^{ 2 }{ LT }^{ -3 } \right]$
0%
$\displaystyle \left[ { MT }^{ -2 } \right]$
0%
$\displaystyle \left[ { LT }^{ -3 } \right]$
0%
$\displaystyle \left[ M{ L }^{ 3 }{ T }^{ -1 } \right]$

Explanation

The given relation is

$P = \dfrac{a- t^2}{bx}$

Thus dimension of

$a = [T^2]$

Dimensions of pressure

$P = [ML^{-1} T^{-2}]$ and that of

$x = [L]$

Thus dimensions of

$b = \dfrac{[T^2]}{[L] \times [M L^{-1} T^{-2}]} = [M^{-1} T^4]$

$\therefore$ Dimensions of

$\dfrac{a}{b} = \dfrac{[T^2]}{[M^{-1} T^4]} = [M T^{-2}]$

The moon is observed from two diametrically opposite points on earth. The angle subtended at the moon by the two directions of observations is

$1^o 54'$ . Given the diameter of the earth to be about

$2.276 \times 10^7 m$ , compute the distance of moon from the earth.

0%
$1 \times 10^8 m$
0%
$6.84 \times 10^8 m$
0%
$6.84 \times 10^{10} m$
0%
$38.4 \times 10^8 m$

Explanation

Here D be the distance between the moon and the earth, d be the diameter of the earth.

Here,

$\theta=1^o54'=1.9^o=\dfrac{\pi}{180}\times 1.9 rad=0.033$ radian

From the figure,

$\tan\theta=d/D$

$\theta$ is very small,

$\theta=d/D$

$D=\dfrac{d}{\theta}=\dfrac{2.276\times 10^7}{0.033}=6.84\times 10^8$ m

The mass of the electron is given

$9.1\times 10^{-31} kg$ . Find the order of magnitude of the number of electrons of

$1 kg$ .

0%
$10^{31}$
0%
$10^{-30}$
0%
$10^{30}$
0%
$10^{-31}$

Explanation

Let n be the required number of electrons.
Thus,

$n\times (9.1\times 10^{-31})=1$
or

$n=1.1\times 10^{30}$
According to rule of order of magnitude, as

$1.1<5$ so it will be taken as 1.
Thus,

$n=10^{30}$

What is the order of magnitude of the distance of a quasar from us if light takes 2.9 billion years to reach us ?

0%
$2.7\times 10^{25} m$
0%
$10^{25} m$
0%
$10^24 m$
0%
$3\times 10^{25} m$

Explanation

Here, time taken

$t=2.9$ billion years

$=2.9\times 10^9$ years

$=2.9\times 10^9\times (365\times 24\times 3600) s=9.14\times 10^{16} s$
Distance

$=$ velocity

$\times$ time
or

$d=(3\times 10^8)\times (9.14\times 10^{16})=2.74\times 10^{25} m$
As

$2.74 <5$ so it will be taken as 1.
Thus, the order of magnitude of distance

$d=1\times 10^{25} m=10^{25} m$

The radius of proton is

$0.6\times 10^{-15} m$ . What is the order of magnitude of it volume?

0%
$10^{-47} m^3$
0%
$10^{-44} m^3$
0%
$10^{-45} m^3$
0%
$10^{-46} m^3$

Explanation

The volume of the proton is

$V=\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\times 3.14\times (0.6\times 10^{-15})^3=9.04\times 10^{-46} m^3$
According to rule of order of magnitude , as

$9.04>5$ so it will be taken 10.
Thus, order of magnitude of volume

$=10\times 10^{-46}=10^{-45} m^3$

What is the order of magnitude of one light year?

0%
$10^{15} m$
0%
$10^{10} m$
0%
$9.2 \times 10^{15} m$
0%
$10^{16} m$

Explanation

One light year is the distance of light at a speed

$3\times 10^8 m/s$ in 1 year.
Thus,

$1$ light year

$= 3\times 10^8 m/s\times 1$ year

$=3\times 10^8 m/s\times (365\times 24\times 3600 s)=9.5\times 10^{15} m$
According to rule of order of magnitude, as

$9.5>5$ so 9.5 will be taken as 10.
Thus, order of magnitude of one light year

$=10\times 10^{15}=10^{16} m$

The sun's angular diameter is measured to be

$1920 ''$ . The distance of the sun from the earth is

$1.496 \times 10^{11} m$ . What is the diameter of the sun?

