NEET 2025: Essential Preparation Tips for Students
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Understanding Units, Dimension and Measurement In physics, to measure physical quantities such as mass, length, and time, we need a standard unit of measurement. For instance, the standard unit for measuring length is the meter. One meter represents a specific length, and when you need to find the length of an object, you simply compare it to how many meters it spans. The process of measuring involves comparing a physical quantity to this predefined standard unit.
Units and Measurements NEET Questions, and answers
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1. If \( \text{Force} = \frac{\alpha}{\beta + \text{Density}} \), what is the dimensional formula of \( [\alpha \beta] \)?
A. \( [M^2L^{-4}T^{-1}] \)
B. \( [M^1L^{-3}T^{-2}] \)
C. \( [M^3L^{-5}T^{-2}] \)
D. \( [M^2L^{-2}T^{-3}] \)
The correct answer is C. \( [M^3L^{-5}T^{-2}] \).
Explanation:
To find the dimensional formula of \( [\alpha \beta] \), let’s analyze the given equation:
Step 1: Dimensional formula of Force
Force is defined as: \( \text{Force} = \text{mass} \times \text{acceleration} \).
The dimensional formula for force is:
\[ [F] = [M^1L^1T^{-2}] \]
Step 2: Dimensional formula of Density
Density is defined as: \( \text{Density} = \frac{\text{mass}}{\text{volume}} \).
The dimensional formula for density is:
\[ [\text{Density}] = [M^1L^{-3}] \]
Step 3: Analyzing the given formula
The formula is \( \text{Force} = \frac{\alpha}{\beta + \text{Density}} \).
- The term \( \beta + \text{Density} \) must have the same dimensional formula as density:
\[ [\beta] = [M^1L^{-3}] \] - Since \( \frac{\alpha}{\beta + \text{Density}} \) has dimensions of force:
\[ [\alpha] = [F] \times [\beta] \]
Step 4: Dimensions of \( \alpha \) and \( \beta \)
Substituting the values:
\[ [\alpha] = [M^1L^1T^{-2}] \times [M^1L^{-3}] \] \[ [\alpha] = [M^2L^{-2}T^{-2}] \] Now, the dimensional formula of \( [\alpha \beta] \) is:
\[ [\alpha \beta] = [\alpha] \times [\beta] \] Substituting values:
\[ [\alpha \beta] = [M^2L^{-2}T^{-2}] \times [M^1L^{-3}] \] \[ [\alpha \beta] = [M^3L^{-5}T^{-2}] \]
Therefore, the correct answer is \( [M^3L^{-5}T^{-2}] \).
Explanation:
To find the dimensional formula of \( [\alpha \beta] \), let’s analyze the given equation:
Step 1: Dimensional formula of Force
Force is defined as: \( \text{Force} = \text{mass} \times \text{acceleration} \).
The dimensional formula for force is:
\[ [F] = [M^1L^1T^{-2}] \]
Step 2: Dimensional formula of Density
Density is defined as: \( \text{Density} = \frac{\text{mass}}{\text{volume}} \).
The dimensional formula for density is:
\[ [\text{Density}] = [M^1L^{-3}] \]
Step 3: Analyzing the given formula
The formula is \( \text{Force} = \frac{\alpha}{\beta + \text{Density}} \).
- The term \( \beta + \text{Density} \) must have the same dimensional formula as density:
\[ [\beta] = [M^1L^{-3}] \] - Since \( \frac{\alpha}{\beta + \text{Density}} \) has dimensions of force:
\[ [\alpha] = [F] \times [\beta] \]
Step 4: Dimensions of \( \alpha \) and \( \beta \)
Substituting the values:
\[ [\alpha] = [M^1L^1T^{-2}] \times [M^1L^{-3}] \] \[ [\alpha] = [M^2L^{-2}T^{-2}] \] Now, the dimensional formula of \( [\alpha \beta] \) is:
\[ [\alpha \beta] = [\alpha] \times [\beta] \] Substituting values:
\[ [\alpha \beta] = [M^2L^{-2}T^{-2}] \times [M^1L^{-3}] \] \[ [\alpha \beta] = [M^3L^{-5}T^{-2}] \]
Therefore, the correct answer is \( [M^3L^{-5}T^{-2}] \).
2. A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. What is the least count of this vernier calipers?
A. 0.2 mm
B. 0.1 mm
C. 0.02 mm
D. 0.25 mm
The correct answer is A. 0.2 mm.
