10th Standard Syllabus & Materials

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TN 10th Tamil நாகரிகம், நாடு, சமூகம் - கவிதை பேழை (செய்யுள்) -முத்தொள்ளாயிரம் Sample Question Papers Study Material - QB365 Set A

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10th Standard

TN 10th Tamil அறம்,தத்துவம், சிந்தனைகவிதை பேழை (செய்யுள்) -அக்கறை Sample Question Papers Study Material - QB365 Set A

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10th Standard

TN 10th Tamil உயிரின்ஓசை - துணைப்பாடம் -பிருமம் Sample Question Papers Study Material - QB365 Set A

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10th Standard

TN 10th Tamil அறம், தத்துவம், சிந்தனை - கவிதை பேழை (செய்யுள்) -தேம்பாவணி * Sample Question Papers Study Material - QB365 Set A

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TN 10th Tamil கலை, அழகியல், புதுமை - உரைநடை உலகம் -பன்முகக் கலைஞர் Sample Question Papers Study Material - QB365 Set A

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TN 10th Tamil மணற்கேணி - இலக்கணம் - இலக்கணம் -பொது Sample Question Papers Study Material - QB365 Set A

State board English medium frequently asked five mark problems in science chapter one

Published on: 28/06/2019

frequently asked five mark problems chapter one

Download Tamil Nadu 10th Standard Science question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

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Std: 10th Standard - Science

chapter: - Model Question paper

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Questions:

Generic 2

2 x 2 = 4

A door is pushed, at a point whose distance from the hinges is 90 cm, with a force of 40 N. Calculate the moment of the force about the hinges.

Calculate the velocity of a moving body of mass 5 kg whose linear momentum is 2.5 kg m s^-1.

Generic 4

1 x 4 = 4

At what height from the centre of the Earth the acceleration due to gravity will be 1/4^th of its value as at the Earth.

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Answers Key:

Generic 2

2 x 2 = 4

Formula: The moment of a force M = F × d
Given: F = 40 N and d = 90 cm = 0.9 m.
Hence, moment of the force = 40 × 0.9 = 36 N m.

Linear momentum = mass × velocity
Velocity = linear momentum / mass. V = 2.5 / 5 = 0.5 m s^-1

Generic 4

1 x 4 = 4

Data: Height from the centre of the Earth,
R' = R + h
The acceleration due to gravity at that height, g' = g/4'
Formula: g = GM /R², gʹ = GM /Rʹ2
\(\frac { g }{ g' } =\left( \frac { R' }{ R } \right) ^{ 2 }={ \left( \frac { R+h }{ R } \right) }^{ 2 }={ \left( 1+\frac { h }{ R } \right) }^{ 2 }\)
\(4=\left( 1+\frac { h }{ R } \right) ^{ 2 },\)
\(2=1+\frac { h }{ R } \quad or\quad h=R.\quad R'=2R\)
From the centre of the Earth, the object is placed at twice the radius of the earth.