Question No. 8

In an experiment to determine the acceleration due to gravity $g$, the formula used for the time period of a periodic motion is $T=2\pi \sqrt{\frac{7(R-r)}{5g}}$. The values of $R$ and $r$ are measured to be $(60\pm 1)$ mm and $(10\pm 1)$ mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?
(A) The error in the measurement of $r$ is 10%
(B) The error in the measurement of $T$ is 3.57%
(C) The error in the measurement of $T$ is 2%
(D) The error in the determined value of $g$ is 11%

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Solution:

Step 1: Error in $r$

Given: $r=(10\pm 1)$ mm
Absolute error, $\mathrm{\Delta }r=1$ mm
Percentage error:

$$$$Δrr×100=110×100=10%

So, option (A) is true.

Step 2: Error in $T$

Time period measurements: 0.52 s, 0.56 s, 0.57 s, 0.54 s, 0.59 s
Least count = 0.01 s
Mean time period:

$$$$Tmean=0.52+0.56+0.57+0.54+0.595=2.785=0.556 s≈0.56 s

Absolute error in each reading (using mean absolute deviation):

$$$$∣0.52−0.56∣=0.04,∣0.56−0.56∣=0.00,∣0.57−0.56∣=0.01,∣0.54−0.56∣=0.02,∣0.59−0.56∣=0.03

Mean absolute error:

$$$$ΔT=0.04+0.00+0.01+0.02+0.035=0.105=0.02 s

Percentage error in $T$:

$$$$ΔTTmean×100=0.020.56×100≈3.57%

So, option (B) is true and option (C) is false.

Step 3: Error in $g$

Given formula:

$$$$T=2π7(R−r)5g

Squaring both sides:

$$$$T2=4π2⋅75⋅R−rg  ⟹  g=28π25⋅R−rT2

Let $D=R-r$. Then:

$$$$g=kDT2,wherek=28π25

Percentage error in $g$:

$$$$Δgg=ΔDD+2ΔTT

Now, $R=(60\pm 1)$ mm, $r=(10\pm 1)$ mm
So, $D=R-r=60-10=50$ mm
$\mathrm{\Delta }D=\mathrm{\Delta }R+\mathrm{\Delta }r=1+1=2$ mm

$$$$ΔDD=250=0.04

From above, $\frac{\mathrm{\Delta }T}{T}=0.0357$
Thus:

$$$$Δgg=0.04+2×0.0357=0.04+0.0714=0.1114

Percentage error in $g$:

$$$$Δgg×100≈11.14%≈11%

So, option (D) is true.

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Final Answer:

$$\boxed{{\displaystyle }}$$A, B, D