Units and Measurements Test

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Units and Measurements Test - 54

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Question 1
1 / -0

Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would be

A
$$[ML^2T^{-3}I^{-1}]$$

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

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

D
$$[ML^2T^{-3}I^{-2}]$$

Solution

Ohm's law: $$R = \dfrac{V}{I}$$
Also, $$W = Vq$$$$\implies V = \dfrac{W}{q}$$ (where, $$q = It$$)
$$\implies$$ $$R = \dfrac{W}{q I} = \dfrac{W}{I^2 t}$$

Dimensions of work done, $$W = [M L^2 T^{-2}]$$
$$\therefore$$ Dimension of resistance, $$R = \dfrac{[ML^2 T^{-2}]}{I^2 T} = [M L^2 T^{-3} I^{-2}]$$
Question 2
1 / -0

In which of the following pairs, the two physical quantities have different dimensions?

A
Planck's constant and angular momentum

B
Impulse and linear momentum

C
Moment of inertia and moment of a force

D
Energy and torque

Solution

(A) : Using $$L = mvr = \dfrac{nh}{2\pi}$$
Thus the angular momentum $$(L)$$ has the same dimensions as that of Planck's constant $$(h)$$

(B) : Impulse $$I = \Delta P$$
Thus impulse $$(I)$$ has the same dimensions as that of linear momentum $$(P)$$

(C) : Using $$\tau = I\alpha$$ $$\implies$$ $$I = \dfrac{\tau}{\alpha}$$
where angular acceleration is not dimensionless.
Thus the dimensions of moment of inertia are different from those of moment of force
( OR torque)

(D) : Dimension of energy $$E = [ML^2 T^{-2}]$$
Dimension of torque $$\tau = Fr = [ML^2 T^{-2}]$$
Hence, torqua and energy have same dimensions.
Question 3
1 / -0

The dimensional formula of farad is

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

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

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

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

Solution

$$\left [C\right ] = \left [\dfrac {Q}{V}\right ] = \left [\dfrac {Q^{2}}{W}\right ] = \left [M^{-1} L^{-2} T^{2} Q^{2}\right ]$$
Question 4
1 / -0

What is the dimensions of impedance?

A
$$[ML^{2} T^{-3} I^{-2}]$$

B
$$[M^{-1}L^{-2} T^{3} I^{2}]$$

C
$$[ML^{3} T^{-3} I^{-2}]$$

D
$$[M^{-1}L^{-3} T^{3} I^{2}]$$

Solution

Impedance is same as resistance but in ac circuit

$$\therefore$$ Dimension of impedance $$= \dfrac {\text {dimension of voltage}}{\text {dimension of current}}$$ $$= \dfrac {\left [V\right ]}{\left [I\right ]} = \dfrac {\left [ML^{2} T^{-3} I^{-1}\right ]}{I} = \left [ML^{2} T^{-3} I^{-2}\right ]$$
Question 5
1 / -0

The comparison of an unknown quantity with a known constant quantity is known as :

A
Unit

B
Scale

C
Magnitude

D
Measurement

Solution

The standard quantity (constant quantity) used for comparison is called unit. Measurement is the comparison of an unknown quantity with a known standard quantity (constant quantity) or unit.
Question 6
1 / -0

On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between.

A
$$\displaystyle \frac { 290 }{ 12.5 } $$ and $$\displaystyle \frac { 290 }{ 11.5 } $$

B
$$\displaystyle \frac { 295 }{ 12 } $$ and $$\displaystyle \frac { 285 }{ 11.5 } $$

C
$$\displaystyle \frac { 285 }{ 12 } $$ and $$\displaystyle \frac { 295 }{ 12 } $$

D
$$\displaystyle \frac { 285 }{ 12.5 } $$ and $$\displaystyle \frac { 295 }{ 11.5 } $$

E
$$\displaystyle \frac { 295 }{ 12.5 } $$ and $$\displaystyle \frac { 285 }{ 11.5 } $$

Solution

Here, the rounding off of numbers is tested.
Distance rounded to nearest 10 miles = 290.
So actual distance covered may be b/w 285 and 295.
Gasoline used rounded to nearest gallon = 12.
So the actual gas used may be b/w 11.5 and 12.5.
Now to get range of of miles/gallon,
least value = the least of distance/the max of gas = 285/12.5
highest value = max of distance/least of gas = 295/11.5
Option D is the correct choice.
Question 7
1 / -0

