MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 15

Out of the following pairs, choose the pair in which the physical quantities do not have identical dimension?

0%
Pressure and Young's modules
0%
Planck's constant and Angular momentum
0%
Impulse and moment of force
0%
Force and rate of change of linear momentum

Explanation

(A) : Young's modulus is given by

$Y = \dfrac{F}{A}.\dfrac{\Delta L}{L}$
This implies Pressure

$(F/A)$ and Young's modulus have same dimensions.
(B) :
From the equation

$L = \dfrac{nh}{2\pi}$
We say that, Planck's constant and angular momentum have same dimensions.
(C) : Using the relation between impulse

$J$ and moment of force

$\tau$ ,

$J = \tau\Delta t$
We say that impulse and moment of force have different dimensions.
(D) :
From the relation

$F = \dfrac{dP}{dt}$
We say that force

$F$ and rate of change of momentum

$\dfrac{dP}{dt}$ have same dimension.

Amol has three objects, X, Y and Z. He puts X into a measuring cylinder with

$30$ mL of water. Then he puts Y and Z one by one into the same measuring cylinder as shown.
Which of the objects has/have a volume less than

$30$ mL?

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X only
0%
Y only
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X and Y only
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X, Y and Z

Assertion: The dimensional formula for relative velocity is same as that of the change in velocity.
Reason: Relative velocity of P w.r.t. Q is the ratio of velocity of P and that of Q.

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Both assertion and reason are true but the reason is the correct explanation of assertion
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Both assertion and reason are true but the reason is not the correct explanation of assertion
0%
Assertion is true but reason is false
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Both the assertion and reason are false
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Reason is true but assertion is false

The Van der Waal's equation of

$'n'$ moles of a real gas is

$\displaystyle \left( P+\frac { a }{ { V }^{ 2 } } \right) \left( V-b \right) =nRT$
Where

$P$ is pressure,

$V$ is volume,

$T$ is absolute temperature,

$R$ is molar gas constant and

$a, b, c$ are Van der Waal constants. The dimensional formula for

$ab$ is:

0%
[ $\displaystyle M{ L }^{ 8 }{ T }^{ -2 }$ ]
0%
[ $\displaystyle M{ L }^{ 6 }{ T }^{ -2 }$ ]
0%
[ $\displaystyle M{ L }^{ 4 }{ T }^{ -2 }$ ]
0%
[ $\displaystyle M{ L }^{ 2 }{ T }^{ -2 }$ ]

Explanation

Since, the dimensions of

$P$ must be same as

$\dfrac { a }{ { V }^{ 2 } }$ ,
Hence,

$\dfrac { [a] }{ L^{ 6 } } =[M{ L }^{ -1 }{ T }^{ -2 }]$

$[a]=[M{ L }^{ 5 }{ T }^{ -2 }]$

Also, dimension of

$b$ must be same as that of

$V$ . Hence,

$[b]={ L }^{ 3 }$ .

Thus,

$[ab]=[M{ L }^{ 8 }{ T }^{ -2 }]$

A block of mass 1kg is placed on a rough horizontal surface. A spring is attached to the block whose other end is joined to a rigid wall, as shown in the figure. A horizontal force is applied to the block so that it remains at rest while the spring is elongated by

$x(x \geq \dfrac{ \mu mg }{ k } )$ . Let

$F_{max}$ and

$F_{min}$ be the maximum and minimum values of force F for which the block remains in equilibrium. For a particular

$x, \ F_{max} -F_{min} = 2N$ . Also shown is the variation of

$F_{max} + F_{min}$ versus

$x$ , the elongation of the spring.

The coefficient of friction between the block and the horizontal surface is :

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$0.1$
0%
$0.2$
0%
$0.3$
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$0.4$

A beam balance of unequal arm length is used by an unscruplus trader. When an object is weighed on left pan. The weight

$i$ found to be

${W}_{1}$ . When the object is weighed on the right plan, the weight is found to be

${W}_{2}$ . If

${W}_{1}\neq {W}_{2}$ , the correct weight of the object is

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$\dfrac{W_{1}+W_{2}}{2}$
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$\sqrt { { W }_{ 1 }{ W }_{ 2 } }$
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$\sqrt { { W }_{ 1 }^{ 2 }{ +W }_{ 2 }^{ 2 } }$
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$\sqrt { { W }_{ 2 }^{ 2 }{ -W }_{ 1 }^{ 2 } }$

Explanation

Let the beam balance be as shown.

