MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 10

While measuring the length of a wooden box, the reading at one end is 1.5 cm and the other end is 4.7 cm. What is the length of the wooden box?

0%
4.7 cm
0%
1.5 cm
0%
3.2 cm
0%
6.2 cm

The position x of a particle at time t is given by

$x = \dfrac{v_{0}}{a}(1 - e^{-at})$ where

$v_{0}$ is a constant and a >The dimensional formula of

$v_{0}$ and a are

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

Explanation

For finding

$v_0=\cfrac{ax}{1-e^{a/t}}\\ \quad=\cfrac{m/s\times m/s^2}{1-e^{-m/s^2\times s}}\\ \quad=\cfrac{m^2}{s^2}\times\cfrac{s}{m}\\ \quad=m/s[M^0LT^{-1}]$

For finding,

$v_0=\cfrac{xa}{1-e^{-at}}\\ \cfrac{v_0}{x}=\cfrac{a}{1-e^{-at}}\\ \cfrac{x}{v_0}=\cfrac{1-e^{-at}}{a}\\a=\cfrac{v_0}{x}\\=\quad \cfrac{m/s}{m}\\a=\cfrac{1}{s}\\a=s^{-1}\\ \quad=[M^0L^0T^{-1}]$

The dimensional formula for permittivity of free space

$(\epsilon_{0})$ in the equation

$F = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q_{1}q_{2}}{r^{2}}$ where, symbols have their usual meaning is

0%
$[M^{1}L^{3}A^{-2}T^{-4}]$
0%
$[M^{-1}L^{-3}T^{4}A^{2}]$
0%
$[M^{-1}L^{-3}A^{-2}T^{-4}]$
0%
$[M^{1}L^{3}A^{2}T^{-4}]$

Explanation

Permittivity of free space

$\epsilon_o = \dfrac{q_1q_2}{4\pi Fr^2}$

Dimensions of

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

Dimensions of

$[q] = [AT]$

Dimensions of

$[r] = [L]$

Thus dimensional formula of

$[\epsilon_o] = \dfrac{[AT][AT]}{[MLT^{-2}][L^2]}$

$\implies$

$[\epsilon_o] = [M^{-1}L^{-3} T^4A^2]$

Four pieces of wooden sticks P, Q, R and S are placed along the length of 15 cm long scale as shown in figure. What is the average length of these sticks?

0%
2.0 cm
0%
2.5 cm
0%
2.6 cm
0%
2.9 cm

Explanation

The length of the different sticks is as follows:

$P = 2.1 cm;\\ Q = 3.0 cm;\\ R = 3.1 cm;\\ S = 2.5 cm$

Average length of these sticks

$=\dfrac{P+Q+R+S}{4}$

$= 2.6 cm$

${ \varepsilon }_{ 0 }$ is the permittivity of free space and

$E$ is the electric field, then the dimensions of

$\cfrac { 1 }{ 2 } { \varepsilon }_{ 0 }{ E }^{ 2 }\quad$ is

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

Explanation

Here,

$\cfrac { 1 }{ 2 } { \varepsilon }_{ 0 }{ E }^{ 2 }\quad$ represetns energy per unit volume
So the dimensions of

$\left[ { \varepsilon }_{ 0 } \right] \left[ { E }^{ 2 } \right] =\cfrac { \left[ Dimensions\quad of\quad Energy \right] }{ \left[ Dimensions\quad of\quad Volume \right] }$

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

For a circuit, L, C and R represent the inductance, capacitance and resistance. Then, the dimensional formula for

$C^2LR$ is

0%
$[MLT]$
0%
$[M^0L^0TI^0]$
0%
$[M^2L^2T^0I^0]$
0%
$[M^0L^0T^3I^0]$

Explanation

The ratio of inductance L to resistance R is time constance T, hence, we have

$\left [ \frac {L}{R} \right ]= T$ and

$\sqrt {LC}=T$
Using these two relation we have

$C^2LR=[C^2L^2]\times \left [ \frac {L}{R} \right ]=[T^4]\times \left [ \frac {1}{T} \right ]=[T^3]$

$R,L$ and

$C$ represent the physical quantities resistance, inductance and capacitance respectively. Which one of the following combination has dimensions of frequency?

0%
$\cfrac { 1 }{ \sqrt { RC } }$
0%
$\cfrac { R }{ L }$
0%
$\cfrac { 1 }{ LC }$
0%
$\cfrac { C }{ L }$

Explanation

$\dfrac{L}{R}$ is defined as the time constant of the L-R circuit. So,

$\dfrac{L}{R}$ has the dimension of time.