Explanation:
To calculate the least count of the vernier calipers, we use the formula:
\[ \text{Least Count (L.C.)} = \text{Value of 1 main scale division} - \text{Value of 1 vernier scale division} \]
Step 1: Value of 1 main scale division
The main scale has marks at 1 mm intervals, so:
\[ \text{Value of 1 main scale division} = 1 \, \text{mm} \]
Step 2: Value of 1 vernier scale division
The vernier scale has 20 divisions matching with 16 main scale divisions. Thus, the total value of 20 vernier scale divisions is equal to the total value of 16 main scale divisions:
\[ 20 \, \text{vernier scale divisions} = 16 \, \text{mm} \] Therefore, the value of 1 vernier scale division is:
\[ \text{Value of 1 vernier scale division} = \frac{16}{20} \, \text{mm} = 0.8 \, \text{mm} \]
Step 3: Calculating the least count
Substituting the values:
\[ \text{Least Count (L.C.)} = 1 \, \text{mm} - 0.8 \, \text{mm} = 0.2 \, \text{mm} \]
Conclusion:
The least count of the vernier calipers is \( 0.2 \, \text{mm} \).
Explanation:
To calculate the least count of the vernier calipers, we use the formula:
\[ \text{Least Count (L.C.)} = \text{Value of 1 main scale division} - \text{Value of 1 vernier scale division} \]
Step 1: Value of 1 main scale division
The main scale has marks at 1 mm intervals, so:
\[ \text{Value of 1 main scale division} = 1 \, \text{mm} \]
Step 2: Value of 1 vernier scale division
The vernier scale has 20 divisions matching with 16 main scale divisions. Thus, the total value of 20 vernier scale divisions is equal to the total value of 16 main scale divisions:
\[ 20 \, \text{vernier scale divisions} = 16 \, \text{mm} \] Therefore, the value of 1 vernier scale division is:
\[ \text{Value of 1 vernier scale division} = \frac{16}{20} \, \text{mm} = 0.8 \, \text{mm} \]
Step 3: Calculating the least count
Substituting the values:
\[ \text{Least Count (L.C.)} = 1 \, \text{mm} - 0.8 \, \text{mm} = 0.2 \, \text{mm} \]
Conclusion:
The least count of the vernier calipers is \( 0.2 \, \text{mm} \).
3. The pitch of a screw gauge is 0.5 mm and there are 50 divisions on its circular scale. When the screw is closed and the plane face of the screw is in contact with the plane face of the stud on the frame, it was observed that the zero of the circular scale is above the pitch line by 5 divisions. What is the zero error in the instrument?
A. 0.025 mm
B. –0.25 mm
C. 0.05 mm
D. –0.05 mm
The correct answer is D. –0.05 mm.
Explanation:
To determine the zero error, let’s analyze the given data:
Step 1: Pitch and least count of the screw gauge
- The pitch of the screw gauge is the distance moved by the screw in one complete rotation, which is given as \( 0.5 \, \text{mm} \).
- The least count (L.C.) of the screw gauge is calculated as:
\[ \text{L.C.} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} \] Substituting the values:
\[ \text{L.C.} = \frac{0.5 \, \text{mm}}{50} = 0.01 \, \text{mm} \]
Step 2: Observed zero error
When the screw is fully closed, the zero of the circular scale is above the pitch line by 5 divisions. This means the zero error is negative because the zero mark of the circular scale lies below the reference line.
- The zero error in terms of the least count is:
\[ \text{Zero Error} = - (\text{Number of divisions above the pitch line}) \times \text{L.C.} \] Substituting the values:
\[ \text{Zero Error} = - (5) \times 0.01 \, \text{mm} = -0.05 \, \text{mm} \]
Conclusion:
The zero error in the screw gauge is \( -0.05 \, \text{mm} \).
Explanation:
To determine the zero error, let’s analyze the given data:
Step 1: Pitch and least count of the screw gauge
- The pitch of the screw gauge is the distance moved by the screw in one complete rotation, which is given as \( 0.5 \, \text{mm} \).
- The least count (L.C.) of the screw gauge is calculated as:
\[ \text{L.C.} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} \] Substituting the values:
\[ \text{L.C.} = \frac{0.5 \, \text{mm}}{50} = 0.01 \, \text{mm} \]
Step 2: Observed zero error
When the screw is fully closed, the zero of the circular scale is above the pitch line by 5 divisions. This means the zero error is negative because the zero mark of the circular scale lies below the reference line.
- The zero error in terms of the least count is:
\[ \text{Zero Error} = - (\text{Number of divisions above the pitch line}) \times \text{L.C.} \] Substituting the values:
\[ \text{Zero Error} = - (5) \times 0.01 \, \text{mm} = -0.05 \, \text{mm} \]
Conclusion:
The zero error in the screw gauge is \( -0.05 \, \text{mm} \).
4. If the unit of force is 1 kN, the unit of length is 1 km, and the unit of time is 100 s in a new system, then what is the new unit of mass?
A. 1000 kg
B. 1 kg
C. 10000 kg
D. 100 kg
The correct answer is C. 10000 kg.
Explanation:
To find the new unit of mass, we use the relationship between force, mass, and acceleration:
\[ F = m \cdot a \] where \( F \) is force, \( m \) is mass, and \( a \) is acceleration.
