MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 8

Write down the number of significant figures in the following value.

$1200 N$

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

Zeroes after non zero digit and before decimal are not counted as significant.

Therefore, (C) has two significant figures in

$1200 N$ .

Dimension of electrical resistance is :

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

Explanation

We know that Power,

$P=I^2R$
Dimensions of Power are that of Force times velocity, i.e.

$MLT^{-2}\times LT^{-1}=ML^2T^{-3}$
Thus,

$ML^2T^{-3}=A^2|R|$ , or

$|R|=ML^2T^{-3}A^{-2}$

In two different systems of units an acceleration is represented by the same number, whilest a velocity is represented by numbers in the ratio

$1\,\colon\,3$ . The ratios of unit of length and time are

0%
$\;\displaystyle\frac{1}{3},\displaystyle\frac{1}{9}$
0%
$\;\displaystyle\frac{1}{9},\displaystyle\frac{1}{3}$
0%
$\;1,1$
0%
$\;\displaystyle\frac{1}{3},\displaystyle\frac{1}{3}$

Explanation

Given,

$\dfrac{[v_1]}{[v_2]}=\dfrac{1}{3}$

Here,

$[a_1]=[a_2]$ or

$[v_1T_1^{-1}]=[v_2T_2^{-1}]$ ( as

$a=\dfrac{dv}{dt}$ )

$\dfrac{T_1}{T_2}=\dfrac{v_1}{v_2}=\dfrac{1}{3}$

$\dfrac{[L_1T_1^{-1}]}{[L_2T_2^{-1}]}=\dfrac{1}{3}$ as

$[v]=[LT^{-1}]$

$\dfrac{L_1}{L_2}=\dfrac{1}{3}\times \dfrac{[T_1]}{[T_2]}=1/3 \times 1/3=1/9$

Which of the statements are correct?

0%
Unit of work is $kgm^2s^{-2}$
0%
force is $kgms^{-2}$
0%
work is $kgms^{-2}$
0%
force is $kgm^2s^{-2}$

Explanation

Work = Force

$\times$ displacement

Force = Mass

$\times$ acceleration (kg. m.s

$^{-2}$ )

So, unit of Work is kg. m/s

$^2$ . m = kg.m

$^2$ .s

$^{-2}$

Option A and B

A force

$\displaystyle F$ is given by

$F = at + bt^2$ , where

$t$ is time. The dimensions of

$a$ and

$b$ are :

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

Explanation

Dimensions of force are

$MLT^{-2}$ . Thus, both

$at$ and

$bt^2$ must have units of force.
Thus,

$|a|T=MLT^{-2}$ or

$|a|=MLT^{-3}$
Also,

$|b|T^2=MLT^{-2}$ or

$|b|=MLT^{-4}$

From the following pairs, choose the pair that does not have identical dimensions ?

0%
Impulse and momentum
0%
Work and torque
0%
Moment of inertia and moment of force
0%
Angular moment and Planck's constant

Explanation

Impulse is change in momentum (

$p_2-p_1$ ).

Thus, they have same dimensions.
Work is dot product of Force and displacement while Torque is the cross product of Force and radius vector.

Thus they have same dimensions (of Fr =

$MLT^{-2}\times L = ML^2T^{-2}$ )
Angular moment and Planck's constant also have the same dimensions because Angular moment = Momentum

$\times$ length. Planck's constant is Momentum

$\times$ Wavelength (because

$p=\dfrac{h}{\lambda}$ )
Thus they have same dimensions.
Moment of Inertia has dimensions

$ML^2$ and moment of force, i.e. Torque has dimensions

$ML^2T^{-2}$ , thus different.

The ratio of the dimensions of Planck's constant and that of inertia has the dimension of ?

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

Explanation

The plank's constant has the dimensions of

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

The dimension of the inertia is

$I=[M^1L^2]$

The ratio of the dimension of the plank's constant and the inertia is

$\dfrac hI=T^{-1}$

It is the same as the dimension of frequency.

