National Level Solutions (SAT) - 2015-16 | NTSE for Class 10 PDF Download

 Page 1


 
     
 
 
 
NTSE STAGE II 
CODE: 13 – 15  
SAT 
HINTS & SOLUTIONS 
 
1. 3 
Sol. Aerobic respiration takes place in mitochondria. 
 
2. 2 
Sol. Cow has a special additional part in their stomach to digest cellulose present in the food. 
 
3. 1 
Sol. In touch me not plant leaflets are closed after contact due to change in Turgon pressure. 
 
4. 3 
Sol. Pancreas is known as ‘mixocrine or dual gland’. 
 
5. 4 
Sol. Placenta provide nutrition to the foetus during pregnancy in human beings. 
 
6. 4 
Sol. Endocrine glands pour their secretions directly in the blood. 
 
7 2 
Sol. Cell of meristematic tissues are actively dividing having dense cytoplasm, thin cell wall and 
no vacuoles or minute vacuole. 
 
8. 3 
Sol. The 4 characteristics present in chordates are 
 – Notocord  
 – Pharyngeal gill slits or pouches 
 – Dorsal tubular nerve cord 
 – Post anal Tail 
 
9. 3 
Sol. In symbiotic relationship between a bacterium and a root of leguminous plants the bacteria 
provide NH 4 to plant and the root provides carbon. 
 
10. 1 
Sol. Biological magnification is the Accumulation of chemicals from lower trophic level to higher 
trophic level. 
 
11. 4 
Sol. Ex – situ conservation– 
Strategies include botanical gardens, zoos, conservation stands, gene, pollen, seed, 
seedling, tissue culture and DNA banks. 
 
12. 2 
Sol. Temperature increases due to the entrapment of infra red radiations. 
 
13. 2 
Sol. A. small pox caused by virus 
 B. cholera caused by bacteria. 
Page 2


 
     
 
 
 
NTSE STAGE II 
CODE: 13 – 15  
SAT 
HINTS & SOLUTIONS 
 
1. 3 
Sol. Aerobic respiration takes place in mitochondria. 
 
2. 2 
Sol. Cow has a special additional part in their stomach to digest cellulose present in the food. 
 
3. 1 
Sol. In touch me not plant leaflets are closed after contact due to change in Turgon pressure. 
 
4. 3 
Sol. Pancreas is known as ‘mixocrine or dual gland’. 
 
5. 4 
Sol. Placenta provide nutrition to the foetus during pregnancy in human beings. 
 
6. 4 
Sol. Endocrine glands pour their secretions directly in the blood. 
 
7 2 
Sol. Cell of meristematic tissues are actively dividing having dense cytoplasm, thin cell wall and 
no vacuoles or minute vacuole. 
 
8. 3 
Sol. The 4 characteristics present in chordates are 
 – Notocord  
 – Pharyngeal gill slits or pouches 
 – Dorsal tubular nerve cord 
 – Post anal Tail 
 
9. 3 
Sol. In symbiotic relationship between a bacterium and a root of leguminous plants the bacteria 
provide NH 4 to plant and the root provides carbon. 
 
10. 1 
Sol. Biological magnification is the Accumulation of chemicals from lower trophic level to higher 
trophic level. 
 
11. 4 
Sol. Ex – situ conservation– 
Strategies include botanical gardens, zoos, conservation stands, gene, pollen, seed, 
seedling, tissue culture and DNA banks. 
 
12. 2 
Sol. Temperature increases due to the entrapment of infra red radiations. 
 
13. 2 
Sol. A. small pox caused by virus 
 B. cholera caused by bacteria. 
 
     
 
 
 C. the carrier organism of malaria is female anopheles mosquito 
 D. deficiency of iron leads to anaemia.  
 
14. 4 
Sol. The phenotype ratio of dihybrid cross is 9:3:3:1 
 ? 
240 1
15
16
?
? 
 This shows wrinkled and green is the recessive character. 
 
