Question

The weights W and $2$W are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled with an acceleration of $10m/s^2$, the tension in the string will be.

• A. $\frac{W}{3}$
• B. $\frac{2W}{3}$
• C. $\frac{4W}{3}$
• D. $\frac{8W}{3}$

Answer 1

Answer: (D)

$\frac{8W}{3}$

$\begin{array}{c}\text{Solution -we know,}\\ \text{weight = mass x gravity}\\ \text{Thus, weight of object}1=\frac{\text{Mass(M1)}}{\text{Gravity(G)}}and\\ \text{weight of object}2=\frac{\text{Mans}\left(M_2\right)}{\text{Gravity}\left(g\right)}\end{array}$
$$\begin{array}{cc}{M}_{2g}-T=M_{2}a-g& -\left(A\right)\\ -M_{1}g+T=M_{1}a-g& -\left(B\right)\end{array}$$
(A)-(B)
$\begin{array}{c}2M_{2}g-2M_{1g}=M_{2}a+M_{1}a\\ a=\frac{2g(M_2-M_1)}{(M_1+M_2)}-\left(C\right)\end{array}$

Now terrsion in string is
$\begin{array}{cc}T& =M_{1}g+M_{1}a+M_{1}g\\ & =2M_{1}g+M_{1}a\end{array}$

Reducing the value of mass of object1 we get,
$=\frac{\omega 1}{g}\sqrt{g}\times 2+\frac{2g(M_2-M_1)}{(M_1+M_2)}]$

$={\omega }_1\left\{\frac{2(M_1+M_2)+2(M_2-M_1)\}}{(M_1+M_2)}\right\}$

$={\omega }_1\left\{\frac{4\left({\omega }_2/g\right)}{\frac{w1}{g}+\frac{{\omega }_2}{g}}\right\}$

$=\frac{4{\omega }_1{\omega }_2}{{\omega }_1+{\omega }_2}[given,\begin{array}{c}{\omega }_1={\omega }_2\\ {\omega }_2=2\omega \end{array}]=\frac{4\times \omega \times 2\omega }{3\omega }=\frac{8\omega }{3}ans$