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A solid ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at the speed $60 \mathrm{~cm}^{-1}$ on the string, when the car is at rest. When the car accelerates on a horizontal road, then speed of the pulse is $66 \mathrm{~cm}^{-1}$. The acceleration of the car is nearly $\left(g=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$6.8 \mathrm{~ms}^{-2}$
When car is at rest, tension in string is $T=m g$.
$$
\begin{aligned}
T & =M g=m v_1^2 \\
V_1=\sqrt{\frac{T}{r}} & =\sqrt{\frac{M g}{\mu}} &..(i)
\end{aligned}
$$
When car is accelerating tension,
$$
\begin{aligned}
T & =\sqrt{M\left(a^2+g^2\right)^{1 / 2}} \\
\therefore \quad v_2 & =\sqrt{\frac{M\left(a^2+g^2\right)^{1 / 2}}{\mu}}
\end{aligned}
$$
Dividing Eq (i) by Eq. (ii), we get
$$
\begin{gathered}
\frac{v_2}{v_1}=\frac{\sqrt{M\left(a^2+g^2\right)^{1 / 2}}}{\sqrt{M g}}=\frac{66}{60} \\
\frac{\left(a^2+g^2\right)^{1 / 2}}{g}=\frac{121}{100}
\end{gathered}
$$
Squaring, we get
$$
\begin{array}{rlrl}
& \Rightarrow & a^2+g^2 & =\left(\frac{121}{100}\right)^2 \\
\Rightarrow & a^2 & =146.41-100 \\
\Rightarrow & a^2 & =46.41
\end{array}
$$
So, $a=6.8 \mathrm{~ms}^{-2}$
$$
\begin{aligned}
T & =M g=m v_1^2 \\
V_1=\sqrt{\frac{T}{r}} & =\sqrt{\frac{M g}{\mu}} &..(i)
\end{aligned}
$$
When car is accelerating tension,
$$
\begin{aligned}
T & =\sqrt{M\left(a^2+g^2\right)^{1 / 2}} \\
\therefore \quad v_2 & =\sqrt{\frac{M\left(a^2+g^2\right)^{1 / 2}}{\mu}}
\end{aligned}
$$
Dividing Eq (i) by Eq. (ii), we get
$$
\begin{gathered}
\frac{v_2}{v_1}=\frac{\sqrt{M\left(a^2+g^2\right)^{1 / 2}}}{\sqrt{M g}}=\frac{66}{60} \\
\frac{\left(a^2+g^2\right)^{1 / 2}}{g}=\frac{121}{100}
\end{gathered}
$$
Squaring, we get
$$
\begin{array}{rlrl}
& \Rightarrow & a^2+g^2 & =\left(\frac{121}{100}\right)^2 \\
\Rightarrow & a^2 & =146.41-100 \\
\Rightarrow & a^2 & =46.41
\end{array}
$$
So, $a=6.8 \mathrm{~ms}^{-2}$
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