Thus,

$\dfrac{R}{L}$ has the dimensions of frequency (because frequency is the reciprocal of time).

The dimensions of angular momentum/ magnetic moment are

0%
$[MA^{-1}T^{-1}]$
0%
$[M^{-1}AT^{-1}]$
0%
$[MAT^{-1}]$
0%
$[MA^{-1}T]$

Explanation

Angular momentum

$= l\omega = [M^{1}L^{2}T^{-1}]$

Magnetic moment

$= [AL^2]$

$\therefore = \dfrac {Angular\ momentum}{Magnetic\ moment}$

$= \dfrac {[M^{1}L^{2}T^{-1}]}{[AL^{2}]} = [MA^{-1}T^{-1}]$ .

The mass of a box measured by a grocer's balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. The total mass of the box is:

0%
2.3 kg
0%
2.34 kg
0%
2.340 kg
0%
2.3403 kg

Explanation

Here, mass of the box,

$m = 2.3 kg$
Mass of one gold piece,

$m_1 = 20.15 g = 0.02015 kg$
Mass of other gold piece,

$m_2 = 20.17 g = 0.02017 kg$

$\therefore$ Total mass =

$m + m_1 +m_2 = 2.3 kg + 0.02015 kg +0.02017 kg = 2.34032 kg$
As the result is correct only upto one place of decimal, therefore, on rounding off, we get Total mass

$= 2.3 kg$

A cube has a side of length 1.2 x

$10^{-2}$ . Its volume upto correct significant figures is

0%
1.7 x $10^{-6}m^3$
0%
1.73 x $10^{-6}m^3$
0%
1.78 x $10^{-6}m^3$
0%
1.732 x $10^{-6}m^3$

Explanation

Here

Length of the cube,

$L =12 x10^{-2} m$

Volume of the cube,

$V= (1.2 \times10^{-2} m)^3= 1.728 \times 10^{-6}m^3$

As the result can have only two significant figures, therefore, on rounding off, we get,

$V. 1.7 \times 10^{-6} m^3$

The radius of a sphere is 1.41 cm. Its volume to an appropriate number of significant figures is then

0%
11.73 $cm^3$
0%
11.736 $cm^3$
0%
11.7 $cm^3$
0%
117 $cm^3$

Explanation

Radius of the sphere,

$r = 1.41 cm$ (

$3$ significant figures)

Volume of the sphere,

$\displaystyle V = \dfrac {4}{3} \pi r^3 = \dfrac{4}{3} \times 3.14 \times (1.41)^3\,cm^3 = 11.736\, cm^3$

Rounded off upto

$3$ significant figures

$= 11.7 cm^3$ .

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 then?

0%
$1.39 \times 10^9 m$ .
0%
$1.39 \times 10^{10} m$ .
0%
$1.39 \times 10^{11} m$ .
0%
$1.39 \times 10^{12} m$ .

Explanation

The Sun’s diameter (d) = ${1.494\times {{10}^{11}}m}$

And Sun’s angular diameter ( $\alpha$ ) = $1920''$

To convert the angular diameter into radians, converting $1920''$ into radians.

$1''=4.85\times {{10}^{-6}}rad$

⸫ $1920''$ = $1920\times 4.85\times {{10}^{-6}}=9.31\times {{10}^{-3}}rad$ = $\alpha$

Now, Sun’s diameter (D) = $\alpha \times d$

D = ( $9.31\times {{10}^{-3}}\times 1.494\times {{10}^{11}}$ )m

⸫ D = $1.39\times {{10}^{9}}m$

$\textbf{Answer:}$

$\textbf{Hence, the correct option is (a)}$ $1.39\times {{10}^{9}}m$

Which of the following pairs of physical quantities have same dimensions?

0%
Force and power
0%
Torque and energy
0%
Torque and power
0%
Force and torque

Explanation

Torque is the moment of force given by:

$\vec\tau=\vec r\times \vec F$

It’s unit is Newton-meter.

The unit of work/energy is also Newton-meter or Joule.

Thus, both of these physical quantities have the same units/dimensions.

$[Torque]=[M][L]^{2}[T]^{−2}=[Energy]$

The number of significant figure In the numbers

$4.8000 \times 10^4$ and

$48000.50$ are respectively then

0%
$5$ and $6.$
0%
$5$ and $7$ .
0%
$2$ and $7.$
0%
$2$ and $6$ .

