Units and Measurements Test

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Units and Measurements Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

The physical quantity having dimensions $$2$$ in length is:

A
Power

B
Acceleration

C
Force constant

D
Stress

Solution

$$Power = ML^{2} T^{-3}$$
$$Acceleration = LT^{-2}$$
$$force \ constant = M^{1}L^{0}T^{-2}$$
$$Stress = F/A = ML^{-1}T^{-2}$$
Hence, option A is correct.
Question 2
1 / -0

The fundamental physical quantities that have the same dimension in the dimensional formula of Torque and Angular Momentum are:

A
mass, time

B
time, length

C
mass, length

D
time, mole

Solution

Torque$$\rightarrow M^{1}L^{2}T^{-2}$$
Angular momentum$$\rightarrow M^{1}L^{2}T^{-1}$$
Mass and length have same dimensions.
Question 3
1 / -0

The physical quantity which has the dimensional formula $${ M }^{ 1 }{ T }^{ -3 }$$ is:

A
surface tension

B
solar constant

C
density

D
compressibility

Solution

Surface tension $$\displaystyle \rightarrow \frac{F}{l}=\frac{MLT^{-2}}{L}=MT^{-2}$$

Solar constant $$=MT^{-3}$$

Density $$=\dfrac{Mass}{Vol}=ML^{-3}$$
Compressibillity $$=\dfrac{1}{Bulk \ Modulus}=\dfrac{\dfrac{dv}{v}}{dp}=\dfrac{L^{3}/L^{3}}{ML^{-1}T^{-2}}=M^{-1}L^{1}T^{2}$$

Hence, option B is correct.
Question 4
1 / -0

The unit of impulse per unit area is same as that of:

A
coefficient of viscosity

B
surface tension

C
bulk modulus

D
none of the above

Solution

$$\displaystyle \frac { Impulse }{ Area } = \frac { f\times t }{ Area } =\frac { [{ M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }]\times [{ T }^{ 1 }] }{ [{ L }^{ 2 }] } =[ M{ L }^{ -1 }{ T }^{ -1 }]$$

$$Viscocity = \dfrac{stress}{\dfrac{dv}{dy}} = [M{ L }^{ -1 }{ T }^{ -1 }]\quad $$

$$Surface \ Tension = \dfrac{f}{l}= \dfrac {[ M{ L }^{ 1 }{ T }^{ -2 } ]}{ [{ L }^{ 1 }] } = { [MT }^{ -2 }]$$

$$Bulk \ Modulus = \dfrac { dP }{ dV/v } =[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 }]\quad $$

Hence, option A is correct.
Question 5
1 / -0

Given $$M$$ is the mass suspended from a spring of force constant $$k$$. The dimensional formula for $$\left[ M/k \right] ^{ 1/2 }$$ is same as that for:

A
frequency

B
time period

C
velocity

D
wavelength

Solution

$$ K = \dfrac{f}{x} = \dfrac { { MLT }^{ -2 } }{ L } ={ MT }^{ -2 }$$

$$\displaystyle \Rightarrow \frac { M }{ K } = \frac { { M }^{ 1 } }{ { M }^{ 1 }{ T }^{ -2 } } = { T }^{ 2 }$$

$$\Rightarrow \sqrt { M/K } \quad =\quad { T }^{ 1 }$$

Hence, option B is correct.
Question 6
1 / -0

$$\dfrac { 1 }{ \sqrt { Capacitance\times Inductance } } $$ has the same unit as:

A
time

B
velocity

C
velocity gradient

D
none of the above

Solution

$$C=Capacitance = { M }^{ -1 }{ L }^{ -2 }{ I }^{ 2 }{ T }^{ Y }$$
$$ I=Inductance =M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }$$

$$\dfrac { 1 }{ \sqrt { CI } } = \dfrac { I }{ \sqrt { \left( { M }^{ -1 }{ L }^{ -2 }{ I }^{ 2 }{ T }^{ 4 } \right) \times \left( M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 } \right) } } $$

$$\displaystyle \frac { 1 }{ \sqrt { CI } } =\frac { 1 }{ T } = { T }^{ -1 }$$

$$Time = { T }^{ 1 }, Velocity = { LT }^{ -1 }$$

$$Velocity _{grad} = \dfrac{dv}{dx} = \dfrac { { LT }^{ -1 } }{ L } = { T }^{ -1 } $$