If a, b, c are the percentage errors in the measurement of A, B and C, then the percentage error in ABC would be approximately

0%
$abc$
0%
$a + b + c$
0%
$ab + bc + ac$
0%
$\displaystyle \frac {a} {b} + \frac {b} {c} + \frac {c} {a}$

Explanation

$\displaystyle \frac{\Delta ABC}{ABC}=\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}$

$=a+b+c$

In the relation

$\displaystyle P = \frac {\alpha} {\beta} e^{-\dfrac {\alpha Z} {k \theta}}$ ,

$P$ is pressure,

$Z$ is distance,

$k$ is Boltzmann constant and

$\theta$ is the temperature. The dimensional formula of

$\beta$ will be :

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

Explanation

Since exponential term is dimensionless,

Dimension of

$\alpha$ ,

$\left[ \alpha \right] =\dfrac { \left[ k \right] \left[ \theta \right] }{ \left[ Z \right] } =\dfrac { (M{ L }^{ 2 }{ T }^{ -2 }{ K }^{ -1 })*(K) }{ L } =ML{ T }^{ -2 }$

Since

$\dfrac { \alpha }{ \beta }$ must have the dimensions of pressure(

$M{ L }^{ -1 }{ T }^{ -2 }$ )

Therefore, dimensions of

$\beta$ ,

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

A unitless quantity ?

0%
Does not exist
0%
Always has a nonzero dimension
0%
Never has a nonzero dimension
0%
May have a nonzero dimension

Two quantities

$A$ and

$B$ are related by

$\dfrac{A}{B} = m$ , where

$m$ is linear density and

$A$ is force. The dimensions of

$B$ will be same as that of

0%
Pressure
0%
Work
0%
Momentum
0%
Latent heat

Explanation

Linear Density, m has dimensions of

$\dfrac{Mass}{Length} = ML^{-1}$ and Force, A, has dimensions

$MLT^{-2}$
Thus dimensions of B are

$\dfrac{MLT^{-2}}{ML^{-1}}=L^2T^{-2}$ , which is the same as Latent Heat (which is heat per unit mass, or

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

Which of the following does not have the dimensions of force ?

0%
Gravitational Potential gradient
0%
Energy gradient
0%
Weight
0%
Rate of change of momentum

Explanation

The gradient of the gravitational potential gradient is the gravitational electric field (i.e. force acting on unit mass).

$\overset { \rightarrow }{ E } =-\nabla \phi$ which has dimensions of

$N/Kg$

Weight is the force exerted by the gravity. So, it will have dimensions of force.

Energy gradient is simply the force vector. So, it will have dimensions of force.

By newton's 2nd law, Rate of change of momentum is the net force. So, it will have dimensions of force.

Hence, Option A is correct.

The velocity

$v$ of a particle at time

$t$ is given by

$\displaystyle v = at + \frac {b} {t + c}$ where

$a, b$ and

$c$ are constants. The dimensions of

$a, b$ and

$c$ are respectively :

0%
$LT^{-2}\, , L$ and $\displaystyle T$
0%
$L^{2}\, , T$ and $\displaystyle LT^{2}$
0%
$LT^{2}\, , LT$ and $\displaystyle L$
0%
$L\, , LT$ and $\displaystyle T^{2}$

Explanation

According to the principle of dimensional homogenity

Dimension of v = dimension of at = dimension of $\displaystyle \frac {b} {t +c}$

Dimension of a = $\displaystyle \frac {[v]} {[t]} = \frac {LT^{-2}} {T} = [LT^{-2}]$

Dimension of b = $\displaystyle [v] [t] = [LT^{-1}T] = [L]$

Dimension of c = $\displaystyle [t] = [T]$

If the unit of length is doubled the numerical value of Area will become ________ times its original value.

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
2
0%
4

Explanation

As unit of length is doubled unit of Area will become four times. So the numerical value of Area will became one fourth Because numerical value

$\displaystyle \propto\frac{1}{unit}$

External and internal diameters of a hollow cylinder are measured to be

$(4.23 \pm 0.01)$ cm and

$(3.89 \pm 0.01)$ cm. The thickness of the wall of the cylinder is

0%
$(0.34 \pm0.03)$ cm
0%
$(0.17 \pm 0.04)$ cm
0%
$(0.17 \pm 0.01)$ cm
0%
$(0.34 \pm 0.02)$ cm

Explanation

External diameter

$d_e = (4.23\pm 0.01) \ cm$

So external radius

$r_e = \dfrac{d_e}{2} = (2.12\pm 0.005) \ cm$

Internal diameter

$d_i = (3.89\pm 0.01) \ cm$

Internal diameter

$r_i = \dfrac{d_i}{2} = (1.95\pm 0.005) \ cm$

Thickness of the wall

$t = r_e-r_i = (2.12-1.95) \pm (0.005+0.005) \ cm =( 0.17\pm 0.01) \ cm$

The pair of quantities that do not have the same dimensions is

0%
Latent heat, specific heat
0%
Gravitational force, Coulomb force
0%
Kinetic energy of a freely falling body, potential energy of compressed spring
0%
Coefficient of friction, number of molecules in a container.