0%
$1 \times 10^9 m$
0%
$1.39 \times 10^9 m$
0%
$13.9 \times 10^9 m$
0%
$139 \times 10^9 m$

Explanation

Hint: Use

$\tan\theta=\frac dD$

Step 1: Note the given values.

$\theta=1920"=\dfrac{1920}{3600}=0.53^o=0.53\times (\pi/180)=0.0093$ radian

and

$D=1.496\times 10^{11}m$

Step 2:Calculating the diameter of the Sun.

$\tan\theta=\frac dD$

$\theta$ is very small so

$\tan\theta\sim \theta$

So, diameter of the sun is

$d=D\theta=1.496\times 10^{11}\times 0.0093=1.39\times 10^9 m$

$\textbf{Hence option B correct}$

The radius of the sun is

$7\times 10^8 m$ and its mass is

$2\times 10^{30} kg$ . What is the order of magnitude of density of the sun?

0%
$1.4\times 10^3 kg/m^3$
0%
$10^7 kg/m^3$
0%
$1.5\times 10^3 kg/m^3$
0%
$10^3 kg/m^3$

Explanation

Given :

$R = 7\times 10^8 \ m$ and

$M = 2\times 10^{30} \ kg$
Volume of sun

$V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi\times (7\times 10^8)^3 = 1.4\times 10^{27} \ m^3$
Density of sun

$\rho = \dfrac{M}{V} = \dfrac{2\times 10^{30}}{1.4\times 10^{27}} = 1.43\times 10^3 \ kg/m^3$
Thus order of magnitude of density of sun is

$10^3 \ kg/m^3$ .

The given rectangular shaped object pictured is being measured. Its left end is perfectly lined up with "0 m" on the measuring tape. Which of the following is the best and most precise length of the green, rectangular shaped object that can be obtained, using the measuring tape?

0%
2.0 m
0%
2.4 m
0%
2.37 m
0%
2.377 m
0%
2.5 m

Explanation

Least count of the given scale,

$LC =\dfrac{1}{10} =0.1$ m

From figure, the end of the green rectangular object is close to

$4th$ line after

$2\ m$ mark.

Thus most precise length of the object,

$L = 2.0 + 0.4 =2.4\ m$

Students A, B and C measure the diameter of a wire using three different screw gauges of least count 0.01 cm, 0.05 cm and 0.001 cm respectively. Each one makes 10 measurements.The measurements will be more precise for

0%
A
0%
B
0%
C
0%
all

The radius of the earth is given

$6.4\times 10^6 m$ . What is the order of magnitude of the size of the earth?

0%
$10^6 m$
0%
$6 \times 10^6 m$
0%
$10^7 m$
0%
$5 \times 10^6 m$

Explanation

When a number rounded to the nearest power of 10, it is called order of magnitude. Here a number which is less than 5 is taken as 1 and a number greater than 5 is taken as 10.
As

$6.4>10,$ so it would be takes as 10.
The order of magnitude of earth's size

$=10\times 10^6=10^7 m$

State whether the given statement is TRUE or FALSE.

The total height of 10 identical coins each of thickness 0.5 cm placed one over the other will be 50 mm.

0%
True
0%
False

Explanation

Thickness of each coin

$t = 0.5 \ cm$

Number of identical coin

$N = 10$

Thus, the total height

$h = tN = 0.5\times 10 = 5 \ cm = 50 \ mm$

So, the given statement is true.

Which of the following is not a characteristic of the standard unit?

0%
It should be of convenient size
0%
It should change with respect to space and time
0%
It should not be perishable
0%
It should be easily reproducible

Match the following:

	Column - I		Column - II
A	Deca	p	$10^3$
B	Hecto	q	$10^6$
C	Kilo	r	$10^1$
D	Mega	s	$10^2$
E	Giga	t	$10^8$
		u	$10^9$

0%
A-r, B-s, C-p, D-q, E-u
0%
A-s, B-r, C-p, D-q, E-u
0%
A-r, B-s, C-u, D-q, E-p
0%
A-q, B-s, C-p, D-r, E-u

Explanation

Here column-I represents the some prefixes and column-II represents the power of ten corresponding prefixes.