And let actual weight be

$W$ .

(1)

$Wr_{L}=W_{1}r_{R}$

(2)

$W_{2}r_{L}=Wr_{R}$

Dividing Equation 1 by Equation 2,

$\cfrac{W}{W_{2}}=\cfrac{W_{1}}{W}$

$\Rightarrow W=\sqrt{W_{1}W_{2}}$

1102553_1015532_ans_b1dab76572924812b9302fbb76017968.png

The dimension of

$\left( \frac { 1 }{ 2 } \right) { \varepsilon }_{ 0 }{ E }^{ 2 }$ is (

${ \varepsilon }_{ 0 }$ : permittivity of free space, Electric field)

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

Explanation

Dimensional formula of permittivity of free space=

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

Dimensional formula of permittivity of electric field=

$[M^{1}L^{1}T^{-3}A^{-1}]$

Dimensional formula of

$\dfrac{1}{2}\epsilon_0 E^2$ =

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

A particle of mass

$m$ collides with a stationary particle and continues to move at an angle of

${45}^{o}$ with respect to the original direction. The second particle also recoils at an angle of

${45}^{o}$ to this direction. The mass of the second particle is (collision is elastic)

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$m$
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$\sqrt { 2 } m$
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$\cfrac { m }{ \sqrt { 2 } }$
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$\cfrac { m }{ 2 }$

Two resistors of resistances $R_1= 150\pm2 \space \Omega$ and $R_2 = 220\pm6 \space \Omega$ areconnected in parallel combination.Calculate the equivalent resistance.

0%
$180\Omega$
0%
$90\Omega$
0%
$18\Omega$
0%
$10\Omega$

Explanation

$\text { Given }-R_{1}=150 \pm 2 \Omega \quad R_{2}=220 \pm 6 \Omega$

$\begin{array}{l} \text { When connect in Parallel we have - } \\ \qquad R_{e q}=\frac{R_{1} \cdot R_{2}}{R_{1}+R_{2}} \Rightarrow R_{e q}=\frac{150 \cdot 220}{150+220}=89.19 \mathrm{~s} \end{array}$

$\begin{array}{l} \text { for calculating error }- \\ \text { we have } \frac{1}{\operatorname{Req}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \end{array}$

$\begin{array}{l} \text { differentiate both Sides } \Rightarrow \\ \qquad-\frac{d\left(R_{e q}\right)}{\left(R_{e q}\right)^{2}}=-\frac{d R_{1}}{\left(R_{1}\right)^{2}}-\frac{d R_{2}}{\left(R_{2}\right)^{2}} \end{array}$

$\Rightarrow d(\text { R}_{eq})=(89.19)^{2} \cdot\left[\frac{2}{(150)^{2}}+\frac{6}{(220)^{2}}\right]$

$\begin{array}{l} d\left(R_{e q}\right)=1.69 \Omega \\ \text { Now we can write } R_{e q}=89.19 \pm 1.69 \Omega \\\Rightarrow \text{90}\Omega \text { can be be the maximum suitable answer. } \end{array}$

Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii

$R_1$ and

$R_2$ respectively. The ratio of the mass of X to that of Y is?

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$(R_1/R_2)^{1/2}$
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$R_2/R_1$
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$(R_1/R_2)^2$
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$R_1/R_2$

Multiple Correct Answers Type
Which of the following pairs have the same dimensions?
(L = inductance, C = capacitance, R = Resistance)

0%
$\dfrac{L}{R}$ and CR
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LR and CR
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$\dfrac{L}{R}$ and $\sqrt{LC}$
0%
RC and $\dfrac{1}{LC}$

Explanation

$\left[ L \right] =\left[ \dfrac { \phi }{ I } \right]$

Where

$\phi$ is the magnetic flux and

$I$ is the current.

