A resistance is to be measured using a meter bridge; the student chooses the standard resistance R to be $100\ \Omega$. The student finds the null point at $l_1=16$ cm. To get the null point near the middle of the wire, the value of R will be

For a meter bridge, the balance condition is given by $\dfrac{X}{R} = \dfrac{l}{100 - l}$, where $X$ is the unknown resistance and $R$ is the standard resistance.

Given $R = 100\ \Omega$ and $l = 16$ cm, we can find $X$:

$\dfrac{X}{100} = \dfrac{16}{100 - 16} = \dfrac{16}{84}$

$X = \dfrac{1600}{84} = \dfrac{400}{21} \approx 19.04\ \Omega$

To obtain the null point near the middle of the wire, the balancing length should be $l' \approx 50$ cm. Using the balance condition again with the new standard resistance $R'$:

$\dfrac{X}{R'} = \dfrac{50}{100 - 50} = 1$

$R' = X \approx 19.04\ \Omega$

Thus, the value of the standard resistance should be chosen as $19\ \Omega$.

Answer: $19\ \Omega$