15. 2 
Sol. Number of neutrons in one molecule of water(H 2O) = 8 
 Number of neutrons in five moles of water  = 8 ? 5 ? 6.022 ? 10
23
 
  = 240.88 ? 10
23
 
  = 2.4088 ? 10
25
 
  ? 2.409 ? 10
25
 
 
16. 4 
Sol. Sodium is highly reactive, so it will form precipitation  of Cu(OH) 2.  
 Iron is more reactive than copper 
 So, it displaces copper from aqueous solution of copper sulphate 
 
44
Fe CuSO FeSO Cu ? ? ? ? ? 
 
17. 2 
Sol. A is suspension as particles are visible to naked eye and settle down. 
 C is solution as beam of light is invisible in it 
 B and D are colloids as particles are invisible and beam of light visible. 
 
18. 2 
Sol. Alpha particles penetrate through thin aluminium foil and scattering cannot be observed. 
 
19. 3 
Sol. Magnesium ribbon is rubbed with sand paper to remove magnesium oxide layer. 
  
20. 2  
Sol. Formation of CaO from CaCO 3, Na2CO3 from NaHCO 3 and Hg from HgO are undergoing 
thermal decomposition. 
 Al from Al2O3 is electrolytic decomposition  
 
21. 1 
Sol. X is amphoteric in nature and electropositive  
 
22. 2 
Sol. Z is sodium chloride, which does not conduct electricity in its pure solid state.   
 
23. 4 
Sol. Covalent bonds in 
4
NH
?
 and ionic bond between 
4
NH
?
 and Cl
–
 
  
24. 3 
Sol. Sulphur cannot be used as reducing agent  
 
25. 4 
Sol. Number of oxygen atoms = 9.033 ? 10
23
 
 Number of moles of oxygen atoms = 
23
23
9.033 10
1.5
6.022 10
?
?
?
 
 Mass of 1.5 mole oxygen atoms = 1.5 ? 16 = 24 
Page 3


 
     
 
 
 
NTSE STAGE II 
CODE: 13 – 15  
SAT 
HINTS & SOLUTIONS 
 
1. 3 
Sol. Aerobic respiration takes place in mitochondria. 
 
2. 2 
Sol. Cow has a special additional part in their stomach to digest cellulose present in the food. 
 
3. 1 
Sol. In touch me not plant leaflets are closed after contact due to change in Turgon pressure. 
 
4. 3 
Sol. Pancreas is known as ‘mixocrine or dual gland’. 
 
5. 4 
Sol. Placenta provide nutrition to the foetus during pregnancy in human beings. 
 
6. 4 
Sol. Endocrine glands pour their secretions directly in the blood. 
 
7 2 
Sol. Cell of meristematic tissues are actively dividing having dense cytoplasm, thin cell wall and 
no vacuoles or minute vacuole. 
 
8. 3 
Sol. The 4 characteristics present in chordates are 
 – Notocord  
 – Pharyngeal gill slits or pouches 
 – Dorsal tubular nerve cord 
 – Post anal Tail 
 
9. 3 
Sol. In symbiotic relationship between a bacterium and a root of leguminous plants the bacteria 
provide NH 4 to plant and the root provides carbon. 
 
10. 1 
Sol. Biological magnification is the Accumulation of chemicals from lower trophic level to higher 
trophic level. 
 
11. 4 
Sol. Ex – situ conservation– 
Strategies include botanical gardens, zoos, conservation stands, gene, pollen, seed, 
seedling, tissue culture and DNA banks. 
 
12. 2 
Sol. Temperature increases due to the entrapment of infra red radiations. 
 
13. 2 
Sol. A. small pox caused by virus 
 B. cholera caused by bacteria. 
 
     
 
 
 C. the carrier organism of malaria is female anopheles mosquito 
 D. deficiency of iron leads to anaemia.  
 
14. 4 
Sol. The phenotype ratio of dihybrid cross is 9:3:3:1 
 ? 
240 1
15
16
?
? 
 This shows wrinkled and green is the recessive character. 
 
