CBSE Class 10 Science Term 2 Sample Paper Solved 2022 Set C

Read and download PDF of CBSE Class 10 Science Term 2 Sample Paper Solved 2022 Set C designed as per the latest curriculum and examination pattern for Class 10 issued by CBSE, NCERT and KVS. The latest Class 10 Science Sample Papers have been provided with solutions so that the students can solve these practice papers and then compare their answers. This will help them to identify mistakes and improvement areas in Science Standard 10 which they need to study more to get better marks in Grade 10 exams. After solving these guess papers also refer to solved Class 10 Science Question Papers available on our website to build strong understanding of the subject

Science Sample Paper Class 10 2022

Students can refer to the below Class 10 Science Sample Paper designed to help students understand the pattern of questions that will be asked in Grade 10 exams. Please download CBSE Class 10 Science Term 2 Sample Paper Solved 2022 Set C

Science Class 10 Sample Paper 2022

Question : An element ‘X’ belongs to 3rd period and group 17 of the periodic table. State its valency. Justify your answer with reason.
OR
Choose from the following :
4Be, 9F, 19K, 20Ca
(a) The element having one electron in the outermost shell.
(b) Two elements of the same group
Answer :  As element X belongs to group 17, it will have 7 electrons in its outermost shell. Moreover, X belongs to period number 3 so, it will have 3 shells.
Electronic configuration of X = 2, 8, 7
Valency of element X
= 8 – (Number of valence electrons)
= 8 – 7 = 1
OR
The electronic configurations of the given elements are :
4Be = 2, 2
9F = 2, 7
19K = 2, 8, 8, 1
20Ca = 2, 8, 8, 2
(a) Potassium (K) has one electron in its outermost shell.
(b) Be and Ca have two electrons in their outermost shells hence, they belong to same group.
 
Question :  What do you mean by biological magnification?
Answer : Biological magnification or biomagnification is the process of increase in amount of some toxic, nonbiodegradable substances such as DDT and heavy metals in successive trophic levels of a food chain. It results in accumulation of highest concentration of these toxins in topmost trophic level.
 
Question : Characters that are transmitted from parents to offsprings during sexual reproduction show
(a) only similarities with parents (b) only variations with parents
(c) both similarities and variations with parents (d) neither similarities nor variations.
Answer : (c) : When germ cells (gametes) from two individuals combine during sexual reproduction to form a new individual, it results in combination of characters of two different individuals. Hence, the offspring resembles parents in some characters as well as differs from them in some other characters i.e., both similarities and variations are exhibited.
 
Question : Mendel crossed a pure recessive wrinkled seeded pea plant with a pure dominant round seeded plant. The first generation of hybrids from the cross should show
(a) 50% wrinkled seeded and 50% round seeded plants
(b) all round seeded plants
(c) 75% round seeded and 25% wrinkled seeded plants
(d) all wrinkled seeded plants.
Answer : (b) : In first generation all plants will have one dominant allele and one recessive allele. Presence of dominant allele in all progenies will produce all round seeded plants in F1 generation.
OR
The crossing of a homozygous tall plant with a dwarf plant would yield plants in the ratio of
(a) two tall and two dwarf
(b) one homozygous tall, one homozygous dwarf and two heterozygous tall
(c) all homozygous dwarf
(d) all heterozygous tall.
Answer : (d) : The crossing of a homozygous tall plant with a dwarf would yield all heterozygous tall plants containing both the alleles. Tallness is a dominant trait over dwarfism.
So even in heterozygous condition all plants are tall.
pi-
 
 
Question : In the given food chain suppose the amount of energy at fourth trophic level is 5 KJ, what will be the energy available at the producer level?
Grass → Grasshopper → Frog → Snake → Hawk
(a) 5 KJ (b) 50 KJ (c) 500 KJ (d) 5000 KJ
Answer : (d) : In the given food chain, if the amount of energy at fourth trophic level is 5 KJ, then 5000 KJ will be the energy available at the producer level. As only 10% of energy is transferred to next higher trophic level, hence, 500 KJ energy will be stored in body of grasshopper and 50 KJ in frog. 5 KJ energy will be present in the body of snake and 0.5 KJ in hawk.

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