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SATYAM SATAPATHY from ODM PUBLIC SCHOOL BHUBANESWAR, BHUBANESWAR
Subject :NSO    Class : Class 7

Ans 1:

Pranav Prakash from SSSVV GAIL Vijaipur, Guna
Class : Class 9
Apropos to the question the mass of the car has a role in the momentum of the car, as Momentum=mv ,therefore option 'C' should be the correct answer.

Ans 2:

ALICE from Delhi Public School, UDAIPUR
Class : Class 9
No b is only

Ans 3:

Jeevika from St Marys , Titardi , Udaipur
Class : Class 7
if you read the solution properly it is quite clear that the answer is b

Ans 4:

Sargah Surya from bgsnps, Bengaluru
Class : Class 4
C is the correct answer. Force is applied to stop the vehicle. The formulae of force is mass * acceleration. Acceleration is final velocity - initial velocity by time. F= M*{(v-u)/t} and its kinetically equation is F=M *[v square - u square/ 2 * distance]. Which ever formula for force has been taken MASS is important. So clearly MASS HAS A ROLE TO PLAY. hence the answer must be C

Ans 5:

Raavi from DPS, Ranchi
Class : Class 8
c is the correct answer and solution is also given in favour of option c. but answer is b

Ans 6:

Anu Aarna Jagadeesan from Don Bosco School of Excellence, Chennai
Class : Class 5

Ans 7:

ALICE from Delhi Public School, UDAIPUR
Class : Class 9
Yes

Ans 8:

ALICE from Delhi Public School, UDAIPUR
Class : Class 9

Ans 9:

ALICE from Delhi Public School, UDAIPUR
Class : Class 9
C only

Ans 10:

TRIYA SOUMIK NATH from SYMBIOSIS PRIMARY SCHOOL, PUNE
Class : Class 6
Though it looks like mass has no role to play in braking distance, actually coeff of friction (COF) between tyre and road surface has inverse relationship with braking distance. COF depends on normal force due to weight of vehicle and therefore indirectly mass plays a role on braking distance. The correct answer should be C

Ans 11:

Soham Dighe from DAV Boys Senior Secondary School Gopalapuram Chennai, Chennai
Class : Class 8
C should be the correct answer. Same has been explained in the solution also.

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Anshuman D from MVM, Hyderabad
Subject :NSO    Class : Class 7

Ans 1:

KHYATI LAKHCHAURA from MVN, FARIDABAD
Class : Class 7
it should be destarched first then keep the setup in sunlight because if we will directly test the presence of starch both the leaves will show blue -black color as there will be starch already present in both of them. So the correct answer should be OPTION A

Ans 2:

priyanka from kendriya vidalaya kathmandu, JHAJJAR
Class : Class 8
option b

Ans 3:

Karan Oberoi from Amanora School, Pune
Class : Class 8
Destarching of leaf is a more effective way so option A should be correct

Ans 4:

priyanka from kendriya vidalaya kathmandu, JHAJJAR
Class : Class 8
option b

Ans 5:

Advika Narkar from The Shishukunj International School, Jhalaria Campus, Indore
Class : Class 8
The leaves should be destarched first, so that there is no presence of starch prior to the experiment in any of the leaves. In that manner, the experiment will be more clear. Therefore, Option A is also a potential answer.

Ans 6:

priyanka from kendriya vidalaya kathmandu, JHAJJAR
Class : Class 8
option b

Ans 7:

Anish Ranjan Senapati from V S VIDYALAYA, KORAPUT
Class : Class 5

Ans 8:

Rohan A Gowda from Jnana Sagara International Public School, Channarayapatna
Class : Class 7
The plant should be destarched first and then used for the experiment To accurately test for the presence of starch in leaves X and Y.

