Science Practical Manual
10th Standard
Practical Sets Index
SET - A
1. DETERMINATION OF WEIGHT OF AN OBJECT USING THE PRINCIPLE OF MOMENTS (PHYSICS)
6. TESTING THE WATER OF HYDRATION OF SALT (CHEMISTRY)
8. PHOTOSYNTHESIS – TEST TUBE AND FUNNEL EXPERIMENT (BIO – BOTANY)
13. IDENTIFICAION OF BLOOD CELLS (BIO-ZOOLOGY)
SET – B
2. DETERMINATION OF FOCAL LENGTH OF A CONVEX LENS (PHYSICS)
5. TESTING THE SOLUBILITY OF THE SALT (CHEMISTRY)
10. TO STUDY THE LAW OF DOMINANCE (BIO – BOTANY)
12. IDENTIFICATION OF MODELS – HUMAN HEART AND HUMAN BRAIN (BIO-ZOOLOGY)
SET – C
1. DETERMINATION OF WEIGHT OF AN OBJECT USING THE PRINCIPLE OF MOMENTS (PHYSICS)
4. IDENTIFY THE DISSOLUTION OF GIVEN SALT WHETHER IT IS EXOTHERMIC OR ENDOTHERMIC (CHEMISTRY)
9. PARTS OF FLOWER (BIO – BOTANY)
13. IDENTIFICAION OF BLOOD CELLS (BIO-ZOOLOGY)
SET – D
3. DETERMINATION OF RESISTIVITY (PHYSICS)
7. TEST THE GIVEN SAMPLE FOR THE PRESENCE OF ACID OR BASE (CHEMISTRY)
11. OBSERVATION OF TRANSVESE SECTION OF DICOT STEM AND DICOT ROOT (BIO – BOTANY)
14. IDENTIFICATION OF ENDOCRINE GLANDS (BIO-ZOOLOGY)
PHYSICS
To determine the weight of an object using the principle of moments
Apparatus required:A metre scale, a knife edge, slotted weights and thread
Principle:Clock wise moment (W1×d1) = Anticlockwise moment (W2×d2)
Formula:$$ W_1 = \frac{W_2 \times d_2}{d_1} \text{ kg} $$
W1 – Unknown weight
W2 – Known weight
d1 – Distance of unknown weight
d2 – Distance of known weight
(i) A metre scale is supported at its centre of gravity by a knife edge and ensure that the scale is in equilibrium position.
(ii) A known weight W2 and unknown weight W1 are suspended from to either side of the scale using the weight hangers.
(iii) Fix the position of one weight hanger and adjust the position of the second weight hanger such that the scale is in equilibrium.
(iv) Measure the distance d1 and d2 of the two weight hangers.
(v) The experiment is repeated for the different positions of the unknown weight. Measure the distances. The reading is tabulated as follows.
| S. No | Known weight W2 (kg) | Distance of known weight d2 (x10-2 m) | Distance of unknown weight d1 (x10-2 m) | W2×d2 (x10-2 kg m) | Unknown weight W1 = (W2×d2)/d1 (kg) |
|---|---|---|---|---|---|
| 1 | 0.1 | 15 | 10 | 1.5 | 0.15 |
| 2 | 0.1 | 22 | 15 | 2.2 | 0.15 |
| Mean | 0.15 | ||||
Using the principle of moments, the weight of the unknown body W1 = 150 x 10-3 kg
To determine the focal length of a convex lens by using
1. Distant object method
2. U - V method
A convex lens, stand, wire gauze object, screen and measuring scale
Formula:$$ f = \frac{uv}{(u+v)} \text{ m} $$
f – focal length of convex lens
u – distance between the object and the lens
v – distance between the image and the lens
1. Distance object method:
Fix the given lens to the stand and place the screen behind the lens. Move the lens back and forth to capture the clear image of the object. Measure the distance between lens and screen. This is focal length (f) of the convex lens.