Explanation

If any number has more than one digit after decimal place are significant. Power of 10 does not affect number of significant figures.

$4.8000 \times 10^{4} = 5$ significant figure

$(4,8,0,0,0)$

$48000.50 = 7$ significant figure

$(4,8,0,0,0,5, 0)$

Four charges are placed at the circumference of the dial of a clock as shown in figure. If the clock has only hour hand, then the resultant force on a positive charge

$q_{0}$ placed at the centre, points in the direction which shows the time as :-

0%
1:30
0%
7:30
0%
4:30
0%
10:30

Explanation

Where

$F=\cfrac{kq^2}{r^2}\\F_{net}=\sqrt{(2F)^2+(2F)^2}$

Between

$6:00$ and

$9:00$

$Time =7:30$

1380499_872988_ans_095664092b8b43a9976eadfadf08d866.png

The numbers 3.845 and 3.835 on rounding off to 3 significant figures will give then

0%
3.85 and 3.84
0%
3.84 and 3.83
0%
3.85 and 3.83
0%
3.84 and 3.84

Which one of the following methods is used to measure distance of a planet or a star from the earth?

0%
Echo method
0%
Parallax method
0%
Triangulation method
0%
None of these

The dimensional formula of electric potential is

0%
$[ML^2T^{-3}A^{-1}]$
0%
$[M^{-1}L^2T^{-2}A]$
0%
$[M^{-1}L^2T^{-2}A^{-1}]$
0%
$[ML^2T^{-2}A]$

Explanation

Electric potential,

$\displaystyle V = \frac{W}{q} = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$

Which of the following relation is dimensionally incorrect?

0%
1 u = 931.5 MeV
0%
1 u = 931.5 MeV/c $^2$
0%
1 u = 1.67 x 10 $^{-27}$ kg
0%
None of these

Explanation

As, we know,

$E = mc^2$
and energy released can be given as:

$E=931MeV$
for,

$m =1u$

$1u = 931.5 MeV/c^{2}$ - Dimensionally correct [Dimension of mass]

$1u=1.67\times 10^{-27}\ kg$ - Dimensionally correct [Dimension of mass]
but,

$931.5 \ MeV$ is having the dimension of energy so it can't be equated to dimension of mass or dimension of

$1u$ .
Hence, dimensionally correct data is given in

$option(A)$
Hence, the correct option will be

$(A)$

The velocity of a particle (v) at an instant t is given by v =at + bt

$^{2}$ . The dimension of b is the

0%
$[L]$
0%
$[LT^{-1}]$
0%
$[LT^{-2}]$
0%
$[LT^{-3}]$

Explanation

$v = at + bt^2$

$[v] = [bt^2]$ or

$[LT^{-1}] = [bT^2]$ or

$[b] = [LT^{-3}]$

The numbers

$2.745$ and

$2.735$ on rounding off to

$3$ significant figures will give:

0%
$2.75$ and $2.74$
0%
$2.74$ and $2.73$
0%
$2.75$ and $2.73$
0%
$2.74$ and $2.74$

Explanation

Hint: While rounding off measurements, we use the following rules by convention:

If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is left unchanged if it is even.
If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one if it is odd.

Step 1: finding the rounded off digits

Let us round off $2.745$ to $3$ significant figures.

Here the digit to be dropped is $5$ , then the preceding digit is left unchanged because it is even.

Hence on rounding off $2.745$ , it would be $2.74$ .

Now consider $2.737$ , here also the digit to be dropped is $5$ , then the preceding digit is raised by one if it is odd.

Hence on rounding off $2.735$ to 3 significant figures, it would be $2.74$ .

If E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively, then the quantity

$\displaystyle (\frac {El^2}{m^5G^2})$ has the dimensions of then

0%
mass
0%
length
0%
time
0%
angle.

Explanation

$\displaystyle [E] = [ML^2T^{-2}], [m]$

$= [M][l] = [ML^2T^{-1}], [G] = [M^{-1}L^3T^{-2}]$

$\displaystyle \therefore \left [ \frac {El^2}{m^5G^2} \right ] = \frac {[ML^2T^{-2}][M^2L^4T^{-2}]}{[M^5][M^{-2}L^6T^{-4}]}$

$= [M^0L^0T^0]$
As angle has no dimensions, therefore

$\displaystyle \frac {El^5}{m^5G^2}$ has the same dimensions as that of angle.