Explanation

Hint:

Dimension of all forms of Energy is same.

Dimension of all forms of force is same.

Constants are Dimensionless.

Explanation for the correct answer:

$\bullet$ Dimension of Latent heat is,

$[M^0 L^2 T^{-2}]$ . And dimension for specific heat is

$[M^0 L^2 T^{-2}K^{-1}]$ . Thus, they have different dimensions. Option A is correct.

$\bullet$ Gravitational force and Coulomb force are two different types of Force. Thus, they have same Dimensions. Option B is incorrect.

$\bullet$ Kinetic Energy of freely falling body and Potential Energy of a compressed spring are two different types of Energy. Thus, they have same Dimensions. Option C is incorrect.

$\bullet$ Coefficient of friction and number of molecules in a container are two different types of Constant. Thus, they have same Dimensions, they are dimensionless. Option D is incorrect.

Thus, Only Option A is correct.

Statement -1: Kinetic energy can be divided by volume
and
Statement -2: Dimensions of kinetic energy and volumeare different

0%
Statement -1 is True Statement -2 is True;

Statement -2 is a correct explanation for Statement -1
0%
Statement -1 is True Statement -2 is True;

Statement -2 is NOT a correct explanation for Statement -1
0%
Statement -1 is True Statement -2 is False .
0%
Statement -1 is False Statement -2 is True;

Explanation

Division of two Quantities with different dimensions is possible.

Hence, the statement I is correct.

Since dimensions of kinetic energy (

$M{ L }^{ 2 }{ T }^{ -2 }$ ) and volume (

${ L }^{ 3 }$ ) are different, the statement II is correct but is not an explanation for statement I.

Hence, option B is correct.

A cube has a side of length

$1.2 \times 10^{-2}m$ . Calculate its volume.

0%
$1.7 \times 10^{-6} m^3$
0%
$1.73 \times 10^{-6} m^3$
0%
$1.70 \times 10^{-6} m^3$
0%
$1.728 \times 10^{-6} m^3$

Explanation

Length of the cube,

$L = 1.2 \times 10^{-2} m$

$\therefore$ Volume of the cube,

$V = L^3 = (1.2 \times 10^{-2})$

$\times (1.2 \times 10^{-2})$

$\times (1.2 \times 10^{-2}) m^3$

$\implies V = 1.728 \times 10^{-6} m^3$

But as the length contains only

$2$ significant digits, thus volume must also contain

$2$ significant digits.

Hence, volume of the cube is

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

Count total number of S.F. in

$6.020\, \times\, 10^{23}$

0%
4
0%
3
0%
6
0%
7

Count total number of S.F. in 300.00

0%
5
0%
6
0%
8
0%
9

Count total number of S.F. in 5.003020

0%
3
0%
20
0%
7
0%
9

What are the dimensions of electrical resistance ?

0%
[ML $\displaystyle ^{2}$ T $\displaystyle ^{-2}$ I $\displaystyle ^{2}]$
0%
[ML $\displaystyle ^{2}$ T $\displaystyle ^{-3}$ I $\displaystyle ^{-2}]$
0%
[ML $\displaystyle ^{2}$ T $\displaystyle ^{-3}$ I $\displaystyle ^{2}]$
0%
[ML $\displaystyle ^{2}$ T $\displaystyle ^{-2}$ I $\displaystyle ^{-2}]$

Explanation

Hint:

Resistance is defined as the opposition offered by any conductor to the flow of current.

Step 1: Write formula for Electrical Resistance.

By, Ohm's Law, we know that

$R = \dfrac{V}{I}$

where

$V$ is potential difference, and

$I$ is current.

Step 2: Find Dimensions of Potential and Current.

