SECTION – E

Q. Nos. 37-39 are source-based/case-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts:

37. The metals produced by various reduction processes are not very pure. They contain impurities, which must be removed to obtain pure metals. The most widely used method for refining impure metals is electrolytic refining.

(i) What is the cathode and anode made of in the refining of copper by this process ?

Answer:

• Cathode: The cathode in the refining of copper is typically made of pure copper metal. As the impure copper dissolves at the anode and pure copper ions migrate towards the cathode, they are deposited onto the pure copper cathode, gradually building up a layer of pure copper.
• Anode: The anode in the refining of copper is made of impure copper. As electric current passes through the electrolytic cell, the impure copper at the anode dissolves into the electrolyte solution, releasing copper ions into the solution.

(ii) Name the solution used in the above process and write its formula.

Answer:

The solution used in the electrolytic refining of copper is typically an aqueous solution of copper sulfate (CuSO₄​)

(iii) (A) How copper gets refined when electric current is passed in the electrolytic cell?

Answer:

In the electrolytic cell used for refining copper, the impure copper anode and a pure copper cathode are immersed in an electrolyte solution of copper sulfate (CuSO₄​).

1. When electric current is passed through the cell, copper ions (Cu²⁺) from the copper sulfate solution migrate towards the cathode.
2. At the cathode (pure copper electrode), copper ions gain electrons and are reduced to form solid copper metal. This deposition of pure copper on the cathode gradually increases its mass.
3. At the anode (impure copper electrode), the impure copper metal dissolves into the electrolyte solution as copper ions (Cu²⁺). This process replenishes the copper ions in the solution and maintains the concentration of copper sulfate.
4. As a result of the electrolysis process, the impurities present in the impure copper anode settle down as anode mud or sludge at the bottom of the electrolytic cell, leaving behind pure copper at the cathode.
5. The pure copper deposited on the cathode is periodically removed, washed, and dried to obtain high-purity copper metal. This electrolytic refining process helps to purify the copper obtained from various reduction processes, ensuring its high quality and purity for industrial applications.

OR

(iii) (B) You have two beakers ‘A’ and ‘B’ containing copper sulphate solution. What would you observe after about 2 hours if you dip a strip of zine in beaker “A’ and a strip of silver in beaker ‘B”? Give reason for your observations in each case.

Answer:

If you dip a strip of zinc in beaker ‘A’ containing copper sulfate solution and a strip of silver in beaker ‘B’ containing copper sulfate solution, the following observations can be made:

Beaker A (Zinc strip):

Reason: This reddish-brown coating is due to the displacement reaction between zinc and copper ions present in the copper sulfate solution. Zinc is more reactive than copper, so it displaces copper from copper sulfate solution, forming solid copper metal. The reaction can be represented as:

Zn (s) + CuSO₄(aq) → ZnSO₄(aq) + Cu (s)

After about 2 hours, you would observe that the zinc strip appears to have a reddish-brown coating on its surface.

The solid copper formed deposits onto the surface of the zinc strip, giving it a reddish-brown appearance.

Beaker B (Silver strip):

Reason: Silver is less reactive than copper, so it cannot displace copper from copper sulfate solution. Therefore, no chemical reaction occurs between silver and copper sulfate solution, and the silver strip remains unchanged.

After about 2 hours, you would observe that there is no visible change in the appearance of the silver strip.

38. Mendel worked out the rules of heredity by working on garden pea using a number of visible contrasting characters. He conducted several experiments by making a cross with one or two pairs of contrasting characters of pea plant. On the basis of his observations he gave some interpretations which helped to study the mechanism of inheritance.

(i) When Mendel crossed pea plants with pure tall and pure short characteristics to produce F₁ progeny, which two observations were made by him in F₁ plants?

Answer:

When Mendel crossed pea plants with pure tall and pure short characteristics to produce F1 progeny, two observations were made:

1. All the F₁ progeny plants were tall.
2. The trait of shortness disappeared in the F1 generation, indicating that the tall trait was dominant over the short trait.

(ii) Write one difference between dominant and recessive trait.

Answer:

The difference between dominant and recessive trait is:

• Dominant trait: A dominant trait is expressed phenotypically when at least one copy of the dominant allele is present in the genotype. It masks the expression of the recessive allele in heterozygous individuals.
• Recessive trait: A recessive trait is expressed phenotypically only when two copies of the recessive allele are present in the genotype. It is masked by the dominant allele in heterozygous individuals.

(iii) (A) In a cross with two pairs of contrasting characters

RRYY X rryy

(round Yellow) (wrinkled green)

Mendel observed 4 types of combinations in F₂ generation. By which method did he obtain F₂ generation? Write the ratio of the parental combinations obtained and what conclusions were drawn from this experiment.

