Units And Measurement

The Vernier caliper problem involves understanding and calculating measurements using Vernier principles. Let's solve this step by step.

1. Least Count Calculation: The least count is the difference between one main scale division (MSD) and one Vernier scale division (VSD).
   • Given: 1 MSD = 1 mm; 10 VSD = 9 MSDs.
   • Hence, 1 VSD = .
   • Least Count = 1 MSD - 1 VSD = 1 mm - 0.9 mm = 0.1 mm.
2. Zero Error Determination: When the jaws touch, the fourth division of the Vernier coincides with a main scale division. This indicates a zero error.
   • Since the fourth Vernier division coincides, zero error = .
3. Measurement of the Diameter: When the bob is placed, the zero of the Vernier lies between 4.1 cm and 4.2 cm, and the sixth Vernier division coincides with a main scale division.
   • Main Scale Reading is between 4.1 cm and 4.2 cm, or between 41 mm and 42 mm.
   • Vernier Scale Reading = .
   • Reading in mm = .
4. Verification: Measurement = 41.2 mm ≤ 42 mm (confirmed range).
5. Conversion to Result Format: Convert mm to cm: .

The diameter of the bob = .

The correct answer is 5 × 10^–6 m
Given : 40 M = 1 cm
⇒ M = 0.025 cm ………..(1)
And also, 50 V = 49 M
⇒ Least count = M – V




⇒ LC = 5 × 10^–6 m

To determine the least count of the travelling microscope, we need to understand how the main scale and vernier scale relate to each other.

1. The main scale of the microscope has 20 divisions per cm. Thus, each division on the main scale is: .
2. The vernier scale has 50 divisions, and 25 of these vernier scale divisions are equal to 24 main scale divisions. Thus, the length of one division on the vernier scale is: .
3. The least count (LC) of the vernier caliper is calculated as the difference between one main scale division and one vernier scale division: .

Therefore, the least count of the travelling microscope is 0.002 cm which corresponds to the correct option.

This means that the smallest measurement that can be accurately read using this microscope is 0.002 cm.

The correct answer is 150.
I_mean =
ΔI_mean =
So %I= ×100= ×100 = %
x= 150
So, value of x is 150.

The correct answer is (D) :

The correct answer is (C) : A is true but R is false.


Statement 'A' is true
But statement R is false are coefficient of viscosity
=

To find the value of , we first need to calculate the mean thickness of the wire. The measured values are 1.22 mm, 1.23 mm, 1.19 mm, and 1.20 mm. The mean ( ) is given by:

Next, we calculate the deviations from the mean:

The absolute deviations (|deviation|) are:

• 0.01 mm
• 0.02 mm
• 0.02 mm
• 0.01 mm

The mean absolute deviation is:

The percentage error is calculated using the formula:

Given that the percentage error is , equate it to 1.24%:

Solving for :

Rounded to the nearest whole number, .

Since 150 falls within the given range of 150—150, the final value is valid.

The correct answer is 180
As the zero error is negative, we will add 0.05 cm to the reading
⇒ Corrected reading = 1.7 cm + 5 × 0.1 mm + 0.05 cm
= 180 × 10^–2 cm

The problem involves finding the diameter of a wire using the measurements obtained from a screw gauge. Let us solve this step-by-step using the observations provided:

1. Pitch of the Screw Gauge:
   • The pitch is defined as the distance moved by the screw on the main scale in one complete rotation of the circular scale.
   • Given that the screw moves 0.5 mm per complete rotation, the pitch is .
2. Least Count of the Screw Gauge:
   • The least count can be calculated using the formula: .
   • Substitute the values to get: .
3. Main Scale Reading (MSR):
   • The main scale reading given is 2.5 mm.
4. Circular Scale Reading (CSR):
   • The 45^th division of the circular scale coincides with the pitch line, so CSR = 45.
5. Total Reading:
   • Total Reading (TR) is calculated as: .
   • Substitute the known values: .
6. Correction for the Negative Error:
   • Negative error provided is 0.03 mm. This error needs to be added to the total reading to get the corrected reading.
   • Corrected Diameter =

Therefore, the diameter of the wire is 2.98 mm, which matches the given correct answer.

The correct answer is (A) : A-IV, B-I, C-III, D-II
A: Torque ⇒ IV : kg m² s^-2
B: Energy density ⇒ I : kg m^-1 s^-2
C: Pressure gradient ⇒ III : kg m^-2 s^-2
D: Impulse ⇒ II : kg ms^-1

M⁰L⁰T^–1 = L^a(ML^–3)^b(MT^–2)^c
M⁰L⁰T^–1 = L^aM^bL^–3b McT^–2c
Equivalent the power of MLT
M 0 = b + c
L 0 = a – 3b
T –1 = – 2c
a = , b = , c =

Step 1: Use the formula for density and apply error propagation principles.- Given m = (0.4±0.01)g, ℓ = (8±0.04)cm, r = (6±0.03)mm.- Density ρ = .- Differentiate logarithmically:- Substitute:

.

.

Step 2: Simplify the calculations.

.

Final Answer: Maximum error in density = 0.04 × 100 = 4%.

and



Therefore, % %
So, the answer is (B): 4.33%

Parallel Resistor Combination and Error Calculation

In parallel combination,

Given and ,

Therefore, .

Now, for error calculation,

Given and ,

Now, the percentage error is,

Conclusion:

The percentage error is approximately 4.33%.

The correct answer is option (A)

Given:

Force ( ), velocity ( ), and time ( ) are fundamental quantities.

Step 1: Dimensional Formula for Density

Let the dimensional formula for density ( ) be:

where and are constants to be determined.

Step 2: Dimensions of Fundamental Quantities

• (force)
• (velocity)
• (time)
• (density)

Step 3: Substitute Dimensions and Equate Powers

Substituting the dimensions into :

Expanding the dimensions on the right-hand side:

Equating the powers of and :

Step 4: Solve the System of Equations

• From , substitute into :
• Substitute and into :

Step 5: Write the Dimensional Formula for Density

Substituting , , and :

Final Answer:

The dimensional formula for density is .

Calculation of Percentage Change in :

Formula:

The percentage change in is given by:

Step 1: Substitute the Given Values:

Substituting the provided values for the terms:

Step 2: Add the Results:

Conclusion:

The percentage change in is 13%.