BIOPHYSICS LABORATORY
TO KNOW HOW TO DETERMINE THE LEAST COUNT OF THE CALIPER USED AND
TO LEARN HOW TO USE THE VERNIER AND MICROMETER CALIPERS CORRECTLY
VERNIER CALIPER
MICROMETER CALIPER
CYLINDER
BEAM BALANCE
SPHERE
CUBE
WASHER
AN AUXILIARY SLIDING SCALE WHICH ENABLES ONE TO INCREASE HIS ACCURACY IN MEASUREMENT.
Total Measurement = Main Scale Reading + (Sequential Vernier Scale Marking × Least Count)
EXAMPLE 1:
EXAMPLE 2:
PRECISE MEASUREMENT OF SMALL LENGTHS AND DIAMETERS.
FOR EVERY ONE REVOLUTION(5O divisions) OF THE SCREW, THE JAW ADVANCES HALF A MILLIMETER.
Total Measurement = Main Scale Reading + (Sequential Micrometer Scale Marking × Least Count)
note: diameter=r/2
DENSITY=mass(g)/volume (cm³)
CUBE= S³
CYLINDER=πr²h
SPHERE: 4/3πr³
WASHER: πrout²-rin²)t
A) 0.1 mm
B) 0.01 mm
C) 0.02 mm
D) 0.5 mm
Answer: C) 0.02 mm
A) 10
B) 20
C) 50
D) 100
Answer: B) 20
A) 0.1 mm
B) 0.01 mm
C) 0.02 mm
D) 0.001 mm
Answer: B) 0.01 mm
A) Main scale
B) Vernier scale
C) Lower jaws
D) Upper jaws
Answer: D) Upper jaws
A) Measures larger objects
B) Easier to use
C) More accurate for smaller dimensions
D) Can measure both internal and external dimensions
Answer: C) More accurate for smaller dimensions
A) Subtract the Vernier reading from the main scale
B) Add the Vernier reading to the main scale
C) Multiply the Vernier reading by the least count
D) Divide the Vernier reading by the least count
Answer: B) Add the Vernier reading to the main scale
A) The number of divisions on the sleeve
B) The distance the spindle moves per full rotation of the thimble
C) The smallest reading on the thimble
D) The length of the thimble
Answer: B) The distance the spindle moves per full rotation of the thimble
A) By closing the jaws completely and checking the alignment of the zero mark on the Vernier scale with the zero on the main scale
B) By rotating the Vernier scale
C) By measuring a standard object
D) By adjusting the main scale
Answer: A) By closing the jaws completely and checking the alignment of the zero mark on the Vernier scale with the zero on the main scale
A) Add the error value to the measured value
B) Subtract the error value from the measured value
C) Multiply the measured value by the error
D) Ignore the error
Answer: A) Add the error value to the measured value
A) To ensure accurate measurement by applying consistent pressure
B) To lock the spindle in place
C) To adjust the zero reading
D) To reset the scale to zero
Answer: A) To ensure accurate measurement by applying consistent pressure
Solution:
Total reading = Main scale reading + (Vernier scale division × Least count)
Main scale reading = 7.5 cm
Vernier division = 9
Least count = 0.02 cm
Total reading = 7.5 cm + (9 × 0.02 cm)
Total reading = 7.5 cm + 0.18 cm
Total reading = 7.68 cm
Answer: 7.68 cm
Solution:
Least count = (Value of 1 main scale division) ÷ (Number of Vernier scale divisions)
Main scale division = 1 mm
20 Vernier scale divisions = 19 mm (given)
Least count = (19 mm ÷ 20)
Least count = 0.95 mm
Now, subtract from 1 main scale division to get the least count:
Least count = 1 mm - 0.95 mm
Least count = 0.05 mm
Answer: 0.05 mm
Solution:
Total reading = Main scale reading + (Thimble reading × Least count)
Main scale reading = 10 mm
Thimble reading = 23
Least count = 0.01 mm
Zero error = -0.02 mm (negative zero error)
Total reading = 10 mm + (23 × 0.01 mm)
Total reading = 10 mm + 0.23 mm
Total reading = 10.23 mm
Correct reading = Total reading - Zero error
Correct reading = 10.23 mm - (-0.02 mm)
Correct reading = 10.25 mm
Answer: 10.25 mm
TO STUDY AND CHECK EXPERIMENTALLY THE GRAPHICAL AND ANALYTICAL METHODS OF DETERMINING THE RESULTANT OF THE GIVEN VECTORS
FORCE TABLE
WEIGHT HANGERS
SET OF SLOTTED MASSES
RULER
PROTRACTOR
A) A vector that represents the sum of two or more vectors.
