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Find the time period of oscillation of a pendulum of length equal to the radius of the earth

1. 2pie(2r/g)^(1/2)

2. 2pie(r/2g)^(1/2)

26 Answers

13
deepanshu001 agarwal ·

WHEN 0.05 MOLE OF NAOH WAS ADDED TO 1 L OF ABUFFER SOLUTION THEN ITS pH CHANGES FROM 5.7 TO 5.85 . THE BUFFER CAPACITY OF THE SOLUTIN IS

1. 0.167
2 0.224
3. 0.3334
4. 0.862

9
Celestine preetham ·

9
Celestine preetham ·

@ #24 yes thats wat u do

@ #25 yes i agree with you , i too used to do these tricks in mock tests but in real xam nowadays papersetters are bcomin cautious

but i feel its necessary to know derivation for fulller understanding
ill post it

1
ANKIT MAHATO ·

celestine ... Time period for a pendulum of infinite length is largest
2π√(R/g) ........

now the time period of R length pendulum should be less than this ... :)
so 2π√(R/2g)

13
deepanshu001 agarwal ·

unitary method is lyk if for change in ph by .15 u nid .05 mols then u r sayin tht for change in ph by 1 u nid .333 mol tht is wat unitary method

9
Celestine preetham ·

k tell me how #7 explains correct ans ?????
im unable to visualise :(

#15 wats unitary method????

13
deepanshu001 agarwal ·

@cele.........thnx

i found the xplanation given by ankit convincin atleast for our purpose

#15 ......how r u using unitary method here ??....even i got the ans but i wasnt convinced

9
Celestine preetham ·

#1
ans is 2 , but deepanshu u said u got it #8 without any valid reason ???
do u want derivation or did u already derive that ??

#8
LR =4a
distance of focii frm directrix =2a so solve
(Q is wrong or ans shud be sin in place of cos as in #20)

#10 is ±1/2 a solution ???? sub and see

method : bring tan-1x to RHs and take tan on both sides
u get x3-3x2-1=0
drawing graph u see it has only one real solution and that also satisfies the original inverse critirea

#11 in eq 1 n (no of moles is a microscopic property ....see definition)

#15 ans = .05/5.85 -5.7 =.33333

1
ANKIT MAHATO ·

i think it should be sin instead of cos ... someone plz verify it !

1
ANKIT MAHATO ·

2a = U^2/2g - U^2/2g(cos2a)
becuse directrix is a line || x axis

4a = U^2/g(1 - cos 2a) = 2U^2/g(sina)^2

1
ANKIT MAHATO ·

oh deep .. yup i thot its ellipse

13
deepanshu001 agarwal ·

@ankit....it is a parabola....

for 3 rd even i got tht .....but the ans is tht the no. of soln is 4

1
ANKIT MAHATO ·

for 3rd ±1/2 is a solution

13
deepanshu001 agarwal ·

cele ...plzz ans this

13
deepanshu001 agarwal ·

YEP ....BHASKAR IT IS....BUT WHY?? HOW ??IS IT A MICROSCOPIC PROPERTY...

13
deepanshu001 agarwal ·

YEP ....BHASKAR IT IS....BUT WHY?? HOW ??IS IT A MICROSCOPIC PROPERTY...

1
Bhaskar Mukherjee ·

yeah i suppose that the ans shld... be 1...

13
deepanshu001 agarwal ·

Which of the following involves microscopic properties?

1. PV=nRT
2. KE=1/2MU^2
3. R1/R2 = (M2/M1)^(1/2)
4. NONE

13
deepanshu001 agarwal ·

The number of solution of the equation tan^{-1}(1-x) + tan^{-1}(x) + tan^{-1}(1+x)
= tan^{-1}(3x)

1. 1
2. 2
3. 3
4. 4

1
ANKIT MAHATO ·

okey working on this one !

13
deepanshu001 agarwal ·

kk....i got it

heres another 1

the lenght of latus rectum of a parabola whose focus is (U^2/2g(sin2a),U^2/2g(cos2a))

and directrix is y= U^2/2g

ans.... 2(U^2)/g((cosa)^2)

1
ANKIT MAHATO ·

now i remembered dude .. we learnt that expression
T = 2π√(l/g)

we also now that time period of infinite length pendulum is 2π√(R/g)

so don't think twice before marking 2. 2π√(R/2g)
.. ans

13
deepanshu001 agarwal ·

thnx,.....

1
ANKIT MAHATO ·

okie ... abi solve kar rahe hai .....

13
deepanshu001 agarwal ·

he....he.....

plzz....go on

1
ANKIT MAHATO ·

yar celestine he kyun .. hum nahi answer kare kya ?

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