15. 2 
Sol. Number of neutrons in one molecule of water(H 2O) = 8 
 Number of neutrons in five moles of water  = 8 ? 5 ? 6.022 ? 10
23
 
  = 240.88 ? 10
23
 
  = 2.4088 ? 10
25
 
  ? 2.409 ? 10
25
 
 
16. 4 
Sol. Sodium is highly reactive, so it will form precipitation  of Cu(OH) 2.  
 Iron is more reactive than copper 
 So, it displaces copper from aqueous solution of copper sulphate 
 
44
Fe CuSO FeSO Cu ? ? ? ? ? 
 
17. 2 
Sol. A is suspension as particles are visible to naked eye and settle down. 
 C is solution as beam of light is invisible in it 
 B and D are colloids as particles are invisible and beam of light visible. 
 
18. 2 
Sol. Alpha particles penetrate through thin aluminium foil and scattering cannot be observed. 
 
19. 3 
Sol. Magnesium ribbon is rubbed with sand paper to remove magnesium oxide layer. 
  
20. 2  
Sol. Formation of CaO from CaCO 3, Na2CO3 from NaHCO 3 and Hg from HgO are undergoing 
thermal decomposition. 
 Al from Al2O3 is electrolytic decomposition  
 
21. 1 
Sol. X is amphoteric in nature and electropositive  
 
22. 2 
Sol. Z is sodium chloride, which does not conduct electricity in its pure solid state.   
 
23. 4 
Sol. Covalent bonds in 
4
NH
?
 and ionic bond between 
4
NH
?
 and Cl
–
 
  
24. 3 
Sol. Sulphur cannot be used as reducing agent  
 
25. 4 
Sol. Number of oxygen atoms = 9.033 ? 10
23
 
 Number of moles of oxygen atoms = 
23
23
9.033 10
1.5
6.022 10
?
?
?
 
 Mass of 1.5 mole oxygen atoms = 1.5 ? 16 = 24 
 
     
 
 
 
2 2 2
2H O 2H O ? ? ? ? 
 Number of moles of hydrogen atoms = 1.5 ? 2 = 3 
 
26. 4 
Sol. C13H26O2, C2H4O2 and C9H18O2 are in the forms of CnH2n+1 COOH and C7H12O2 is not in this 
form. 
 
27. 3 
Sol. Soap foam appears white as it reflects light of all wavelengths  
 
28. 2 
Sol. 
18 19
(n 4.8 10 ) 1.6 10
1.12
1
?
? ? ? ?
? 
 ? n = 2.2 × 10
18
 
 
29. 3 
Sol. A solenoid of finite length carrying current produces magnetic field like bar magnet.   
 
30. 4 
Sol. 
30
i 3A
10
?? ; V – VA = 10 × 1 = 10 
 i 1 = 1A (upper branch) V – V B = 2 × 10 = 20 
 i 2 = 2A (lower branch) V A – V B = 10 V  
 
31. 1 
Sol. Magnetic force does not affect the magnitude of velocity. Because magnetic force always act 
perpendicular to velocity.  
 
32. 3 
Sol. Force due to electric field changes magnitude of velocity and hence momentum. 
 
33. 3 
Sol. ? = (i1 – r1) + (i2 – r2)  
 60° = ( ? i – 30°) + ( ? e – 30°) (since  r 1 = r 2 = 30°) 
 ? ?i + ?e = 120° 
 2 ?i = 120°   (since, ?i = ?e) 
 ?i = ?e = 60° 
 
? i ? e 
60° 
60° 
30° 
30° 
60° 
60° 
 
 
34. 1 
Sol. SPEAR should be thrown at actual object which is below the image. 
 LASER will follow refraction of light hence it will bend. So it should be sent towards virtual 
image.  
35. 2 
Sol. Sin ? = ? sin (90° – ?) 
 tan ? = ? = 3 
 ? = 60°,   r = 90° – ? = 30° 
 
? 
? 
90° 
90°– ? =  r 
 
 
36. 1 
Page 4


 
     
 
 
 
NTSE STAGE II 
CODE: 13 – 15  
SAT 
HINTS & SOLUTIONS 
 
1. 3 
Sol. Aerobic respiration takes place in mitochondria. 
 
2. 2 
Sol. Cow has a special additional part in their stomach to digest cellulose present in the food. 
 
3. 1 
Sol. In touch me not plant leaflets are closed after contact due to change in Turgon pressure. 
 
