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PHXI02:UNITS AND MEASUREMENTS

367503 The least count of a stop watch is 1/5 second. The time of 20 oscillations of a pendulum is measured to be 25 seconds. How much will be the percentage error in the measurement of time ?

1 \(0.6\,\% \)

2 \(0.4\,\% \)

3 \(0.2\,\% \)

4 \(0.8\,\% \)

PHXI02:UNITS AND MEASUREMENTS

367504 If \(X = \frac{A}{B}\) and \(\Delta X,\Delta A\) and \(\;\Delta B\) are maximum absolute errors in \(X\), \(A\) and \(B\) respectively, then the maximum fractional error in \(X\) is given by

1 \(\Delta X = \Delta A\; - \Delta B\)

2 \(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A}\; + \frac{{\Delta B}}{B}\)

3 \(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A}\; - \frac{{\Delta B}}{B}\)

4 \(\Delta X = \Delta A\; + \Delta B\)

PHXI02:UNITS AND MEASUREMENTS

367505 Two resistors \(A\) and \(B\) have values \((3.0 \pm 0.1)k\Omega \) and \((9.0 \pm 0.3)k\Omega \) respectively. If they are connected in parallel, the percentage error in the equivalent resistance is

1 \(2\,\% \)

2 \(3.3\,\% \)

3 \(10\,\% \)

4 \(1\,\% \)

Explanation:

\({A = 3.0k\,\Omega ,\Delta A = 0.1k\,\Omega }\)
\({B = 9.0k\,\Omega ,\Delta B = 0.3k\,\Omega }\)
Now, equivalent parallel resistance \(R_{P}\) is given by,
\({\frac{1}{{{R_p}}} = \frac{1}{{\;A}} + \frac{1}{{\;B}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (1)}\)
\({\therefore \quad {R_p} = \frac{{AB}}{{A + B}} = \frac{{3 \times 9}}{{3 + 9}}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {R_p} = 2.25\,k\,\Omega }\)
Differentiating equation (1), we get,
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{ - \Delta {R_P}}}{{R_p^2}} = \frac{{ - \Delta A}}{{{A^2}}} - \frac{{\Delta B}}{{{B^2}}}}\)
\({\therefore \quad \Delta {R_P} = \Delta A{{\left( {\frac{{{R_P}}}{{\;A}}} \right)}^2} + \Delta B{{\left( {\frac{{{R_P}}}{B}} \right)}^2}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = (0.1){{\left( {\frac{{2.25}}{{3.0}}} \right)}^2} + (0.3){{\left( {\frac{{2.25}}{{9.0}}} \right)}^2}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 0.05625 + 0.01875}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \Delta {R_P} = 0.075\,\,k\Omega }\)
\(\therefore\) Percentage error in equivalent resistance is,
\({\frac{{\Delta {R_P}}}{{{R_P}}} \times 100 = \frac{{0.075}}{{2.25}} \times 100}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 3.33\% }\)

PHXI02:UNITS AND MEASUREMENTS

367506 In an experiment, the percentage of error occurred in the measurement of physical quantities \(A\), \(B\), \(C\) and \(D\) are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement \(X\), where \(x = \frac{{{A^2}{B^{1/2}}}}{{{C^{1/3}}{D^3}}}\) , will be

1 \(\left( {\frac{3}{{13}}} \right)\% \)

2 \(16\% \)

3 \( - 10\% \)

4 \(10\% \)

NEET - 2019

PHXI02:UNITS AND MEASUREMENTS

367503 The least count of a stop watch is 1/5 second. The time of 20 oscillations of a pendulum is measured to be 25 seconds. How much will be the percentage error in the measurement of time ?

1 \(0.6\,\% \)

2 \(0.4\,\% \)

3 \(0.2\,\% \)

4 \(0.8\,\% \)

PHXI02:UNITS AND MEASUREMENTS

367504 If \(X = \frac{A}{B}\) and \(\Delta X,\Delta A\) and \(\;\Delta B\) are maximum absolute errors in \(X\), \(A\) and \(B\) respectively, then the maximum fractional error in \(X\) is given by

1 \(\Delta X = \Delta A\; - \Delta B\)

2 \(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A}\; + \frac{{\Delta B}}{B}\)

3 \(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A}\; - \frac{{\Delta B}}{B}\)

4 \(\Delta X = \Delta A\; + \Delta B\)

PHXI02:UNITS AND MEASUREMENTS

367505 Two resistors \(A\) and \(B\) have values \((3.0 \pm 0.1)k\Omega \) and \((9.0 \pm 0.3)k\Omega \) respectively. If they are connected in parallel, the percentage error in the equivalent resistance is