B) A vector that is always perpendicular to other vectors.
C) A vector that represents the difference between two vectors.
D) A vector with zero magnitude.
Answer: A) A vector that represents the sum of two or more vectors.
A) Parallel
B) Anti-parallel
C) Equal in magnitude but opposite in direction
D) At right angles
Answer: C) Equal in magnitude but opposite in direction
A) Equal to the magnitude of one of the vectors
B) Twice the magnitude of one of the vectors
C) √2 times the magnitude of one of the vectors
D) Half the magnitude of one of the vectors
Answer: C) √2 times the magnitude of one of the vectors
A) It is always horizontal
B) It depends on the magnitudes and directions of the vectors
C) It is always vertical
D) It is always along one of the original vectors
Answer: B) It depends on the magnitudes and directions of the vectors
A) Law of Sines
B) Law of Cosines
C) Pythagorean Theorem
D) Newton's First Law
Answer: B) Law of Cosines
A) Tip-to-tail
B) Tail-to-tail
C) Parallel to each other
D) Head-to-head
Answer: A) Tip-to-tail
A) The vector sum
B) The difference of the vectors
C) The scalar product of the vectors
D) A unit vector
Answer: A) The vector sum
A) It is always larger than the magnitude of each individual vector
B) It is the maximum when two vectors are in the same direction
C) It is the minimum when two vectors are in opposite directions
D) It depends on the angle between the vectors
Answer: A) It is always larger than the magnitude of each individual vector
A) Zero
B) Equal to the sum of the two vectors
C) Equal to the difference of the two vectors
D) Twice the magnitude of one of the vectors
Answer: C) Equal to the difference of the two vectors
A) Adding their magnitudes directly
B) Subtracting their magnitudes
C) Using the Pythagorean theorem
D) Multiplying their magnitudes
Answer: C) Using the Pythagorean theorem
IN THIS ACTIVITY, THE MOVING OBJECT IS A STEEL BALL AND THE MEDIUM IS OIL WITHIN A GLASS TUBE
WE WISH TO VERIFY THE RELATIONSHIP OF DISTANCE VS. TIME FOR THE METAL BALL MOVING AT A CONSTANT SPEED BY GRAPHICAL ANALYSIS
SLOW BALL APPARATUS AND MILLESECONDS TIMER
A) Speed that changes over time.
B) Speed that increases uniformly over time.
C) Speed that decreases uniformly over time.
D) Speed that remains constant over time.
Answer: D) Speed that remains constant over time.
A) The object’s velocity is changing.
B) The object covers equal distances in unequal intervals of time.
C) The object covers equal distances in equal intervals of time.
D) The object accelerates at a constant rate.
Answer: C) The object covers equal distances in equal intervals of time.
A) 120 km
B) 180 km
C) 200 km
D) 240 km
Answer: B) 180 km
A) Zero
B) Constant and non-zero
C) Increasing
D) Decreasing
Answer: A) Zero
A) A curved line
B) A straight horizontal line
C) A straight sloping line
D) A vertical line
Answer: C) A straight sloping line
A) 5 seconds
B) 10 seconds
C) 15 seconds
D) 20 seconds
Answer: B) 10 seconds
A) Speedometer
B) Stopwatch
C) Thermometer
D) Ammeter
Answer: B) Stopwatch
A) 5 m/s
B) 8 m/s
C) 10 m/s
D) 15 m/s
Answer: C) 10 m/s
A) A car speeding up on the highway.
B) A cyclist moving at a steady pace on a flat road.
C) A rocket accelerating into space.
D) A ball falling freely under gravity.
Answer: B) A cyclist moving at a steady pace on a flat road.
A) Increasing speed readings over time.
B) Decreasing speed readings over time.
C) Constant speed readings at regular time intervals.
D) Speed readings fluctuating at random intervals.
Answer: C) Constant speed readings at regular time intervals.
TO DETERMINE THE ACCELERATION BY GRAPHICAL AND THEORETICAL ANALYSIS
LINEAR AIR TRACK SET UP
MILLISECONDS TIMER
A) Acceleration that changes over time.
B) Acceleration that remains constant over time.
C) Acceleration that increases with distance.
D) Acceleration that decreases over time.
Answer: B) Acceleration that remains constant over time.