Step 1: Dimensional relationship
The dimensional formula for force is:
\[ [F] = [M][L][T^{-2}] \] From this, mass can be expressed as:
\[ [M] = \frac{[F] \cdot [T^2]}{[L]} \]
Step 2: Substituting the new units
- The new unit of force is \( 1 \, \text{kN} = 1000 \, \text{N} \).
- The new unit of length is \( 1 \, \text{km} = 1000 \, \text{m} \).
- The new unit of time is \( 100 \, \text{s} \).
Substituting these values:
\[ [M] = \frac{1000 \cdot (100)^2}{1000} \] Simplify the expression:
\[ [M] = \frac{1000 \cdot 10000}{1000} = 10000 \, \text{kg} \]
Conclusion:
The new unit of mass in this system is \( 10000 \, \text{kg} \).
Explanation:
To find the new unit of mass, we use the relationship between force, mass, and acceleration:
\[ F = m \cdot a \] where \( F \) is force, \( m \) is mass, and \( a \) is acceleration.
Step 1: Dimensional relationship
The dimensional formula for force is:
\[ [F] = [M][L][T^{-2}] \] From this, mass can be expressed as:
\[ [M] = \frac{[F] \cdot [T^2]}{[L]} \]
Step 2: Substituting the new units
- The new unit of force is \( 1 \, \text{kN} = 1000 \, \text{N} \).
- The new unit of length is \( 1 \, \text{km} = 1000 \, \text{m} \).
- The new unit of time is \( 100 \, \text{s} \).
Substituting these values:
\[ [M] = \frac{1000 \cdot (100)^2}{1000} \] Simplify the expression:
\[ [M] = \frac{1000 \cdot 10000}{1000} = 10000 \, \text{kg} \]
Conclusion:
The new unit of mass in this system is \( 10000 \, \text{kg} \).
5. The percentage error in measuring the mass (\( m \)) and radius (\( R \)) of a solid sphere are 3% and 2%, respectively. What is the error in the measurement of its moment of inertia about its diameter?
A. 5%
B. 7%
C. 8%
D. 4%
The correct answer is B. 7%.
Explanation:
The moment of inertia (\( I \)) of a solid sphere about its diameter is given by:
\[ I = \frac{2}{5} m R^2 \] Here, \( m \) is the mass, and \( R \) is the radius.
Step 1: Percentage error formula
The percentage error in a quantity is given by the sum of the percentage errors of its constituent terms multiplied by their respective powers:
\[ \Delta I\% = \Delta m\% + 2 \cdot \Delta R\% \]
Step 2: Substituting the given values
- Percentage error in mass (\( \Delta m\% \)) = 3%
- Percentage error in radius (\( \Delta R\% \)) = 2%
Substituting these values into the formula:
\[ \Delta I\% = 3\% + 2 \cdot 2\% = 3\% + 4\% = 7\% \]
Conclusion:
The percentage error in the measurement of the moment of inertia is \( 7\% \).
Explanation:
The moment of inertia (\( I \)) of a solid sphere about its diameter is given by:
\[ I = \frac{2}{5} m R^2 \] Here, \( m \) is the mass, and \( R \) is the radius.
Step 1: Percentage error formula
The percentage error in a quantity is given by the sum of the percentage errors of its constituent terms multiplied by their respective powers:
\[ \Delta I\% = \Delta m\% + 2 \cdot \Delta R\% \]
Step 2: Substituting the given values
- Percentage error in mass (\( \Delta m\% \)) = 3%
- Percentage error in radius (\( \Delta R\% \)) = 2%
Substituting these values into the formula:
\[ \Delta I\% = 3\% + 2 \cdot 2\% = 3\% + 4\% = 7\% \]
Conclusion:
The percentage error in the measurement of the moment of inertia is \( 7\% \).
6. If energy (\( E \)), velocity (\( v \)), and force (\( F \)) are taken as fundamental quantities, what are the dimensions of mass (\( M \))?
A. \( [E][V]^{-2} \)
B. \( [E][V]^{-2}[F]^{0} \)
C. \( [F][V]^{-1} \)
D. \( [F][V]^{-2} \)
The correct answer is B. \( [E][V]^{-2}[F]^{-1} \).
