eksp pd1

8/12/2019 eksp pd1

1/5

INCLINED PLANES

Aim of the experiment:

To find the acceleration of motion

Apparatus/Materials:Trolley, protractor, wooden blocks, cellophane tape, tickertimer,

ticker tape, power supply, friction-compensated runway

Setup:

Procedure:

1. The apparatus is set up as per the diagram, and the inclined angle of the plane is

measured using a protractor. An initial angle of 5 is used.

2. The ticker-timer is started up and at the same time the trolley is released to slide downthe plane.

3. The acceleration of trolley calculate from tape chart.

Results:

Calculate acceleration

8/12/2019 eksp pd1

2/5

INERTIA

Hypothesis:

The larger the mass, the bigger the inertia


To study the effect of mass on the inertia of an object

Variables:

Manipulated: Mass, m

Responding: Period of oscillation, TConstant: Stiffness of the inertia balance

Apparatus/Materials: Inertia balance, masses for the inertia balance, G-clamp,

Stopwatch

Setup:

Procedure:

1. The inertia balance is set up by clamping it onto one end of the table as shown in the

figure above.

2. One mass is placed into the inertia balance. The inertia balance is displaced to one

side so that it oscillates in a horizontal plane.

3. The time for 10 complete oscillations is measured using a stopwatch. This step is

repeated. The average of 10 oscillations is calculated. Then, the period of oscillation

is determined.

4. Steps 2 and 3 are repeated using two and three masses on the inertia balance.5. A graph of T2 versus number of masses, n is drawn.

Results:

Graph of T2 versus m:

Discussion:

The graph of T2 versus m shows a straight line passing through the origin. This means

that the period of oscillation increases with the mass of the load; that is, an object with a

large mass has a large inertia.

Conclusion:

Objects with a large mass have a large inertia. This is the reason why it is difficult to setan object of large mass in motion or to stop it. The hypothesis is valid.

8/12/2019 eksp pd1

3/5

HOOKES LAW

Hypothesis:The bigger the weight, the longer the spring extension


To determine the relationship between the weight and the spring extension

Variables:

Manipulated: Weight of the load

Responding: Spring extensionConstant: Spring constant

Apparatus and Materials:Spring, pin, weights, plasticine, retort stand, metre rule

Setup:

Procedure:

1. The apparatus is setup as shown in the diagram.

2. The length of the spring without any weights, l0 is measured using the metre rule with

the pin as reference.

3. A 50 g weight is hung from the bottom of the spring. The new length of the spring, l

is measured. The spring extension is ll0.

4. Step 4 is repeated with weights 100 g, 150 g, 200 g, and 250 g.

Results:

Original length of spring = l0 = __________ cm

Analysis:

A graph of spring extension,x against weight,F is plotted.

Thex-F graph is a linear graph which passes through the origin. This shows that the

extension of the spring is directly proportional to the stretching force.

Conclusion:Hypothesis proven.

8/12/2019 eksp pd1

4/5

PRESSURE IN LIQUIDS

Experiment 1: Water pressure and depthHypothesis:

Water pressure increases with depth


To find the relationship between the pressure in a liquid according to its depth

Variables:

Manipulated: Depth of liquidResponding: Pressure in liquid

Constant: Density of liquidApparatus and Materials:Measuring cylinder, thistle funnel, rubber tube,

manometer, metre rule

Setup:

Procedure:

1. Apparatus is set up as shown in the diagram.

2. The measuring cylinder is completely filled with water.

3. The thistle funnel is lowered into the water to a depth of 10.0 cm. The manometer

reading is measured. The difference in the liquid heights in the manometer represent

the pressure reading.

4. Step 3 is repeated with values of depth 20.0 cm, 30.0 cm, 40.0 cm and 50.0 cm.

Results:

Analysis:

A graph of pressure against depth is drawn.

Conclusion:

It is observed that the manometer reading increases as the depth of the thistle funnelincreases. This shows that the pressure increases with the depth of the liquid.

Hypothesis proven.

8/12/2019 eksp pd1

5/5

ARCHIMEDES PRINCIPLE

Hypothesis:The buoyant force on an object in a liquid is equal to the weight of the liquid

displaced


To find the relationship between the buoyant force acting upon an object in a liquid

and the weight of the liquid displaced

Variables:Manipulated: Weight of the object

Responding: Buoyant force / Weight of liquid displaced

Constant: Density of liquid used

Apparatus and Materials:Eureka tin, spring balance, stone, thread, beaker, triple

beam balance

Setup:

Procedure:

1. A beaker is weighed with the triple beam balance and its mass, m1 is recorded.

2. The Eureka tin is filled with water right up to the level of the overflow hole. Thebeaker is placed beneath the spout to catch any water that flows out.

3. A stone is suspended from the spring balance with thread and its weight in air, W1 is

read from the spring balance.

4. The stone is lowered into the Eureka tin until it is completely immersed in water

without touching the bottom of the Eureka tin. The water will overflow into the

beaker.

5. The spring balance reading, W2 is recorded.

6. The beaker with water is weighed with the triple beam balance, and the mass, m2 is

recorded.

Results:Weight of stone in air = W1=

Weight of stone in water = W2=..

Buoyant force acting on the stone = W2W1= ..

Weight of the empty beaker = m1g=.

Weight of the beaker and displaced water = m2g=.

Weight of the displaced water = (m2m1)g=.

It is found that W2W1= (m2m1)g=

Discussion:

The loss of weight of the stone immersed in water is due to the buoyant force of the water

acting upon it.

From the results, it is found that the loss in weight of the stone is equal to the weight of

water displaced.

Conclusion:

Buoyant force on the stone = Weight of the water displaced by the stone

Hypothesis proven.

eksp pd1

Documents