Dimensional formula of

$\Delta Q$ , heat supplied to the system is

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

Explanation

Heat supplied to a system is in the form of energy.

$\therefore$ Its dimensional formula is

$\displaystyle [ML^2T^{-2}]$ .

The dimensions of Planck's constant are the same as that of the

0%
linear impulse
0%
work
0%
linear momentum
0%
angular

Explanation

Energy of a photon,

$E = h\upsilon$

where h is the Planck's constant and

$\upsilon$ is the frequency.

$\displaystyle \therefore [h] = \frac {[E]}{[v]} = \frac {[ML^2T^{-2}]}{[T^{-1}]}= [ML^2T^{-1}]$

Angular momentum = Moment of inertia x Angular velocity (Angular momentum)

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

The dimensional formula of physical quantity is

$[M^aL^bT^c]$ . Then that physical quantity is

0%
surface tension if a = 1, b = 1, c = -2
0%
force if a = 1, b = 1, c = 2
0%
angular frequency if a = 0, b = 0, c = -1
0%
spring constant if a = 1, b = -1, c = -2

Explanation

1) Surface Tension

$[S] = \dfrac{Force [F]}{Length [L]}$

So,

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

$\therefore$ Surface Tension

$= S = [M^1 L^0 T^{-2}]$

2) Force

$[F] =$ mass (m)

$\times$ acceleration (a)

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

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

3) Angular frequency

$= w = \dfrac{red [d \phi ]}{\sec [dt]}$

$w = [M^0 L^0 T^{-1}]$

4) Spring constant

$[K] = \dfrac{Force \ applied [F]}{Extension \ of \ spring [x]}$

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

$K = [M^1 T^{-2}]$

$K = [M^1 L^0 T^{-2}]$

$\therefore$ Option C is correct.

The displacement of a progressive wave is represented by y = A sin(

$\omega$ t - kx) where x is distance and t is time. The dimensions of

$\displaystyle \frac{\omega}{k}$ are same as those of the

0%
velocity
0%
wave number
0%
wavelength
0%
frequency

Explanation

$y = A sin (\omega t - kx)$
As

$(\omega t - kx)$ represents an angle which is dimensionless, therefore

$\displaystyle [\omega] = \frac{1}{[t]} = [T^{-1}]$ and

$\displaystyle [k] = \frac{1}{[x]} = [L^{-1}]$

$\displaystyle \left[ \frac{\omega}{k} \right] = \frac{[T^{-1}]}{[L^{-1}]} = [LT^{-1}]$ = velocity

The mean length of an object is 5 cm. Which of the following measurements is most accurate. ?

0%
4.9 cm
0%
4.805 cm
0%
5.25 cm
0%
5.4 cm

Explanation

To signify the mean length of an object comprising

$5$ cm in dimension, it is better to present it with

$4.9$ cm can be easily rounded up to

$5$ cm.

Whereas, other options are either less than or greater than the given option.

The physical quantity to be measured more accurately in the experimental determination of Young's modulus of a wire is

0%
Radius
0%
Length
0%
Force
0%
Extension

Explanation

In the experimental determination of Young's modulus, the extension

$\delta L$ in the wire is measured and using this Young's modulus is calculated by the relation :

$Y = \dfrac{F \ L}{A \ \delta L}$

The dimensional formula for electric intensity is

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

Explanation

Since

$E=\cfrac { F }{ q } ,\left[ E \right] =\cfrac { \left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 } \right] }{ \left[ AT \right] } =\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ 3 }{ A }^{ -1 } \right]$

Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass

$M$ is attached to each of the free ends at the centre of the rods.

0%
Both the rods will elongate but there shall be no perceptible change in shape
0%
The steel rod will elongate and change shape but the rubber rod will only elongate
0%
The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse
0%
The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre

Explanation

According to the diagram shown below in which a man

$AT$ is attached at the center of each rod, then both rods will be elongated. But due to different elastic properties of the material the steel rod will elongate without making any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

1229513_937250_ans_ecc39dce33894931acb8dd7dc15b6bbc.PNG

The dimensional formula of electric flux is:

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

Explanation

Correct option is (B) $\left[\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$

Hint: Electric flux is the measure of the electric field lines crossing the surface.

