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# A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration ‘a’ directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration ‘a’ (relative to car) vertically. The tension in the string is equal to (assume $\theta$ remain constant):A. $m\sqrt {\left( {{g^2} + {a^2}} \right)}$B. $m\sqrt {\left( {{g^2} + {a^2}} \right)} - ma$C. $m\sqrt {\left( {{g^2} + {a^2}} \right)} + ma$D. $m\left( {g + a} \right)$

Last updated date: 20th Jun 2024
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Hint:-Inertia of a body is the state of the being in a state of rest or motion until an external force is applied on the body. The pseudo force on the bob will be on the opposite side of the motion of the car due to the motion of the car.
Formula used: Weight of the mass ‘m’ is given by,
$W = mg$
W is weight and m is the mass and g is the acceleration due to gravity.

Complete step-by-step solution
It is given that a bob is hanging over a pulley inside a car through a string the second end of the string is in the hand of a person standing in the car the car is moving with constant acceleration ‘a’ directed horizontally. Other end of the string is pulled with constant acceleration ‘a’ (relative to car) vertically. We need to find the tension in the string if the $\theta$ is constant.
The force on the bob will be in the opposite direction of the motion of the car and also a tension in the string is present.
The forces on the bob of mass m is given by,

It can be seen in the figure that in the y’-y’ direction,
$\Rightarrow ma\cos \theta = mg\sin \theta$
$\Rightarrow \tan \theta = \dfrac{a}{g}$………eq. (1)
Since the string is pulled with an acceleration a therefore the equation formed will be,
$\Rightarrow T - \left( {ma\sin \theta + mg\cos \theta } \right) = ma$………eq. (2)
The value of $\sin \theta$ and $\cos \theta$ can be calculated by equation (1).

The value of $\sin \theta$ and $\cos \theta$ is given by,
$\sin \theta = \dfrac{a}{{\sqrt {{a^2} + {g^2}} }}$and$\cos \theta = \dfrac{g}{{\sqrt {{a^2} + {g^2}} }}$.
Replace the values into equation (1) and equation (2) we get.
$\Rightarrow T - \left( {ma\sin \theta + mg\cos \theta } \right) = ma$
$\Rightarrow T - ma\left( {\dfrac{a}{{\sqrt {{a^2} + {g^2}} }}} \right) - mg\left( {\dfrac{g}{{\sqrt {{a^2} + {g^2}} }}} \right) = ma$
$\Rightarrow T - \left( {\dfrac{{m{a^2}}}{{\sqrt {{a^2} + {g^2}} }}} \right) - \left( {\dfrac{{m{g^2}}}{{\sqrt {{a^2} + {g^2}} }}} \right) = ma$
$\Rightarrow T = ma + \left( {\dfrac{{m{a^2}}}{{\sqrt {{a^2} + {g^2}} }}} \right) + \left( {\dfrac{{m{g^2}}}{{\sqrt {{a^2} + {g^2}} }}} \right)$
$\Rightarrow T = ma + \left( {\dfrac{{m{a^2} + m{g^2}}}{{\sqrt {{a^2} + {g^2}} }}} \right)$
$\Rightarrow T = m\left[ {a + \left( {\dfrac{{{a^2} + {g^2}}}{{\sqrt {{a^2} + {g^2}} }}} \right)} \right]$
$\Rightarrow T = m\left( {a + \sqrt {{a^2} + {g^2}} } \right)$
$\Rightarrow T = m\sqrt {{a^2} + {g^2}} + ma$

The tension in the string is equal to$T = m\sqrt {{a^2} + {g^2}} + ma$. The correct answer for this problem is option C.

Note:- The pseudo force is the force that comes into play when the force on a body is calculated on the bases of a non-inertial frame of reference. The bob is in equilibrium in the direction y’-y’, therefore the forces can be balanced in this direction.