Explanation:
To express the dimensions of mass (\( M \)) in terms of energy (\( E \)), velocity (\( v \)), and force (\( F \)), we start with the dimensional formula for force and energy:
Step 1: Dimensional formula for energy
Energy (\( E \)) has the dimensional formula:
\[ [E] = [M][L]^2[T]^{-2} \]
Step 2: Dimensional formula for force
Force (\( F \)) has the dimensional formula:
\[ [F] = [M][L][T]^{-2} \]
Step 3: Dimensional formula for velocity
Velocity (\( v \)) has the dimensional formula:
\[ [V] = [L][T]^{-1} \]
Step 4: Relating \( M \), \( E \), \( V \), and \( F \)
We assume the dimensions of \( M \) as:
\[ [M] = [E]^a [V]^b [F]^c \] Substituting the dimensional formulas of \( E \), \( V \), and \( F \):
\[ [M] = ([M][L]^2[T]^{-2})^a ([L][T]^{-1})^b ([M][L][T]^{-2})^c \] Equating the powers of \( [M] \), \( [L] \), and \( [T] \), we get:
- For \( [M] \): \( 1 = a + c \)
- For \( [L] \): \( 0 = 2a + b + c \)
- For \( [T] \): \( 0 = -2a - b - 2c \)
Solving these equations simultaneously:
- From \( a + c = 1 \), we get \( c = 1 - a \)
- Substituting \( c = 1 - a \) into the other equations, we solve to get:
\[ a = 1, \, b = -2, \, c = 0 \]
Step 5: Final expression for \( [M] \)
Substituting the values of \( a \), \( b \), and \( c \):
\[ [M] = [E]^1 [V]^{-2} [F]^{0} \]
Conclusion:
The dimensions of mass (\( M \)) in terms of \( E \), \( v \), and \( F \) are \( [E][V]^{-2}[F]^{0} \).
Explanation:
To express the dimensions of mass (\( M \)) in terms of energy (\( E \)), velocity (\( v \)), and force (\( F \)), we start with the dimensional formula for force and energy:
Step 1: Dimensional formula for energy
Energy (\( E \)) has the dimensional formula:
\[ [E] = [M][L]^2[T]^{-2} \]
Step 2: Dimensional formula for force
Force (\( F \)) has the dimensional formula:
\[ [F] = [M][L][T]^{-2} \]
Step 3: Dimensional formula for velocity
Velocity (\( v \)) has the dimensional formula:
\[ [V] = [L][T]^{-1} \]
Step 4: Relating \( M \), \( E \), \( V \), and \( F \)
We assume the dimensions of \( M \) as:
\[ [M] = [E]^a [V]^b [F]^c \] Substituting the dimensional formulas of \( E \), \( V \), and \( F \):
\[ [M] = ([M][L]^2[T]^{-2})^a ([L][T]^{-1})^b ([M][L][T]^{-2})^c \] Equating the powers of \( [M] \), \( [L] \), and \( [T] \), we get:
- For \( [M] \): \( 1 = a + c \)
- For \( [L] \): \( 0 = 2a + b + c \)
- For \( [T] \): \( 0 = -2a - b - 2c \)
Solving these equations simultaneously:
- From \( a + c = 1 \), we get \( c = 1 - a \)
- Substituting \( c = 1 - a \) into the other equations, we solve to get:
\[ a = 1, \, b = -2, \, c = 0 \]
Step 5: Final expression for \( [M] \)
Substituting the values of \( a \), \( b \), and \( c \):
\[ [M] = [E]^1 [V]^{-2} [F]^{0} \]
Conclusion:
The dimensions of mass (\( M \)) in terms of \( E \), \( v \), and \( F \) are \( [E][V]^{-2}[F]^{0} \).
7. In the relation \( y = r \sin(\omega t - kx) \), what are the dimensions of \( \frac{\omega}{k} \)?
A. \( [L][T]^{-1} \)
B. \( [L][T]^2 \)
C. \( [T] \)
D. \( [L] \)
The correct answer is A. \( [L][T]^{-1} \).
Explanation:
The given equation is:
\[ y = r \sin(\omega t - kx) \] Here:
- \( \omega \) is the angular frequency.
- \( k \) is the wave number.
- \( t \) is time.
- \( x \) is position.
Step 1: Dimensional formula of \( \omega \)
Angular frequency \( \omega \) is related to time (\( t \)) as follows:
\[ \omega t \, \text{(dimensionless)} \implies [\omega] = [T]^{-1} \]
Step 2: Dimensional formula of \( k \)
Wave number \( k \) is related to position (\( x \)) as follows:
\[ kx \, \text{(dimensionless)} \implies [k] = [L]^{-1} \]
Step 3: Dimensional formula of \( \frac{\omega}{k} \)
The quantity \( \frac{\omega}{k} \) has dimensions:
\[ \left[\frac{\omega}{k}\right] = \frac{[T]^{-1}}{[L]^{-1}} = [L][T]^{-1} \]
Conclusion:
The dimensions of \( \frac{\omega}{k} \) are \( [L][T]^{-1} \).
Explanation:
The given equation is:
\[ y = r \sin(\omega t - kx) \] Here:
- \( \omega \) is the angular frequency.
- \( k \) is the wave number.
- \( t \) is time.
- \( x \) is position.
