Units and Measurements Test

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Units and Measurements Test - 51

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Question 1
1 / -0

Dimension of electrical resistance is :

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

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

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

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

Solution

We know that Power, $$P=I^2R$$
Dimensions of Power are that of Force times velocity, i.e. $$MLT^{-2}\times LT^{-1}=ML^2T^{-3}$$
Thus, $$ML^2T^{-3}=A^2|R|$$, or $$|R|=ML^2T^{-3}A^{-2}$$
Question 2
1 / -0

The velocity $$v$$ of a particle at time$$ t$$ is given by $$\displaystyle v = at + \frac {b} {t + c} $$ where $$a, b$$ and $$c$$ are constants. The dimensions of $$a, b$$ and $$c$$ are respectively :

A
$$ LT^{-2}\, , L $$ and $$\displaystyle T $$

B
$$ L^{2}\, , T $$ and $$\displaystyle LT^{2} $$

C
$$ LT^{2}\, , LT $$ and $$\displaystyle L $$

D
$$ L\, , LT $$ and $$\displaystyle T^{2} $$

Solution

According to the principle of dimensional homogenity
Dimension of v = dimension of at = dimension of $$\displaystyle \frac {b} {t +c} $$
Dimension of a = $$\displaystyle \frac {[v]} {[t]} = \frac {LT^{-2}} {T} = [LT^{-2}] $$
Dimension of b = $$\displaystyle [v] [t] = [LT^{-1}T] = [L] $$
Dimension of c = $$\displaystyle [t] = [T] $$
Question 3
1 / -0

Which of the following does not have the dimensions of force ?

A
Gravitational Potential gradient

B
Energy gradient

C
Weight

D
Rate of change of momentum

Solution

The gradient of the gravitational potential gradient is the gravitational electric field (i.e. force acting on unit mass). $$\overset { \rightarrow }{ E } =-\nabla \phi $$ which has dimensions of $$N/Kg$$
Weight is the force exerted by the gravity. So, it will have dimensions of force.
Energy gradient is simply the force vector. So, it will have dimensions of force.
By newton's 2nd law, Rate of change of momentum is the net force. So, it will have dimensions of force.
Hence, Option A is correct.
Question 4
1 / -0

Two quantities $$A$$ and $$B$$ are related by $$\dfrac{A}{B} = m$$, where $$m$$ is linear density and $$A$$ is force. The dimensions of $$B$$ will be same as that of

A
Pressure

B
Work

C
Momentum

D
Latent heat

Solution

Linear Density, m has dimensions of $$\dfrac{Mass}{Length} = ML^{-1}$$ and Force, A, has dimensions $$MLT^{-2}$$
Thus dimensions of B are $$\dfrac{MLT^{-2}}{ML^{-1}}=L^2T^{-2}$$, which is the same as Latent Heat (which is heat per unit mass, or $$\dfrac{ML^2T^{-2}}{M}=L^2T^{-2}$$)
Question 5
1 / -0

A unitless quantity ?

A
Does not exist

B
Always has a nonzero dimension

C
Never has a nonzero dimension

D
May have a nonzero dimension

Solution

If a quantity is unitless, it can not have dimensions, because dimensions imply units (in terms of M, L, T etc.).
But the converse is not true since a quantity can be dimensionless and still have units (eg. Angular displacement).
Question 6
1 / -0

If the unit of length is doubled the numerical value of Area will become ________ times its original value.

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

B
$$\dfrac{1}{4}$$

C
2

D
4

Solution

As unit of length is doubled unit of Area will become four times. So the numerical value of Area will became one fourth Because numerical value $$\displaystyle \propto\frac{1}{unit}$$
Question 7
1 / -0

External and internal diameters of a hollow cylinder are measured to be $$(4.23 \pm 0.01)$$ cm and $$(3.89 \pm 0.01)$$ cm. The thickness of the wall of the cylinder is

A
$$(0.34 \pm0.03)$$ cm

B
$$(0.17 \pm 0.04)$$ cm

C
$$(0.17 \pm 0.01)$$ cm

D
$$(0.34 \pm 0.02)$$ cm

Solution

External diameter $$d_e = (4.23\pm 0.01) \ cm$$
So external radius $$r_e = \dfrac{d_e}{2} = (2.12\pm 0.005) \ cm$$
Internal diameter $$d_i = (3.89\pm 0.01) \ cm$$
Internal diameter $$r_i = \dfrac{d_i}{2} = (1.95\pm 0.005) \ cm$$
Thickness of the wall $$t = r_e-r_i = (2.12-1.95) \pm (0.005+0.005) \ cm =( 0.17\pm 0.01) \ cm$$
Question 8
1 / -0

The pair of quantities that do not have the same dimensions is

A
Latent heat, specific heat

B
Gravitational force, Coulomb force

C
Kinetic energy of a freely falling body, potential energy of compressed spring

D
Coefficient of friction, number of molecules in a container.