4. 3 
Sol. Pancreas is known as ‘mixocrine or dual gland’. 
 
5. 4 
Sol. Placenta provide nutrition to the foetus during pregnancy in human beings. 
 
6. 4 
Sol. Endocrine glands pour their secretions directly in the blood. 
 
7 2 
Sol. Cell of meristematic tissues are actively dividing having dense cytoplasm, thin cell wall and 
no vacuoles or minute vacuole. 
 
8. 3 
Sol. The 4 characteristics present in chordates are 
 – Notocord  
 – Pharyngeal gill slits or pouches 
 – Dorsal tubular nerve cord 
 – Post anal Tail 
 
9. 3 
Sol. In symbiotic relationship between a bacterium and a root of leguminous plants the bacteria 
provide NH 4 to plant and the root provides carbon. 
 
10. 1 
Sol. Biological magnification is the Accumulation of chemicals from lower trophic level to higher 
trophic level. 
 
11. 4 
Sol. Ex – situ conservation– 
Strategies include botanical gardens, zoos, conservation stands, gene, pollen, seed, 
seedling, tissue culture and DNA banks. 
 
12. 2 
Sol. Temperature increases due to the entrapment of infra red radiations. 
 
13. 2 
Sol. A. small pox caused by virus 
 B. cholera caused by bacteria. 
 
     
 
 
 C. the carrier organism of malaria is female anopheles mosquito 
 D. deficiency of iron leads to anaemia.  
 
14. 4 
Sol. The phenotype ratio of dihybrid cross is 9:3:3:1 
 ? 
240 1
15
16
?
? 
 This shows wrinkled and green is the recessive character. 
 
15. 2 
Sol. Number of neutrons in one molecule of water(H 2O) = 8 
 Number of neutrons in five moles of water  = 8 ? 5 ? 6.022 ? 10
23
 
  = 240.88 ? 10
23
 
  = 2.4088 ? 10
25
 
  ? 2.409 ? 10
25
 
 
16. 4 
Sol. Sodium is highly reactive, so it will form precipitation  of Cu(OH) 2.  
 Iron is more reactive than copper 
 So, it displaces copper from aqueous solution of copper sulphate 
 
44
Fe CuSO FeSO Cu ? ? ? ? ? 
 
17. 2 
Sol. A is suspension as particles are visible to naked eye and settle down. 
 C is solution as beam of light is invisible in it 
 B and D are colloids as particles are invisible and beam of light visible. 
 
18. 2 
Sol. Alpha particles penetrate through thin aluminium foil and scattering cannot be observed. 
 
19. 3 
Sol. Magnesium ribbon is rubbed with sand paper to remove magnesium oxide layer. 
  
20. 2  
Sol. Formation of CaO from CaCO 3, Na2CO3 from NaHCO 3 and Hg from HgO are undergoing 
thermal decomposition. 
 Al from Al2O3 is electrolytic decomposition  
 
21. 1 
Sol. X is amphoteric in nature and electropositive  
 
22. 2 
Sol. Z is sodium chloride, which does not conduct electricity in its pure solid state.   
 
23. 4 
Sol. Covalent bonds in 
4
NH
?
 and ionic bond between 
4
NH
?
 and Cl
–
 
  
24. 3 
Sol. Sulphur cannot be used as reducing agent  
 
25. 4 
Sol. Number of oxygen atoms = 9.033 ? 10
23
 
 Number of moles of oxygen atoms = 
23
23
9.033 10
1.5
6.022 10
?
?
?
 
 Mass of 1.5 mole oxygen atoms = 1.5 ? 16 = 24 
 
     
 
 
 
2 2 2
2H O 2H O ? ? ? ? 
 Number of moles of hydrogen atoms = 1.5 ? 2 = 3 
 
26. 4 
Sol. C13H26O2, C2H4O2 and C9H18O2 are in the forms of CnH2n+1 COOH and C7H12O2 is not in this 
form. 
 
27. 3 
Sol. Soap foam appears white as it reflects light of all wavelengths  
 
28. 2 
Sol. 
18 19
(n 4.8 10 ) 1.6 10
1.12
1
?
? ? ? ?
? 
 ? n = 2.2 × 10
18
 
 
29. 3 
Sol. A solenoid of finite length carrying current produces magnetic field like bar magnet.   
 