Hence, option C is correct.
Question 7
1 / -0

Dimensions of solar constant are:

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

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

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

D
$$\left[ { M }^{ 1 }{ T }^{ -3 } \right] $$

Solution

Solar constant(S) is defined as energy due to sun falling on an unit area per unit time .
$$\Rightarrow \displaystyle S=\left ( \frac{E}{A} \right )\times \frac{1}{T}=\left ( \frac{ML^{2}}{T^{2}} \right )\times \frac{1}{L^{2}}\times \frac{1}{T}=MT^{-3}$$
Question 8
1 / -0

If the unit of work is $$100J$$, the unit of power is $$1KW$$, the unit of time in second is:

A
$$10^{ -1 }$$

B
$$10$$

C
$$10^{ -2 }$$

D
$$10^{ -3 }$$

Solution

Power $$=1000=M^{1}L^{2}T^{-3}$$
Work $$=100=M^{1}L^{2}T^{-2}$$
Work $$=power \times time$$
$$\displaystyle time=\frac{work}{power}=T^{1}=\frac{100}{1000}=\frac{1}{10}$$
Hence, Option A is correct.
Question 9
1 / -0

According to theory of significant figures $$\left( 2.0 \right) ^{ 10 }$$ is :

A
1024

B
$$1.024\times { 10 }^{ 10 }$$

C
$$1.0\times { 10 }^{ 3 }$$

D
1000

Solution

$$\left ( 2.0 \right )^{10}=1024$$
But since 2.0 contains 2 significant figure.
So, $$1024=1.0\times 10^{3}$$
Question 10
1 / -0

Dimensions of '$$ohm$$' are same as that of:
(Given : $$h$$ is Planck's constant and $$e$$ is charge)

A
$$\dfrac { h }{ e } $$

B
$$\dfrac { { h }^{ 2 } }{ e } $$

C
$$\dfrac { h }{ { e }^{ 2 } } $$

D
$$\dfrac { { h }^{ 2 } }{ { e }^{ 2 } } $$

Solution

$$a = ohm = ML^{2}I^{-2}T^{-3}$$
$$b = h = ML^{2}T^{-1}$$
$$c = e = I^{1}T^{1}$$
$$a = b^{\alpha } c^{\beta }$$
$$ML^{2}I^{-2}T^{-3} = (ML^{2}T^{-1})^{\alpha } (I^{1}T^{1})^{\beta} $$
$$\Rightarrow \beta = -2, \alpha = 1$$
$$ \Rightarrow ohm = \dfrac{h}{e^{2}}$$

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Units and Measurements Test - 25

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Units and Measurements Test - 25

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SHARING IS CARING

Question 1

Power

Acceleration

Force constant

Stress

Solution

Question 2

mass, time

time, length

mass, length

time, mole

Solution

Question 3

surface tension

solar constant

density

compressibility

Solution

Question 4

coefficient of viscosity

surface tension

bulk modulus

none of the above

Solution

Question 5

frequency

time period

velocity

wavelength

Solution

Question 6

time

velocity

velocity gradient

none of the above

Solution

Question 7

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

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

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

$$\left[ { M }^{ 1 }{ T }^{ -3 } \right] $$

Solution

Question 8

$$10^{ -1 }$$

$$10$$

$$10^{ -2 }$$

$$10^{ -3 }$$

Solution

Question 9

1024

$$1.024\times { 10 }^{ 10 }$$

$$1.0\times { 10 }^{ 3 }$$

1000

Solution

Question 10

$$\dfrac { h }{ e } $$

$$\dfrac { { h }^{ 2 } }{ e } $$

$$\dfrac { h }{ { e }^{ 2 } } $$

$$\dfrac { { h }^{ 2 } }{ { e }^{ 2 } } $$

Solution

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