Question 1: Palette 1/12

DPT MCQ

Concept: Dimensional Analysis

If Gravitational force is given by $F = \frac{G M_1 M_2}{r^2}$, then find the dimensional formula (D.F.) of Gravitational constant $G$.

A

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

B

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

C

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

D

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

Question 2: Palette 2/12

DPT MCQ

Concept: The dimensional formula of a physical quantity is derived from its defining formula. For the coefficient of viscosity ($\eta$), rearranging the formula for viscous force gives $\eta = \frac{F}{6 \pi r v}$. Since $6 \pi$ is a dimensionless constant, the dimensions of $\eta$ depend on force, radius, and velocity.

Find the dimensional formula of the coefficient of viscosity $\eta$, given that the viscous force $F$ is given by $F = 6 \pi \eta r v$, where $r$ is radius and $v$ is velocity.

A

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

B

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

C

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

D

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

Question 3: Palette 3/12

DPT MCQ

Concept: Specific heat capacity is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius or Kelvin. The relation between heat given ($\Delta Q$), mass ($M$), specific heat capacity ($S$), and rise in temperature ($\Delta T$) is given by:
$\Delta Q = M S \Delta T$

What is the dimensional formula of specific heat capacity ($S$)?

A

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

B

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

C

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

D

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

Question 4: Palette 4/12

DPT MCQ

Concept: Electric charge $q$ is defined as the product of electric current $I$ and time $t$. Current is a fundamental physical quantity with dimension $[A]$, and time has dimension $[T]$.

Find the dimensional formula of charge if the formula of charge is equal to current $\times$ time.

A

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

B

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

C

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

D

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

Question 5: Palette 5/12

DPT MCQ

Concept: Electric potential ($V$) is defined as the work done ($W$) per unit electric charge ($q$). Using the relationship $V = \frac{W}{q}$, the dimensions of potential can be derived from the dimensional formulas of work done and electric charge.

Find the dimensional formula of electric potential or voltage.

A

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

B

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

C

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

D

$[M^0 L^2 T^{-3} A^{-1}]$

Question 6: Palette 6/12

DPT MCQ

Concept: Resistance ($R$) is defined by Ohm's law as $R = \frac{V}{I}$, where $V$ is potential difference (Voltage) and $I$ is electric current. Potential difference is the work done per unit charge, $V = \frac{W}{q}$, and charge is given by $q = I \cdot t$.

Find the dimensional formula of resistance.

A

$[\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-1}]$

B

$[\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-2}]$

C

$[\text{M}^1\text{L}^3\text{T}^{-3}\text{A}^{-2}]$

D

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

Question 7: Palette 7/12

DPT MCQ

Concept: According to Ohm's law, resistance $R$ is given by $R = \frac{V}{I}$, where $V$ is electric potential and $I$ is electric current. Electric potential is work done per unit charge, $V = \frac{W}{q}$, and charge is current multiplied by time, $q = I \cdot t$. Combining these gives $R = \frac{W}{I^2 \cdot t}$.

Find the dimensional formula of resistance.

A

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

B

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

C

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

D

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

Question 8: Palette 8/12

DPT MCQ

Concept: Resistivity $\rho$ is calculated by rearranging the formula $R = \rho \frac{l}{\text{Area}}$ to get $\rho = \frac{R \cdot \text{Area}}{l}$. The dimensional formula of resistivity is obtained by combining the dimensions of resistance, area, and length.

If $R = \rho \frac{l}{\text{Area}}$, where $R$ is resistance, $l$ is length, and $\rho$ is resistivity, find the dimensional formula of resistivity.

A

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

B

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

C

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

D

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

Question 9: Palette 9/12

DPT MCQ

Concept: By the principle of homogeneity of dimensions, the dimensions of both sides of a physical equation must be equal. Therefore, $[\text{Kinetic Energy}] = [\alpha] \cdot [s]^2$, which gives $[\alpha] = \frac{[\text{Kinetic Energy}]}{[s]^2}$.

Kinetic energy of a particle moving along an elliptical trajectory is given by $K = \alpha s^2$, where $s$ is the distance travelled by the particle. Determine the dimensions of $\alpha$.

A

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

B

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

C

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

D

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

Question 10: Palette 10/12

DPT MCQ

Concept: Dimensional Analysis and Principle of Homogeneity of Dimensions.

If the centripetal force $F$ acting on a particle depends on its mass $M$, velocity $v$, and radius $r$ as $F = \frac{M v^n}{r}$, find the value of the exponent $n$ using dimensional analysis.

A

1

B

2

C

3

D

4

Question 11: Palette 11/12

DPT MCQ

Concept: Dimensional Analysis and Principle of Homogeneity of Dimensions.

If the formula for the orbital velocity of a satellite is given by $v = \sqrt{\frac{G M}{r^n}}$ where $v$ is orbital velocity, $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $r$ is the radius of the orbit, find the value of the exponent $n$ for the formula to be dimensionally correct.

A

1

B

2

C

3

D

0

Question 12: Palette 12/12

DPT MCQ

Concept: Dimensional Analysis and Principle of Homogeneity of Dimensions.

If the kinetic energy $KE$ of a body depends on the applied force $F$, velocity $v$, and time $t$ according to the relation $KE = F v^n t$, find the value of the variable exponent $n$ using dimensional analysis.

A

0

B

1

C

2

D

3