The main scale of a Vernier caliper has 20 divisions per cm, and 50...

First, find the length of one main scale division (MSD):
$1\text{ MSD} = \dfrac{1}{20}\text{ cm} = 0.05\text{ cm}$

Given $50\text{ VSD} = 49\text{ MSD}$, the least count (LC) is:
$\text{LC} = 1\text{ MSD} - \dfrac{49}{50}\text{ MSD} = \dfrac{1}{50}\text{ MSD} = \dfrac{0.05}{50}\text{ cm} = 0.001\text{ cm}$

When the jaws are closed, the Vernier zero is to the left of the main scale zero, indicating a negative zero error. The $40^{\text{th}}$ division coincides, so the zero error is:
$\text{Zero Error} = -(N - n) \times \text{LC}$
$\text{Zero Error} = -(50 - 40) \times 0.001\text{ cm} = -0.010\text{ cm}$

During the measurement, the main scale reading (MSR) is the reading just before the Vernier zero, which is $2.4\text{ cm}$. The Vernier scale reading (VSR) is:
$\text{VSR} = 24 \times \text{LC} = 24 \times 0.001\text{ cm} = 0.024\text{ cm}$

The measured reading is:
$\text{Measured Reading} = \text{MSR} + \text{VSR} = 2.4\text{ cm} + 0.024\text{ cm} = 2.424\text{ cm}$

The true diameter is obtained by subtracting the zero error from the measured reading:
$\text{True Reading} = \text{Measured Reading} - \text{Zero Error}$
$\text{True Reading} = 2.424\text{ cm} - (-0.010\text{ cm}) = 2.434\text{ cm}$

Answer: $2.434\text{ cm}$