Deca

$=10^1$ , hecto =

$10^2$ , kilo =

$10^3$ , mega

$=10^6$ , giga

$=10^9$

Thus, A-r, B-s, C-p, D-q, E-u

A star has a parallax angle p of 0.723 arcseconds. What is the distance of the star?

0%
1.38 parsecs
0%
2.38 parsecs
0%
3.38 parsecs
0%
4.38 parsecs

Explanation

Relationship between a star's distance and its parallax angle:

$d=\dfrac{1}{p}$

The distance $d$ is measured in parsecs and the parallax angle $p$ is measured in arcseconds.

Hence, $d=\dfrac{1}{0.723}=1.38 parsecs$

Two stars

$S_1$ and

$S_2$ are located at distances

$d_1$ and

$d_2$ respectively. Also if

$d_1>d_2$ then which of the following statements is true?

0%
The parallax of $S_1$ and $S_2$ are same.
0%
The parallax of $S_1$ is twice as that of $S_2$ .
0%
The parallax of $S_1$ is greater than parallax of $S_2$
0%
The parallax of $S_2$ is greater than parallax of $S_1$

Explanation

$Answer:-$ D

Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines.Due to foreshortning, nearby objects have a larger parallax than more distant objects when observed from different positions, so parallax can be used to determine distances.

Hence parallax of $S_2$ is greater than that of $S_1$

Astronomersuse the principle of parallax to measure distances to the closer stars. Here, the term "parallax" is the semi-angle of inclination between two sight-lines to the star, as observed when the Earth is on opposite sides of the Sun in its orbit.

If the angular distance between the stars turns out to be approximately 1100 arcseconds, or 0.30 degrees. The moon appears to shift 0.3 degrees when we observe it from two vantage points 2360 km apart, then find the distance of the moon from the surface of the earth.Given angular diameter of moon is 0.5 degrees.

0%
450642 km
0%
450392 km
0%
325684 km
0%
480264 km

Explanation

The parallax angle in radians is given by:

$\theta_{rad}=\theta_{deg}\times \dfrac{\pi}{180}$

$=0.3\times \dfrac{\pi}{180}rad$

Thus the distance between moon and surface of earth is:

$D=\dfrac{d}{\theta}$

$=\dfrac{2360km}{0.3\times \dfrac{\pi}{180}}$

$=450642km$

A star is

$1.45\ parsec$ light years away. How much parallax would this star show when viewed from two locations of the earth six months apart in its orbit around the sun?

0%
$2\ Parsec$
0%
$0.725\ Parsec$
0%
$1.45\ Parsec$
0%
$2.9\ Parsec$

Explanation

One light year

$=$ speed of light

$\times$ one year

$1 ly=3\times 10^8\times (24\times 3600)=94608\times 10^{11} m$

So,

$4.29 ly=4.29\times (94608\times 10^{11})=4.058\times 10^{16} m$

$1$ parsec

$=3.08\times 10^{16} m$

(Parsec is a unit of length used to measure large distances to objects outside our Solar System)

Thus,

$4.29 ly=\dfrac{4.058\times 10^{16}}{3.08\times 10^{16}}=1.32$ parsec

Now angular displacement ,

$\theta=\dfrac{d}{D}$

where

$d=$ diameter of earth's orbit

$= 3\times 10^{11} m$ and

$D=$ distance of star from the earth

$=4.058\times 10^{16} m$

So,

$\theta=\dfrac{3\times 10^{11}}{4.058\times 10^{16}}=7.39\times 10^{-6}$ rad

$1 sec=4.85\times 10^{-6} rad$ so,

$7.39\times 10^{-6} rad=\dfrac{7.39\times 10^{-6}}{4.85\times 10^{-6}}=1.52 sec$

Which of the following is a possible dimensionless quantity?

0%
Velocity gradient
0%
Pressure gradient
0%
Displacement gradient
0%
Force gradient

Explanation

Displacement gradient in x-direction is given by

$\dfrac{dr}{dx}$
Since

$r$ and

$x$ have same dimension, this results the displacement gradient to be dimensionless quantity.