Step 1: Dimensional formula of \( \omega \)
Angular frequency \( \omega \) is related to time (\( t \)) as follows:
\[ \omega t \, \text{(dimensionless)} \implies [\omega] = [T]^{-1} \]
Step 2: Dimensional formula of \( k \)
Wave number \( k \) is related to position (\( x \)) as follows:
\[ kx \, \text{(dimensionless)} \implies [k] = [L]^{-1} \]
Step 3: Dimensional formula of \( \frac{\omega}{k} \)
The quantity \( \frac{\omega}{k} \) has dimensions:
\[ \left[\frac{\omega}{k}\right] = \frac{[T]^{-1}}{[L]^{-1}} = [L][T]^{-1} \]
Conclusion:
The dimensions of \( \frac{\omega}{k} \) are \( [L][T]^{-1} \).
8. Given below are two statements: one is labelled as Assertion (A) and the other as Reason (R).
Assertion (A): In a Vernier calliper, if a positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading.
Reason (R): The zero error in a Vernier calliper might occur due to manufacturing defects or rough handling.
In the light of the above statements, choose the correct answer:
Assertion (A): In a Vernier calliper, if a positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading.
Reason (R): The zero error in a Vernier calliper might occur due to manufacturing defects or rough handling.
In the light of the above statements, choose the correct answer:
A. Both (A) and (R) are correct and (R) is the correct explanation of (A)
B. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
C. (A) is true but (R) is false
D. (A) is false but (R) is true
The correct answer is A. Both (A) and (R) are correct and (R) is the correct explanation of (A).
Explanation:
Step 1: Understanding zero error and its impact
- A positive zero error occurs when the zero mark of the Vernier scale lies to the right of the zero mark of the main scale.
- This leads to an additional reading being added to the measurement, making the reading appear greater than the actual value.
Step 2: Cause of zero error
- Zero error can occur due to defects during manufacturing or rough handling of the instrument, such as dropping it or improper usage.
Step 3: Linking (A) and (R)
- Assertion (A) correctly describes the effect of positive zero error on measurements.
- Reason (R) correctly explains the possible causes of zero error in Vernier callipers.
- Since the reason aligns with the assertion, (R) is the correct explanation of (A).
Conclusion:
Both (A) and (R) are correct, and (R) is the correct explanation of (A).
Explanation:
Step 1: Understanding zero error and its impact
- A positive zero error occurs when the zero mark of the Vernier scale lies to the right of the zero mark of the main scale.
- This leads to an additional reading being added to the measurement, making the reading appear greater than the actual value.
Step 2: Cause of zero error
- Zero error can occur due to defects during manufacturing or rough handling of the instrument, such as dropping it or improper usage.
Step 3: Linking (A) and (R)
- Assertion (A) correctly describes the effect of positive zero error on measurements.
- Reason (R) correctly explains the possible causes of zero error in Vernier callipers.
- Since the reason aligns with the assertion, (R) is the correct explanation of (A).
Conclusion:
Both (A) and (R) are correct, and (R) is the correct explanation of (A).
9. One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of the vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is:
A. 0.01 cm
B. 0.02 cm
C. 0.05 cm
D. 0.005 cm
The correct answer is B. 0.02 cm.
Explanation:
Step 1: Main scale division (MSD)
- 1 cm on the main scale is divided into 10 equal parts.
- Therefore, 1 MSD = \( \frac{1 \, \text{cm}}{10} = 0.1 \, \text{cm} \).
Step 2: Length of 1 vernier scale division (VSD)
- 10 vernier scale divisions = 8 main scale divisions.
- Length of 1 VSD = \( \frac{8 \, \text{MSD}}{10} = \frac{8 \times 0.1}{10} = 0.08 \, \text{cm} \).
Step 3: Least count (LC)
The least count is the difference between 1 MSD and 1 VSD:
\[ \text{LC} = \text{MSD} - \text{VSD} = 0.1 \, \text{cm} - 0.08 \, \text{cm} = 0.02 \, \text{cm}. \]
Conclusion:
The least count of the vernier callipers is \( 0.02 \, \text{cm} \).
Explanation:
Step 1: Main scale division (MSD)
- 1 cm on the main scale is divided into 10 equal parts.
- Therefore, 1 MSD = \( \frac{1 \, \text{cm}}{10} = 0.1 \, \text{cm} \).
Step 2: Length of 1 vernier scale division (VSD)
- 10 vernier scale divisions = 8 main scale divisions.
- Length of 1 VSD = \( \frac{8 \, \text{MSD}}{10} = \frac{8 \times 0.1}{10} = 0.08 \, \text{cm} \).
Step 3: Least count (LC)
The least count is the difference between 1 MSD and 1 VSD:
\[ \text{LC} = \text{MSD} - \text{VSD} = 0.1 \, \text{cm} - 0.08 \, \text{cm} = 0.02 \, \text{cm}. \]
Conclusion:
The least count of the vernier callipers is \( 0.02 \, \text{cm} \).