Step1: Electric flux $\phi=\int \vec{E} \cdot \vec{s}$

The dimension of $\phi=$ dimension fo $\mathrm{E} \times$ dimension of $\mathrm{s}$

$[\phi]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right][\mathrm{AT}]^{-1}\left[\mathrm{~L}^{2}\right]=\left[\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$

If E, M, J and G denote energy, mass, angular momentum and gravitational constant respectively, then

$\frac{EJ^2}{M^5 G^2}$ has the dimensions of

0%
length
0%
mass
0%
time
0%
angle

Explanation

Dimensional formula of:

$E : [ML^{2}T^{-2}]\\$

$J : [ML^{2}T^{-1}]\\$

$G : [M^{-1}L^{3}T^{-2}]\\$

Thus

$(E J ^{2})/(M^{5} G^{2}) = \dfrac{[ML^{2}T^{-2}][ML^{2}T^{-1}]^{2}}{M^{5} \times [M^{-1}L^{3}T^{-2}]^{2}} \\$

$= \dfrac{[M^{3}L^{6}T^{-4}]}{[M^{3}L^{6}T^{-4}]}\\$

$= [M^{0} L^{0} T^{0}]$

The above quantity is dimensionless. Angle is also a dimensionless quantity while others are having some dimensions.

Thus the correct option is D.

With due regard to significant figures, add the following:
a. 953 and 0.324
b. 953 and 0.625
c. 953.0 and 0.324
d. 953.0 and 0.374

0%
a. $953$ ; b. $954$ c. $953.3$ d. $953.4$
0%
a. $953$ ; b. $955$ c. $953.3$ d. $953.4$
0%
a. $952$ ; b. $954$ c. $953.3$ d. $953.4$
0%
a. $953$ ; b. $954$ c. $953.3$ d. $952.4$

Explanation

$953+0.324=953.324\approx \boxed { 953 }$

$953+0.625=953.625\approx \boxed { 954 }$

$953.0+0.324=953.324\approx \boxed { 953.3 }$

$953.0+0.374=953.374\approx \boxed { 953.4 }$

A physical quantity

$x$ depends on quantities

$y$ and

$z$ as follows:

$x = Ay + B \tan (C_z)$ , where

$A, B$ and

$C$ are constants. Which of the followings do not have the same dimensions?

0%
$x$ and $B$
0%
$C$ and $z^{-}$
0%
$y$ and $B/A$
0%
$x$ and $A$

Explanation

Given

$x=Ay+Btan\left( { C }_{ z } \right)$

Each term in a equation has same dimensions.

$\Rightarrow Dim\left( x \right) =Dim\left( Ay \right) =Dim\left( B \right)$

$\Rightarrow$

$x$ and

$A$ cannot have same dimension.

$L$ and

$R$ denote inductance and resistance, respectively, then the dimensions of

$L/R$ are

0%
$M^1L^0T^0Q^{-1}$
0%
$M^0L^0TQ^{0}$
0%
$M^0L^1T^{-1}Q^{0}$
0%
$M^{-1}LT^0Q^{-1}$

Explanation

We know

$E=\dfrac { 1 }{ 2 } L{ i }^{ 2 }\quad \Rightarrow \quad L=\dfrac { 2E }{ { i }^{ 2 } }$

$\Rightarrow$ dimension of

$L=\dfrac { { MLT }^{ -2 } }{ { Q }^{ 2 } } .{ T }^{ 2 }={ MLT }^{ 0 }{ Q }^{ -2 }$

Also we know

$E={ i }^{ 2 }RT\Rightarrow R=\dfrac { E }{ { i }^{ 2 }t }$

dimension of

$R=\dfrac { { MLT }^{ -2 } }{ { Q }^{ 2 } } .{ T }^{ -1 }={ MLT }^{ -1 }{ Q }^{ -2 }$

$\therefore$

$dim\left( \dfrac { L }{ R } \right) =\dfrac { { MLT }^{ 0 }{ Q }^{ -2 } }{ { MLT }^{ -1 }{ Q }^{ -2 } } =\left[ T \right]$

In the relation

$\dfrac{dy}{dt} = 2 \omega \sin (\omega t + \phi_0)$ , the dimensional formula for

$\omega t + \phi_0$ is

0%
$MLT$
0%
$MLT^0$
0%
$ML^0T^0$
0%
$M^0L^0T^0$

Explanation

The given ralation is:

$\dfrac { dy }{ dt } =2wsin\left( wt+{ \phi }_{ 0 } \right)$

It is the equation for the velocity of the wave.