Answer:

Mendel obtained the F₂ generation by allowing the F₁ plants resulting from the cross between the parental plants with two pairs of contrasting characters to self-pollinate.

Ratio of the parental combinations obtained: In the F₂ generation, Mendel observed four types of combinations:

1. Round yellow (RRYY)
2. Round green (RRYy)
3. Wrinkled yellow (rrYY)
4. Wrinkled green (rrYy)

OR

(iii) (B) Justify the statement :

“It is possible that a trait is inherited but may not be expressed.”

Answer:

The statement “It is possible that a trait is inherited but may not be expressed” can be justified by the concept of dominant and recessive alleles and the phenomenon of incomplete dominance and codominance in genetics.

1. Recessive Alleles: In cases where a trait is controlled by a recessive allele, it may not be expressed phenotypically if the individual possesses at least one dominant allele for that trait. Recessive alleles are only expressed phenotypically when the individual is homozygous for the recessive allele. In heterozygous individuals, the dominant allele masks the expression of the recessive allele, resulting in the trait not being expressed.
2. Incomplete Dominance: In incomplete dominance, neither allele is completely dominant over the other. Instead, the heterozygous phenotype is an intermediate blend of the phenotypes associated with the two homozygous genotypes. In such cases, the trait may not be fully expressed in heterozygous individuals, resulting in a phenotype that differs from both the homozygous dominant and homozygous recessive phenotypes.
3. Codominance: In codominance, both alleles contribute equally to the phenotype of the heterozygous individual. As a result, both alleles are expressed fully, without one dominating over the other. However, if the trait associated with one of the alleles is not observable or detectable, it may seem as though the trait is not expressed, even though it is inherited.
4. Environmental Factors: In some cases, the expression of a trait may be influenced by environmental factors. Even though the trait is inherited genetically, its expression may be influenced or modified by external environmental conditions, leading to variability in the phenotypic expression of the trait.

39. Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors:

(i) In which one of the above cases the mirror will form a diminished image of the object ? Justify your answer.

Answer:

For a concave mirror to form a diminished image of the object, the object distance must be greater than the focal length of the mirror. This situation occurs when the object is placed beyond the focal point of the mirror.

From the given data:

• Case 1: Mirror A has a focal length of 20 cm and an object distance of 45 cm. Object distance > Focal length.
• Case 2: Mirror B has a focal length of 15 cm and an object distance of 30 cm. Object distance > Focal length.
• Case 3: Mirror C has a focal length of 30 cm and an object distance of 20 cm. Object distance < Focal length.

Therefore, in Case 2 (Mirror B), the mirror will form a diminished image of the object because the object distance is greater than the focal length of the mirror.

(ii) List two properties of the image formed in case 2.

Answer:

Two properties of the image formed in Case 2 (Mirror B):

1. Real: Since the object distance is greater than the focal length, the image formed by Mirror B will be real.
2. Inverted: Concave mirrors always produce inverted images, irrespective of the position of the object. Therefore, the image formed by Mirror B will be inverted

(iii) (A) What is the nature and size of the image formed by mirror C ? Draw ray diagram to justify your answer.

Answer:

Nature: The nature of the image formed by Mirror C can be determined using the mirror formula:

1/f = 1/u + 1/v

Where:

• f = focal length of the mirror (30 cm)
• v = image distance
• u = object distance (-20 cm)

Given that the object distance u is negative (since it is measured on the same side as the incident light), the image distance v will also be negative, indicating that the image is formed on the same side as the object. This implies that the image formed is a virtual image.

Size: Since the object distance u is less than the focal length f, the image formed will be magnified. The magnification can be calculated using the magnification formula: Magnification (m) = −Magnification (m) = u − v​

OR

(iii) (B) An object is placed at a distance of 18 cm from the pole of a concave mirror of focal length 12 cm. Find the position of the image formed in this case.

Answer:

To find the position of the image formed by a concave mirror when an object is placed at a distance of 18 cm from the pole and the focal length of the mirror is 12 cm, we can use the mirror formula:

1/f = 1/u + 1/v

Given:

f = −12 cm (negative because it’s a concave mirror)

u = −18 cm (negative because the object is placed on the same side as the incident light)

Substituting the given values into the mirror formula:

1/−12 =1/v + 1/−18

Solving for v, we get :

1/v = -1/36

v = −36

Hence, the position of the image formed by the concave mirror when the object is placed at a distance of 18 cm from the pole is 36 cm on the same side as the object. Therefore, the image is virtual and magnified.

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