The dimensions of Stefan's constant are

A
$$[M^0LT^{-3}K^{-4}]$$

B
$$[MLT^{-3}K^{-3}]$$

C
$$[ML^2T^{-3}K^{-4}]$$

D
$$[ML^0T^{-3}K^{-4}]$$

Solution

Rate of heat radiated is given by, $$R = \sigma A e T^4$$
where, $$\sigma$$ is the Stefan's constant, $$A$$ is the area and $$T$$ is the temperature
$$e$$ is the emissivity of the body
$$\implies \sigma = \dfrac{R}{AeT^4}$$
Dimension of power, $$R = \dfrac{Energy}{Time} = [ML^2T^{-3}]$$
Emissivity is dimensionless.
Dimension of area, $$A = [M^0 L^2 T^0]$$
Dimension of temperature, $$T = [K]$$
$$\therefore$$ Dimension of, $$\sigma = \dfrac{[ML^2T^{-3}]}{[M^0 L^2T^0] \times [K^4]}$$

$$\implies $$ Dimension of Stefan's constant $$\sigma = [M L^0 T^{-3} K^{-4}]$$
Question 8
1 / -0

Dimensions of coefficient of viscosity is

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

B
$$\left [ML^{-3} T^{-4}\right ]$$

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

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

Solution
Question 9
1 / -0

The dimension of angular momentum is

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

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

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

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

Solution

Dimension of angular momentum
$$L = r\times p$$
$$= [L] [MLT^{-1}] $$
$$= [M^1L^{2} T^{-1}]$$
Question 10
1 / -0

The energy (E), angular momentum (L) and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula of Planck's constant (h) is

A
Zero

B
$$-1$$

C
$$\dfrac{5}{3}$$

D
$$1$$

Solution

$$h={ G }^{ x }{ L }^{ y }{ E }^{ z }$$
$$\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -1 } \right]= { \left[ { M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right] }^{ x }{ \left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -1 } \right] }^{ y }{ \left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right] }^{ z }$$

Comparing the power, we get
$$1=-x+y+z$$ ...(i)
$$2=3x+2y+2z$$ ...(ii)
$$-1=-2x-y-2z$$ ...(iii)
On solving equations (i), (ii) and (iii), we get
$$x=0$$

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Units and Measurements Test - 54

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Units and Measurements Test - 54

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Question 1

$$[ML^2T^{-3}I^{-1}]$$

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

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

$$[ML^2T^{-3}I^{-2}]$$

Solution

Question 2

Planck's constant and angular momentum

Impulse and linear momentum

Moment of inertia and moment of a force

Energy and torque

Solution

Question 3

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

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

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

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

Solution

Question 4

$$[ML^{2} T^{-3} I^{-2}]$$

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

$$[ML^{3} T^{-3} I^{-2}]$$

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

Solution

Question 5

Unit

Scale

Magnitude

Measurement

Solution

Question 6

$$\displaystyle \frac { 290 }{ 12.5 } $$ and $$\displaystyle \frac { 290 }{ 11.5 } $$

$$\displaystyle \frac { 295 }{ 12 } $$ and $$\displaystyle \frac { 285 }{ 11.5 } $$

$$\displaystyle \frac { 285 }{ 12 } $$ and $$\displaystyle \frac { 295 }{ 12 } $$

$$\displaystyle \frac { 285 }{ 12.5 } $$ and $$\displaystyle \frac { 295 }{ 11.5 } $$

$$\displaystyle \frac { 295 }{ 12.5 } $$ and $$\displaystyle \frac { 285 }{ 11.5 } $$

Solution

Question 7

$$[M^0LT^{-3}K^{-4}]$$

$$[MLT^{-3}K^{-3}]$$

$$[ML^2T^{-3}K^{-4}]$$

$$[ML^0T^{-3}K^{-4}]$$

Solution

Question 8

$$\left [MT^{2}\right ]$$

$$\left [ML^{-3} T^{-4}\right ]$$

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

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

Solution

Question 9

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

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

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

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

Solution

Question 10

Zero

$$-1$$

$$\dfrac{5}{3}$$

$$1$$

Solution

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