Dimention of current is given as,

$[I] = I^1$

Since we know that Potential is given as,

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

where,

$W$ is Work and

$q$ is charge

Dimention of potential is given as,

$[V] = \dfrac{[W]}{[q]}$

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

$\Rightarrow[V] = [ML^2T^{-3}I^{-1}]$

Step 3: Calculate Dimension for Resistance.

Dimension for Resistance is given as,

$[R] = \dfrac{[V]}{[I]}$

$\Rightarrow [R] = \dfrac{[ML^2 T^{-3}I^{-1}]}{[I]}$

$\Rightarrow [R] = [ML^2 T^{-3}I^{-2}]$

Thus, Dimension for Electrical Resistance is, $[R] = [ML^2 T^{-3}I^{-2}]$ .

Option B is correct.

If a conductance of a conductor (G) is

$\displaystyle \frac { { I }^{ 2 }t }{ W }$ , where I is current, t is time and W is work done then write the unit of conductance expressed in terms of fundamental units.

0%
$\dfrac { { A }^{ 2 }{ s }^{ 3 } }{ { kg\quad m }^{ 2 } }$
0%
$\dfrac { { A }^{ 3 }{ s }^{ 2 } }{ { kg\quad m }^{ 2 } }$
0%
$\dfrac { { A }^{ 2 }{ s }^{ 3 } }{ { kg\quad m }^{ 1 } }$
0%
$\dfrac { { A }^{ 3}{ s }^{ 2 } }{ { kg\quad m }^{1 } }$

Explanation

Fundamental unit of

$I$ is

$A$ and that of

$t$ is

$s$ .

The fundamental unit of work done

$W$ is

$Joules$ and it is given as

$kg \ m^2/s^2$ in the SI system.

Thus unit of conductance

$= \dfrac{I^2 t}{W} = \dfrac{A^2 \ s}{kg \ m^2/s^2} = \dfrac{A^2 \ s^3}{kg \ m^2}$

An ice cube has a stone of 500 g placed on its top, is floating in water with its lateral sides placed vertically. It displaces 5 kg of water. Suddenly, the stone slips into water. Because of this ice cube rises by

$\displaystyle \frac{1}{10}th$ of its length above the water level. What is the density of the ice cube (in

$kg/m^3$ )?

0%
960
0%
840
0%
900
0%
820

What is the dimensional formula of universal gravitational 'G'? The gravitational force of attraction between two objects of masses

$\displaystyle { m }_{ 1 }\quad and\quad { m }_{ 2 }$ seperated by a distance d is given by

$\displaystyle F=\frac { G{ m }_{ 1 }{ m }_{ 2 } }{ { d }_{ 2 } }$ . Where 'G' is the universal gravitational constant.

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

Explanation

Gravitational constant can be written as

$G = \dfrac{Fr^2}{m_1m_2}$

So, its dimensions are

$[G] = \dfrac{[F][r^2]}{[m_1][m_2]}$

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

The pair of quantities that do not have the same dimensions is:

0%
Latent heat, specific heat
0%
Gravitational force, Coulomb force
0%
Kinetic energy of a freely falling body, potential energy of a compressed spring
0%
Coefficient of friction, number of molecules in a container

Explanation

$\textbf{Explanation:}$

$\bullet$ The quantity that has same unit will have same dimension representation in terms of MLTAK where M is mass, L is length, T is time , A is ampere and K is for kelvin (temperature).

$\bullet$ Gravitational force and coulomb force are types of force which can be measured in Newton. Both have same unit. So, there dimensions will be same as well.

$\bullet$ Similarly, coefficient of friction and number of molecules in a container both are unitless and kinetic energy of freely falling body and potential energy of compressed spring both are measure in joule. So, they have same dimensions.

$\bullet$ But latent heat is energy required to change the state of a unit mass while specific heat is energy required to change to change unit mass by 1°. So, unit of latent heat is joule/kg while unit of specific heat is joule/°c. As unit of both quantity are different, their dimensions will be different.

$Answer:$

Hence, option A is the correct answer.

The distance between two places is 25 km. Write the magnitude of the measurement.

0%
km
0%
distance
0%
25
0%
two

Explanation

The distance between the two places is

$25$ km where

$25$ represents the magnitude of measurement and km represents the unit.