10. A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 6 mm and the zero of the circular scale division is 26 divisions above the reference level of the screw gauge. It has a zero error of 0.003 cm. The correct diameter of the ball is:
A. 0.629 cm
B. 0.626 cm
C. 0.623 cm
D. 0.633 cm
The correct answer is C. 0.623 cm.
Explanation:
Step 1: Understanding the given data
- Least count (LC) of screw gauge = 0.001 cm.
- Main scale reading (MSR) = 6 mm = 0.6 cm.
- Circular scale reading (CSR) = 26 divisions.
- Zero error = 0.003 cm.
Step 2: Calculating the measured diameter
- The reading from the circular scale is:
\[ \text{CSR} = \text{Number of divisions} \times \text{Least count} = 26 \times 0.001 \, \text{cm} = 0.026 \, \text{cm}. \] - The total measured diameter is:
\[ \text{Measured Diameter} = \text{MSR} + \text{CSR} = 0.6 \, \text{cm} + 0.026 \, \text{cm} = 0.626 \, \text{cm}. \] Step 3: Correcting for the zero error
- Since the zero error is positive (0.003 cm), the corrected diameter is:
\[ \text{Corrected Diameter} = \text{Measured Diameter} - \text{Zero Error} = 0.626 \, \text{cm} - 0.003 \, \text{cm} = 0.623 \, \text{cm}. \]
Conclusion:
The correct diameter of the ball is \( 0.623 \, \text{cm} \).
Explanation:
Step 1: Understanding the given data
- Least count (LC) of screw gauge = 0.001 cm.
- Main scale reading (MSR) = 6 mm = 0.6 cm.
- Circular scale reading (CSR) = 26 divisions.
- Zero error = 0.003 cm.
Step 2: Calculating the measured diameter
- The reading from the circular scale is:
\[ \text{CSR} = \text{Number of divisions} \times \text{Least count} = 26 \times 0.001 \, \text{cm} = 0.026 \, \text{cm}. \] - The total measured diameter is:
\[ \text{Measured Diameter} = \text{MSR} + \text{CSR} = 0.6 \, \text{cm} + 0.026 \, \text{cm} = 0.626 \, \text{cm}. \] Step 3: Correcting for the zero error
- Since the zero error is positive (0.003 cm), the corrected diameter is:
\[ \text{Corrected Diameter} = \text{Measured Diameter} - \text{Zero Error} = 0.626 \, \text{cm} - 0.003 \, \text{cm} = 0.623 \, \text{cm}. \]
Conclusion:
The correct diameter of the ball is \( 0.623 \, \text{cm} \).
11. In an experiment to find out the thickness of a sheet using a screw gauge, the following observations were noted:
(a) Screw moves 1 mm on the main scale in two complete rotations.
(b) Total divisions on the circular scale = 50.
(c) Main scale reading is 1.5 mm.
(d) 35 divisions on the circular scale coincide with the pitch line.
(e) The instrument has a negative zero error of 0.02 mm.
Calculate the thickness of the sheet using the above data.
(a) Screw moves 1 mm on the main scale in two complete rotations.
(b) Total divisions on the circular scale = 50.
(c) Main scale reading is 1.5 mm.
(d) 35 divisions on the circular scale coincide with the pitch line.
(e) The instrument has a negative zero error of 0.02 mm.
Calculate the thickness of the sheet using the above data.
A. 1.52 mm
B. 1.58 mm
C. 1.87 mm
D. 1.50 mm
The correct answer is C. 1.87 mm.
Explanation:
Step 1: Find the least count of the screw gauge
- Screw moves 1 mm on the main scale in two complete rotations.
- Therefore, the screw moves 0.5 mm on the main scale per rotation.
- There are 50 divisions on the circular scale.
- Hence, the least count of the screw gauge is:
\[ \text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} = \frac{0.5 \, \text{mm}}{50} = 0.01 \, \text{mm}. \]
Step 2: Reading from the main scale and the circular scale
- Main scale reading (MSR) = 1.5 mm.
- Circular scale reading (CSR) = 35 divisions.
- Therefore, the CSR is:
\[ \text{CSR} = 35 \times 0.01 \, \text{mm} = 0.35 \, \text{mm}. \]
Step 3: Total measured thickness
- The total measured thickness is:
\[ \text{Total Thickness} = \text{MSR} + \text{CSR} = 1.5 \, \text{mm} + 0.35 \, \text{mm} = 1.85 \, \text{mm}. \]
Step 4: Correcting for the negative zero error
- The zero error is negative (-0.02 mm), so we subtract it from the total thickness:
\[ \text{Corrected Thickness} = 1.85 \, \text{mm} + 0.02 \, \text{mm} = 1.87 \, \text{mm}. \]
Conclusion:
The corrected thickness of the sheet is \( \boxed{1.87 \, \text{mm}} \).
