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TN 12th Computer Applications மின்னணு தரவு பரிமாற்றம் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிக பாதுகாப்பு அமைப்புகள் Sample Question Papers Study Material - QB365 Set A

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12th Standard

TN 12th Computer Applications மின்னணு செலுத்தல் முறைகள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிகம் Sample Question Papers Study Material - QB365 Set A

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12th Standard

TN 12th Computer Applications திறந்த மூல கருத்துருக்கள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications வலையமைப்பு வடமிடல் Sample Question Papers Study Material - QB365 Set A

12th Standard English Medium Maths Subject Application of Differential Calculus Creative 3 Mark Questions with Solution Part - I

Published on: 19/06/2021

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Std: 12th Standard - Maths

chapter: - Creative Questions

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Questions:

Generic 3

5 x 3 = 15

Obtain Maclaurin’s series for \(\frac { 1 }{ 1+x } \)

A cylindrical hole 4 mm in diameter and 12 mm deep in a metal block is reboared to increase the diameter to 4.12mm. Estimate the amount of metal removed

Find the equation of normal to the curve y⁴=ax²at(a,a)

The side of a square is equal to the diameter of a circle. If the side and radius change at the same rate then find the ratio of the change of their areas.

A ball is thrown vertically upwards, moves according to the law s = 13.8 t - 4.9 t² where s
is in metres and t is in seconds.
(i) Find the acceleration at t = 1
(ii) Find velocity at t = 1
(iii) Find the maximum height reached by the ball?

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Answers Key:

Generic 3

5 x 3 = 15

1−

2.89

4x + 3y = 7a

2:π

s = 13.8 t - 4.9 t²
v = \(\frac { ds }{ dt } \) =13.8 - 4.9 (2t)
=13.8-9.8t
When t = 1, v = 13.8 - 9.8(1)
4 m/sec.
Acceleration = \(\frac { d^{ 2 }x }{ { dt }^{ 2 } } \) = -9.8 m/sec²
At maximum height, v = 0
∴ 13.8 - 9.8 t = 0
⇒ 13.8 = 9.8 t
⇒ t = \(\frac { 13.8 }{ 9.8 } \) = 1.40 sec
At t = 1.4 sec,
distance (s) = 13.8 (1.40) - 4.9 (1.40)²
= 19.32 - 9.604 = 9.716 m