$5\times 10^7 \mu s$ is equivalent to ____

0%
0.5 s
0%
5 s
0%
50 s
0%
500 s

Explanation

We know that

$1\mu s = 10^{-6} \ s$
Thus,

$5\times 10^{7}\mu s = 5\times 10^{7}\mu s\times \dfrac{10^{-6} \ s}{\mu s} = 50 \ s$

The dimension of

$\dfrac{h}{mv}$ is: (

$h$

$=$ planks constant,

$m$ = mass,

$v$ =velocity)

0%
$M$
0%
$L$
0%
$T$
0%
none of the above

Explanation

Debroglie wavelength is given by:

$\lambda=\dfrac{h}{mv}$
Wavelength has dimension of length

$[L]$ .

The dimensional formula of coefficient of viscosity is

0%
$[MLT^{-1}]$
0%
$[M^{-1}L^{2}T^{-2}]$
0%
$[ML^{-1}T^{-1}]$
0%
none

Explanation

Coefficient of viscosity is given by

$\eta =\dfrac{F}{A}\dfrac{dz}{dv}$
Dimension of force

$F = [MLT^{-2}]$
Dimension of Area

$A = [L^2]$
Dimension of velocity gradient

$\dfrac{dv}{dz} = [T^{-1}]$
Thus dimension of coefficient of viscosity

$\eta = \dfrac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1}T^{-1}]$

A physical quantity

$Q$ is found to depend on observables

$x, y$ and

$z$ , obeying relation

$Q=\dfrac{x^{2/5}z^{3}}{y}$ . The percentage error in the measurements of

$x, y$ and

$z$ are

$1\%$ ,

$2\%$ and

$4\%$ respectively. What is percentage error in the quantity

$Q$ will be :

0%
$5\%$
0%
$4.5\%$
0%
$11\%$
0%
$7.75\%$

Explanation

The given quantity

$Q = \dfrac{x^3y^2}{z}$

Percentage error in Q

$\dfrac{\Delta Q}{Q} \times 100 = 3\dfrac{\Delta x}{x}\times 100$

$+2\dfrac{\Delta y}{y} \times 100 + \dfrac{\Delta z}{z}\times 100$

Percentage error in

$x,y$ and

$z$ are

$1\%$ ,

$2\%$ and

$4\%$ respectively.

$\implies$

$\dfrac{\Delta Q}{Q} \times 100 = 3(1) + 2(2)+1(4) = 11$ %

The equation of state of a gas is given by

$\begin{pmatrix}P+\dfrac{a}{v^3}\end{pmatrix} (V-b^2) = cT$ , where P, V, T are pressure, volume and temperature respectively, and a, b, c are constants. The dimensions of a and b are respectively

0%
$ML^8T^{-2}$ and $L^{3/2}$
0%
$ML^5T^{-2}$ and $L^{3}$
0%
$ML^5T^{-2}$ and $L$
0%
$ML^6T^{-2}$ and $L^{3/2}$

Explanation

The dimensions of

$\dfrac{a}{v^3}$ must be that of pressure.

Dimension of pressure are

$[ML^{-1}T^{-2}]$

Hence dimension of

$a$ is

$[ML^{-1+3*3}T^{-2}]=[ML^8T^{-2}]$

The dimension of

$b^2$ must be that of volume.

Hence

$[b^2]=[L^3]$

$\implies [b]=[L^{3/2}]$

The ratio of the dimensions of Planck constant and that of moment of inertia has the dimensions of

0%
Time
0%
Frequency
0%
Angular momentum
0%
Velocity

Explanation

Step 1: Dimension formula of Planck constant:

Energy of photon is given by:

$E=h \nu$
Where

$h$ is Planck's constant and

$\nu$ is frequency of photon

Now, Planck's constant will be

$h = \dfrac{E}{\nu}$

Dimension of energy

$[E] = [ML^2T^{-2}]$

Dimension of frequency

$[ \nu ]=[T^{-1}]$ {frequency is reciprocal of time period)

Dimensions of Planck's constant

$[h]= [ML^2T^{-2}]/[T^{-1}] = [ML^2T^{-1}]$

Step 1: Dimension formula of moment of inertia:

The moment of inertia

$(I)$ is given by:

$I=MR^2$
Where

$M$ is mass of the obect and

$R$ is distance.