A) Zero
B) Constant and directed along the incline
C) Increasing
D) Decreasing
Answer: B) Constant and directed along the incline
A) Distance
B) Speed
C) Time
D) Mass
Answer: B) Speed
A) A straight line on a distance-time graph
B) A straight line on a velocity-time graph
C) A horizontal line on an acceleration-time graph
D) A curved line on a distance-time graph
Answer: D) A curved line on a distance-time graph
A) The mass of the object
B) The length of the inclined plane
C) The angle of inclination of the plane
D) The color of the object
Answer: C) The angle of inclination of the plane
A) 2 m/s
B) 4 m/s
C) 6 m/s
D) 8 m/s
Answer: D) 8 m/s
A) The speed of the object
B) The angle of inclination of the plane
C) The height of the plane
D) The time taken by the object to reach the bottom
Answer: B) The angle of inclination of the plane
A) It will be halved.
B) It will double.
C) It will remain the same.
D) It will quadruple.
Answer: B) It will double.
A) Curved
B) Horizontal line
C) Straight sloping line
D) Vertical line
Answer: C) Straight sloping line
A) Meter stick
B) Stopwatch
C) Spring balance
D) Thermometer
Answer: B) Stopwatch
IN THIS ACTIVITY, YOU ARE TO CALCULATE THE VALUE Of g BY MEASURING THE TIMES, t, TAKEN TO FALL VARIOUS DISTANCES, D.
FREE FALL APPARATUS
A) Uniform motion
B) Uniformly accelerated motion
C) Non-uniform motion
D) Circular motion
Answer: B) Uniformly accelerated motion
A) 9.8 m/s²
B) 10 m/s²
C) 9.81 m/s²
D) 11 m/s²
Answer: A) 9.8 m/s²
A) 2.0 seconds
B) 3.0 seconds
C) 4.0 seconds
D) 5.0 seconds
Answer: B) 3.0 seconds
A) At the moment of release
B) At the maximum height
C) Just before hitting the ground
D) At halfway up its trajectory
Answer: B) At the maximum height
A) It increases
B) It decreases
C) It becomes zero
D) It changes direction
Answer: C) It becomes zero
A) The speed remains constant.
B) The speed increases.
C) The speed decreases.
D) The speed fluctuates.
Answer: B) The speed increases.
A) 9.8 m/s
B) 14.6 m/s
C) 5 m/s
D) 19.6 m/s
Answer: B) 14.6 m/s
A) Normal force
B) Gravitational force
C) Frictional force
D) Tension force
Answer: B) Gravitational force
A) It has no effect.
B) It increases their acceleration.
C) It decreases their acceleration.
D) It makes them fall faster.
Answer: C) It decreases their acceleration.
A) Heavier objects fall faster.
B) Lighter objects fall faster.
C) All objects fall with the same acceleration regardless of mass.
D) Mass has no effect on free fall.
Answer: C) All objects fall with the same acceleration regardless of mass.
TO DETERMINE BY GRAPHICAL ANALYSIS THE RELATIONSHIP BETWEEN THE FORCE AND THE ACCELERATION OF AN OBJECT WITH CONSTANT MASS
LINEAR AIR TRACK SET UP
SET OF MASSES
MILLISECONDS TIMER
A) F=m+aF = m + aF=m+a
B) F=maF = maF=ma
C) F=m/aF = m/aF=m/a
D) F=a/mF = a/mF=a/m
Answer: B) F=maF = maF=ma
A) 2 m/s²
B) 4 m/s²
C) 5 m/s²
D) 10 m/s²
Answer: B) 4 m/s²
A) The force of gravity acting on an object.
B) The mass of an object.
C) The resistance to motion.
D) The acceleration of an object.
Answer: A) The force of gravity acting on an object.
A) The net force is zero.
B) The net force is positive.
C) The net force is negative.
D) The net force is constant.
Answer: A) The net force is zero.
A) 1000 N
B) 1500 N
C) 3000 N
D) 3500 N
Answer: C) 3000 N
A) 1.0 m/s²
B) 2.0 m/s²
C) 3.0 m/s²
D) 4.0 m/s²
Answer: C) 3.0 m/s²
A) The mass of the object.
B) The net force acting on the object.
C) The distance traveled by the object.
D) The direction of the net force.
Answer: C) The distance traveled by the object.
A) 6 N
B) 8 N
C) 10 N
D) 14 N
Answer: B) 8 N
A) Mass of the object
B) Acceleration
C) Force
D) Velocity
Answer: A) Mass of the object
A) The acceleration remains the same.
B) The acceleration is halved.
C) The acceleration is doubled.
D) The acceleration is quadrupled.
Answer: C) The acceleration is doubled.
