The velocity of water waves $v$ may depend upon their wavelength $\lambda $, the density of water $\rho $ and the acceleration due to gravity $g$. The method of dimensions gives the relation between these quantities as
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- 1The period of oscillation of a simple pendulum is $T=2\pi \sqrt {\frac{l}{g}} $. Measured value of $L$ is $20.0\; cm$ known to $1\; mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90\ s$ using a wrist watch of $1\; s$ resolution. The accuracy in the determination of $g$ is ........ $\%$View Solution
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$(a)$ Screw moves $0.5\,mm$ on main scale in one complete rotation
$(b)$ Total divisions on circular scale $=50$
$(c)$ Main scale reading is $2.5\,mm$
$(d)$ $45^{\text {th }}$ division of circular scale is in the pitch line
$(e)$ Instrument has $0.03 \;mm$ negative error
Then the diameter of wire is $...........\,mm$
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Student No. Length of pendulum $(cm)$ No. of oscillations $(n)$ Total time for oscillations Time period $(s)$ $1.$ $64.0$ $8$ $128.0$ $16.0$ $2.$ $64.0$ $4$ $64.0$ $16.0$ $3.$ $20.0$ $4$ $36.0$ $9.0$ (Least count of length $=0.1 \,{m}$, least count for time $=0.1\, {s}$ )
If $E_{1}, E_{2}$ and $E_{3}$ are the percentage errors in $'g'$ for students $1,2$ and $3$ respectively, then the minimum percentage error is obtained by student no. ....... .
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($B$) $l=\sqrt{\left(\frac{\varepsilon k_B T}{n q^2}\right)}$
($C$)$l=\sqrt{\left(\frac{q^2}{\varepsilon n^{2 / 3} k_B T}\right)}$
($D$) $l=\sqrt{\left(\frac{q^2}{\varepsilon n^{1 / 3} k_B T}\right)}$
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$(B)$ Error $\Delta \mathrm{T}$ in measuring $\mathrm{T}$, the time period, is $1$ second
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$(D)$ Percentage error in the determination of $g$ is $2.5 \%$
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