Arguments of trignometric functions are dimension less

$\Rightarrow \left( wt+\phi \right)$ has dimension

$\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$ .

The following observations were taken for determining the surface tension of water by capillary tube method : diameter of capillary ,

$D = 1.25 \times 10^{-2} m$ . Taking

$g = 9.80 m s^{-2}$ and using the relation

$T = (rgh/2) \times 10^3 Nm^{-1}$ . what is the possible error in measurement of surface tension T?

0%
$2.4\%$
0%
$15\%$
0%
$1.6\%$
0%
$0.15\%$

Write the dimensions of

$a \times b$ in the relation.

$E = \dfrac{b - x^2}{at}$ , where

$E$ is the energy,

$x$ is the displacement, and

$t$ is the time.

0%
$ML^2T$
0%
$M^{-1}L^2T^1$
0%
$ML^2T^{-2}$
0%
$MLT^{-2}$

Explanation

$E=\dfrac { b-{ x }^{ 2 } }{ at }$

$E$ is energy

$x$ is displacement

$t$ time

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

$x=\left[ L \right]$

$t=\left[ T \right]$

$b$ will have unit same

${ x }^{ 2 }$

$b=\left[ { L }^{ 2 } \right]$

$a=\left[ \dfrac { b-{ x }^{ 2 } }{ Et } \right] =\dfrac { \left[ { L }^{ 2 } \right] }{ \left[ { ML }^{ 2 }{ T }^{ -2 }\times T \right] } =\left[ { M }^{ -1 }T{ L }^{ 0 } \right]$

$a\times b=\left[ { L }^{ 2 } \right] \left[ { M }^{ -1 }{ L }^{ 0 }T \right]$

$=\left[ { M }^{ -1 }{ L }^{ 2 }T \right]$

The percentage errors in the measurement of mass and speed are

$2\%$ and

$3\%$ , respectively. How much will be the maximum error in the estimation of KE obtained by measuring mass and speed?

0%
$5\%$
0%
$1\%$
0%
$8\%$
0%
$11\%$

Explanation

We know

$K=\dfrac { 1 }{ 2 } { mV }^{ 2 }$

$\dfrac { \Delta K }{ K } =\dfrac { \Delta M }{ M } +\dfrac { 2\Delta V }{ V }$

$\dfrac { \Delta K }{ K }=2$ %

$+2\times3$ %

$=(2+6)$ %

$=8$ %

If frequency

$F$ , velocity

$V$ , and density

$D$ are considered fundamental units, the dimensional formula for momentum will be

0%
$DVF^2$
0%
$DV^2F^{-1}$
0%
$D^2V^2F^2$
0%
$DV^4F^{-3}$

Explanation

Momentum

$=[F]^a [V]^b [D]^c$

$\implies [MLT^{-1}]=[T^{-1}]^a [LT^{-1}]^b[ML^{-3}]^c$

Comparing both sides,

$c=1$

$b-3c=1$

$\therefore b=4$

Also,

$a+b=1$

$\therefore a=-3$

Therefore,

Momentum

$=[F]^{-3}[V]^4[D]^1=DV^4F^{-3}$

If the velocity of light

$C$ , the universal gravitational constant

$G$ , and Planck's constant

$h$ are chosen as fundamental units, the dimension of mass in this system are

0%
$h^{1/2}C^{1/2}G^{-1/2}$
0%
$h^{-1}C^{-1}G$
0%
$hCG^{-1}$
0%
$hCG$

Explanation

Speed of light

$C=\left[ { LT }^{ -1 } \right]$

gravitational constant

$G=\left[ { M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right]$

planck's constant

$h={ ML }^{ 2 }{ T }^{ -1 }$

Let

$M={ C }^{ x }{ G }^{ y }{ h }^{ z }$

$\left[ M \right] ={ \left[ { LT }^{ -1 } \right] }^{ x }{ \left[ { M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right] }^{ y }{ \left[ { ML }^{ 2 }{ T }^{ -1 } \right] }^{ z }$

$\left[ M \right] =\left[ { M }^{ -y+z }{ L }^{ x+3y+2z }{ T }^{ -x-2y-z } \right]$