TO DETERMINE BY GRAPHICAL ANALYSIS THE RELATIONSHIP BETWEEN THE MASS (m) AND THR ACCELERATION (a) OF AN OBJECT WITH THE FORCE ACTING ON IT BEING KEPT CONSTANT.
LINEAR AIR TRACK SET UP
SET OF MASSES
MILLISECONDS TIMER
A) The net force acting on the object is zero.
B) The object must be in free fall.
C) The mass of the object is increasing.
D) The velocity of the object is changing.
Answer: D) The velocity of the object is changing.
A) The feather
B) The hammer
C) Both hit the ground at the same time
D) It depends on the mass of each object
Answer: C) Both hit the ground at the same time
A) They experience the same net force.
B) They experience different net forces.
C) The object with greater acceleration must have a smaller mass.
D) The object with smaller acceleration must have a larger mass.
Answer: B) They experience different net forces.
A) Weight is mass multiplied by acceleration.
B) Weight is mass divided by acceleration due to gravity.
C) Weight is the mass times the gravitational acceleration (W=mgW = mgW=mg).
D) Weight is independent of mass.
Answer: C) Weight is the mass times the gravitational acceleration (W=mgW = mgW=mg).
A) 5 N
B) 10 N
C) 15 N
D) 20 N
Answer: D) 10 N
A) 6 m/s²
B) 7.2 m/s²
C) 8 m/s²
D) 9.8 m/s²
Answer: B) 8 m/s²
A) 3 m/s²
B) 4 m/s²
C) 5 m/s²
D) 6 m/s²
Answer: C) 5 m/s²
A) The acceleration doubles.
B) The acceleration is halved.
C) The acceleration remains the same.
D) The acceleration quadruples.
Answer: A) The acceleration doubles.
A) 30 N
B) 40 N
C) 50 N
D) 10 N
Answer: A) 30 N
A) 5 N
B) 10 N
C) 15 N
D) 25 N
Answer: D) 25 N
TO UNDERSTAND THE CONCEPT OF ENERGY AND ITS CONVERSION
CURVE TRACK
BOND PAPER
MARBLE
PAPER TAPE
METER STICK
CARBON PAPER
A) Energy can be created or destroyed.
B) Energy can be transformed from one form to another, but the total energy remains constant.
C) Energy can only exist in kinetic or potential forms.
D) Energy is always lost as heat during conversion.
Answer: B) Energy can be transformed from one form to another, but the total energy remains constant.
A) 49 J
B) 98 J
C) 196 J
D) 245 J
Answer: B) 98 J
(Calculation: PE=mgh=2 kg×9.8 m/s2×5 m=98 JPE = mgh = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5 \, \text{m} = 98 \, \text{J}PE=mgh=2kg×9.8m/s2×5m=98J)
A) Potential energy to kinetic energy.
B) Kinetic energy to thermal energy.
C) Chemical energy to kinetic energy.
D) Thermal energy to mechanical energy.
Answer: C) Chemical energy to kinetic energy.
A) It is lost as heat.
B) It is converted to kinetic energy.
C) It increases.
D) It stays the same.
Answer: B) It is converted to kinetic energy.
A) Electrical energy to mechanical energy.
B) Electrical energy to light and thermal energy.
C) Thermal energy to electrical energy.
D) Mechanical energy to chemical energy.
Answer: B) Electrical energy to light and thermal energy.
A) Kinetic energy to thermal energy.
B) Thermal energy to electrical energy.
C) Mechanical energy to electrical energy.
D) Chemical energy to electrical energy.
Answer: C) Mechanical energy to electrical energy.
A) 50 J
B) 100 J
C) 150 J
D) 200 J
Answer: D) 200 J
(Calculation: Total Mechanical Energy =KE+PE=150 J+50 J=200 J= KE + PE = 150 \, \text{J} + 50 \, \text{J} = 200 \, \text{J}=KE+PE=150J+50J=200J)
A) It increases.
B) It decreases.
C) It remains constant.
D) It is converted to sound energy.
Answer: C) It remains constant.
A) 2000 J
B) 4000 J
C) 8000 J
D) 200000 J
Answer: C) 200000 J
(Calculation: KE=12mv2=12×1000 kg×(20 m/s)2=200000 JKE = \frac{1}{2} mv^2 = \frac{1}{2} \times 1000 \, \text{kg} \times (20 \, \text{m/s})^2 = 200000 \, \text{J}KE=21mv2=21×1000kg×(20m/s)2=200000J)
A) Potential energy to kinetic energy.
B) Kinetic energy to potential energy.
C) Thermal energy to kinetic energy.
D) Chemical energy to potential energy.
Answer: A) Potential energy to kinetic energy.