$\left[ \phi \right] =\left[ B \right] \left[ { L }^{ 2 } \right] =\left[ { MT }^{ 2 }{ A }^{ -1 } \right] { L }^{ 2 }$

$\Rightarrow \left[ \phi \right] =\left[ { ML }^{ 2 }{ T }^{ -2 }{ A }^{ -1 } \right]$

So,

$\left[ L \right] =\left[ { MT }^{ -2 }{ L }^{ 2 }{ A }^{ -2 } \right]$

Now for

$R$

$\left[ R \right] =\left[ \dfrac { V }{ I } \right] =\left[ \dfrac { { ML }^{ 2 }{ T }^{ -3 }{ A }^{ -1 } }{ { A }^{ -1 } } \right]$

$\left[ R \right] =\left[ { ML }^{ 2 }{ T }^{ -3 }{ A }^{ -2 } \right]$

Now,

$\left[ \dfrac { L }{ R } \right] =\left[ \dfrac { { MT }^{ 2 }{ L }^{ 2 }{ A }^{ -2 } }{ { ML }^{ 2 }{ T }^{ -3 }{ A }^{ -2 } } \right]$

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

$\left[ C \right] =\left[ \dfrac { AT }{ V } \right] =\left[ \dfrac { AT }{ { ML }^{ 2 }{ T }^{ -3 }{ A }^{ -1 } } \right] =\left[ { M }^{ -1 }{ L }^{ -2 }{ T }^{ 4 }{ A }^{ 2 } \right]$

$\left[ CR \right] =\left[ T \right]$

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

${ \left[ LC \right] }^{ 1/2 }=\left[ T \right]$

$\left[ RC \right] =T$

A body of mass

$m_1$ , moving with a uniform velocity of 50

$m{s^{ - 1}}$ collides with another body of mass

$m_2$ at rest and then two together start moving with a velocity 40

$m{s^{ - 1}}$ . The ratio of their masses

$\left( {\frac{{{m_1}}}{{{m_2}}}} \right)$ is

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1 : 3
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2 : 3
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2 : 5
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4 : 1

A body is dropped from a height h. If it acquired a momentum p, Then the mass of the body is

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$\frac{P}{{\sqrt {2gh} }}$
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$\frac{{{P^2}}}{{2gh}}$
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$\frac{{2gh}}{{{P}}}$
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$\sqrt {\frac{{2gh}}{p}}$

Two particle X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii

$R_1$ and

$R_2$ respectively. The ratio of the mass X to the Y is:

0%
${\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^{\frac{1}{2}}}$
0%
${\frac{{{R_2}}}{{{R_1}}}}$
0%
${\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^2}$
0%
${\frac{{{R_1}}}{{{R_2}}}}$

Which of the following quantity is dimension less?

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Gravitational Constant
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Plank's Constant
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Power of lens
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None of these

van der Waal's constant

$'a'$ has the dimensions of

0%
$Mol \ {L}^{-1}$
0%
$Atm\ {L}^{2}\ {mol}^{-2}$
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$L\ {mol}^{-1}$
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$Atm\ L\ {mol}^{-2}$

The length of two rods measured by a meter scale are 60.0 cm and 20.0 cm respectively. The least count of meter scale is 1 mm. Then

0%
First measurement is more accurate
0%
Second measurement is more accurate
0%
Both measurement are equally accurate
0%
First measurement is more is more precise

Explanation

$\begin{aligned} \text { Given - } & \text { Length of Rod } 1 \text { measured }=60.0 \mathrm{~cm} \\ & \text { Length of Rod } 2 \text { measured }=20.0 \mathrm{~cm} \\ & \text { Least count of Scale }=1 \mathrm{~mm}=0.1 \mathrm{~cm} \end{aligned}$