Dimensions of moment of inertia

$[I] = [ML^2]$

Step 3: Dimension formula of their ratio:

$\therefore$ Dimensions of

$\dfrac{h}{I}$

$=\dfrac{[ML^2T^{-1}]}{[ML^2]}=[T^{-1}]$

Thus ratio of Planck's constant and moment of inertia has dimensions of frequency.

Option B is correct.

$ML^{-1}T^{-2}$ is the dimensional formula of

0%
force
0%
coefficient of friction
0%
modulus of elasticity
0%
energy

Explanation

Modulus of elasticity is given by

$Y = \dfrac{F}{A}.\dfrac{L}{\Delta L}$
Dimension of force

$F = [MLT^{-2}]$
Dimension of Area

$A = [L^2]$
Dimension of length

$L = [L]$
Thus dimension of modulus of elasticity

$Y = \dfrac{[MLT^{-2}][L]}{[L^2][L]} = [ML^{-1}T^{-2}]$

Parallax angles _______

$0.01/ arcsec$ are very difficult to measure from Earth.

0%
more than
0%
less than
0%
equal to
0%
greater than or equal to

Explanation

Parallax effect depends upon the path of light that travels from object to the observer. For distant objects observed from Earth, the light that reaches Earth has to go through a number of layers in Earth's atmosphere, during which it refracts and disperses and hence decreases the accuracy of the method. Resultantly parallax angles less than

$0.01/arcsec$ are very difficult to measure from Earth.

If a star is

$5.2\times 10^{16}\ m$ away. What is the parallax angle in degrees?

0%
$1.67 \times 10^{-4}$ degrees
0%
$1.67 \times 10^{-5}$ degrees
0%
$0.67 \times 10^{-4}$ degrees
0%
$2.3 \times 10^{-4}$ degrees

Explanation

Given :

$1$ AU

$= 1.5\times 10^11$ m

$d = 5.2\times 10^{15}$ m

Parallax angle:

$\alpha = \dfrac{1 AU}{d} =\dfrac{1.5\times 10^{11}}{5.2\times 10^{16}} = 0.288\times 10^{-5}$ radians

$\implies$

$\alpha = \dfrac{180}{\pi} \times 0.288\times 10^{-5} = 1.67\times 10^{-4}$ degrees

Two quantities X and Y have different dimension. Which mathematical operation given below is physically meaningful?

0%
$X+Y$
0%
$X-Y$
0%
$\dfrac{X}{Y}$
0%
None of these

Four students measure the height of a tower. Each student uses different methods and each measures the height many times. The data for each are plotted. The measurement with the highest precision is:

0%
I
0%
II
0%
III
0%
IV

The dimensions formula of ( velocity

$)^2 /$ radius are the same of that of :

0%
Planck's constant
0%
Gravitational constant
0%
Dielectric constant
0%
None of these

Explanation

Dimensions formula of ( velocity

$)^2 /$ radius

$= \dfrac {[M^0 LT^{-1}]^2 }{[M^0 LT^0]} = [LT^{-2}]$

Thus, it has the dimensions of acceleration

If the dimensions of length are expressed as a

$G^x \cdot C^y \cdot h^z$ , where G, C and h are the universal gravitational constant, speed of light and plank constant respectively, then value of x,y,z will be :

0%
$x=1, y=-\dfrac{3}{2}, z=\dfrac{1}{2}$
0%
$x=\dfrac{1}{2}, y=\dfrac{1}{2}, z=\dfrac{3}{2}$
0%
$x=\dfrac{1}{2}, y=-\dfrac{3}{2}, z=\dfrac{1}{2}$
0%
$x=-\dfrac{3}{2}, y=\dfrac{1}{2}, z=\dfrac{1}{2}$

Explanation

Length

$\propto G^xC^yh^z$

$[L]=[M^{-1}L^3T^{-2}]^x[LT^{-1}]^y[ML^2T^{-1}]^z$
On comparing the power of M, L and T in both sides, we get

$-x+z=0$ ....(i)

$3x+y+2z= 1$ ....(ii)
and

$-2x-y-z=0$ ....(iii)
By solving Eqs. (i), (ii) and (iii) we get

$x=\dfrac{1}{2}; y=-\dfrac{3}{2}$ and

$z=\dfrac{1}{2}$

Someone is weighing some box as 25 ____. Fill in the blank with appropriate answer.