30. 4 
Sol. 
30
i 3A
10
?? ; V – VA = 10 × 1 = 10 
 i 1 = 1A (upper branch) V – V B = 2 × 10 = 20 
 i 2 = 2A (lower branch) V A – V B = 10 V  
 
31. 1 
Sol. Magnetic force does not affect the magnitude of velocity. Because magnetic force always act 
perpendicular to velocity.  
 
32. 3 
Sol. Force due to electric field changes magnitude of velocity and hence momentum. 
 
33. 3 
Sol. ? = (i1 – r1) + (i2 – r2)  
 60° = ( ? i – 30°) + ( ? e – 30°) (since  r 1 = r 2 = 30°) 
 ? ?i + ?e = 120° 
 2 ?i = 120°   (since, ?i = ?e) 
 ?i = ?e = 60° 
 
? i ? e 
60° 
60° 
30° 
30° 
60° 
60° 
 
 
34. 1 
Sol. SPEAR should be thrown at actual object which is below the image. 
 LASER will follow refraction of light hence it will bend. So it should be sent towards virtual 
image.  
35. 2 
Sol. Sin ? = ? sin (90° – ?) 
 tan ? = ? = 3 
 ? = 60°,   r = 90° – ? = 30° 
 
? 
? 
90° 
90°– ? =  r 
 
 
36. 1 
 
     
 
 
Sol. 
1 1 1
v u f
??     
 ?  v = –300  ;  u = ? 
 ? 
1 1 1
f 300
??
??
     
? f = –300 
 
2
nd
 case 
1 1 1
300 50 d
??
? ? ?
  ?  d = 60 cm 
 
37. 4    
Sol. From conservation of energy 
 
22
1
11
mv mgh mv
22
?? 
 v 1 is the speed before hitting ground.    
 
38. 3 
Sol. A
hh
g A 2 g mg
24
? ? ? ? ? ? ? …(i) 
 Ah 1 × 2 ?h = mg …(ii) 
  
 
m 
A 
h/2 
h/4 
? 
2 ? 
 
 From (i) and (ii) 
 
1
hh
A g A 2 g Ah 2 g
24
? ? ? ? ? ? ? ? 
 ? 
1
h
h
2
? 
 
h 1 
2 ? 
 
 
39. 3     
Sol. F net = 0 
 
1 2 3
P P P 0 ? ? ? 
 ? 
22
3 1 2
V V V ?? 
 
MV 3 
MV 2 
MV 1 
 
 
40. 1 
Sol. F × x 1 = KE 
 F × x2 = KE 
 ? x1 = x2 
 
 
41. 1 
Sol. 
1
N 13Q 3 ?? 
 
2
N 21 Q 11 ?? 
 Number lies between 500 and 600. So the only number is 536 
 ? Remainder by 19 = 4   
 
42. 4 
Sol. 0.34 0.34 0.68787......... 0.687 ? ? ? 
 
43. 1 
Page 5


 
     
 
 
 
NTSE STAGE II 
CODE: 13 – 15  
SAT 
HINTS & SOLUTIONS 
 
1. 3 
Sol. Aerobic respiration takes place in mitochondria. 
 
2. 2 
Sol. Cow has a special additional part in their stomach to digest cellulose present in the food. 
 
3. 1 
Sol. In touch me not plant leaflets are closed after contact due to change in Turgon pressure. 
 
4. 3 
Sol. Pancreas is known as ‘mixocrine or dual gland’. 
 
5. 4 
Sol. Placenta provide nutrition to the foetus during pregnancy in human beings. 
 
6. 4 
Sol. Endocrine glands pour their secretions directly in the blood. 
 
7 2 
Sol. Cell of meristematic tissues are actively dividing having dense cytoplasm, thin cell wall and 
no vacuoles or minute vacuole. 
 