TO KNOW HOW TO DETERMINE THE LEAST COUNT OF THE CALIPER USED AND
TO LEARN HOW TO USE THE VERNIER AND MICROMETER CALIPERS CORRECTLY
VERNIER CALIPER
MICROMETER CALIPER
CYLINDER
BEAM BALANCE
SPHERE
CUBE
WASHER
AN AUXILIARY SLIDING SCALE WHICH ENABLES ONE TO INCREASE HIS ACCURACY IN MEASUREMENT.
Total Measurement = Main Scale Reading + (Sequential Vernier Scale Marking × Least Count)
EXAMPLE 1:
EXAMPLE 2:
PRECISE MEASUREMENT OF SMALL LENGTHS AND DIAMETERS.
FOR EVERY ONE REVOLUTION(5O divisions) OF THE SCREW, THE JAW ADVANCES HALF A MILLIMETER.
Total Measurement = Main Scale Reading + (Sequential Micrometer Scale Marking × Least Count)
note: diameter=r/2
DENSITY=mass(g)/volume (cm³)
CUBE= S³
CYLINDER=πr²h
SPHERE: 4/3πr³
WASHER: πrout²-rin²)t
A) 0.1 mm
B) 0.01 mm
C) 0.02 mm
D) 0.5 mm
Answer: C) 0.02 mm
A) 10
B) 20
C) 50
D) 100
Answer: B) 20
A) 0.1 mm
B) 0.01 mm
C) 0.02 mm
D) 0.001 mm
Answer: B) 0.01 mm
A) Main scale
B) Vernier scale
C) Lower jaws
D) Upper jaws
Answer: D) Upper jaws
A) Measures larger objects
B) Easier to use
C) More accurate for smaller dimensions
D) Can measure both internal and external dimensions
Answer: C) More accurate for smaller dimensions
A) Subtract the Vernier reading from the main scale
B) Add the Vernier reading to the main scale
C) Multiply the Vernier reading by the least count
D) Divide the Vernier reading by the least count
Answer: B) Add the Vernier reading to the main scale
A) The number of divisions on the sleeve
B) The distance the spindle moves per full rotation of the thimble
C) The smallest reading on the thimble
D) The length of the thimble
Answer: B) The distance the spindle moves per full rotation of the thimble
A) By closing the jaws completely and checking the alignment of the zero mark on the Vernier scale with the zero on the main scale
B) By rotating the Vernier scale
C) By measuring a standard object
D) By adjusting the main scale
Answer: A) By closing the jaws completely and checking the alignment of the zero mark on the Vernier scale with the zero on the main scale
A) Add the error value to the measured value
B) Subtract the error value from the measured value
C) Multiply the measured value by the error
D) Ignore the error
Answer: A) Add the error value to the measured value
A) To ensure accurate measurement by applying consistent pressure
B) To lock the spindle in place
C) To adjust the zero reading
D) To reset the scale to zero
Answer: A) To ensure accurate measurement by applying consistent pressure
Solution:
Total reading = Main scale reading + (Vernier scale division × Least count)
Main scale reading = 7.5 cm
Vernier division = 9
Least count = 0.02 cm
Total reading = 7.5 cm + (9 × 0.02 cm)
Total reading = 7.5 cm + 0.18 cm
Total reading = 7.68 cm
Answer: 7.68 cm
Solution:
Least count = (Value of 1 main scale division) ÷ (Number of Vernier scale divisions)
Main scale division = 1 mm
20 Vernier scale divisions = 19 mm (given)
Least count = (19 mm ÷ 20)
Least count = 0.95 mm
Now, subtract from 1 main scale division to get the least count:
Least count = 1 mm - 0.95 mm
Least count = 0.05 mm
Answer: 0.05 mm
Solution:
Total reading = Main scale reading + (Thimble reading × Least count)
Main scale reading = 10 mm
Thimble reading = 23
Least count = 0.01 mm
Zero error = -0.02 mm (negative zero error)
Total reading = 10 mm + (23 × 0.01 mm)
Total reading = 10 mm + 0.23 mm
Total reading = 10.23 mm
Correct reading = Total reading - Zero error
Correct reading = 10.23 mm - (-0.02 mm)
Correct reading = 10.25 mm
Answer: 10.25 mm
TO STUDY AND CHECK EXPERIMENTALLY THE GRAPHICAL AND ANALYTICAL METHODS OF DETERMINING THE RESULTANT OF THE GIVEN VECTORS
FORCE TABLE
WEIGHT HANGERS
SET OF SLOTTED MASSES
RULER
PROTRACTOR
A) A vector that represents the sum of two or more vectors.
