Student I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different length of the pendulum and/or record time for different number of oscillations. The observations area shown in the table. Least count for length = 0.1 cm Least count for time = 0.1 s .tg border-collapse:collapse;border-spacing:0; .tg tdfont-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black; .tg thfont-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black; .tg .tg-s6z2text-align:center .tg .tg-baqhtext-align:center;vertical-align:top Student Length of the pendulum (cm) No. of oscillations (n) Total time for(n) oscillations (s) Time periods (s) I 64.0 8 128.0 16.0 II 64.0 4 64.0 16.0 III 20.0 4 36.0 9.0 If E I , E II and E III are the percentage errors in g, i.e., (\frac\triangle gg\times100) for students I, II and III, respectively,then

Units and Measurements PYQ — JEE Advanced 2008 · Shift 1 · Q47 with Solution

Student I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different length of the pendulum and/or record time for different number of oscillations. The observations area shown in the table.

Least count for length = 0.1 cm

Least count for time = 0.1 s

Student	Length of the pendulum (cm)	No. of oscillations (n)	Total time for(n) oscillations (s)	Time periods (s)
I	64.0	8	128.0	16.0
II	64.0	4	64.0	16.0
III	20.0	4	36.0	9.0

If E_I, E_II and E_III are the percentage errors in g, i.e., $\left(\frac{\triangle g}g\times100\right)$ for students I, II and III, respectively,then

A
EI = 0
B
EI is minimum
C
EI = EII
D
EII is maximum

View written solutionFree

Correct answer: B

Formula for $g$ using a simple pendulum

For a simple pendulum,

T = 2\pi \sqrt{\frac{L}{g}}

g = \frac{4\pi^2 L}{T^2}

Hence, the fractional error in $g$ is

\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}

Therefore, percentage error is

E = \left(\frac{\Delta L}{L} + 2\frac{\Delta T}{T}\right)\times 100

Error in length measurement

Least count in length = $0.1\text{ cm}$ . So we take

\Delta L = 0.1\text{ cm}

Error in time period measurement

If total time $t$ for $n$ oscillations is measured, then

T = \frac{t}{n}

So,

\Delta T = \frac{\Delta t}{n}

where least count of stopwatch is $0.1\text{ s}$ , hence

\Delta t = 0.1\text{ s}

Thus,

\frac{\Delta T}{T} = \frac{\Delta t/n}{T} = \frac{\Delta t}{nT} = \frac{\Delta t}{t}

So we can directly use

\frac{\Delta T}{T} = \frac{0.1}{t}

For Student I

Given:

$L = 64.0\text{ cm}$
$n=8$
total time $t = 128.0\text{ s}$
$T = 16.0\text{ s}$

Fractional error in length:

\frac{\Delta L}{L} = \frac{0.1}{64.0}

Fractional error in time period:

\frac{\Delta T}{T} = \frac{0.1}{128.0}

So,

\frac{\Delta g}{g} = \frac{0.1}{64} + 2\left(\frac{0.1}{128}\right)

But

2\left(\frac{0.1}{128}\right)=\frac{0.1}{64}

Hence,

\frac{\Delta g}{g} = \frac{0.1}{64}+\frac{0.1}{64}=\frac{0.2}{64}=0.003125

Thus,

E_I = 0.3125\%

So, option A ( $E_I=0$ ) is false.

For Student II

Given:

$L = 64.0\text{ cm}$
$n=4$
total time $t = 64.0\text{ s}$
$T = 16.0\text{ s}$

Fractional error in length:

\frac{\Delta L}{L} = \frac{0.1}{64}

Fractional error in time period:

\frac{\Delta T}{T} = \frac{0.1}{64}

So,

\frac{\Delta g}{g} = \frac{0.1}{64}+2\left(\frac{0.1}{64}\right)=\frac{0.3}{64}=0.0046875

Thus,

E_{II}=0.46875\%

Clearly,

E_{II} > E_I

So option C ( $E_I=E_{II}$ ) is false.

For Student III

Given:

$L = 20.0\text{ cm}$
$n=4$
total time $t = 36.0\text{ s}$
$T = 9.0\text{ s}$

Fractional error in length:

\frac{\Delta L}{L} = \frac{0.1}{20.0}=0.005

Fractional error in time period:

\frac{\Delta T}{T} = \frac{0.1}{36.0}

So,

\frac{\Delta g}{g} = \frac{0.1}{20}+2\left(\frac{0.1}{36}\right)

=0.005+0.005555\ldots =0.010555\ldots

Thus,

E_{III}\approx 1.0556\%

Hence,

E_{III} > E_{II} > E_I

So option D ( $E_{II}$ is maximum) is false.

