Units and Measurements Test

Result

Units and Measurements Test - 59

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The dimensional formula for Young's modulus is :

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

B
$$ [ M^{0} LT^{-2} ] $$

C
$$ [ MLT^{-2} ] $$

D
$$ [ ML^{2} T^{-2} ] $$

Solution

Young's modulus
$$ Y = \dfrac {stress}{strain} $$
Dimension of $$ Y = \dfrac {ML^{-1} T^{-2} }{M^0L^0T^0} $$
$$ = [ ML^{-1} T^{-2} ] $$
Question 2
1 / -0

The dimensional formula of modulus of elasticity is

A
$$\displaystyle \left[ { ML }^{ -1 }{ T }^{ -2 } \right] $$

B
$$\displaystyle \left[ { M }^{ 0 }{ LT }^{ -2 } \right] $$

C
$$\displaystyle \left[ { MLT }^{ -2 } \right] $$

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

Solution

Modules of elasticity $$\displaystyle =\frac { stress }{ strain } $$
Stress $$\displaystyle =\frac { \left[ F \right] }{ \left[ A \right] } =\frac { \left[ { MLT }^{ -2 } \right] }{ \left[ { L }^{ 2 } \right] } =\left[ { ML }^{ -1 }{ T }^{ -2 } \right] $$
Strain is a dimensionless quantity.
So, modules of elasticity = $$\displaystyle \left[ { ML }^{ -1 }{ T }^{ -2 } \right] $$
Question 3
1 / -0

The quantity which has the same dimensions as that of gravitational potential is

A
latent heat

B
impulse

C
angular acceleration

D
planck's constant

E
specific heat capacity

Solution

Dimensions of gravitational potential
$$=\dfrac{Dimension \,of\, work}{Dimension \,of\, mass}$$
$$=\dfrac{[ML^2T^{-2}]}{[M]}=[L^2T^{-2}]$$
and dimension of latent heat
$$=\dfrac{Dimension \,of \,[Q]}{Dimension \,of\, [M]}$$
$$=\dfrac{[ML^2T^{-2}]}{[M]}=[L^2T^{-2}]$$
Question 4
1 / -0

The frequency of vibration of the string is given by
$$v=\dfrac {p}{2l}\left [ \dfrac {F}{m} \right ]^{1/2}$$
Here, p is the number of segments in the string and l is the length. The dimensional formula for m will be

A
$$[M^0LT^{-1}]$$

B
$$[ML^0T^{-1}]$$

C
$$[ML^{-1}T^{0}]$$

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

Solution

$$v=\dfrac{p}{2l}\left [ \dfrac{F} {M}\right ]^{1/2}$$
Squaring the equation on either side, we have
$$v^2=\dfrac{p^2}{4l^2}\left ( \dfrac{F} {M}\right )$$
$$m=\dfrac {P^2F}{4l^2v^2}$$
$$[m]=\dfrac {[MLT^{-2}]}{[L^2][T^{-1}]^2}=[ML^{-1}T^0]$$
Question 5
1 / -0

The physical quantity that does not have the dimensional formula $$[ML^{-1}T^{-2}]$$ is

A
force

B
pressure

C
stress

D
modulus of elasticity

E
energy density

Solution

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

$$Pressure=Force/Area$$
$$P=[MLT^{-2}][L^{-2}]=[ML^{-1}T^{-2}]$$

$$Stress$$ is force per unit area therefore it's dimensional formula will be same as pressure.

$$Modulus \ of \ elasticity=Fl/A \triangle l=[MLT^{-2}] [L][L^{-2}][L^{-1}]=[ML^{-1}T^{-2}]$$

Hence $$D$$ is correct answer.
Question 6
1 / -0

The dimensional formula of $$\dfrac {1}{\mu_{0}\epsilon_{0}}$$ is

A
$$[M^{0}LT^{-2}]$$

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

C
$$[M^{0}LT^{-1}]$$

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

Solution

Velocity of $$EM$$ waves, $$v = \dfrac {1}{\sqrt {\mu_{0}\epsilon_{0}}}$$
$$\Rightarrow = \dfrac {1}{\mu_{0}\epsilon_{0}} = v^{2}$$
$$\therefore$$ Dimensional formula of $$\dfrac {1}{\mu_{0}\epsilon_{0}} = [M^{0}LT^{-1}]^{2}$$
$$= [M^{0}L^{2}T^{-2}]$$.
Question 7
1 / -0