B) A vector that is always perpendicular to other vectors.
C) A vector that represents the difference between two vectors.
D) A vector with zero magnitude.
Answer: A) A vector that represents the sum of two or more vectors.
A) Parallel
B) Anti-parallel
C) Equal in magnitude but opposite in direction
D) At right angles
Answer: C) Equal in magnitude but opposite in direction
A) Equal to the magnitude of one of the vectors
B) Twice the magnitude of one of the vectors
C) √2 times the magnitude of one of the vectors
D) Half the magnitude of one of the vectors
Answer: C) √2 times the magnitude of one of the vectors
A) It is always horizontal
B) It depends on the magnitudes and directions of the vectors
C) It is always vertical
D) It is always along one of the original vectors
Answer: B) It depends on the magnitudes and directions of the vectors
A) Law of Sines
B) Law of Cosines
C) Pythagorean Theorem
D) Newton's First Law
Answer: B) Law of Cosines
A) Tip-to-tail
B) Tail-to-tail
C) Parallel to each other
D) Head-to-head
Answer: A) Tip-to-tail
A) The vector sum
B) The difference of the vectors
C) The scalar product of the vectors
D) A unit vector
Answer: A) The vector sum
A) It is always larger than the magnitude of each individual vector
B) It is the maximum when two vectors are in the same direction
C) It is the minimum when two vectors are in opposite directions
D) It depends on the angle between the vectors
Answer: A) It is always larger than the magnitude of each individual vector
A) Zero
B) Equal to the sum of the two vectors
C) Equal to the difference of the two vectors
D) Twice the magnitude of one of the vectors
Answer: C) Equal to the difference of the two vectors
A) Adding their magnitudes directly
B) Subtracting their magnitudes
C) Using the Pythagorean theorem
D) Multiplying their magnitudes
Answer: C) Using the Pythagorean theorem
IN THIS ACTIVITY, THE MOVING OBJECT IS A STEEL BALL AND THE MEDIUM IS OIL WITHIN A GLASS TUBE
WE WISH TO VERIFY THE RELATIONSHIP OF DISTANCE VS. TIME FOR THE METAL BALL MOVING AT A CONSTANT SPEED BY GRAPHICAL ANALYSIS
SLOW BALL APPARATUS AND MILLESECONDS TIMER
A) Speed that changes over time.
B) Speed that increases uniformly over time.
C) Speed that decreases uniformly over time.
D) Speed that remains constant over time.
Answer: D) Speed that remains constant over time.
A) The object’s velocity is changing.
B) The object covers equal distances in unequal intervals of time.
C) The object covers equal distances in equal intervals of time.
D) The object accelerates at a constant rate.
Answer: C) The object covers equal distances in equal intervals of time.
A) 120 km
B) 180 km
C) 200 km
D) 240 km
Answer: B) 180 km
A) Zero
B) Constant and non-zero
C) Increasing
D) Decreasing
Answer: A) Zero
A) A curved line
B) A straight horizontal line
C) A straight sloping line
D) A vertical line
Answer: C) A straight sloping line
A) 5 seconds
B) 10 seconds
C) 15 seconds
D) 20 seconds
Answer: B) 10 seconds
A) Speedometer
B) Stopwatch
C) Thermometer
D) Ammeter
Answer: B) Stopwatch
A) 5 m/s
B) 8 m/s
C) 10 m/s
D) 15 m/s
Answer: C) 10 m/s
A) A car speeding up on the highway.
B) A cyclist moving at a steady pace on a flat road.
C) A rocket accelerating into space.
D) A ball falling freely under gravity.
Answer: B) A cyclist moving at a steady pace on a flat road.
A) Increasing speed readings over time.
B) Decreasing speed readings over time.
C) Constant speed readings at regular time intervals.
D) Speed readings fluctuating at random intervals.
Answer: C) Constant speed readings at regular time intervals.
TO DETERMINE THE ACCELERATION BY GRAPHICAL AND THEORETICAL ANALYSIS
LINEAR AIR TRACK SET UP
MILLISECONDS TIMER
A) Acceleration that changes over time.
B) Acceleration that remains constant over time.
C) Acceleration that increases with distance.
D) Acceleration that decreases over time.
Answer: B) Acceleration that remains constant over time.