$-y+z=1,\quad x+3y+2z=0,\quad -x-2y-z=0$

On solving, we get

$x=\dfrac { 1 }{ 2 } ;\quad y=\dfrac { -1 }{ 2 } ;\quad z=\dfrac { 1 }{ 2 }$

So, dimension of

$M=\left[ { C }^{ 1/2 }{ G }^{ -1/2 }{ h }^{ 1/2 } \right]$

Force F is given in terms of time t and distance x by

$F = A \sin Ct + B \cos Dx$ . Then the dimensions of

$\dfrac {A}{B}$ and

$\dfrac{C}{D}$ are:

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

Explanation

$\textbf{Step 1: Dimensions of C and D}$

$F = A \sin \, Ct + B \cos \, Dx$

Since the argument of trigonometric functions is the angle, which is dimensionless

$\therefore [Ct] = M^{0} L^{0} T^{0}$

$\Rightarrow[CT^1] = M^{0} L^{0} T^{0}$

$\Rightarrow [C] = T^{-1}$

$....(1)$

Similarly

$[Dx] = M^{0} L^{0} T^{0}$

$\Rightarrow$

$[DL^{1}] = M^{0} L^{0} T^{0}$

$\Rightarrow$

$[D] = L^{-1}$

$....(2)$

$\textbf{Step 2: Using principle of homogeneity}$

Since the value of trigonometric functions is dimensionless,

$\therefore$ Using the principle of homogeneity, Dimension of

$A$ and

$B$ is same as Force

$F$

$[A] = [B] = MLT^{-2}$

$....(3)$

$\textbf{Step 3: Final calculation}$

From equation

$(3)$

$\dfrac{A}{B} = M^{0} L^{0} T^{0}$

From equation

$(1)$ and

$(2)$

$\dfrac{C}{D} = M^{0} L T^{-1}$

Hence Option

$C$ is correct.

The number of significant figures in

$5.69 \times 10^{15}$ kg is

0%
$1$
0%
$2$
0%
$3$
0%
$18$

Explanation

No. of significant figures in

$\underbrace { 5.69 } \times { 10 }^{ 15 }$ Kg is

$3$ .

Digits that cannot meaning to measurement and measurement resolution.

The radius of the earth is

$6.37 \times 10^6 m$ and its mass is

$5.975 \times 10^{24} kg$ . Find the earth's average density to appropriate significant figures.

0%
$5 \times 10^3 kg m^{-3}$
0%
$5.52 \times 10^3 kg m^{-3}$
0%
$2 \times 10^3 kg m^{-3}$
0%
$5.52 \times 10^3 kg m^{-4}$

Explanation

$\begin{array}{l}\text { radius }=6.37 \times 10^{6}\mathrm{~m} \\\text { Volume }=\frac{4}{3} \pi r^{3}\\V=\frac{4}{3} \times \pi \times\left(6.37 \times 10^{6}\right)^{3} \\\text { Mass }=5.975 \times 10^{24}\mathrm{~kg} \\\text { density }(\rho)=\frac{\operatorname{mass}(\mathrm{m})}{V}\end{array}$

$\begin{aligned}\rho &=\frac{5.975 \times 10^{24}}{4\times\pi\times\left(6.37\times10^{6}\right)^{3}}\\&=5.52\times10^{3}\mathrm{kgm}^{-3}\end{aligned}$

In the relation

$y = r sin (\omega t - kx)$ , the dimensions of

$\omega/k$ are

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

Explanation

$y=rsin\left( wt-kx \right)$

$\left( wt-kr \right)$ is angle so it is dimensionless.

$\left[ wt \right] =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] \quad \quad \left[ kx \right] =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$

$w=\dfrac { 1 }{ t }$

$\left[ w \right] =\left[ { T }^{ -1 } \right]$

$k=\dfrac { 1 }{ x }$

$\dfrac { w }{ k } =\dfrac { \left[ { T }^{ -1 } \right] }{ \left[ { L }^{ -1 } \right] } =\left[ { M }^{ 0 }{ LT }^{ -1 } \right]$

$k=\left[ { L }^{ -1 } \right]$

The time dependence of a physical quantity

$P$ is given by

$P = P_0e^{-\alpha t^2}$ , where

$\alpha$ is a constant and t is time. Then constant

$\alpha$ is / has

0%
Dimensionless
0%
Dimensions of $T^{-2}$
0%
Dimensions of P
0%
Dimensions of $T^{2}$

Explanation

$P={ P }_{ 0 }{ e }^{ -{ \alpha t }^{ 2 } }$

${ \alpha t }^{ 2 }$ must be dimension less in

${ e }^{ -{ \alpha t }^{ 2 } }$

$\Rightarrow \quad dim\left( { \alpha t }^{ 2 } \right) ={ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }\Rightarrow dim\left( \alpha \right) =\dfrac { { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } }{ { T }^{ 2 } }$