Explanation:
Step 1: Find the least count of the screw gauge
- Screw moves 1 mm on the main scale in two complete rotations.
- Therefore, the screw moves 0.5 mm on the main scale per rotation.
- There are 50 divisions on the circular scale.
- Hence, the least count of the screw gauge is:
\[ \text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} = \frac{0.5 \, \text{mm}}{50} = 0.01 \, \text{mm}. \]
Step 2: Reading from the main scale and the circular scale
- Main scale reading (MSR) = 1.5 mm.
- Circular scale reading (CSR) = 35 divisions.
- Therefore, the CSR is:
\[ \text{CSR} = 35 \times 0.01 \, \text{mm} = 0.35 \, \text{mm}. \]
Step 3: Total measured thickness
- The total measured thickness is:
\[ \text{Total Thickness} = \text{MSR} + \text{CSR} = 1.5 \, \text{mm} + 0.35 \, \text{mm} = 1.85 \, \text{mm}. \]
Step 4: Correcting for the negative zero error
- The zero error is negative (-0.02 mm), so we subtract it from the total thickness:
\[ \text{Corrected Thickness} = 1.85 \, \text{mm} + 0.02 \, \text{mm} = 1.87 \, \text{mm}. \]
Conclusion:
The corrected thickness of the sheet is \( \boxed{1.87 \, \text{mm}} \).
12. The length, breadth, and thickness of a block are measured as 205.4 cm, 7.0 cm, and 0.43 cm respectively. Which one of the following measurements is the most accurate?
A. Length
B. Breadth
C. Thickness
D. None of these
The correct answer is C. Thickness.
Explanation:
The accuracy of a measurement is determined by the Least count in the measured value. The more the LC in the measurement, the more accurate it is considered to be.
- The length is measured as 205.4 cm, which has 0.1 LC.
- The breadth is measured as 7.0 cm, which has 0.1 LC.
- The thickness is measured as 0.43 cm, which has 0.01 LC.
Therefore, the measurement with the more least count and thus the most accurate measurement is the Thickness 0.43 mm.
Explanation:
The accuracy of a measurement is determined by the Least count in the measured value. The more the LC in the measurement, the more accurate it is considered to be.
- The length is measured as 205.4 cm, which has 0.1 LC.
- The breadth is measured as 7.0 cm, which has 0.1 LC.
- The thickness is measured as 0.43 cm, which has 0.01 LC.
Therefore, the measurement with the more least count and thus the most accurate measurement is the Thickness 0.43 mm.
13. A person is sitting in a moving train and is facing the engine. He tosses up a coin which falls behind him. He concludes that the train is moving:
(A) forward with increasing speed.
(B) forward with decreasing speed.
(C) backward with increasing speed.
(D) backward with decreasing speed.
Select the alternative:
(A) forward with increasing speed.
(B) forward with decreasing speed.
(C) backward with increasing speed.
(D) backward with decreasing speed.
Select the alternative:
A. Only (A)
B. Only (A) and (C)
C. Only (A) and (D)
D. Only (B) and (C)
The correct answer is A. Only (A).
Explanation:
The person in the train faces the engine and tosses a coin. If the coin falls behind him, it implies that the motion of the train is not uniform.
- In a uniform motion (constant speed), the coin would fall vertically downward to the person's hand.
- However, if the coin falls behind, it suggests that the train is moving with either increasing speed.
- Therefore, the train is most likely moving forward but increasing speed (option A).
Explanation:
The person in the train faces the engine and tosses a coin. If the coin falls behind him, it implies that the motion of the train is not uniform.
- In a uniform motion (constant speed), the coin would fall vertically downward to the person's hand.
- However, if the coin falls behind, it suggests that the train is moving with either increasing speed.
- Therefore, the train is most likely moving forward but increasing speed (option A).
14. A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by:
A. αβt
B. (αβ)t
C. (αβt) / (α + β)
D. (α + β)t
The correct answer is C. (αβt) / (α + β).
Explanation:
To find the maximum velocity, let the time spent in accelerating be \( t_1 \) and the time spent in decelerating be \( t_2 \). Since the total time elapsed is \( t \), we have: \[ t_1 + t_2 = t \] During acceleration, the final velocity \( v_{\text{max}} \) after time \( t_1 \) is given by: \[ v_{\text{max}} = \alpha t_1 \] During deceleration, the car comes to rest, so the final velocity is 0 after time \( t_2 \), and the velocity at the start of deceleration is \( v_{\text{max}} \): \[ v_{\text{max}} = \beta t_2 \] From the second equation, we can express \( t_2 \) in terms of \( v_{\text{max}} \) and \( \beta \): \[ t_2 = \frac{v_{\text{max}}}{\beta} \] Substituting \( t_2 \) into the total time equation \( t_1 + t_2 = t \), we get: \[ t_1 = t - t_2 = t - \frac{v_{\text{max}}}{\beta} \] Now, from the equation \( v_{\text{max}} = \alpha t_1 \), we can solve for \( v_{\text{max}} \): \[ v_{\text{max}} = \alpha \left( t - \frac{v_{\text{max}}}{\beta} \right) \] Simplifying this, we find: \[ v_{\text{max}} \left( 1 + \frac{\alpha}{\beta} \right) = \alpha t \] Therefore, the maximum velocity is: \[ v_{\text{max}} = \frac{\alpha \beta t}{\alpha + \beta} \] Hence, the correct expression for the maximum velocity is \( v_{\text{max}} = \frac{\alpha \beta t}{\alpha + \beta} \).