1 \(2\,\% \)

2 \(3.3\,\% \)

3 \(10\,\% \)

4 \(1\,\% \)

Explanation:

\({A = 3.0k\,\Omega ,\Delta A = 0.1k\,\Omega }\)
\({B = 9.0k\,\Omega ,\Delta B = 0.3k\,\Omega }\)
Now, equivalent parallel resistance \(R_{P}\) is given by,
\({\frac{1}{{{R_p}}} = \frac{1}{{\;A}} + \frac{1}{{\;B}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (1)}\)
\({\therefore \quad {R_p} = \frac{{AB}}{{A + B}} = \frac{{3 \times 9}}{{3 + 9}}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {R_p} = 2.25\,k\,\Omega }\)
Differentiating equation (1), we get,
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{ - \Delta {R_P}}}{{R_p^2}} = \frac{{ - \Delta A}}{{{A^2}}} - \frac{{\Delta B}}{{{B^2}}}}\)
\({\therefore \quad \Delta {R_P} = \Delta A{{\left( {\frac{{{R_P}}}{{\;A}}} \right)}^2} + \Delta B{{\left( {\frac{{{R_P}}}{B}} \right)}^2}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = (0.1){{\left( {\frac{{2.25}}{{3.0}}} \right)}^2} + (0.3){{\left( {\frac{{2.25}}{{9.0}}} \right)}^2}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 0.05625 + 0.01875}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \Delta {R_P} = 0.075\,\,k\Omega }\)
\(\therefore\) Percentage error in equivalent resistance is,
\({\frac{{\Delta {R_P}}}{{{R_P}}} \times 100 = \frac{{0.075}}{{2.25}} \times 100}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 3.33\% }\)

PHXI02:UNITS AND MEASUREMENTS

367506 In an experiment, the percentage of error occurred in the measurement of physical quantities \(A\), \(B\), \(C\) and \(D\) are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement \(X\), where \(x = \frac{{{A^2}{B^{1/2}}}}{{{C^{1/3}}{D^3}}}\) , will be

1 \(\left( {\frac{3}{{13}}} \right)\% \)

2 \(16\% \)

3 \( - 10\% \)

4 \(10\% \)

NEET - 2019

PHXI02:UNITS AND MEASUREMENTS

367503 The least count of a stop watch is 1/5 second. The time of 20 oscillations of a pendulum is measured to be 25 seconds. How much will be the percentage error in the measurement of time ?

1 \(0.6\,\% \)

2 \(0.4\,\% \)

3 \(0.2\,\% \)

4 \(0.8\,\% \)

PHXI02:UNITS AND MEASUREMENTS

367504 If \(X = \frac{A}{B}\) and \(\Delta X,\Delta A\) and \(\;\Delta B\) are maximum absolute errors in \(X\), \(A\) and \(B\) respectively, then the maximum fractional error in \(X\) is given by

1 \(\Delta X = \Delta A\; - \Delta B\)

2 \(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A}\; + \frac{{\Delta B}}{B}\)

3 \(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A}\; - \frac{{\Delta B}}{B}\)

4 \(\Delta X = \Delta A\; + \Delta B\)

PHXI02:UNITS AND MEASUREMENTS

367505 Two resistors \(A\) and \(B\) have values \((3.0 \pm 0.1)k\Omega \) and \((9.0 \pm 0.3)k\Omega \) respectively. If they are connected in parallel, the percentage error in the equivalent resistance is

1 \(2\,\% \)

2 \(3.3\,\% \)

3 \(10\,\% \)

4 \(1\,\% \)

Explanation:

\({A = 3.0k\,\Omega ,\Delta A = 0.1k\,\Omega }\)
\({B = 9.0k\,\Omega ,\Delta B = 0.3k\,\Omega }\)
Now, equivalent parallel resistance \(R_{P}\) is given by,
\({\frac{1}{{{R_p}}} = \frac{1}{{\;A}} + \frac{1}{{\;B}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (1)}\)
\({\therefore \quad {R_p} = \frac{{AB}}{{A + B}} = \frac{{3 \times 9}}{{3 + 9}}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {R_p} = 2.25\,k\,\Omega }\)
Differentiating equation (1), we get,
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{ - \Delta {R_P}}}{{R_p^2}} = \frac{{ - \Delta A}}{{{A^2}}} - \frac{{\Delta B}}{{{B^2}}}}\)
\({\therefore \quad \Delta {R_P} = \Delta A{{\left( {\frac{{{R_P}}}{{\;A}}} \right)}^2} + \Delta B{{\left( {\frac{{{R_P}}}{B}} \right)}^2}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = (0.1){{\left( {\frac{{2.25}}{{3.0}}} \right)}^2} + (0.3){{\left( {\frac{{2.25}}{{9.0}}} \right)}^2}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 0.05625 + 0.01875}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \Delta {R_P} = 0.075\,\,k\Omega }\)
\(\therefore\) Percentage error in equivalent resistance is,
\({\frac{{\Delta {R_P}}}{{{R_P}}} \times 100 = \frac{{0.075}}{{2.25}} \times 100}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 3.33\% }\)