$\begin{array}{l} \text { Both the Readings are equally accurate because } \\ \text { the meauswred units are accurate upto the least } \\ \text { count }=0.1 \mathrm{~cm} \\ \text { and hence both are equaly accurate. } \end{array}$

$\begin{array}{l} \text { Percentage error in Rod } 1=\frac{0.1}{60}=0.16 \% \\ \text { Percentage error in Rod } 2=\frac{0.1}{20}=0.5 \% \\ \text { since } \% \text { eeror in Rod } 1<\% \text { error in Rod } 2 \\ \text { Hence Reading of Rod } 1 \text { is more precise. } \end{array}$

The linear mass density i.e. mass per unit length of a rod of length L is given by

$\rho$ =

$\rho_{0}$ (1 +

$\dfrac{x}{L}$ ), where

$\rho_{0}$ is constant , x is distance from the left end. Find the c.o.m from the left end.

0%
$\dfrac{5L}{9}$
0%
$\dfrac{4L}{9}$
0%
$\dfrac{5L}{8}$
0%
$\dfrac{3L}{8}$

Figure shows a weighing machine kept in a lift is moving upwards with acceleration of

$5$

$m/s^2$ . A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown by weighing machine and spring balance is

$15$ kg and

$45$ kg respectively.
Answer the following question. Assume that the weighing machine can measure weight by having negligible deformation due to block, while the spring balance requires larger expansion(take

$g=10m/s^2$ ).
Mass of the object in kg and the normal force acting on the block due to weighing machine are?

0%
$60$ kg, $450$ N
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$40$ kg, $150$ N
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$80$ kg, $400$ N
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$10$ kg, Zero

Two isotopes of an element

$X$ are present in the ratio of

$1:2$ , having mass number

$M$ and (

$M+0.5$ ) respectively. Find the mean mass number of

$X$

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$M + (2/3)$
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$M + (1/3)$
0%
$M + (1/4)$
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$M + 3$

The dimension[ M L T] may correspond to

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Work done by a force
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Linear momentum
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Pressure
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Energy per unit volume
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None of these

A pressure

$O$ , velocity of light

$C$ and acceleration due to gravity g are chosen as fundamental units, then dimensional formula of mass is

0%
$PC^{3}\ g^{-4}$
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$PC^{-4}\ g^{3}$
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$PC^{4}\ g^{-3}$
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$PC^{4}\ g^{3}$

A student perform an experiment for determination of

$(g=\frac{4\pi ^2l}{T^2}),l \cong 1\;m$ , and he commits an error of

$\delta l$ . For

$T$ he takes the time of

$n$ oscillations with the stop watch of least count

$\Delta T$ and he commits a human error of

$0.1\;s$ . For which the following data, the measurement of

$g$ will be most accurate ?

0%
$\Delta L=.5,\Delta T=0.1,n=20$
0%
$\Delta L=.5,\Delta T=0.1,n=50$
0%
$\Delta L=.5,\Delta T=0.01,n=20$
0%
$\Delta L=.0.1,\Delta T=0.05,n=50$

The dimension of magnetic field in

$M,L,T$ and

$C$ (Colulomb) is given as

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$MLT^{-1}C^{-1}$
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$MLT^{2}C^{2}$
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$MT^{-1}C^{-1}$
0%
$MT^{-2}C^{-1}$

What is dimension of

$\frac {d^2x}{dy^2}$ where x is pressure and y is energy incident over unit area in unit time of a surface?

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

The equation of state for a real gas such as hydrogen, oxygen, etc. is called the Van der Waal's equation which reads

$\left(P+\dfrac{a}{V^{2}}\right)(V-b)=nRT$
Where a and b are constant of a gas. the dimensional formula of constant a is:

0%
$ML^{5}T^{-2}$
0%
$ML^{5}T^{-1}$
0%
$ML^{-1}T^{-1}$
0%
$\alpha$ being a constant, is dimensionless

Four particles of mass

$m_1=2$ m,

$m_2=4$ m,

$m_3=$ m and

$m_4$ are placed at four corners of a square. What should be the value of

$m_4$ so that the centre of mass of all the four particles are exactly at the centre of the square?