2. U - V method:
Fix the lens into the stand and place the wire gauze object at the specified distance to the left side of the lens. Measure the distance between the lens and the object (u). place the screen on the right side of the lens and capture the clear image on the screen. Measure the distance between the lens and the image (v). Repeat the same procedure by changing ‘u’ and tabulate the observations.
Observation:Focal length of the convex lens (by distance object method) is (f) = 10 cm
(2f) = 20 cm
| S. No | Position of the object | Distance of the Object (cm) | Distance of the Image (cm) | Focal length of the convex lens f = uv/(u+v) cm |
|---|---|---|---|---|
| 1 | u>2f | 22 | 19 | 10.20 |
| 2 | u=2f | 20 | 21 | 10.24 |
| 3 | u<2f | 18 | 24 | 10.29 |
| Mean | 10.24 | |||
The focal length of the given convex lens
1. By distance object method f = 10.00 x 10-2 m
2. By u-v method f = 10.24 x 10-2 m
To determine the resistivity of the material of the given coil of wire
Apparatus required:A coil wire, battery, key, ammeter, voltmeter, rheostat, a metre scale and screw gauge
Formula:$$ Resistivity \space \rho = \left( \frac{RA}{L} \right) \Omega m $$
A – Area of the cross section of the wire
L – Length of the coil wire
R – Resistance of the coil wire
(i) According to the picture the circuit should be installed. Close the key and hence the circuit is closed.
(ii) The potential difference should be noted in the table for the change of the rheostat and for different current measurements.
(iii) Measure the diameter of the wire using a screw gauge.
(iv) Measure the length of the coil using meter scale.
(I) To find the resistance
| S. No | Ammeter reading – I (A) | Volt meter reading – V (V) | Resistance R= V/I (Ω) |
|---|---|---|---|
| 1 | 1.0 | 1.5 | 1.50 |
| 2 | 1.5 | 2.2 | 1.47 |
| Mean | 1.49 | ||
(II) To find the diameter of the wire using screw gauge
Least Count: 0.01 mm Zero Error: No Error
| S. No | Pitch scale reading PSR (mm) | Head scale coincidence (HSC) | Head scale reading HSR=(PSR×LC) ± ZE (mm) | Corrected reading PSR+HSR (mm) |
|---|---|---|---|---|
| 1 | 0 | 39 | 0.39 | 0.39 |
| 2 | 0 | 37 | 0.37 | 0.37 |
| Mean | 0.38 | |||
Radius of the wire r = diameter/2 = 0.19×10-3 m
Area of the cross section of the wire $$ A = \pi r^2 = 0.11 \times 10^{-6} m^2 $$
Length of the wire L = 33 x 10-2 m
Resistivity of the wire $$ \rho = \left( \frac{RA}{L} \right) = 5 \times 10^{-7} \Omega m $$
The resistivity of the material of the wire = 5 ×10-7 Ωm
CHEMISTRY
To test the dissolution of given salt is exothermic or endothermic
Material required:Two beakers, Thermometer, stirrer and two samples
Principle:If the reaction liberates the heat, then it is called exothermic
If the reaction absorbs the heat, then it is called endothermic
(i) Take 50 ml of water in two beakers and label them as A and B. Note the temperature of the water from the beaker A and B.
(ii) Then, add 5 g of sample A into the beaker A and stir well until it dissolved completely. Record final temperature of the solution.
(iii) Now, repeat the same for sample B. Record the observation
| S. No | Sample | Temperature before addition of sample (oC) | Temperature after addition of sample (oC) | Inference |
|---|---|---|---|---|
| 1 | A | 25 | 45 | Temperature increases |
| 2 | B | 25 | 20 | Temperature decreases |
From the inferences made
The dissolution of sample A is exothermic.
The dissolution of sample B is endothermic.
To test the solubility of the given salt based on the saturation and unsaturation of the solution at a given temperature
Materials required:A 250 ml beaker, 100 ml measuring jar, a stirrer, distilled water and salt
Principle:A solution in which more solute can be dissolved in the solvent at a given temperature is called unsaturated solution
A solution in which no more solute can be dissolved in the solvent at a given temperature is called saturated solution
(i) In a 250 ml beaker pour 100 ml water using measuring jar. To this water add 25 g salt from the first packet. stir the content very well.