Solution

Hint:
Dimension of all forms of Energy is same.
Dimension of all forms of force is same.
Constants are Dimensionless.

Explanation for the correct answer:
$$\bullet$$ Dimension of Latent heat is, $$[M^0 L^2 T^{-2}]$$. And dimension for specific heat is $$[M^0 L^2 T^{-2}K^{-1}]$$. Thus, they have different dimensions. Option A is correct.
$$\bullet$$ Gravitational force and Coulomb force are two different types of Force. Thus, they have same Dimensions. Option B is incorrect.
$$\bullet$$ Kinetic Energy of freely falling body and Potential Energy of a compressed spring are two different types of Energy. Thus, they have same Dimensions. Option C is incorrect.
$$\bullet$$ Coefficient of friction and number of molecules in a container are two different types of Constant. Thus, they have same Dimensions, they are dimensionless. Option D is incorrect.

Thus, Only Option A is correct.
Question 9
1 / -0

Statement -1: Kinetic energy can be divided by volume
and
Statement -2: Dimensions of kinetic energy and volumeare different

A
Statement -1 is True Statement -2 is True;

Statement -2 is a correct explanation for Statement -1

B
Statement -1 is True Statement -2 is True;

Statement -2 is NOT a correct explanation for Statement -1

C
Statement -1 is True Statement -2 is False .

D
Statement -1 is False Statement -2 is True;

Solution

Division of two Quantities with different dimensions is possible.
Hence, the statement I is correct.
Since dimensions of kinetic energy ($$M{ L }^{ 2 }{ T }^{ -2 }$$) and volume ($${ L }^{ 3 }$$) are different, the statement II is correct but is not an explanation for statement I.
Hence, option B is correct.
Question 10
1 / -0

In the relation $$\displaystyle P = \frac {\alpha} {\beta} e^{-\dfrac {\alpha Z} {k \theta}} $$, $$P$$ is pressure, $$Z$$ is distance, $$k$$ is Boltzmann constant and $$ \theta $$ is the temperature. The dimensional formula of $$ \beta $$ will be :

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

B
$$ [ML^2T] $$

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

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

Solution

Since exponential term is dimensionless,
Dimension of $$\alpha$$, $$\left[ \alpha \right] =\dfrac { \left[ k \right] \left[ \theta \right] }{ \left[ Z \right] } =\dfrac { (M{ L }^{ 2 }{ T }^{ -2 }{ K }^{ -1 })*(K) }{ L } =ML{ T }^{ -2 }$$
Since $$\dfrac { \alpha }{ \beta } $$ must have the dimensions of pressure($$M{ L }^{ -1 }{ T }^{ -2 }$$)
Therefore, dimensions of $$\beta$$, $$\left[ \beta \right] =\dfrac { ML{ T }^{ -2 } }{ M{ L }^{ -1 }{ T }^{ -2 } } ={ L }^{ 2 }$$

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Units and Measurements Test - 51

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Units and Measurements Test - 51

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Question 1

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

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

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

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

Solution

Question 2

$$ LT^{-2}\, , L $$ and $$\displaystyle T $$

$$ L^{2}\, , T $$ and $$\displaystyle LT^{2} $$

$$ LT^{2}\, , LT $$ and $$\displaystyle L $$

$$ L\, , LT $$ and $$\displaystyle T^{2} $$

Solution

Question 3

Gravitational Potential gradient

Energy gradient

Weight

Rate of change of momentum

Solution

Question 4

Pressure

Work

Momentum

Latent heat

Solution

Question 5

Does not exist

Always has a nonzero dimension

Never has a nonzero dimension

May have a nonzero dimension

Solution

Question 6

$$\dfrac{1}{2}$$

$$\dfrac{1}{4}$$

2

4

Solution

Question 7

$$(0.34 \pm0.03)$$ cm

$$(0.17 \pm 0.04)$$ cm

$$(0.17 \pm 0.01)$$ cm

$$(0.34 \pm 0.02)$$ cm

Solution

Question 8

Latent heat, specific heat

Gravitational force, Coulomb force

Kinetic energy of a freely falling body, potential energy of compressed spring

Coefficient of friction, number of molecules in a container.

Solution

Question 9

Statement -1 is True Statement -2 is True; Statement -2 is a correct explanation for Statement -1

Statement -1 is True Statement -2 is True; Statement -2 is NOT a correct explanation for Statement -1

Statement -1 is True Statement -2 is False .

Statement -1 is False Statement -2 is True;

Solution

Question 10

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

$$ [ML^2T] $$

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

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

Solution

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Statement -1 is True Statement -2 is True;

Statement -2 is a correct explanation for Statement -1

Statement -1 is True Statement -2 is True;

Statement -2 is NOT a correct explanation for Statement -1