Dimensional Analysis

where is dimensionless constant

Hence, Correct answer is option (B) : .

To find the relation for distance and time between two different systems of units, we will use the given relationships for velocity and acceleration between the two systems.

The relationship for velocity between the two systems is given by:

The relationship for acceleration between the two systems is given by:

We know that velocity is the rate of change of distance with respect to time . Thus, we can write:

Similarly, acceleration is the rate of change of velocity with respect to time . So:

For the system of units 1 and 2, the formulas for length and time can be derived by connecting the given equations for velocity and acceleration.

Equate the expressions of acceleration from both systems using the relation :

Similarly, equate the expressions of velocity:

Solving these two equations, we get:

and

Which simplifies to:

and

Thus, the correct relation for distance and time in the two systems is:

and

This matches the given correct option.

[S]= ×[ΔT]
and, L=
⇒ They have different dimensions
The correct option is (D): Specific heat capacity and Latent heat

The correct answer is 5





Therefore , the frequency of ac source will be 5 × 10² Hz

Step 1: Express Mass in Terms of Fundamental Quantities

Let be the mass, and let its dimensions in terms of , , and be given by:

Step 2: Write the Dimensional Equation

The dimensions of (Planck’s constant) are . The dimensions of (speed of light) are . The dimensions of (gravitational constant) are . Substituting these dimensions into the equation from Step 1, we get:

Step 3: Equate the Exponents

Equating the exponents of , , and on both sides, we get the following system of equations:

Step 4: Solve the System of Equations

Adding the second and third equations, we get:

Also, from the first equation, . Adding these two equations gives , so . Since , we have . Substituting and into the third equation gives:

Thus, , , and .

Step 5: Write the Dimensions of Mass

Therefore, the dimensions of mass in the new system are:

Conclusion:

The correct answer is option (3).

To determine the dimensional formula for latent heat, we need to understand what latent heat refers to. Latent heat is the heat absorbed or released by a substance during a phase change (such as melting or boiling) without a change in temperature. It is usually expressed in energy per unit mass.

The formula for latent heat (L) is:
where is the total heat absorbed or released, and is the mass.

The dimensional formula for heat energy is: . Mass has the dimensional formula .

Since , the dimensional formula of latent heat will be calculated as:

After simplifying, we get:

Therefore, the correct dimensional formula for latent heat is: .

To determine the dimensions of , we need to understand the dimensions of each component:

1. The permittivity of free space, denoted by , has dimensions of . This is derived from the relation for Coulomb's law, or by using the formula for capacitance , where A is area and d is the distance.
2. The electric field has dimensions of . This arises from the definition , where F is the force with dimensions and is the charge with dimensions .

Now, let's calculate the dimensions of :

• has the dimensions of .
• Thus, has dimensions .

Performing dimensional multiplication, we have:

• Mass:
• Length:
• Time:
• Current:

Thus, the dimensions of are .

The correct answer is therefore: .

Dimensional analysis:

This corresponds to the dimensions of force per unit mass. Therefore, Option (2) is correct.

We need to find the dimensional formula of angular impulse.

Concept Used:

Angular impulse is defined as the product of torque and time. Torque is the rotational analogue of force, given by . The dimensional formula can be derived from this relationship.

Step-by-Step Solution:

Step 1: Write the definition of angular impulse.

Step 2: Find the dimensional formula of torque.

Torque ( ) = Force × Perpendicular Distance

The dimensional formula of force is , and distance is .

Step 3: Find the dimensional formula of time.

Step 4: Combine the dimensions to get the dimensional formula for angular impulse.

Thus, the dimensional formula of angular impulse is [M L² T⁻¹].

From the given equation , we have the following dimensions for each variable: Now, .
Thus, the dimensions of correspond to the dimension of compressibility.

To match the items from List-I with those in List-II, let's identify the dimensional formulas for each physical quantity.

1. Permeability of free space (A):
   The dimensional formula is .
   Thus, (A) matches with (III).
2. Magnetic field (B):
   The dimensional formula is .
   Thus, (B) matches with (II).
3. Magnetic moment (C):
   The dimensional formula is .
   Thus, (C) matches with (IV).
4. Torsional constant (D):
   The dimensional formula is .
   Thus, (D) matches with (I).

Hence, the correct matching is:

(A)-(III), (B)-(II), (C)-(IV), (D)-(I).