Instrument used to measure the temperature is

0%
protractor
0%
stethoscope
0%
thermometer
0%
none of these

Count total number of S.F. in

$1.60\, \times\, 10^{- 19}$

0%
3
0%
5
0%
2`
0%
22

What is the dimensional formula of gravitational constant?

0%
$\left [ML^{2} T^{-2}\right ]$
0%
$\left [ML^{-1} T^{-1}\right ]$
0%
$\left [M^{-1}L^{3} T^{-2}\right ]$
0%
none of these

Explanation

$F = \dfrac {G.M.M}{L^{2}} = MLT^{-2}$

$G = \dfrac {MLT^{-2} L^{2}}{M^{2}} = M^{-1} L^{3} T^{-2}$

The dimensional formula of magnetic flux is

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

Explanation

$F = Bqv \rightarrow F = \dfrac {\phi}{A} qv$

$\phi = \dfrac {F . A}{qv} = \dfrac {MLT^{-2} L^{2}}{ATLT^{-1}} = ML^{+2} A^{-1} T^{-2}$

Write the dimensional formula of Force.

0%
$MLT$
0%
$MLT^{-2}$
0%
$MLT^2$
0%
None of the above

Explanation

Force

$=$ mass

$\times$ acceleration

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

Which one of the following defines the mass of a body?

0%
Total weight of the body
0%
Gravitational force exerted by the body
0%
Matter contained in the body
0%
Gravitational force exerted on the body
0%
None of these

Using mass

$(M)$ , length

$(L)$ , time

$(T)$ and current

$(A)$ as fundamental quantities, the dimensions of permeability are

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

Explanation

$B = \mu ni; F = Bqv$

$F = \mu ni qv \Rightarrow \mu = \dfrac {F}{niqv}$

$\mu = \dfrac {MLT^{-2}}{\dfrac {1}{L} A.AT.LT^{-1}}$ [n is no. of turns per unit length]

$= MLA^{-2} T^{-2}$

The relation between force

$F$ and density

$d$ is

$F = \dfrac{x}{\sqrt{d}}$ . The dimensions of

$x$ are

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

Explanation

Given,

$F = \dfrac{x}{\sqrt{d}}$
Substituting dimensions

$ML{T}^{-2} = \dfrac{x}{\sqrt{M{L}^{-3}}}$

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

Dimensions of electrical resistance are

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

The dimensional formula for torque is

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

Using mass (

$M$ ), length (

$L$ ), time (

$T$ ) and current (

$A$ ) as fundamental quantities, the dimensional formula of permittivity is

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

Explanation

$F = \dfrac {1}{4\pi \epsilon_{0}} \cdot \dfrac {q_{1} q_{2}}{r^{2}} \Rightarrow \epsilon_{0} = \dfrac {q_{1}q_{2}}{4\pi Fr^{2}}$

$\dfrac {AT \cdot AT}{MLT^{-2} L^{2}} = \dfrac {A^{2}T^{2}}{ML^{3} T^{-2}} = M^{-1} L^{-3} A^{2} T^{4}$

The dimensional formula of farad is

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

Explanation

$\left [C\right ] = \left [\dfrac {Q}{V}\right ] = \left [\dfrac {Q^{2}}{W}\right ] = \left [M^{-1} L^{-2} T^{2} Q^{2}\right ]$

What is the dimensions of impedance?

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

Explanation

Impedance is same as resistance but in ac circuit

$\therefore$ Dimension of impedance $= \dfrac {\text {dimension of voltage}}{\text {dimension of current}}$ $= \dfrac {\left [V\right ]}{\left [I\right ]} = \dfrac {\left [ML^{2} T^{-3} I^{-1}\right ]}{I} = \left [ML^{2} T^{-3} I^{-2}\right ]$

The comparison of an unknown quantity with a known constant quantity is known as :

0%
Unit
0%
Scale
0%
Magnitude
0%
Measurement

On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between.

