In This post you will get answer of Q.
A solid ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at the speed $60, cm ^{-1}$ on the string, when the car is at rest. When the car accelerates on a horizontal road, then speed of the pulse is $66, cm ^{-1}$. The acceleration of the car is nearly $left(g=10, ms ^{-2}right)$
Solution:
When car is at rest, tension in string is $T=m g$.
$T=M g=m v_{1}^{2}$
$V_{1}=sqrt{frac{T}{r}}=sqrt{frac{M g}{mu}}$ …(i)
When car is accelerating tension,
$T =sqrt{Mleft(a^{2}+g^{2}right)^{1 / 2}}$
$therefore v_{2} =sqrt{frac{Mleft(a^{2}+g^{2}right)^{1 / 2}}{mu}}$
Dividing Eq (i) by Eq. (ii), we get
$frac{v_{2}}{v_{1}}=frac{sqrt{Mleft(a^{2}+g^{2}right)^{1 / 2}}}{sqrt{M g}}=frac{66}{60}$
$frac{left(a^{2}+g^{2}right)^{1 / 2}}{g}=frac{121}{100}$
Squaring, we get
$Rightarrow a^{2}+g^{2}=left(frac{121}{100}right)^{2}$
$Rightarrow a^{2}=146.41-100$
$Rightarrow a^{2}=46.41$
So, $a=6.8, ms ^{-2}$
Solution:
When car is at rest, tension in string is $T=m g$.
$T=M g=m v_{1}^{2}$
$V_{1}=sqrt{frac{T}{r}}=sqrt{frac{M g}{mu}}$ …(i)
When car is accelerating tension,
$T =sqrt{Mleft(a^{2}+g^{2}right)^{1 / 2}}$
$therefore v_{2} =sqrt{frac{Mleft(a^{2}+g^{2}right)^{1 / 2}}{mu}}$
Dividing Eq (i) by Eq. (ii), we get
$frac{v_{2}}{v_{1}}=frac{sqrt{Mleft(a^{2}+g^{2}right)^{1 / 2}}}{sqrt{M g}}=frac{66}{60}$
$frac{left(a^{2}+g^{2}right)^{1 / 2}}{g}=frac{121}{100}$
Squaring, we get
$Rightarrow a^{2}+g^{2}=left(frac{121}{100}right)^{2}$
$Rightarrow a^{2}=146.41-100$
$Rightarrow a^{2}=46.41$
So, $a=6.8, ms ^{-2}$