Ans 9:

kartick Somasundaram from cps, chennai
Class : Class 3
option B

Ans 10:

Derek from Arsha vidya Mandir, Chennai
Class : Class 7
It should be destarched first

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Ashwak from Cambridge, Jalandhar
Subject :NSO    Class : Class 3

Ans 1:

SRIRAM SRINIVASAN PAKU from KENDRIYA VIDHALAYA ASHOK NAGAR, CHENNAI
Class : Class 6

Ans 2:

Haris from Deens academy , Banglore
Class : Class 5
all the questions answer is b

Ans 3:

Yatindra G from SBOA, Chennai
Class : Class 4
B!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Ans 4:

Haris from Deens academy , Banglore
Class : Class 5
b

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Arnav kaul from Chirec, Hyderabad
Subject :NSO    Class : Class 5

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

Ans 2:

Shreetej from Pearson public school, Pune
Class : Class 5
d is correct answer

Ans 3:

Shreetej from Pearson public school, Pune
Class : Class 5
D is the answer

Ans 4:

Shubhank Mittal from Lotus Valley International School, Noida
Class : Class 5

Ans 5:

Sanskriti Nagar from Rose Bud Academy, khagaria
Class : Class 5
D

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Arnav kaul from Chirec, Hyderabad
Subject :NSO    Class : Class 5

Ans 1:

Soham from DPS Whitefield, Bangalore
Class : Class 4
wrong answer

Ans 2:

CHIRAG KHORIA from EVERGREEN PUBLIC SCHOOL, MAYNAGURI
Class : Class 5
tough one,i think the explanation is:S is lens and that part is not colored and t is either eye lids or cornea and neither of them is folded skin.acoording to me

Ans 3:

niuv from meridian, Hyderabad
Class : Class 6

Ans 4:

Anya the NSO prodigy gymnast from Gopalan International School, Bengaluru
Class : Class 4
very tough though the ans is c

Ans 5:

Mehar from NCS, Vasco
Class : Class 5
please explain

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Arnav kaul from Chirec, Hyderabad
Subject :NSO    Class : Class 5

Ans 1:

Adyasha Senapati from Gopalan International School, Bangalore
Class : Class 6
What's the Correct answer?

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Arnav kaul from Chirec, Hyderabad
Subject :NSO    Class : Class 5

Ans 1:

Dushyant Singh from Shri Ram Centennial School, Indore
Class : Class 6

Ans 2:

Dushyant Singh from Shri Ram Centennial School, Indore
Class : Class 6

Ans 3:

Dushyant Singh from Shri Ram Centennial School, Indore
Class : Class 6
both answers b and c are not advisable

Ans 4:

Dushyant Singh from Shri Ram Centennial School, Indore
Class : Class 6

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Arnav kaul from Chirec, Hyderabad
Subject :NSO    Class : Class 5

Ans 1:

PARUL GHARSE from Vibgyor, Mumbai
Class : Class 6

Ans 2:

Vallabh Agrawalla from Sai International School, Bhubaneswar
Class : Class 8
After unscrambling the four jumbled words. We get, Sedimentation, Chlorination, Distillation and Filtration. Filtration and Sedimentation are methods of removing insoluble impurities. Distillation is the only option that is used to remove soluble impurities.

Ans 3:

Devina kashyap from TPS, Mumbai
Class : Class 4

Ans 4:

Devina kashyap from TPS, Mumbai
Class : Class 4

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shreyash rawat from podar international school, pune
Subject :NSO    Class : Class 4

Ans 1:

Kanav Dhingra from DPS sec45, Gurgaon
Class : Class 10
option B

Ans 2:

Advika Narkar from The Shishukunj International School, Jhalaria Campus, Indore
Class : Class 8
The pattern followed is 6 7=137 9=16So,8 4=12

Ans 3:

Advika Narkar from The Shishukunj International School, Jhalaria Campus, Indore
Class : Class 8
i.e. 6 plus 7 = 13 and 9 plus 7 = 16 . SO, 8 plus 4 is 12.

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Diti Sharma from Amity International School, Noida
Subject :NSO    Class : Class 8

Ans 1:

Arnav Raj Sharma from The gulibilium school, Gurgaon
Class : Class 9
Please explain

Ans 2:

Pratyay Anand from Ryan International School, NAVI MUMBAI
Class : Class 9
i dont understand how ?

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