$\therefore$

$dim\left( \alpha \right) ={ M }^{ 0 }{ L }^{ 0 }{ T }^{ -2 }$

The position x of a particle at time t is given by

$x = \dfrac{V_0}{a}(1 - e^{-at})$ , where

$V_0$ is constant and

$a > 0$ . The dimensions of

$V_0$ and

$a$ are

0%
$M^0LT^{-1}$ and $T^{-1}$
0%
$M^0LT^0$ and $T^{-1}$
0%
$M^0LT^{-1}$ and $LT^{-2}$
0%
$M^0LT^{-1}$ and $T$

Explanation

$x=\dfrac { { V }_{ 0 } }{ a } \left( 1-{ e }^{ -at } \right)$

at should be dimensions

$\Rightarrow Dim\left( a \right) =\dfrac { 1 }{ Dim\left( T \right) } ={ T }^{ -1 }$

$x$ is position

$\Rightarrow dim\left( x \right) ={ M }^{ 0 }L{ T }^{ 0 }=dim\left( \dfrac { { V }_{ 0 } }{ a } \right)$

$\therefore$

$dim\left( { V }_{ 0 } \right) ={ M }^{ 0 }L{ T }^{ -1 }$

The mass of the liquid flowing per second per unit area of cross section of the tube if proportional to

$P^x$ and

$v^y$ , where

$P$ is the pressure difference and

$v$ is the velocity then the relation between

$x$ and

$y$ is

0%
$x = y$
0%
$x = - y$
0%
$y^2 = x$
0%
$y = -x^2$

Explanation

Given

$m\alpha { P }^{ x }\times { V }^{ y }$

$\left[ { ML }^{ -2 }{ T }^{ 0 } \right] ={ \left[ { ML }^{ -1 }{ T }^{ -2 } \right] }^{ x }{ \left[ { M }^{ 0 }{ LT }^{ 1 } \right] }^{ y }$

$x=1$ } Comparing coefficients

$-2=-x+y$ and

$-1=-2x-y$

$-2=-4x-2y$

$0=\left( -4x-2y \right) -\left( -x+y \right)$

$0=-3x-3y$

$\Rightarrow$

$3x=-3y$

$\Rightarrow$

$\boxed { x=-y }$ .

The Van der Waal's equation of state for some gases can be expressed as:

$(P + \dfrac{a}{V^2})(V - b) = RT$
where P is the pressure, V is the molar volume, and T is the absolute temperature of the given sample of gas and a,b and R are constants.
The dimensions of a are

0%
$ML^5T^{-2}$
0%
$ML^{-1}T^{-2}$
0%
$L^3$
0%
none of the above

Explanation

$\left[ P \right] =\left[ \dfrac { a }{ { V }^{ 2 } } \right]$

$\left[ a \right] =\left[ P \right] \left[ { V }^{ 2 } \right]$

$\left[ a \right] =\left[ \dfrac { Force }{ Area } \times { \left( Volume \right) }^{ 2 } \right]$

$\left[ a \right] =\left[ \dfrac { { MLT }^{ -2 }\times { L }^{ 6 } }{ { L }^{ 2 } } \right]$

$\left[ a \right] =\left[ { ML }^{ 5 }{ T }^{ -2 } \right]$

In a given system of units, 1 unit of mass = 2 kg. 1 unit of length = 5 m and 1 unit of time = 5 sec. then in this system, 1 N represents:

0%
5/2 units of force
0%
2/5 units of force
0%
2 units of force
0%
1/2 units of force

Explanation

1N can be represented as

$kgm/s^2$

$1N=\dfrac { [M][L] }{ { [T] }^{ 2 } } \\ =\dfrac { (1Kg)(1m) }{ { (1s) }^{ 2 } } \\ =\dfrac { (\dfrac { 1 }{ 2 } unit\quad of\quad mass)(\dfrac { 1 }{ 5 } unit\quad of\quad length) }{ { (\dfrac { 1 }{ 5 } unit\quad of\quad time) }^{ 2 } } \\ =\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 5 } \times \dfrac { 5 }{ 1 } \times \dfrac { 5 }{ 1 } unit\quad of\quad force\\ =\dfrac { 5 }{ 2 } unit\quad of\quad force.$

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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