0%
$2$ m
0%
$8$ m
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$6$ m
0%
None of these

Let

$[{\varepsilon _o}]$ denotes the dimensional formula of the permittivity of the vacuum. If

$M$ = mass,

$L$ = Length,

$T$ = time and

$A$ = electric current, then:

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

The accuracy in the measurement of the diameter of hydrogen atom as 1.06 x

$10^{-10}$ m is

0%
0.01
0%
106 x $10^{-10}$
0%
$\frac{1}{106}$
0%
0.01 x $10^{-10}$

A uniform rod of mass M and length L leans against a frictionless wall, with quarter of its length hanging over a corner as shown. Friction at corner is sufficient to keep the rod at rest. Then the ratio of magnitude of normal reaction on rod by wall and the magnitude of normal reaction on rod by corner is?

0%
$\dfrac{1}{2\sin\theta}$
0%
$\dfrac{2}{\sin\theta}$
0%
$\dfrac{1}{2\cos\theta}$
0%
$\dfrac{2}{\cos\theta}$

$V$ is the volume of a liquid flowing per second through a capillary tube of length

$l$ and radius

$r$ , under a pressure difference

$(p)$ . If the velocity

$(v)$ , mass

$(M)$ and time

$(T)$ are taken as the fundamental quantities, then the dimensional formula for

$\eta$ in the relation

$V = \dfrac {\pi p r^{4}}{8\eta l}$ is

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

Dimensions of magnetic flux density is-

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

The figure shows the position-time (x-t) graph of one-dimenstional motion of a body of mass

$0.4kg$ . The magnitude of each impulse is

0%
$0.2Ns$
0%
$0.4Ns$
0%
$0.8Ns$
0%
$1.6Ns$

Explanation

$\text { impulse }=\Delta p=m \Delta v$

$u=\dfrac{2-0}{2-0}=1$

$v=\dfrac{2-4}{2-0}=-1$

$\Delta v=1-(-1)$

$=2$

$\begin{aligned}\Delta p &=0.4 \times 2\\&=0.8\mathrm{NS}\end{aligned}$

1991627_1324822_ans_b91655acd3dc446f92d1e32d6fe6f13b.PNG

The dimensions of the quantity

$\dfrac {L}{RCV}$ are-

0%
$M^0L^0T^1A^1$
0%
$M^0L^0T^0A^{-1}$
0%
$M^0L^0T^0A^1$
0%
$M^0L^0T^{-1}A^{-1}$

The dimensions of emissive power are:

0%
$[M^1 L^{-2}T^{-3} ]$
0%
$[M^1 L^2T^{-3} ]$
0%
$[M^1L^0T^{-3} ]$
0%
$[M^1L^0T6{-2} ]$

The dimensional formula
for the physical quantity $\cfrac {B \mu_0 \epsilon_0}{E}$ is : (E=intensity
of electric field, B-magnetic induction and symbols have their usual meaning)

0%
$L^oMT$
0%
$L^1MT^{-1}$
0%
$L^{-3}MT^3$
0%
$L^{\cfrac {1}{2}}MT^{-\cfrac {1}{2}}$

The moon's distance from the earth is 360000 km and its diameter subtends an angle of 42' at the eye of the observer. The diameter of the moon in kilometers is :

0%
4400
0%
1000
0%
3600
0%
8800

The dimensional formula of

$\left( \dfrac { slug }{ barn } \right)$ is

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

The diameter and height of a cylinder are measured by a meter scale to be

$12.6\pm 0.1$ cm and

$34.2\pm 0.1$ cm ,respectively. What will be the value of its volume in appropriate significant figures?

