Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is ____________ ms^{$$-$$1}. (Take g = 10 ms^{$$-$$2})

Your Input ________

Correct Answer is **6**

by energy conservation $$mgl = {1 \over 2}I{\omega ^2} = {1 \over 2}{{m{l^2}{\omega ^2}} \over 3}$$

$$ \Rightarrow \omega = \sqrt {{{6g} \over l}} $$

As we know the relation between the linear speed and angular speed,

$$v = \omega r = \omega l = \sqrt {6gl} $$

$$v = \sqrt {6 \times 10 \times .6} $$ = 6 m/s

Hence, the speed of the free end of the rod when it passes through its lowest position is 6 m/s.

2

Numerical

In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias $$\left( {{{{I_1}} \over {{I_2}}}} \right)$$ will be x : 1. The value of x will be _____________.

Your Input ________

Correct Answer is **9**

$${1 \over 2}{I_1}{({\omega _1})^2} = {1 \over 2}{I_2}{({\omega _2})^2}$$

$${I_1}{\left( {{v \over {3R}}} \right)^2} = {I_2}{\left( {{v \over R}} \right)^2}$$

$${{{I_1}} \over {{I_2}}} = {9 \over 1}$$

3

Numerical

A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10 s is ____________ $$\pi$$ $$\times$$ 10^{$$-$$1} Nm.

Your Input ________

Correct Answer is **4**

$$\tau = {{\Delta L} \over {\Delta t}} = {{I({\omega _f} - {\omega _i})} \over {\Delta t}}$$

$$\tau = {{{{m{R^2}} \over 2} \times [0 - \omega ]} \over {\Delta t}}$$

$$ = {{10 \times {{(20 \times {{10}^{ - 2}})}^2}} \over 2} \times {{600 \times \pi } \over {30 \times 10}}$$

$$ = 0.4\pi = 4\pi \times {10^{ - 2}}$$

$$\tau = {{{{m{R^2}} \over 2} \times [0 - \omega ]} \over {\Delta t}}$$

$$ = {{10 \times {{(20 \times {{10}^{ - 2}})}^2}} \over 2} \times {{600 \times \pi } \over {30 \times 10}}$$

$$ = 0.4\pi = 4\pi \times {10^{ - 2}}$$

4

Numerical

A particle of mass 'm' is moving in time 't' on a trajectory given by

$$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$

Where $$\alpha$$ and $$\beta$$ are dimensional constants.

The angular momentum of the particle becomes the same as it was for t = 0 at time t = ____________ seconds.

$$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$

Where $$\alpha$$ and $$\beta$$ are dimensional constants.

The angular momentum of the particle becomes the same as it was for t = 0 at time t = ____________ seconds.

Your Input ________

Correct Answer is **10**

$$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$

$$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$$

$$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$$

$$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\widehat i + 5\beta \widehat j]$$

$$\overrightarrow L = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$$

At t = 0, $$\overrightarrow L = \overrightarrow 0 $$

$$50\alpha \beta {t^2} - 100\alpha \beta ({t^2} - 5t) = 0$$

t $$-$$ 2 (t $$-$$ 5) = 0

t = 10 sec

$$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$$

$$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$$

$$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\widehat i + 5\beta \widehat j]$$

$$\overrightarrow L = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$$

At t = 0, $$\overrightarrow L = \overrightarrow 0 $$

$$50\alpha \beta {t^2} - 100\alpha \beta ({t^2} - 5t) = 0$$

t $$-$$ 2 (t $$-$$ 5) = 0

t = 10 sec

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