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A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $g=10 \mathrm{~ms}^{-2}$, find thework done (in Joule) by string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.
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Verified Answer
The correct answer is:
8
$$
\text { } \begin{aligned}
\text { Now, } B & =\frac{\mu_0}{4 \pi} \frac{I}{12 x / 5}\left[\sin 37^{\circ}+\sin 53^{\circ}\right] \\
& =7\left(\frac{\mu_0 I}{48 \pi x}\right) \\
a & =\frac{\text { Net pulling force }}{\text { Total mass }} \\
& =\frac{0.72 g-0.36 g}{0.72+0.36}=\frac{g}{3} \\
s & =\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{g}{3}\right)(1)^2=\frac{g}{6}
\end{aligned}
$$
$$
\begin{aligned}
& T-0.36 g=0.36 a=0.36 \frac{g}{3} \\
& \therefore \quad T=0.48 g \\
& \text { Now, } W_T=T S \cos 0^{\circ} \text { (on } 3.6 \mathrm{~kg} \text { mass) } \\
& =(0.48 g)\left(\frac{g}{6}\right)(1)=0.08\left(g^2\right) \\
& =0.08(10)^2=8 \mathrm{~J} \\
&
\end{aligned}
$$
\text { } \begin{aligned}
\text { Now, } B & =\frac{\mu_0}{4 \pi} \frac{I}{12 x / 5}\left[\sin 37^{\circ}+\sin 53^{\circ}\right] \\
& =7\left(\frac{\mu_0 I}{48 \pi x}\right) \\
a & =\frac{\text { Net pulling force }}{\text { Total mass }} \\
& =\frac{0.72 g-0.36 g}{0.72+0.36}=\frac{g}{3} \\
s & =\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{g}{3}\right)(1)^2=\frac{g}{6}
\end{aligned}
$$
$$
\begin{aligned}
& T-0.36 g=0.36 a=0.36 \frac{g}{3} \\
& \therefore \quad T=0.48 g \\
& \text { Now, } W_T=T S \cos 0^{\circ} \text { (on } 3.6 \mathrm{~kg} \text { mass) } \\
& =(0.48 g)\left(\frac{g}{6}\right)(1)=0.08\left(g^2\right) \\
& =0.08(10)^2=8 \mathrm{~J} \\
&
\end{aligned}
$$
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