(ii) Add the next packet containing 11 g salt followed by constant stirring.
(iii) Now add third packet containing 1 g salt. Record the observations.
| S. No | Amount of salt added (g) | Observation (Dissolved/ Undissolved) | Inference (Unsaturated/ Saturated / Super saturated) |
|---|---|---|---|
| 1 | 25 | Dissolved | Unsaturated |
| 2 | 36 (25 + 11) | Dissolved | Saturated |
| 3 | 37 (25 + 11 + 1) | Undissolved | Super saturated |
From the above observation, it is inferred that the amount of salt required for saturation is 36 g (or) 36 x 10-3 kg
To check whether the given sample of salt possesses “Water of Hydration” or not.
Materials required:Crystalline copper sulphate salt, test tube, tongs and spirit lamp
Principle:Some salts crystallize with water molecules. This is called hydrated salt.
Procedure:| S. No | Experiment | Observation | Inference |
|---|---|---|---|
| 1 | A pinch of crystalline copper sulphate taken in a test tube and heated for sometime | Water droplets are seen on the inner walls of the test tube | The water of hydration is present |
In the given sample of salt water of hydration is present
To check whether the given sample of salt possesses “Water of Hydration” or not.
Materials required:Crystalline copper sulphate salt, test tube, tongs and spirit lamp
Principle:Some salts crystallize with water molecules. This is called hydrated salt.
Procedure:| S. No | Experiment | Observation | Inference |
|---|---|---|---|
| 1 | A pinch of crystalline copper sulphate taken in a test tube and heated for sometime | Water droplets are not seen on the inner walls of the test tube | The water of hydration is absent |
In the given sample of salt water of hydration is absent
To identify the presence of an acid or a base in a given sample
Materials required:Test-tube, test-tube stand, glass rod, indicators and the given sample
Principle:| Indicator | Acid | Base |
|---|---|---|
| Phenolphthalein | Colourless | Pink colour |
| Methyl orange | Pink colour | Yellow colour |
| Sodium carbonate | Brisk effervescence | No brisk effervescence |
| S. No | Experiment | Observation | Inference |
|---|---|---|---|
| 1 | Take 5 ml of the solution in a test tube and add few drops of Phenolphthalein in it | No change in colour | Presence of acid |
| 2 | Take 5 ml of the solution in a test tube and add few drops of Methyl orange in it | Turns pink in colour | Presence of acid |
| 3 | Take 5 ml of the solution in a test tube and add pinch of sodium carbonate in it | Brisk effervescence | Presence of acid |
The given test solution contains acid
To identify the presence of an acid or a base in a given sample
Materials required:Test-tube, test-tube stand, glass rod, indicators and the given sample
Procedure:| S. No | Experiment | Observation | Inference |
|---|---|---|---|
| 1 | Take 5 ml of the solution in a test tube and add few drops of Phenolphthalein in it | Turns pink in colour | Presence of base |
| 2 | Take 5 ml of the solution in a test tube and add few drops of Methyl orange in it | Turns yellow in colour | Presence of base |
| 3 | Take 5 ml of the solution in a test tube and add pinch of sodium carbonate in it | No brisk effervescence | Presence of base |
The given test solution contains base
BIO – BOTANY
To prove that oxygen is evolved during photosynthesis
Materials required:Test tube, funnel, beaker, pond water and Hydrilla plant
Procedure:(i) Take a few twigs of Hydrilla plant in a beaker containing pond water.
(ii) Place an inverted funnel over the plant.
(iii) Invert a test tube filled with water over the stem of the funnel.
(iv) Keep the apparatus in the sunlight for few hours.
It is noted that water gets displaced down from the test tube.
Inference:Take the test tube and keep the burning stick near the mouth of the test tube. Increased the flame will appear.
This test proves that oxygen is released during photosynthesis.
