1equation is correct yaar !!
A student performs an experiment for determination of g= \frac{4\pi^2l}{T^2} , l≈1 m,
he commits an error of Δl. For T he takes the time of n oscillations with a astop watch of least count ΔT and hes a human error of 0.1 sec.
Which of the followingdata the measurement of g will be most accurate
\begin{matrix} A) &\Delta l & \Delta T sec & n & \text{Amplitude of oscillation} \\ B) & 5mm & 0.2 sec & 10 & 5mm\\ C) & 5mm & 0.2 sec & 20 & 5mm\\ D) & 5mm & 0.1 sec & 20 & 1mm\\ E) & 1mm & 0.1 sec & 50 & 1mm \end{matrix}
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28 Answers
1increasing the no. of trials increases accuracy is correct ...
also small amplitude means small angular displacement θ .. now for small θ .. sinθ ≈ θ approximation holds true .. .. and in derivation this approximation is used somewhere .. which u can figure out .. [50]
11bhaiyya one doubt
\Delta g/g=\Delta l/l+2\Delta T/T
is that equation rite bhaiyya?
how to take consideration for n and amplitude
for n , increasing number of trials increases accuracy( say if im rite)
what about amplitude how to reason?
21hmmmm....... YEAH!!!
thnx!! [1]
Jus today I found it in a buk [11th std. ISC txt buk [3] ]
62@tapan
to solve this question take derivative
and divide by original
or more simply
just take log of the whole thing
and take derivative! :)
21PL. teasch me HOW TO SOLVE THIS TYPE OF QUESTIONS!!!!
I jus came across an identiacal IIT Questn
21DOUBT :
In wat condition of "n" and amplitude of oscillation wud the answer hav bn sumthin oder than E had ∂T and ∂L in the table remained same.............
Ankit/Abhi : can u pl. brief upon the points wich u took into consideration
62yes abhi and ankit are right..
others make sure u understandthis one.. :)
21BTW Virang jus rite the 3 columns in of ur ans rather than writing OPTIONs
Abhi wer do u get these BIG smilyes?
11Let Nishant Bhaiyya now decide which is rite and which is wrong
11The answer is 5mm 0.2 10
The t = 1 /10
T = 0.1
E = (0.005 + 2 *0.2/0.1)*100
E = (4.005)*100
E = 400.5%
For 5mm 0.2 20
T = 1/20
T = 0.05
Therefore
E = (0.005 + 2*0.2/0.05)*100
E = 800.05%
For 5mm 0.1 20
T = 0.05
E = (0.005 + 2* 0.1/0.05) *100
E = 400.05
For 1mm 0.1 50
T = 0.02
E = (0.001 + 2* 0.1/0.002)*100
E = 10000.001%
21Wat's amplitude of oscillation ????
matlab wich forula [2] [17]