Two weights W1 and W2 are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up with an acceleration g, the tension in the string will be

The correct option is 4W1W2/(W1+W2).

To find the tension in the string when the pulley is accelerating upwards, we can analyze the system from the frame of reference of the accelerating pulley.

Step 1: Relate weights to masses
Let $W_1$ and $W_2$ be the weights of the two suspended masses. Their respective masses $m_1$ and $m_2$ are related to their weights by:
$m_1=\frac{W_1}{g}$
and
$m_2=\frac{W_2}{g}$
where $g$ is the acceleration due to gravity.

Step 2: Find the effective acceleration due to gravity
The pulley is being pulled upwards with an acceleration of $a=g$ relative to the ground.
Since the pulley's frame of reference is non-inertial (accelerating upwards), we apply a downward pseudo force on both masses in this frame. The pseudo acceleration is equal in magnitude and opposite in direction to the pulley's acceleration.
Therefore, the effective acceleration due to gravity $g_{\text{eff}}$ acting downwards in the pulley's frame is:
$g_{\text{eff}}=g+a=g+g=2g$

Step 3: Determine the tension in the string
For a standard Atwood machine with an effective gravitational acceleration $g_{\text{eff}}$, the tension $T$ in the string is given by the formula:
$T=\frac{2m_1{m}_2}{m_1+m_2}{g}_{\text{eff}}$

Substituting $g_{\text{eff}}=2g$ into the tension formula:
$T=\frac{2m_1{m}_2}{m_1+m_2}\left(2g\right)=\frac{4m_1{m}_{2}g}{m_1+m_2}$

Step 4: Express tension in terms of weights
Now, substitute the expressions for masses in terms of weights ($m_1=W_1/g$ and $m_2=W_2/g$):
$T=\frac{4\left(\frac{W_1}{g}\right)\left(\frac{W_2}{g}\right)g}{\frac{W_1}{g}+\frac{W_2}{g}}$
Simplifying the numerator and denominator by cancelling out the common factor of $g$:
$T=\frac{4\left(\frac{W_1{W}_2}{g}\right)}{\frac{W_1+W_2}{g}}=\frac{4W_1{W}_2}{W_1+W_2}$

Thus, the tension in the string is 4W1W2/(W1+W2).