Units and Measurements Test

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Units and Measurements Test - 84

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Weekly Quiz Competition

Question 1
1 / -0

In the relation, $$\displaystyle P=\frac{\alpha }{\beta }e^{{\alpha z}/{k\theta }}$$ $$P$$ is pressure, $$Z$$ is distance, $$K$$ is Boltzmann constant and $$\displaystyle \theta $$ is the temperature. The dimensions of $$\displaystyle \beta $$ will be

A
$$\displaystyle \left [ M^{\circ}L^{2}T^{0} \right ]$$

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

C
$$\displaystyle \left [ ML^{\circ}T^{~1} \right ]$$

D
$$\displaystyle \left [ M^{\circ}L^{2}T^{-1} \right ]$$

Solution

$$P=\dfrac { \alpha }{ \beta } { e }^{ \dfrac { \alpha z }{ k\theta } }$$

The dimension of power of e is zero.
$$\dfrac { \alpha z }{ k\theta } =\quad \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] $$

Thus unit of: $$\alpha =\dfrac { k\theta }{ z } $$
Unit of boltzman's constant K is $$K=\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }K^{ -1 } \right] $$

By putting thes values we get,
Unit of $$\alpha =\dfrac { \left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }K^{ -1 } \right] \left[ { K }^{ 1 } \right] }{ \left[ { L }^{ 1 } \right] } =\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }K^{ 0 } \right] $$

Therefore unit of $$P=\dfrac { \alpha }{ \beta } $$

Unit of $$\beta =\dfrac { \alpha }{ P } $$

$$\beta =\dfrac { \left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }K^{ 0 } \right] }{ \left[ { { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } } \right] } =\left[ { { M }^{ 0 }{ L }^{ 2 }{ T }^{ 0 } } \right] $$
Question 2
1 / -0

If $$E, M, J$$ and $$G$$ respectively denote energy, mass, angular momentum and universal gravitational constant, the quantity, which has the same dimensions as the dimensions of $$\frac{EJ^2}{M^5G^2}$$ is:

A
time

B
angle

C
mass

D
length

Solution

Dimensions of the given quantity are $$\dfrac { M{ L }^{ 2 }{ T }^{ -2 }{ [M{ L }^{ 2 }{ T }^{ -1 }] }^{ 2 } }{ { M }^{ 5 }{ [{ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }] }^{ 2 } } =[{ M }^{ 0 }L^{ 0 }{ T }^{ 0 }]$$

Since, the given quantity is dimensionless, among the given options, it can only have dimensions same as angle.

Question 3

1 / -0

Match the following physical quantities with their respective dimensional formula:

(a) Angular Momentum	(e) $$\displaystyle \left[ { ML }^{ 2 }{ T }^{ -3 } \right] $$
(b) Impulse	(f) $$\displaystyle \left[ { ML }^{ 2 }{ T }^{ -1 } \right] $$
(c) Pressure	(g) $$\displaystyle \left[ ML{ T }^{ -1 } \right] $$
(d) Power	(h) $$\displaystyle \left[ { ML }^{ -1 }{ T }^{ -2 } \right] $$

a-f, b-g, c-h, d-e

a-f, b-g, c-e, d-h

a-g, b-h, c-f, d-e

a-g, b-h, c-e, d-f

Solution

Angular momentum has formula $$J=mvr$$. Hence its dimensions are $$[J]=[M{ L }^{ 2 }{ T }^{ -1 }]$$

Impulse has a formula $$I=Ft$$. Hence its dimensions are $$[I]=[M{ L }{ T }^{ -1 }]$$

Pressure has a formula $$P=\dfrac { F }{ A } $$. Hence its dimensions are $$[P]=[ML^{-1}T^{-2}]$$

Power has a formula $$W=Fv$$. Hence its dimensions are $$[W]=[ML^{2}T^{-3}]$$

Question 4
1 / -0

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

A
Pressure and Young's modules

B
Planck's constant and Angular momentum

C
Impulse and moment of force

D
Force and rate of change of linear momentum

Solution

(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.
Question 5
1 / -0

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?

A
X only

B
Y only

C
X and Y only

D
X, Y and Z
Question 6
1 / -0

Directions For Questions

These questions consist of two statement, each printed as assertion and reason. While answering these question you are required to cheese any one of the following five responses.

...view full instructions

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.