8. 3 
Sol. The 4 characteristics present in chordates are 
 – Notocord  
 – Pharyngeal gill slits or pouches 
 – Dorsal tubular nerve cord 
 – Post anal Tail 
 
9. 3 
Sol. In symbiotic relationship between a bacterium and a root of leguminous plants the bacteria 
provide NH 4 to plant and the root provides carbon. 
 
10. 1 
Sol. Biological magnification is the Accumulation of chemicals from lower trophic level to higher 
trophic level. 
 
11. 4 
Sol. Ex – situ conservation– 
Strategies include botanical gardens, zoos, conservation stands, gene, pollen, seed, 
seedling, tissue culture and DNA banks. 
 
12. 2 
Sol. Temperature increases due to the entrapment of infra red radiations. 
 
13. 2 
Sol. A. small pox caused by virus 
 B. cholera caused by bacteria. 
 
     
 
 
 C. the carrier organism of malaria is female anopheles mosquito 
 D. deficiency of iron leads to anaemia.  
 
14. 4 
Sol. The phenotype ratio of dihybrid cross is 9:3:3:1 
 ? 
240 1
15
16
?
? 
 This shows wrinkled and green is the recessive character. 
 
15. 2 
Sol. Number of neutrons in one molecule of water(H 2O) = 8 
 Number of neutrons in five moles of water  = 8 ? 5 ? 6.022 ? 10
23
 
  = 240.88 ? 10
23
 
  = 2.4088 ? 10
25
 
  ? 2.409 ? 10
25
 
 
16. 4 
Sol. Sodium is highly reactive, so it will form precipitation  of Cu(OH) 2.  
 Iron is more reactive than copper 
 So, it displaces copper from aqueous solution of copper sulphate 
 
44
Fe CuSO FeSO Cu ? ? ? ? ? 
 
17. 2 
Sol. A is suspension as particles are visible to naked eye and settle down. 
 C is solution as beam of light is invisible in it 
 B and D are colloids as particles are invisible and beam of light visible. 
 
18. 2 
Sol. Alpha particles penetrate through thin aluminium foil and scattering cannot be observed. 
 
19. 3 
Sol. Magnesium ribbon is rubbed with sand paper to remove magnesium oxide layer. 
  
20. 2  
Sol. Formation of CaO from CaCO 3, Na2CO3 from NaHCO 3 and Hg from HgO are undergoing 
thermal decomposition. 
 Al from Al2O3 is electrolytic decomposition  
 
21. 1 
Sol. X is amphoteric in nature and electropositive  
 
22. 2 
Sol. Z is sodium chloride, which does not conduct electricity in its pure solid state.   
 
23. 4 
Sol. Covalent bonds in 
4
NH
?
 and ionic bond between 
4
NH
?
 and Cl
–
 
  
24. 3 
Sol. Sulphur cannot be used as reducing agent  
 
25. 4 
Sol. Number of oxygen atoms = 9.033 ? 10
23
 
 Number of moles of oxygen atoms = 
23
23
9.033 10
1.5
6.022 10
?
?
?
 
 Mass of 1.5 mole oxygen atoms = 1.5 ? 16 = 24 
 
     
 
 
 
2 2 2
2H O 2H O ? ? ? ? 
 Number of moles of hydrogen atoms = 1.5 ? 2 = 3 
 
26. 4 
Sol. C13H26O2, C2H4O2 and C9H18O2 are in the forms of CnH2n+1 COOH and C7H12O2 is not in this 
form. 
 
27. 3 
Sol. Soap foam appears white as it reflects light of all wavelengths  
 
28. 2 
Sol. 
18 19
(n 4.8 10 ) 1.6 10
1.12
1
?
? ? ? ?
? 
 ? n = 2.2 × 10
18
 
 
29. 3 
Sol. A solenoid of finite length carrying current produces magnetic field like bar magnet.   
 
30. 4 
Sol. 
30
i 3A
10
?? ; V – VA = 10 × 1 = 10 
 i 1 = 1A (upper branch) V – V B = 2 × 10 = 20 
 i 2 = 2A (lower branch) V A – V B = 10 V  
 
31. 1 
Sol. Magnetic force does not affect the magnitude of velocity. Because magnetic force always act 
perpendicular to velocity.  
 