To dissect and display the parts of given flower and draw labelled sketches
Material required:Flower, needle and paper
Procedure:With the help of the needle dissect the different whorls of the flower
Diagram:
Floral parts:
Accessory organ
(i) Calyx
(ii) Corolla
Reproductive organ
(i) Androecium (Male part)
(ii) Gynoecium (Female part)
Parts of the given flower were dissected and submitted to sight.
To study the law of dominance by using model/ picture/ photograph. To find out the genotype ratio and phenotype ratio in pea plant using checker board.
Material required:Colour chalk pieces or Graph sheets
Procedure:Depict parental generation and the gametes using colour chalk pieces.
Phenotypic ratio 3:1
Genotypic ratio 1:2:1
Using the model, the law of dominance and the monohybrid cross study were found.
To identify the given slide with the help of microscope
Material required:Slides and Microscope
Identification:The given slide is identified as T.S of Dicot Stem
Reasons:(i) Vascular bundles are arranged in a ring
(ii) Conjoint, collateral, end arch and open vascular bundle
(iii) Ground tissue differentiated
(iv) 3 to 6 layer of collenchyma tissues present in hypodermis
The given slide was identified as T.S of the Dicot stem.
To identify the given slide with the help of microscope
Material required:Slides and Microscope
Identification:The given slide is identified as T.S of Dicot Root
Reasons:(i) Radial vascular bundle
(ii) 2 to 4 xylem presents
(iii) Cambium present
(iv) Cortex is made up of parenchymatous cells
The given slide was identified as T.S of the Dicot root.
BIO – ZOOLOGY
To identify the given models, draw a labelled diagram and write a note on it.
Material required:Models (Human Heart and Human Brain)
Identification:The given model is identified as L.S of human heart
Notes:(i) The human heart made up of cardiac muscle
(ii) The heart has four chambers
(iii) The heart pumps blood to all parts of the body
(iv) The heart is covered by pericardium
The given model was identified as the L.S of the human heart.
To identify the given models, draw a labelled diagram and write a note on it.
Material required:Models (Human Heart and Human Brain)
Identification:The given model is identified as L.S of human brain
Notes:(i) The brain is enclosed in the cranial cavity
(ii) It is the controlling centre of the all body activities
(iii) It is covered by three connective tissue membrane
(iv) The brain is divided into three parts
The given model was identified as the L.S of the human brain.
To identify the given slides, draw a labelled diagram and write a note on it.
Material required:Permanent slides of blood cells and Microscope
Identification:The given slide is identified as Red Blood Cell
Notes:(i) They are biconcave and disc shaped
(ii) Mature mammalian RBC’s do not have nucleus
(iii) Haemoglobin is a respiratory pigment which gives red colour
The given slide was identified as the Red Blood Cell.
To identify the given slides, draw a labelled diagram and write a note on it.
Material required:Permanent slides of blood cells and Microscope
Identification:The given slide is identified as White Blood Cell
Notes:(i) They are colourless and they have amoeboid shaped.
(ii) They have nucleus.
(iii) They protect the body from diseases.
The given slide was identified as the White Blood Cell.
To identify the endocrine glands, its location hormone secreted and functions
Material required:Flag labelled Endocrine gland models
Identification:The flag labelled endocrine gland is identified as Thyroid gland
Location:Thyroid gland is bilobed gland located in the neck region on either side of the trachea.
Hormones secreted:Tri iodothyronine (T3) and Thyroxine (T4)
Functions:(i) It regulates metabolism
(ii) It increases the body temperature
(iii) It is required for normal growth and development
(iv) It is also known as personality hormone
The given model was identified as the thyroid gland.
To identify the endocrine glands, its location hormone secreted and functions
Material required:Flag labelled Endocrine gland models
Identification:The flag labelled endocrine gland is identified as Islets of Langerhans in the pancreas
Location:Islets of Langerhans are seen embedded in the pancreas which is located in the abdominal region
Hormones secreted:(i) α cells secrete glucagon
(ii) β cells secrete insulin
(i) Insulin converts glucose into glycogen
(ii) Glucagon converts glycogen into glucose
(iii) Insulin and Glucagon maintain the blood sugar level
The given model was identified as the Islets of Langerhans in the pancreas.