In a metre bridge experiment null point is obtained at 40 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then the new position of the null point from the same end, if one decides to balance a resistance of 3X against Y, will be close to :

In a meter bridge experiment, we have a wire of length 100 cm. The null point occurs when the ratio of the resistances equals the ratio of the lengths of the wire segments. Initially, resistance X is balanced against resistance Y, and the null point is at 40 cm from the end where X is connected. Since X is less than Y, we use the formula:

$$\frac{X}{Y} = \frac{l_1}{100 - l_1}$$

Here, $$ l_1 = 40 $$ cm. Substituting the values:

$$\frac{X}{Y} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3}$$

So, we have $$ \frac{X}{Y} = \frac{2}{3} $$, which means $$ X = \frac{2}{3} Y $$.

Now, we need to balance a resistance of 3X against Y. Let the new null point be at $$ l $$ cm from the same end where X was connected. The balance condition becomes:

$$\frac{3X}{Y} = \frac{l}{100 - l}$$

Substitute $$ \frac{X}{Y} = \frac{2}{3} $$ into the equation:

$$\frac{3X}{Y} = 3 \times \frac{X}{Y} = 3 \times \frac{2}{3} = 2$$

So, the equation simplifies to:

$$2 = \frac{l}{100 - l}$$

To solve for $$ l $$, multiply both sides by $$ 100 - l $$:

$$2 \times (100 - l) = l$$

Which gives:

$$200 - 2l = l$$

Now, bring all terms involving $$ l $$ to one side:

$$200 = l + 2l$$

$$200 = 3l$$

Dividing both sides by 3:

$$l = \frac{200}{3} \approx 66.67 \text{ cm}$$

This value is approximately 67 cm. Comparing with the options, 67 cm corresponds to option C.

Hence, the correct answer is Option C.