Explanation:
To find the maximum velocity, let the time spent in accelerating be \( t_1 \) and the time spent in decelerating be \( t_2 \). Since the total time elapsed is \( t \), we have: \[ t_1 + t_2 = t \] During acceleration, the final velocity \( v_{\text{max}} \) after time \( t_1 \) is given by: \[ v_{\text{max}} = \alpha t_1 \] During deceleration, the car comes to rest, so the final velocity is 0 after time \( t_2 \), and the velocity at the start of deceleration is \( v_{\text{max}} \): \[ v_{\text{max}} = \beta t_2 \] From the second equation, we can express \( t_2 \) in terms of \( v_{\text{max}} \) and \( \beta \): \[ t_2 = \frac{v_{\text{max}}}{\beta} \] Substituting \( t_2 \) into the total time equation \( t_1 + t_2 = t \), we get: \[ t_1 = t - t_2 = t - \frac{v_{\text{max}}}{\beta} \] Now, from the equation \( v_{\text{max}} = \alpha t_1 \), we can solve for \( v_{\text{max}} \): \[ v_{\text{max}} = \alpha \left( t - \frac{v_{\text{max}}}{\beta} \right) \] Simplifying this, we find: \[ v_{\text{max}} \left( 1 + \frac{\alpha}{\beta} \right) = \alpha t \] Therefore, the maximum velocity is: \[ v_{\text{max}} = \frac{\alpha \beta t}{\alpha + \beta} \] Hence, the correct expression for the maximum velocity is \( v_{\text{max}} = \frac{\alpha \beta t}{\alpha + \beta} \).
15. A particle is projected with speed 40 m/s at an angle of 53° from the horizon from the ground. After what time will the velocity of the particle be perpendicular to the initial velocity?
A. 15 s
B. 10 s
C. 5 s
D. 20 s
The correct answer is C. 5 s.
Explanation:
To find the time at which the velocity becomes perpendicular to the initial velocity, we need to analyze the horizontal and vertical components of the velocity.
The initial velocity is \( u = 40 \, \text{m/s} \) at an angle \( \theta = 53^\circ \), so: - Horizontal component: \( u_x = u \cos \theta = 40 \times \cos 53^\circ = 40 \times 0.6 = 24 \, \text{m/s} \) - Vertical component: \( u_y = u \sin \theta = 40 \times \sin 53^\circ = 40 \times 0.8 = 32 \, \text{m/s} \) The velocity will be perpendicular to the initial velocity when the vertical component of the velocity becomes zero. This happens at the time of flight when the particle reaches its maximum height. The vertical velocity decreases due to gravity at the rate of \( g = 9.8 \, \text{m/s}^2 \). The time taken to reach the maximum height can be found using the equation: \[ v_y = u_y - g t = 0 \] Solving for \( t \): \[ t = \frac{u}{gsin \ theta} = \frac{40}{10X0.8} = \ 5 sec \, \text{s} \] Thus, the time at which the velocity becomes perpendicular to the initial velocity is 5 seconds.
Explanation:
To find the time at which the velocity becomes perpendicular to the initial velocity, we need to analyze the horizontal and vertical components of the velocity.
The initial velocity is \( u = 40 \, \text{m/s} \) at an angle \( \theta = 53^\circ \), so: - Horizontal component: \( u_x = u \cos \theta = 40 \times \cos 53^\circ = 40 \times 0.6 = 24 \, \text{m/s} \) - Vertical component: \( u_y = u \sin \theta = 40 \times \sin 53^\circ = 40 \times 0.8 = 32 \, \text{m/s} \) The velocity will be perpendicular to the initial velocity when the vertical component of the velocity becomes zero. This happens at the time of flight when the particle reaches its maximum height. The vertical velocity decreases due to gravity at the rate of \( g = 9.8 \, \text{m/s}^2 \). The time taken to reach the maximum height can be found using the equation: \[ v_y = u_y - g t = 0 \] Solving for \( t \): \[ t = \frac{u}{gsin \ theta} = \frac{40}{10X0.8} = \ 5 sec \, \text{s} \] Thus, the time at which the velocity becomes perpendicular to the initial velocity is 5 seconds.
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