A) Zero
B) Constant and directed along the incline
C) Increasing
D) Decreasing
Answer: B) Constant and directed along the incline
A) Distance
B) Speed
C) Time
D) Mass
Answer: B) Speed
A) A straight line on a distance-time graph
B) A straight line on a velocity-time graph
C) A horizontal line on an acceleration-time graph
D) A curved line on a distance-time graph
Answer: D) A curved line on a distance-time graph
A) The mass of the object
B) The length of the inclined plane
C) The angle of inclination of the plane
D) The color of the object
Answer: C) The angle of inclination of the plane
A) 2 m/s
B) 4 m/s
C) 6 m/s
D) 8 m/s
Answer: D) 8 m/s
A) The speed of the object
B) The angle of inclination of the plane
C) The height of the plane
D) The time taken by the object to reach the bottom
Answer: B) The angle of inclination of the plane
A) It will be halved.
B) It will double.
C) It will remain the same.
D) It will quadruple.
Answer: B) It will double.
A) Curved
B) Horizontal line
C) Straight sloping line
D) Vertical line
Answer: C) Straight sloping line
A) Meter stick
B) Stopwatch
C) Spring balance
D) Thermometer
Answer: B) Stopwatch
IN THIS ACTIVITY, YOU ARE TO CALCULATE THE VALUE Of g BY MEASURING THE TIMES, t, TAKEN TO FALL VARIOUS DISTANCES, D.
FREE FALL APPARATUS
A) Uniform motion
B) Uniformly accelerated motion
C) Non-uniform motion
D) Circular motion
Answer: B) Uniformly accelerated motion
A) 9.8 m/s²
B) 10 m/s²
C) 9.81 m/s²
D) 11 m/s²
Answer: A) 9.8 m/s²
A) 2.0 seconds
B) 3.0 seconds
C) 4.0 seconds
D) 5.0 seconds
Answer: B) 3.0 seconds
A) At the moment of release
B) At the maximum height
C) Just before hitting the ground
D) At halfway up its trajectory
Answer: B) At the maximum height
A) It increases
B) It decreases
C) It becomes zero
D) It changes direction
Answer: C) It becomes zero
A) The speed remains constant.
B) The speed increases.
C) The speed decreases.
D) The speed fluctuates.
Answer: B) The speed increases.
A) 9.8 m/s
B) 14.6 m/s
C) 5 m/s
D) 19.6 m/s
Answer: B) 14.6 m/s
A) Normal force
B) Gravitational force
C) Frictional force
D) Tension force
Answer: B) Gravitational force
A) It has no effect.
B) It increases their acceleration.
C) It decreases their acceleration.
D) It makes them fall faster.
Answer: C) It decreases their acceleration.
A) Heavier objects fall faster.
B) Lighter objects fall faster.
C) All objects fall with the same acceleration regardless of mass.
D) Mass has no effect on free fall.
Answer: C) All objects fall with the same acceleration regardless of mass.
TO DETERMINE BY GRAPHICAL ANALYSIS THE RELATIONSHIP BETWEEN THE FORCE AND THE ACCELERATION OF AN OBJECT WITH CONSTANT MASS
LINEAR AIR TRACK SET UP
SET OF MASSES
MILLISECONDS TIMER
A) F=m+aF = m + aF=m+a
B) F=maF = maF=ma
C) F=m/aF = m/aF=m/a
D) F=a/mF = a/mF=a/m
Answer: B) F=maF = maF=ma
A) 2 m/s²
B) 4 m/s²
C) 5 m/s²
D) 10 m/s²
Answer: B) 4 m/s²
A) The force of gravity acting on an object.
B) The mass of an object.
C) The resistance to motion.
D) The acceleration of an object.
Answer: A) The force of gravity acting on an object.
A) The net force is zero.
B) The net force is positive.
C) The net force is negative.
D) The net force is constant.
Answer: A) The net force is zero.
A) 1000 N
B) 1500 N
C) 3000 N
D) 3500 N
Answer: C) 3000 N
A) 1.0 m/s²
B) 2.0 m/s²
C) 3.0 m/s²
D) 4.0 m/s²
Answer: C) 3.0 m/s²
A) The mass of the object.
B) The net force acting on the object.
C) The distance traveled by the object.
D) The direction of the net force.
Answer: C) The distance traveled by the object.
A) 6 N
B) 8 N
C) 10 N
D) 14 N
Answer: B) 8 N
A) Mass of the object
B) Acceleration
C) Force
D) Velocity
Answer: A) Mass of the object
A) The acceleration remains the same.
B) The acceleration is halved.
C) The acceleration is doubled.
D) The acceleration is quadrupled.
Answer: C) The acceleration is doubled.