Number of particles is given by $$n=-D\cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 2 } } $$ crossing a unit area perpendicular to X-axis in unit time, where $${ n }_{ 1 }$$ and $${ n }_{ 2 }$$ are number of particles per unit volume for the value of $$x$$ meant to $${x}_{2}$$ and $${x}_{1}$$. Find dimensions of $$D$$ called as diffusion constant :

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

B
$$\left[ { M }^{ 0 }{ L }^{ 2 }{ T }^{ -4 } \right] $$

C
$$\left[ { M }^{ 0 }{ L }^{ }{ T }^{ -3 } \right] \quad $$

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

Solution

Given, number of particle passing from unit area in unit time $$=n$$
$$=\cfrac { Number\quad of\quad particle }{ A\times t } =\cfrac { \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] }{ \left[ { L }^{ 2 } \right] \left[ T \right] } =\left[ { L }^{ -2 }{ T }^{ -1 } \right] $$
Now from the given formula $$\left[ D \right] =\cfrac { \left[ n \right] \left[ { x }_{ 2 }-{ x }_{ 1 } \right] }{ \left[ { n }_{ 2 }-{ n }_{ 1 } \right] } =\left[ { L }^{ 2 }{ T }^{ -1 } \right] $$
Question 8
1 / -0

If $${ \varepsilon }_{ 0 }$$ 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

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

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

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

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

Solution

Here, $$\cfrac { 1 }{ 2 } { \varepsilon }_{ 0 }{ E }^{ 2 }\quad $$ represetns energy per unit volume
So the dimensions of
$$\left[ { \varepsilon }_{ 0 } \right] \left[ { E }^{ 2 } \right] =\cfrac { \left[ Dimensions\quad of\quad Energy \right] }{ \left[ Dimensions\quad of\quad Volume \right] } $$
$$=\dfrac { \left[ M{ L }^{ 2 }{ T }^{ -2 } \right] }{ \left[ { L }^{ 3 } \right] } =\left[ M{ L }^{ -1 }{ T }^{ -2 } \right] $$
Question 9
1 / -0

For a circuit, L, C and R represent the inductance, capacitance and resistance. Then, the dimensional formula for $$C^2LR$$ is

A
$$[MLT]$$

B
$$[M^0L^0TI^0]$$

C
$$[M^2L^2T^0I^0]$$

D
$$[M^0L^0T^3I^0]$$

Solution

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
$$C^2LR=[C^2L^2]\times \left [ \frac {L}{R} \right ]=[T^4]\times \left [ \frac {1}{T} \right ]=[T^3]$$
Question 10
1 / -0

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

A
$$\cfrac { 1 }{ \sqrt { RC } } $$

B
$$\cfrac { R }{ L } $$

C
$$\cfrac { 1 }{ LC } $$

D
$$\cfrac { C }{ L } $$

Solution

$$\dfrac{L}{R}$$ 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).

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Units and Measurements Test - 59

Result

Units and Measurements Test - 59

Score

-

Rank

-

SHARING IS CARING

Question 1

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

$$ [ M^{0} LT^{-2} ] $$

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

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

Solution

Question 2

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

$$\displaystyle \left[ { M }^{ 0 }{ LT }^{ -2 } \right] $$

$$\displaystyle \left[ { MLT }^{ -2 } \right] $$

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

Solution

Question 3

latent heat

impulse

angular acceleration

planck's constant

specific heat capacity

Solution

Question 4

$$[M^0LT^{-1}]$$

$$[ML^0T^{-1}]$$

$$[ML^{-1}T^{0}]$$

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

Solution

Question 5

force

pressure

stress

modulus of elasticity

energy density

Solution

Question 6

$$[M^{0}LT^{-2}]$$

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

$$[M^{0}LT^{-1}]$$

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

Solution

Question 7

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

$$\left[ { M }^{ 0 }{ L }^{ 2 }{ T }^{ -4 } \right] $$

$$\left[ { M }^{ 0 }{ L }^{ }{ T }^{ -3 } \right] \quad $$

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

Solution

Question 8

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

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

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

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

Solution

Question 9

$$[MLT]$$

$$[M^0L^0TI^0]$$

$$[M^2L^2T^0I^0]$$

$$[M^0L^0T^3I^0]$$

Solution

Question 10

$$\cfrac { 1 }{ \sqrt { RC } } $$

$$\cfrac { R }{ L } $$

$$\cfrac { 1 }{ LC } $$

$$\cfrac { C }{ L } $$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.