0%
$4264\pm 81{ cm }^{ 3 }$
0%
$4260\pm 80{ cm }^{ 3 }$
0%
$4264.4\pm 81.0{ cm }^{ 3 }$
0%
$4300\pm 80{ cm }^{ 3 }$

Explanation

$\begin{aligned}\text { Given } \rightarrow & \text{ diameter }(d)=12.6 \pm 0.1 \mathrm{~cm} \\& \text { height }(h)=34.2 \pm 0.1 \mathrm{~cm}\end{aligned}$

$\text { Volume }=\pi \dfrac{d^{2}}{4} h=\dfrac{22}{7}\times \dfrac{12.6 \times 12.6 \times 34.2}{4}=4260\mathrm{~cm}^{3}$

$\begin{array}{l}\text { Taking log both side }\\\log V=\log \left(\pi \frac{d^{2}}{4}h\right)\\\log V=\log \pi+2 \log d-\log4+\operatorname{logh} \\\text { Differentiate both side } \\\frac{\delta V}{V}=0+2 \frac{\delta d}{d}-0+\frac{\delta h}{h}\end{array}$

$\begin{aligned}\delta V &=\left(\frac{2 \times 0.1}{12.6}+\frac{0.1}{34.2}\right) V \\&=\left(\frac{0.2}{12.6}+\frac{0.1}{34.2}\right){4260} \\&=8\mathrm{~cm}^{3} \\\therefore \text { Volume } &=4260 \pm 80 \mathrm{~cm}^{3}\end{aligned}$

The periodic time

$(T)$ of a simple pendulum of length

$(L)$ is given by

$T = 2\pi \sqrt {\dfrac {L}{g}}$ . What is the dimensional formula of

$T = \sqrt {\dfrac {g}{L}}$ ?

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

The dimensional formula for

$\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}$ is identical to that of

0%
$\frac{\mathrm{B}^{2}}{2 \mu_{0}}$
0%
$\frac{1}{2} B^{2} \mu_{0}$
0%
$\frac{\mu_{0}^{2}}{2 B}$
0%
$\frac{1}{2} B \mu_{0}^{2}$

If force

$F ,$ acceleration

$A$ and time T are are chosen asfundamental quantities, the dimensiogs of energy in the system are

0%
$F A T ^ { 2 }$
0%
$F ^ { - 2 } A ^ { - 1 } T$
0%
$F ^ { 0 } A T ^ { 2 }$
0%
$F ^ { - 1 } A ^ { 2 } T ^ { 0 }$

A quantity

$Q = \frac { E L ^ { 2 } } { m ^ { 5 }{G}^{2} }$ where

$E, I, G$ and

$m$ are energy, angular momentum, gravitational constant and mass respectively. What are the dimensional of Q ?

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

The term

$( 1 / 2 )$ pv

$^ { 2 }$ occurs in Bernaullis equations, with p being the density of a fluid and v its speed The dimensions of this term are

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

The dimensional formula for Stefan's constant is

0%
$M^0L^0T^{-2}K^{-4}$
0%

$M^1L^0T^{-3}K^{-4}$
0%
$M^1L^1T^{-2}K^{-2}$
0%
$M^0L^2T^{-3}K^{-3}$

A glass cube of edge 1

${cm}$ and

$\mu=1.5$ has a spot at the centre .The area of the cube face that must be covered to prevent the spot from being seen is (in

${cm}^{2} )$

0%
$\sqrt{5} \pi$
0%
$5 \pi$
0%
$\frac{\pi}{\sqrt{5}}$
0%
$\frac{\pi}{5}$

A force acts for

$0.5$ s on a body of mass

$1.5$ kg initially at rest. When the force ceases to act, the body is found to cover a distance of

$5$ m in

$2$ s. The magnitude of the applied force is

0%
$5.0$ N
0%
$7.5$ N
0%
$10$ N
0%
$12.5$ N

The dimensions of (force constant

\ mass)

$^ { 1 / 2 }$ are the same as that

0%
acceleration
0%
angular acceleration
0%
angular velocity
0%
none of the above

Dimensional formula for electromotive force is same as that for

0%
potential
0%
current
0%
force
0%
energy

The dimensions of

$\dfrac { mass }{ Force\quad constant\quad of\quad SHM }$ are same as that of :

0%
time
0%
${ \left( time \right) }^{ 2 }$
0%
acceleration
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$\dfrac { 1 }{ acceleration }$

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 15 - 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 15 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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