PHXI02:UNITS AND MEASUREMENTS

367506 In an experiment, the percentage of error occurred in the measurement of physical quantities \(A\), \(B\), \(C\) and \(D\) are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement \(X\), where \(x = \frac{{{A^2}{B^{1/2}}}}{{{C^{1/3}}{D^3}}}\) , will be

1 \(\left( {\frac{3}{{13}}} \right)\% \)

2 \(16\% \)

3 \( - 10\% \)

4 \(10\% \)

NEET - 2019

PHXI02:UNITS AND MEASUREMENTS

367503 The least count of a stop watch is 1/5 second. The time of 20 oscillations of a pendulum is measured to be 25 seconds. How much will be the percentage error in the measurement of time ?

1 \(0.6\,\% \)

2 \(0.4\,\% \)

3 \(0.2\,\% \)

4 \(0.8\,\% \)

PHXI02:UNITS AND MEASUREMENTS

367504 If \(X = \frac{A}{B}\) and \(\Delta X,\Delta A\) and \(\;\Delta B\) are maximum absolute errors in \(X\), \(A\) and \(B\) respectively, then the maximum fractional error in \(X\) is given by

1 \(\Delta X = \Delta A\; - \Delta B\)

2 \(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A}\; + \frac{{\Delta B}}{B}\)

3 \(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A}\; - \frac{{\Delta B}}{B}\)

4 \(\Delta X = \Delta A\; + \Delta B\)

PHXI02:UNITS AND MEASUREMENTS

367505 Two resistors \(A\) and \(B\) have values \((3.0 \pm 0.1)k\Omega \) and \((9.0 \pm 0.3)k\Omega \) respectively. If they are connected in parallel, the percentage error in the equivalent resistance is

1 \(2\,\% \)

2 \(3.3\,\% \)

3 \(10\,\% \)

4 \(1\,\% \)

Explanation:

\({A = 3.0k\,\Omega ,\Delta A = 0.1k\,\Omega }\)
\({B = 9.0k\,\Omega ,\Delta B = 0.3k\,\Omega }\)
Now, equivalent parallel resistance \(R_{P}\) is given by,
\({\frac{1}{{{R_p}}} = \frac{1}{{\;A}} + \frac{1}{{\;B}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (1)}\)
\({\therefore \quad {R_p} = \frac{{AB}}{{A + B}} = \frac{{3 \times 9}}{{3 + 9}}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {R_p} = 2.25\,k\,\Omega }\)
Differentiating equation (1), we get,
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{ - \Delta {R_P}}}{{R_p^2}} = \frac{{ - \Delta A}}{{{A^2}}} - \frac{{\Delta B}}{{{B^2}}}}\)
\({\therefore \quad \Delta {R_P} = \Delta A{{\left( {\frac{{{R_P}}}{{\;A}}} \right)}^2} + \Delta B{{\left( {\frac{{{R_P}}}{B}} \right)}^2}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = (0.1){{\left( {\frac{{2.25}}{{3.0}}} \right)}^2} + (0.3){{\left( {\frac{{2.25}}{{9.0}}} \right)}^2}}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 0.05625 + 0.01875}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \Delta {R_P} = 0.075\,\,k\Omega }\)
\(\therefore\) Percentage error in equivalent resistance is,
\({\frac{{\Delta {R_P}}}{{{R_P}}} \times 100 = \frac{{0.075}}{{2.25}} \times 100}\)
\({{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = 3.33\% }\)

PHXI02:UNITS AND MEASUREMENTS

367506 In an experiment, the percentage of error occurred in the measurement of physical quantities \(A\), \(B\), \(C\) and \(D\) are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement \(X\), where \(x = \frac{{{A^2}{B^{1/2}}}}{{{C^{1/3}}{D^3}}}\) , will be

1 \(\left( {\frac{3}{{13}}} \right)\% \)

2 \(16\% \)

3 \( - 10\% \)

4 \(10\% \)

NEET - 2019