0%
$\displaystyle \frac { 290 }{ 12.5 }$ and $\displaystyle \frac { 290 }{ 11.5 }$
0%
$\displaystyle \frac { 295 }{ 12 }$ and $\displaystyle \frac { 285 }{ 11.5 }$
0%
$\displaystyle \frac { 285 }{ 12 }$ and $\displaystyle \frac { 295 }{ 12 }$
0%
$\displaystyle \frac { 285 }{ 12.5 }$ and $\displaystyle \frac { 295 }{ 11.5 }$
0%
$\displaystyle \frac { 295 }{ 12.5 }$ and $\displaystyle \frac { 285 }{ 11.5 }$

The dimensions of Stefan's constant are

0%
$[M^0LT^{-3}K^{-4}]$
0%
$[MLT^{-3}K^{-3}]$
0%
$[ML^2T^{-3}K^{-4}]$
0%
$[ML^0T^{-3}K^{-4}]$

Explanation

Rate of heat radiated is given by,

$R = \sigma A e T^4$

where,

$\sigma$ is the Stefan's constant,

$A$ is the area and

$T$ is the temperature

$e$ is the emissivity of the body

$\implies \sigma = \dfrac{R}{AeT^4}$

Dimension of power,

$R = \dfrac{Energy}{Time} = [ML^2T^{-3}]$

Emissivity is dimensionless.

Dimension of area,

$A = [M^0 L^2 T^0]$

Dimension of temperature,

$T = [K]$

$\therefore$ Dimension of,

$\sigma = \dfrac{[ML^2T^{-3}]}{[M^0 L^2T^0] \times [K^4]}$

$\implies$ Dimension of Stefan's constant

$\sigma = [M L^0 T^{-3} K^{-4}]$

Dimensions of coefficient of viscosity is

0%
$\left [MT^{2}\right ]$
0%
$\left [ML^{-3} T^{-4}\right ]$
0%
$\left [ML^{-1} T^{-2}\right ]$
0%
$\left [ML^{-1} T^{-1}\right ]$

Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would be

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

Explanation

Ohm's law:

$R = \dfrac{V}{I}$

Also,

$W = Vq$

$\implies V = \dfrac{W}{q}$ (where,

$q = It$ )

$\implies$

$R = \dfrac{W}{q I} = \dfrac{W}{I^2 t}$

Dimensions of work done,

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

$\therefore$ Dimension of resistance,

$R = \dfrac{[ML^2 T^{-2}]}{I^2 T} = [M L^2 T^{-3} I^{-2}]$

The energy (E), angular momentum (L) and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula of Planck's constant (h) is

0%
Zero
0%
$-1$
0%
$\dfrac{5}{3}$
0%
$1$

Explanation

$h={ G }^{ x }{ L }^{ y }{ E }^{ z }$

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

Comparing the power, we get

$1=-x+y+z$ ...(i)

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

$-1=-2x-y-2z$ ...(iii)
On solving equations (i), (ii) and (iii), we get

$x=0$

The dimension of angular momentum is

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

Explanation

Dimension of angular momentum

$L = r\times p$

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

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

The physical quantity which has the dimensional formula

$\displaystyle \left[ { M }^{ }{ T }^{ -3 } \right]$ is:

0%
Surface tension
0%
Density
0%
Solar constant
0%
Compressibility

Explanation

Surface tension,

$S = \dfrac{F}{l}$

Dimensional formula of force,

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

$\therefore$ Dimensional formula for surface tension,

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

Dimensional formula for density,

$\rho = \dfrac{M}{V} = \dfrac{M}{L^3} = [ML^{-3}]$

Solar constant

$= \dfrac{\sigma R^2_s T_s^4}{R^2_{se}}$ where all symbols have their usual meaning

Dimension of Stefan's constant,

$\sigma = [MT^{-3} K^{-4}]$

$\therefore$ Dimensional formula for solar constant

$= \dfrac{[MT^{-3} K^{-4}] \times L^2 \times K^4}{L^2} = [M T^{-3}]$

Compressibility,

$k = -\dfrac{\Delta V}{V} \dfrac{1}{\Delta P}$

$\therefore$ Dimensional formula for compressibility,

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

If accuracy of weighing machine is better than 1/2 percent, a student record his own mass. Choose the option that explains the recording most accurately.

0%
6.43 kg
0%
60 kg
0%
64.3 kg
0%
600kg
0%
643 kg

Ampere-hour is the unit of

0%
Quantity of charge
0%
Potential
0%
Energy
0%
Current

Explanation

Charge = current

$\times$ time

Coulomb = ampere

$\times$ hour

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.