Final comparison of options

A: $E_I=0$ ❌
B: $E_I$ is minimum ✅
C: $E_I=E_{II}$ ❌
D: $E_{II}$ is maximum ❌

Therefore, the correct option is:

\boxed{B}

Comparison with stored answer

Stored correct answer: $B$

My derived answer: $B$

So, the derived answer agrees with the stored answer.

More from Units and Measurements

Least count for length = 0.1 cm

Least count for time = 0.1 s

Student	Length of the pendulum (cm)	No. of oscillations (n)	Total time for(n) oscillations (s)	Time periods (s)
I	64.0	8	128.0	16.0
II	64.0	4	64.0	16.0
III	20.0	4	36.0	9.0

If E_I, E_II and E_III are the percentage errors in g, i.e., $\left(\frac{\triangle g}g\times100\right)$ for students I, II and III, respectively,then

A
EI = 0
B
EI is minimum
C
EI = EII
D
EII is maximum

View written solutionFree

Correct answer: B

Formula for $g$ using a simple pendulum

For a simple pendulum,

T = 2\pi \sqrt{\frac{L}{g}}

g = \frac{4\pi^2 L}{T^2}

Hence, the fractional error in $g$ is

\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}

Therefore, percentage error is

E = \left(\frac{\Delta L}{L} + 2\frac{\Delta T}{T}\right)\times 100

Error in length measurement

Least count in length = $0.1\text{ cm}$ . So we take

\Delta L = 0.1\text{ cm}

Error in time period measurement

If total time $t$ for $n$ oscillations is measured, then

T = \frac{t}{n}

So,

\Delta T = \frac{\Delta t}{n}

where least count of stopwatch is $0.1\text{ s}$ , hence

\Delta t = 0.1\text{ s}

Thus,

\frac{\Delta T}{T} = \frac{\Delta t/n}{T} = \frac{\Delta t}{nT} = \frac{\Delta t}{t}

So we can directly use

\frac{\Delta T}{T} = \frac{0.1}{t}

For Student I

Given:

$L = 64.0\text{ cm}$
$n=8$
total time $t = 128.0\text{ s}$
$T = 16.0\text{ s}$

Fractional error in length:

\frac{\Delta L}{L} = \frac{0.1}{64.0}

Fractional error in time period:

\frac{\Delta T}{T} = \frac{0.1}{128.0}

So,

\frac{\Delta g}{g} = \frac{0.1}{64} + 2\left(\frac{0.1}{128}\right)

But

2\left(\frac{0.1}{128}\right)=\frac{0.1}{64}

Hence,

\frac{\Delta g}{g} = \frac{0.1}{64}+\frac{0.1}{64}=\frac{0.2}{64}=0.003125

Thus,

E_I = 0.3125\%

So, option A ( $E_I=0$ ) is false.

For Student II

Given:

$L = 64.0\text{ cm}$
$n=4$
total time $t = 64.0\text{ s}$
$T = 16.0\text{ s}$

Fractional error in length:

\frac{\Delta L}{L} = \frac{0.1}{64}

Fractional error in time period:

\frac{\Delta T}{T} = \frac{0.1}{64}

So,

\frac{\Delta g}{g} = \frac{0.1}{64}+2\left(\frac{0.1}{64}\right)=\frac{0.3}{64}=0.0046875

Thus,

E_{II}=0.46875\%

Clearly,

E_{II} > E_I

So option C ( $E_I=E_{II}$ ) is false.

For Student III

Given:

$L = 20.0\text{ cm}$
$n=4$
total time $t = 36.0\text{ s}$
$T = 9.0\text{ s}$

Fractional error in length:

\frac{\Delta L}{L} = \frac{0.1}{20.0}=0.005

Fractional error in time period:

\frac{\Delta T}{T} = \frac{0.1}{36.0}

So,

\frac{\Delta g}{g} = \frac{0.1}{20}+2\left(\frac{0.1}{36}\right)

=0.005+0.005555\ldots =0.010555\ldots

Thus,

E_{III}\approx 1.0556\%

Hence,

E_{III} > E_{II} > E_I

So option D ( $E_{II}$ is maximum) is false.

Final comparison of options

A: $E_I=0$ ❌
B: $E_I$ is minimum ✅
C: $E_I=E_{II}$ ❌
D: $E_{II}$ is maximum ❌

Therefore, the correct option is:

\boxed{B}

Comparison with stored answer

Stored correct answer: $B$

My derived answer: $B$

So, the derived answer agrees with the stored answer.

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