TO DETERMINE BY GRAPHICAL ANALYSIS THE RELATIONSHIP BETWEEN THE MASS (m) AND THR ACCELERATION (a) OF AN OBJECT WITH THE FORCE ACTING ON IT BEING KEPT CONSTANT.
LINEAR AIR TRACK SET UP
SET OF MASSES
MILLISECONDS TIMER
A) The net force acting on the object is zero.
B) The object must be in free fall.
C) The mass of the object is increasing.
D) The velocity of the object is changing.
Answer: D) The velocity of the object is changing.
A) The feather
B) The hammer
C) Both hit the ground at the same time
D) It depends on the mass of each object
Answer: C) Both hit the ground at the same time
A) They experience the same net force.
B) They experience different net forces.
C) The object with greater acceleration must have a smaller mass.
D) The object with smaller acceleration must have a larger mass.
Answer: B) They experience different net forces.
A) Weight is mass multiplied by acceleration.
B) Weight is mass divided by acceleration due to gravity.
C) Weight is the mass times the gravitational acceleration (W=mgW = mgW=mg).
D) Weight is independent of mass.
Answer: C) Weight is the mass times the gravitational acceleration (W=mgW = mgW=mg).
A) 5 N
B) 10 N
C) 15 N
D) 20 N
Answer: D) 10 N
A) 6 m/s²
B) 7.2 m/s²
C) 8 m/s²
D) 9.8 m/s²
Answer: B) 8 m/s²
A) 3 m/s²
B) 4 m/s²
C) 5 m/s²
D) 6 m/s²
Answer: C) 5 m/s²
A) The acceleration doubles.
B) The acceleration is halved.
C) The acceleration remains the same.
D) The acceleration quadruples.
Answer: A) The acceleration doubles.
A) 30 N
B) 40 N
C) 50 N
D) 10 N
Answer: A) 30 N
A) 5 N
B) 10 N
C) 15 N
D) 25 N
Answer: D) 25 N
TO UNDERSTAND THE CONCEPT OF ENERGY AND ITS CONVERSION
CURVE TRACK
BOND PAPER
MARBLE
PAPER TAPE
METER STICK
CARBON PAPER
A) Energy can be created or destroyed.
B) Energy can be transformed from one form to another, but the total energy remains constant.
C) Energy can only exist in kinetic or potential forms.
D) Energy is always lost as heat during conversion.
Answer: B) Energy can be transformed from one form to another, but the total energy remains constant.
A) 49 J
B) 98 J
C) 196 J
D) 245 J
Answer: B) 98 J
(Calculation: PE=mgh=2 kg×9.8 m/s2×5 m=98 JPE = mgh = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5 \, \text{m} = 98 \, \text{J}PE=mgh=2kg×9.8m/s2×5m=98J)
A) Potential energy to kinetic energy.
B) Kinetic energy to thermal energy.
C) Chemical energy to kinetic energy.
D) Thermal energy to mechanical energy.
Answer: C) Chemical energy to kinetic energy.
A) It is lost as heat.
B) It is converted to kinetic energy.
C) It increases.
D) It stays the same.
Answer: B) It is converted to kinetic energy.
A) Electrical energy to mechanical energy.
B) Electrical energy to light and thermal energy.
C) Thermal energy to electrical energy.
D) Mechanical energy to chemical energy.
Answer: B) Electrical energy to light and thermal energy.
A) Kinetic energy to thermal energy.
B) Thermal energy to electrical energy.
C) Mechanical energy to electrical energy.
D) Chemical energy to electrical energy.
Answer: C) Mechanical energy to electrical energy.
A) 50 J
B) 100 J
C) 150 J
D) 200 J
Answer: D) 200 J
(Calculation: Total Mechanical Energy =KE+PE=150 J+50 J=200 J= KE + PE = 150 \, \text{J} + 50 \, \text{J} = 200 \, \text{J}=KE+PE=150J+50J=200J)
A) It increases.
B) It decreases.
C) It remains constant.
D) It is converted to sound energy.
Answer: C) It remains constant.
A) 2000 J
B) 4000 J
C) 8000 J
D) 200000 J
Answer: C) 200000 J
(Calculation: KE=12mv2=12×1000 kg×(20 m/s)2=200000 JKE = \frac{1}{2} mv^2 = \frac{1}{2} \times 1000 \, \text{kg} \times (20 \, \text{m/s})^2 = 200000 \, \text{J}KE=21mv2=21×1000kg×(20m/s)2=200000J)
A) Potential energy to kinetic energy.
B) Kinetic energy to potential energy.
C) Thermal energy to kinetic energy.
D) Chemical energy to potential energy.
Answer: A) Potential energy to kinetic energy.