0%
Kilogram
0%
Meter
0%
Seconds
0%
Joules

The Quantum Hall Resistance

$R_{H}$ is a fundamental constant with dimensions of resistance. If

$h$ is Planck's constant and

$e$ the electron charge, then the dimension of

$R_{H}$ is the same as.

0%
$e^{2}/h$
0%
$h/e^{2}$
0%
$h^{2}/e$
0%
$e/h^{2}$

Explanation

From the fundamental relation of energy of a photon, we have

$E=h\nu$ ,

$[h] = \textrm{[Energy][Time]}$ and

$[e] = [Charge]$

$\textrm{[Energy] = [Charge][Volt]}$

Thus,

$\textrm{[Volt]} =\frac{[Energy]}{\textrm{[Charge]}}=\frac{[h]}{[e][Time]}$

Similarly,

$\textrm{[Current]} = \frac{\textrm{[Charge]}}{\textrm{[Time]}}=\frac{[e]}{\textrm{[Time]}}$

From Ohm's Law,

$R = \frac{V}{I}$ , we have

$[R_H]=\frac{\textrm{[Volt]}}{\textrm{[Current]}}=\frac{[h]}{[e]\textrm{[Time]}} \times \frac{\textrm{[Time]}}{[e]} = [\frac{h}{e^2}]$

Thus, the dimensions of

$R_H$ match with the dimensions of

$\frac{h}{e^2}$

The dimensional formula for Young's modulus is :

0%
$[ ML^{-1} T^{-2} ]$
0%
$[ M^{0} LT^{-2} ]$
0%
$[ MLT^{-2} ]$
0%
$[ ML^{2} T^{-2} ]$

Explanation

Young's modulus

$Y = \dfrac {stress}{strain}$
Dimension of

$Y = \dfrac {ML^{-1} T^{-2} }{M^0L^0T^0}$

$= [ ML^{-1} T^{-2} ]$

The quantity which has the same dimensions as that of gravitational potential is

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latent heat
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impulse
0%
angular acceleration
0%
planck's constant
0%
specific heat capacity

Explanation

Dimensions of gravitational potential

$=\dfrac{Dimension \,of\, work}{Dimension \,of\, mass}$

$=\dfrac{[ML^2T^{-2}]}{[M]}=[L^2T^{-2}]$
and dimension of latent heat

$=\dfrac{Dimension \,of \,[Q]}{Dimension \,of\, [M]}$

$=\dfrac{[ML^2T^{-2}]}{[M]}=[L^2T^{-2}]$

The frequency of vibration of the string is given by

$v=\dfrac {p}{2l}\left [ \dfrac {F}{m} \right ]^{1/2}$
Here, p is the number of segments in the string and l is the length. The dimensional formula for m will be

0%
$[M^0LT^{-1}]$
0%
$[ML^0T^{-1}]$
0%
$[ML^{-1}T^{0}]$
0%
$[M^0L^0T^{0}]$

Explanation

$v=\dfrac{p}{2l}\left [ \dfrac{F} {M}\right ]^{1/2}$
Squaring the equation on either side, we have

$v^2=\dfrac{p^2}{4l^2}\left ( \dfrac{F} {M}\right )$

$m=\dfrac {P^2F}{4l^2v^2}$

$[m]=\dfrac {[MLT^{-2}]}{[L^2][T^{-1}]^2}=[ML^{-1}T^0]$

The physical quantity that does not have the dimensional formula

$[ML^{-1}T^{-2}]$ is

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force
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pressure
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stress
0%
modulus of elasticity
0%
energy density

Explanation

$F=mass \times acceleration$

$F= [M][LT^{-2}]=[MLT^{-2}]$

$Pressure=Force/Area$

$P=[MLT^{-2}][L^{-2}]=[ML^{-1}T^{-2}]$

$Stress$ is force per unit area therefore it's dimensional formula will be same as pressure.

$Modulus \ of \ elasticity=Fl/A \triangle l=[MLT^{-2}] [L][L^{-2}][L^{-1}]=[ML^{-1}T^{-2}]$

Hence

$D$ is correct answer.