A
Both assertion and reason are true but the reason is the correct explanation of assertion

B
Both assertion and reason are true but the reason is not the correct explanation of assertion

C
Assertion is true but reason is false

D
Both the assertion and reason are false

E
Reason is true but assertion is false

Solution

Relative velocity which is vector subtraction of two velocities will also be a vector of the form of velocity so, its dimensional formula will remain unchanged. Relative velocity is measured not by calculating ratio but by calculating difference.
Question 7
1 / -0

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:

A
[$$\displaystyle M{ L }^{ 8 }{ T }^{ -2 }$$]

B
[$$\displaystyle M{ L }^{ 6 }{ T }^{ -2 }$$]

C
[$$\displaystyle M{ L }^{ 4 }{ T }^{ -2 }$$]

D
[$$\displaystyle M{ L }^{ 2 }{ T }^{ -2 }$$]

Solution

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, $$[ab]=[M{ L }^{ 8 }{ T }^{ -2 }]$$
Question 8
1 / -0

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 :

A
$$ 0.1 $$

B
$$ 0.2 $$

C
$$ 0.3 $$

D
$$ 0.4 $$
Question 9
1 / -0

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

A
$$\dfrac{W_{1}+W_{2}}{2}$$

B
$$\sqrt { { W }_{ 1 }{ W }_{ 2 } }$$

C
$$\sqrt { { W }_{ 1 }^{ 2 }{ +W }_{ 2 }^{ 2 } }$$

D
$$\sqrt { { W }_{ 2 }^{ 2 }{ -W }_{ 1 }^{ 2 } }$$

Solution

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,
$$\cfrac{W}{W_{2}}=\cfrac{W_{1}}{W}$$
$$\Rightarrow W=\sqrt{W_{1}W_{2}}$$
Question 10
1 / -0

The dimension of $$\left( \frac { 1 }{ 2 } \right) { \varepsilon }_{ 0 }{ E }^{ 2 }$$ is ($${ \varepsilon }_{ 0 }$$: permittivity of free space, Electric field)

A
$${MLT}^{-1}$$

B
$${ML}^{2}{T}^{-2}$$

C
$${ML}^{-1}{T}^{-2}$$

D
$${ML}^{2}{T}^{-1}$$

Solution

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 $$\dfrac{1}{2}\epsilon_0 E^2$$=$$[M^{1}L^{-1}T^{-2}]$$

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Units and Measurements Test - 84

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Units and Measurements Test - 84

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Question 1

$$\displaystyle \left [ M^{\circ}L^{2}T^{0} \right ]$$

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

$$\displaystyle \left [ ML^{\circ}T^{~1} \right ]$$

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

Solution

Question 2

time

angle

mass

length

Solution

Question 3

a-f, b-g, c-h, d-e

a-f, b-g, c-e, d-h

a-g, b-h, c-f, d-e

a-g, b-h, c-e, d-f

Solution

Question 4

Pressure and Young's modules

Planck's constant and Angular momentum

Impulse and moment of force

Force and rate of change of linear momentum

Solution

Question 5

X only

Y only

X and Y only

X, Y and Z

Question 6

Both assertion and reason are true but the reason is the correct explanation of assertion

Both assertion and reason are true but the reason is not the correct explanation of assertion

Assertion is true but reason is false

Both the assertion and reason are false

Reason is true but assertion is false

Solution

Question 7

[$$\displaystyle M{ L }^{ 8 }{ T }^{ -2 }$$]

[$$\displaystyle M{ L }^{ 6 }{ T }^{ -2 }$$]

[$$\displaystyle M{ L }^{ 4 }{ T }^{ -2 }$$]

[$$\displaystyle M{ L }^{ 2 }{ T }^{ -2 }$$]

Solution

Question 8

$$ 0.1 $$

$$ 0.2 $$

$$ 0.3 $$

$$ 0.4 $$

Question 9

$$\dfrac{W_{1}+W_{2}}{2}$$

$$\sqrt { { W }_{ 1 }{ W }_{ 2 } }$$

$$\sqrt { { W }_{ 1 }^{ 2 }{ +W }_{ 2 }^{ 2 } }$$

$$\sqrt { { W }_{ 2 }^{ 2 }{ -W }_{ 1 }^{ 2 } }$$

Solution

Question 10

$${MLT}^{-1}$$

$${ML}^{2}{T}^{-2}$$

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

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

Solution

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