32. 3 
Sol. Force due to electric field changes magnitude of velocity and hence momentum. 
 
33. 3 
Sol. ? = (i1 – r1) + (i2 – r2)  
 60° = ( ? i – 30°) + ( ? e – 30°) (since  r 1 = r 2 = 30°) 
 ? ?i + ?e = 120° 
 2 ?i = 120°   (since, ?i = ?e) 
 ?i = ?e = 60° 
 
? i ? e 
60° 
60° 
30° 
30° 
60° 
60° 
 
 
34. 1 
Sol. SPEAR should be thrown at actual object which is below the image. 
 LASER will follow refraction of light hence it will bend. So it should be sent towards virtual 
image.  
35. 2 
Sol. Sin ? = ? sin (90° – ?) 
 tan ? = ? = 3 
 ? = 60°,   r = 90° – ? = 30° 
 
? 
? 
90° 
90°– ? =  r 
 
 
36. 1 
 
     
 
 
Sol. 
1 1 1
v u f
??     
 ?  v = –300  ;  u = ? 
 ? 
1 1 1
f 300
??
??
     
? f = –300 
 
2
nd
 case 
1 1 1
300 50 d
??
? ? ?
  ?  d = 60 cm 
 
37. 4    
Sol. From conservation of energy 
 
22
1
11
mv mgh mv
22
?? 
 v 1 is the speed before hitting ground.    
 
38. 3 
Sol. A
hh
g A 2 g mg
24
? ? ? ? ? ? ? …(i) 
 Ah 1 × 2 ?h = mg …(ii) 
  
 
m 
A 
h/2 
h/4 
? 
2 ? 
 
 From (i) and (ii) 
 
1
hh
A g A 2 g Ah 2 g
24
? ? ? ? ? ? ? ? 
 ? 
1
h
h
2
? 
 
h 1 
2 ? 
 
 
39. 3     
Sol. F net = 0 
 
1 2 3
P P P 0 ? ? ? 
 ? 
22
3 1 2
V V V ?? 
 
MV 3 
MV 2 
MV 1 
 
 
40. 1 
Sol. F × x 1 = KE 
 F × x2 = KE 
 ? x1 = x2 
 
 
41. 1 
Sol. 
1
N 13Q 3 ?? 
 
2
N 21 Q 11 ?? 
 Number lies between 500 and 600. So the only number is 536 
 ? Remainder by 19 = 4   
 
42. 4 
Sol. 0.34 0.34 0.68787......... 0.687 ? ? ? 
 
43. 1 
 
     
 
 
Sol. ? ? ? ?
2
P x k x 1 ?? 
 ? ? ? ?
2
P 2 k 1 2 ? ? ? ?  
 k2 ?? 
 ? ? ? ?
2
P 2 2 2 1 ? ? ? 
 = 18   
 
44. 4 
Sol. x y 2 ?? 
 kx y 3 ?? 
 
5 3 2k
x , y
1 k 1 k
?
??
??
 
 1 k 0 ? ? ? and 3 2k 0 ?? 
 k1 ?? and 
3
k
2
? 
 
 
 
45. 4 
Sol. 
10 8
9
a 2a
2a
?
 
 
10 10 8 8
99
22
22
? ? ? ? ? ? ?
?
? ? ?
 
 
? ? ? ?
? ?
8 2 8 2
99
22
2
? ? ? ? ? ? ?
?
? ? ?
 
 
? ? ? ?
? ?
88
99
66
2
? ? ? ? ?
?
? ? ?
 
 
? ?
? ?
99
99
6
2
? ? ?
?
? ? ?
 
 = 3  
 
46. 2 
Sol. 
1 2 r
S S ............. S ? ? ? 
 
? ? ? ?
2
n
n 1 2 ....... r 1 3 5 .................. 2r 1
2
?? ? ? ? ? ? ? ? ? ? ?
??
 
 
? ? ? ?
n
1 3 5 ....... 2r 1
2
? ? ? ? ? 
 
? ?
2 2 2
r. r 1
n r nr
.n
2 2 2
?
? ? ? 
 
? ? nr 1 nr
2
?
? 
 
47. 2