The dimensional formula of

$\dfrac {1}{\mu_{0}\epsilon_{0}}$ is

0%
$[M^{0}LT^{-2}]$
0%
$[M^{0}L^{-2}T^{-2}]$
0%
$[M^{0}LT^{-1}]$
0%
$[M^{0}L^{2}T^{-2}]$

Explanation

Velocity of

$EM$ waves,

$v = \dfrac {1}{\sqrt {\mu_{0}\epsilon_{0}}}$

$\Rightarrow = \dfrac {1}{\mu_{0}\epsilon_{0}} = v^{2}$

$\therefore$ Dimensional formula of

$\dfrac {1}{\mu_{0}\epsilon_{0}} = [M^{0}LT^{-1}]^{2}$

$= [M^{0}L^{2}T^{-2}]$ .

The dimension of

$\displaystyle \frac{a}{b}$ in the equation

$\displaystyle p = \frac{a-t^2}{bx}$ where p is pressure, x is distance and t is time is

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$\displaystyle [LT^{-3}]$
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$\displaystyle [ML^3 T^{-1}]$
0%
$\displaystyle [M^2 L T^{-3}]$
0%
$\displaystyle [MT^{-2}]$

Explanation

Given equation

$\displaystyle P= \frac{a-t^2}{bx} = \frac{a}{bx} - \frac{t^2}{bx}$

Applying the principal of homogeneity, we have

$\displaystyle [pressure]= \left[ \frac{force}{area} \right] = \left[ \frac{a}{bx} \right]$

$\displaystyle [ML^{-1}T^{-2}] = \left[ \frac{a}{b} \right] \left[ \frac{1}{L} \right]$

$\displaystyle \left[ \frac{a}{b} \right] = [MT^{-2}]$

Number of particles is given by

$n=-D\cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 2 } }$ crossing a unit area perpendicular to X-axis in unit time, where

${ n }_{ 1 }$ and

${ n }_{ 2 }$ are number of particles per unit volume for the value of

$x$ meant to

${x}_{2}$ and

${x}_{1}$ . Find dimensions of

$D$ called as diffusion constant :

0%
$\left[ { M }^{ 0 }{ L }^{ }{ T }^{ 2 } \right]$
0%
$\left[ { M }^{ 0 }{ L }^{ 2 }{ T }^{ -4 } \right]$
0%
$\left[ { M }^{ 0 }{ L }^{ }{ T }^{ -3 } \right] \quad$
0%
$\left[ { M }^{ 0 }{ L }^{ 2 }{ T }^{ -1 } \right]$

Explanation

Given, number of particle passing from unit area in unit time

$=n$

$=\cfrac { Number\quad of\quad particle }{ A\times t } =\cfrac { \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] }{ \left[ { L }^{ 2 } \right] \left[ T \right] } =\left[ { L }^{ -2 }{ T }^{ -1 } \right]$

Now from the given formula

$\left[ D \right] =\cfrac { \left[ n \right] \left[ { x }_{ 2 }-{ x }_{ 1 } \right] }{ \left[ { n }_{ 2 }-{ n }_{ 1 } \right] } =\left[ { L }^{ 2 }{ T }^{ -1 } \right]$

Express 0.006006 into scientific notation in three significant digits:

0%
$6.01 \times 10^{-3}$
0%
$6.0006 \times 10^{-3}$
0%
$6.00 \times 10^{-3}$
0%
$6.0 \times 10^{-3}$

Explanation

The number

$0.006006$ has 6 significant figures.

We need to round off the given number to convert into 3 significant figure. Thus, the correct answer is

$6.01\times 10^{-3}$ as it has 3 significant figure including rounding off.

The dimensions of capacitance are :

0%
$\displaystyle \left[ { ML }^{ -2 }{ Q }^{ -2 }{ T }^{ 2 } \right]$
0%
$\displaystyle \left[ { M }^{ -1 }{ L }^{ 2 }{ T }^{ -2 }{ Q }^{ 2 } \right]$
0%
$\displaystyle \left[ { M }^{ -1 }{ L }^{ 2 }{ T }^{ -2 }{ Q }^{ -2 } \right]$
0%
$\displaystyle \left[ { M }^{ -1 }{ L }^{ -2 }{ T }^{ 2 }{ Q }^{ 2 } \right]$

Explanation

$\textbf{Step 1-Formula of Capacitance}$

If charge on capacitor is C and potential on it is V.

