Units and Measurements Test

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Units and Measurements Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Dimension of specific heat is:

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

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

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

D
$${ M^{-1}L }^{ 2 }{ T }^{ -2 }{ K }^{ -1 }$$

Solution

$$Q=ms\delta\theta$$
Dimension for specific heat will be $$=\dfrac{ML^2T^{-2}}{MK}$$$$=M^0L^2T^{-2}K^{-1}$$
Question 2
1 / -0

The dimensional formula for latent heat is :

A
$${ M^1LT }^{ -2 }$$

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

C
$${ M }^{ 0 }{ L }^{ 2 }{ T }^{ -2 }$$

D
$${ M^1L^1T }^{ -1 }$$

Solution

$$\textbf {Hint :}$$ Use latent heat $$L$$ =$$\dfrac{Q}{m}$$
$${\textbf{Explanation:}}$$
Latent heat, $$L$$ =$$\dfrac{Q}{m}=\dfrac{Energy\ released\ or\ absorbed}{Mass}$$
Dimensional formula of $$L$$ $$=\dfrac{M^{1}L^{2}T^{-2}}{M^{1}}$$
$$=M^{0}L^{2}T^{-2}$$

$${\textbf{Correct option: C}}$$
Question 3
1 / -0

Dimensional formula of magnetic field of induction (B) is:
Given Force on a charged particle is given by: $$F = qvBsin \theta$$ where $$\theta$$ is angle between velocity and magnetic field.

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

B
$$MLT^{2}I^{-1}$$

C
$$MT^{-2}I^{-1}$$

D
$$MT^{2}I^{1}$$

Solution

It is given:
$$F=qvBsin \theta$$
Hence, $$B= \dfrac{F}{qvsin\theta}$$
Dimensional formula of magnetic field is:
$$[B] = [F/qv]$$ as $$sin \theta$$ is dimensionless
$$[B] = [MLT^{-2}][I^{-1}T^{-1}][L^{-1}T]$$
$$[B] = [MT^{-2}I^{-1}]$$
Question 4
1 / -0

Match the following
Column-I Column-II
A. Pressure E. $$ML^{-1} T^{-2}$$
B. Stress F. $$Nm^{-2}$$
C. Energy per unit
volume
G. $$M^{0} L^{0} T^{0}$$
D. Strain H. $$Jm^{-3}$$

A
A - E, F, H : B - E, F, H : C - E, F, H : D - G

B
A - E, F : B - E,H : C - E, F, H : D - G

C
A - E, H : B - E, F : C - E, F, H : D - G

D
A - E, F, H : B - E, F : C - E, F, H : D - G

Solution

Stress $$= Pressure = \dfrac{force}{area}=\dfrac{N}{M^{2}}=\dfrac{M^{1}L^{1}T^{-2}}{L^{2}}=M^{1}L^{-1}T^{-2}$$

Energy per unit volume $$= \dfrac{M^{1}L^{2}T^{-2}}{L^{3}}=M^{1}L^{-1}T^{-2}$$

Strain $$= \dfrac{\Delta l}{l}=M^{0}L^{0}T^{0}$$
Hence, A, B, & C each has E, F, & H options dimensionally
while D has only G.
Hence, option A is correct.
Question 5
1 / -0

If $$L$$ is the inductance, $$'i'$$ is current in the circuit, $$\dfrac { 1 }{ 2 } { Li }^{ 2 }$$ has the dimensions of:

A
Work

B
Power

C
Pressure

D
Force

Solution

$$L \rightarrow M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }$$
$$ I\rightarrow I$$
$${ LI }^{ 2 } = (M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }){ I }^{ 2 }= M{ L }^{ 2 }{ T }^{ -2 } \rightarrow (unit \ of \ work)$$
$$Power = f\times v = ML{ T }^{ -2 } \times L{ T }^{ -1 }\quad =\quad M{ L }^{ 2 }{ T }^{ -3 }$$
$$ Pressure = f/A = \dfrac { ML{ T }^{ -2 } }{ { L }^{ 2 } } = { ML }^{ -1 }{ T }^{ -2 }$$
$$force = { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }$$
Question 6
1 / -0

The units of force, velocity and energy are 100 dyne, 10 $${cm s }^{ -1 }$$ and 500 erg respectively. The units of mass, length and time are:

A
5 g, 5 cm, 5 s

B
5 g, 5 cm, 0.5 s

C
0.5 g, 5 cm, 5 s

D
5 g, 0.5cm, 5 s

Solution

Force $$= 100 dyne= M^{1}L^{1}T^{-2}$$
$$velocity = 10 \dfrac{cm}{s}=LT^{-1}$$
$$Energy = 500 erg =M^{1}L^{2}T^{-2}\Rightarrow \dfrac{energy}{force}=L^{1}=\dfrac{500}{100}=5cm$$
$$\Rightarrow \dfrac{energy}{(velocity)^{2}}=M^{1}=\dfrac{500}{(10)^{2}} =5gm \Rightarrow T=0.5\ sec$$
Question 7
1 / -0

The unit of an electrical parameter whose formula is $$[M^{1}L^{2}T^{-3}A^{-2}]$$ is:

A
ohm

B
ampere

C
volt

D
newton

Solution

$$ V = IR$$

So, $$ [R] = [\dfrac{V}{I}]$$$$ = [\dfrac{W}{QI} ]= [\dfrac{W}{I^{2} T}] = [\dfrac{F \times S}{I^{2} \times T}] = [\dfrac{m \times a \times S}{I^{2} \times T}]$$$$ = [\dfrac{M(\dfrac{L}{T^{2}}) \times L}{A^{2} \times T}]$$

$$\Rightarrow [\Omega]= [ML^{2}T^{-3}A^{-2}]$$
Question 8
1 / -0

If $$e, { \epsilon}_{0}, h$$ and $$c$$ respectively represents electric charge, permittivity of free space, Planck's constant and speed of light. The dimension of $$\dfrac{e^2 }{ 4\pi \epsilon_0 hc} $$ is :

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

B
$$[M^1 L^0 T^0] $$

C
$$[M^0 L^1 T^0] $$

D
$$[M^0 L^0 T^1] $$

Solution

Planck's constant (h) has joule-seconds unit, dimensions are $$=[ML^2T^{-1}]$$
Electron charge (e) will have $$[TI]$$
Dimension for $${ \varepsilon }_{ 0 }$$ is $$[M^{-1}L^{-3}T^{4}I^{2}]$$
Dimensions for $$\dfrac { e^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }hc } $$ will be $$[M^0L^0T^0]$$.
Question 9
1 / -0

The product of energy and time is called action. The dimensional formula for action is same as that for

A
$$force \times velocity$$

B
$$impulse \times distance$$

C
$$power$$

D
$$angular \times energy$$

Solution

$$ Energy   \times   Time = \left ( M^{1}L^{2}T^{-2} \right )\times \left ( T^{1} \right )= M^{1}L^{2}T^{-1}$$

$$ Force \times   Velocity = \left ( M^{1}L^{1}T^{-2} \right )\times \left ( L^{1}T^{-1} \right ) = M^{1}L^{2}T^{-3}$$

$$ Impulse \times   distance = \left ( M^{1}L^{1}T^{-1} \right )\times \left ( L^{1} \right )   = M^{1}L^{2}T^{-1}$$

$$ Power = M^{1}L^{2}T^{-3}$$

$$ Angular \ Energy = M^{1}L^{1}T^{-2} $$
Question 10
1 / -0

A highly rigid cubical block $$A$$ of small mass $$M$$ and side $$L$$ is fixed rigidly on to another cubical block of same dimensions and of low modulus of rigidity $$\eta$$ such that the lower face of $$A$$ completely covers the upper face $$B$$. The lower face of $$B$$ is rigidly held on a horizontal surface. A small force $$F$$ is applied perpendicular to one of the side face of $$A$$. After the force is withdrawn, block $$A$$ executes small oscillations, the time period of which is given by :

A
$$2\pi \sqrt{M\eta L}$$

B
$$2\pi \sqrt{\dfrac{M\eta}{ L}}$$

C
$$2\pi \sqrt{\dfrac{ML}{\eta}}$$

D
$$2\pi \sqrt{\dfrac{M}{L\eta}}$$

Solution

Modulus of rigidity $$\eta =\dfrac { E }{ A\theta } $$

We know, $$A={ L }^{ 2 }$$

For small $$\theta$$,
$$\theta =\dfrac { x }{ L } $$

Restoring force $$F=-\eta AB$$

$$F=-\eta Lx$$

Acceleration, $$a=-\dfrac { \eta L }{ M } x\qquad \rightarrow (1)\\$$
$$ a=\dfrac { F }{ m } $$

Here, $$\alpha \propto -x\\$$

$$\therefore$$here we have SHM,

Hence the time period is given by

$$T=2x\sqrt { \left| \dfrac { x }{ a } \right| } \\ T=2\pi \sqrt { \dfrac { M }{ \eta L } } $$

where,
$$a$$=acceleration and $$x$$=small deformation.

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Re-Attempt Weekly Quiz Competition

Column-I	Column-II
A. Pressure	E. $$ML^{-1} T^{-2}$$
B. Stress	F. $$Nm^{-2}$$
C. Energy per unit volume	G. $$M^{0} L^{0} T^{0}$$
D. Strain	H. $$Jm^{-3}$$

Units and Measurements Test - 44

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Units and Measurements Test - 44

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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

$${ M^1LT }^{ -2 }$$

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

$${ M }^{ 0 }{ L }^{ 2 }{ T }^{ -2 }$$

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

Solution

Question 3

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

$$MLT^{2}I^{-1}$$

$$MT^{-2}I^{-1}$$

$$MT^{2}I^{1}$$

Solution

Question 4

A - E, F, H : B - E, F, H : C - E, F, H : D - G

A - E, F : B - E,H : C - E, F, H : D - G

A - E, H : B - E, F : C - E, F, H : D - G

A - E, F, H : B - E, F : C - E, F, H : D - G

Solution

Question 5

Work

Power

Pressure

Force

Solution

Question 6

5 g, 5 cm, 5 s

5 g, 5 cm, 0.5 s

0.5 g, 5 cm, 5 s

5 g, 0.5cm, 5 s

Solution

Question 7

ohm

ampere

volt

newton

Solution

Question 8

$$[M^0 L^0 T^0] $$

$$[M^1 L^0 T^0] $$

$$[M^0 L^1 T^0] $$

$$[M^0 L^0 T^1] $$

Solution

Question 9

$$force \times velocity$$

$$impulse \times distance$$

$$power$$

$$angular \times energy$$

Solution

Question 10

$$2\pi \sqrt{M\eta L}$$

$$2\pi \sqrt{\dfrac{M\eta}{ L}}$$

$$2\pi \sqrt{\dfrac{ML}{\eta}}$$

$$2\pi \sqrt{\dfrac{M}{L\eta}}$$

Solution

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