The correct answer is: (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

The acceleration due to gravity on the surface of the Earth is given by the formula: where:
- is the gravitational constant,
- is the mass of the Earth,
- is the radius of the Earth. If the diameter is reduced to one third of its original value, the new radius becomes: Since mass remains unchanged, the new acceleration due to gravity is: Thus, the acceleration due to gravity increases by a factor of 9.
Final Answer: .

The problem requires us to find the dimensional formulas for the physical quantities in List-I and match them with the correct dimensions provided in List-II.

Concept Used:

Dimensional analysis involves expressing physical quantities in terms of fundamental dimensions: Mass (M), Length (L), and Time (T). To find the dimension of a quantity, we use its definition or a formula relating it to other quantities with known dimensions.

Key dimensional formulas used in derivations:

• Force
• Area
• Pressure
• Energy/Work
• Power

Step-by-Step Solution:

Step 1: (A) Coefficient of viscosity ( )

According to Newton's law of viscosity, the viscous force is given by . We can rearrange this to find the dimensions of .

This matches with (I) in List-II.

Step 2: (B) Intensity of wave (I)

The intensity of a wave is defined as the power transmitted per unit area.

This matches with (II) in List-II.

Step 3: (C) Pressure gradient

Pressure gradient is the rate of change of pressure with respect to distance.

This matches with (III) in List-II.

Step 4: (D) Compressibility (K)

Compressibility is the reciprocal of the Bulk Modulus (B). The Bulk Modulus has the same dimensions as pressure.

This matches with (IV) in List-II.

Final Result:

Based on the derivations, the correct matching is:

• (A) Coefficient of viscosity → (I)
• (B) Intensity of wave → (II)
• (C) Pressure gradient → (III)
• (D) Compressibility → (IV)

Therefore, the correct matching is (A)-(I), (B)-(II), (C)-(III), (D)-(IV).

Let's match the physical quantities in List-I with their corresponding dimensions in List-II.

1. Mass density is defined as mass per unit volume. The dimension formula is:
   .
2. Impulse is defined as the change in momentum. Its dimension is the same as momentum, which is:
   .
3. Power is defined as work done per unit time. Its dimension is:
   .
4. Moment of inertia is represented as , hence its dimension is:
   .

Based on the above analysis, we can match them as:

• (A) Mass density - (IV)
• (B) Impulse - (II)
• (C) Power - (I)
• (D) Moment of inertia - (III)

Thus, the correct answer is (A)-(IV), (B)-(II), (C)-(I), (D)-(III).

- The dimension of is .
- The dimension of should match the dimension of , i.e., .
- For the numerator , dimensions of both terms must be consistent.
- has dimensions of , as gives , which balances the dimension of velocity.
- has dimensions of , as it has to balance the dimension of velocity when multiplied by .
- The denominator has dimensions of , so must have dimensions .
Thus, the dimension of , , and is .

Modulus of Elasticity (Young's Modulus) =
Strain is dimensionless, so the dimension of Modulus of Elasticity = Dimension of Stress.
Stress =

Torque =

The measured quantity is Modulus of Elasticity per unit Torque.

Comparing the dimensions:

Since we need to compare it with , we take the magnitude of the value .
Therefore, the value of is:

Step 1: Understanding the Concept:
We use dimensional analysis to match units to their expressions in terms of fundamental constants .
Step 2: Detailed Explanation:
Let .
Dimensions: .
For Length:
.
.
.
.
So .
Expression: . (Matches A-I).
Similarly for Time: . (Matches B-II).
For Mass: . (Matches C-III).
Step 3: Final Answer:
The correct matching is A-I, B-II, C-III.

Given equation:

where:

• is pressure,
• is density,
• is speed,
• and are quantities for which we need to find the dimensions of .

Step-by-Step Solution:

1. Dimensions of Pressure :

Pressure is force per unit area. The dimensions of force are , and area has dimensions . Therefore, the dimensions of pressure are:

2. Dimensions of :

Since the sum is equal to a constant, the dimensions of both terms must be the same. Therefore, the dimensions of must be .

Let’s assume the dimensions of and are and , respectively. Then, the dimensions of are:

For the dimensions to be consistent with , we must have:

• (for mass),
• (for length),
• (for time).

3. Dimensions of :

From the above, we can solve for the dimensions of and . After solving, we find that:

Final Answer: The dimensions of are .