then capacitance

$C=\dfrac{Q}{V}$

$\textbf{Step 2- Dimensions of charge and potential}$

Dimension of charge [Q]={AT]

Potential

$V=\dfrac{W}{Q}$

So dimension of potential is

$[V]=\dfrac{[ML^2T^{-2}]}{Q}$

$[V]=[ML^2T^{-2}Q^{-1}]$

$\textbf{Step 3-Dimensions of capacitance}$

$[C]=\dfrac{[Q]}{[ML^2T^{-2}Q^{-1}]}$

$[C]=[M^{-1}L^{-2}T^2Q^{2}]$

Hence the correct answer is (D).

$L,C$ and

$R$ denote the inductance, capacitance and resistance respectively, the dimensional formula for

${C}^{2}LR$ is :

0%
$\left[ { M }^{ }{ L }^{ -2 }{ T }^{ -1 }{ I }^{ 0 } \right]$
0%
$\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 3 }{ I }^{ 0 } \right]$
0%
$\left[ { M }^{ -1 }{ L }^{ -2 }{ T }^{ 6 }{ I }^{ 2 } \right]$
0%
$\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 2 }{ I }^{ 0 } \right]$

Explanation

Given

$\left[ { C }^{ 2 }LR \right] =\left[ { C }^{ 2 }{ L }^{ 2 }\cfrac { R }{ L } \right] =\left[ { (LC) }^{ 2 }\cfrac { R }{ L } \right]$

ad we know that frequency of

$LC$ circuits

$f=\cfrac { 1 }{ 2\pi } \cfrac { 1 }{ \sqrt { LC } }$ Here the dimension of

$LC$ is equal to

$\left[ { T }^{ 2 } \right]$

$\left[ \cfrac { L }{ R } \right]$ gives the time constant of

$L-R$ circuit, so that the dimension of

$\left[ \cfrac { L }{ R } \right]$ is equal to

$\left[ T \right]$

Hence the required dimensions

$\left[ \cfrac { L }{ R } \right] \quad \left[ { (LC) }^{ 2 }\cfrac { R }{ L } \right] ={ \left[ { T }^{ 2 } \right] }^{ 2 }\left[ { T }^{ -1 } \right] =\left[ { T }^{ 3 } \right]$

The dimensional formula of modulus of elasticity is

0%
$\displaystyle \left[ { ML }^{ -1 }{ T }^{ -2 } \right]$
0%
$\displaystyle \left[ { M }^{ 0 }{ LT }^{ -2 } \right]$
0%
$\displaystyle \left[ { MLT }^{ -2 } \right]$
0%
$\displaystyle \left[ { ML }^{ 2 }{ T }^{ -2 } \right]$

Explanation

Modules of elasticity

$\displaystyle =\frac { stress }{ strain }$
Stress

$\displaystyle =\frac { \left[ F \right] }{ \left[ A \right] } =\frac { \left[ { MLT }^{ -2 } \right] }{ \left[ { L }^{ 2 } \right] } =\left[ { ML }^{ -1 }{ T }^{ -2 } \right]$
Strain is a dimensionless quantity.

So, modules of elasticity =

$\displaystyle \left[ { ML }^{ -1 }{ T }^{ -2 } \right]$

The dimensions of

$\dfrac {\alpha}{\beta}$ in the equation

$F = \dfrac {\alpha - t^{2}}{\beta v^{2}}$ , where

$F$ is force,

$v$ is velocity and

$t$ is time, is :

0%
$[MLT^{-1}]$
0%
$[ML^{-1}T^{-2}]$
0%
$[ML^{3}T^{-4}]$
0%
$[ML^{2}T^{-4}]$

Explanation

The given equation is

$F = \dfrac{\alpha - t^2}{\beta v^2}$

Dimension of

$\alpha$ is same as that of square of time i.e.

$\alpha = [T^2]$

Dimensoin of

$\beta$ is

$\beta = \dfrac{\alpha - t^2}{Fv^2} = \dfrac{[T^2]}{[MLT^{-2}][L^2T^{-2}]} = [M^{-1}L^{-3}T^6]$

Thus

$\dfrac{\alpha}{\beta} = \dfrac{[T^2]}{[M^{-1}L^{-3}T^6]} = [ML^3T^{-4}]$

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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