Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The position vector of the centre of mass $$\overrightarrow r {\,_{cm}}\,$$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is :

A

$${\overrightarrow r _{cm}}\, = {{11} \over 8}L\,\,\widehat x + {8 \over 8}L\widehat y$$

B

$${\overrightarrow r _{cm}}\, = {5 \over 8}L\,\,\widehat x + {{13} \over 8}L\widehat y$$

C

$${\overrightarrow r _{cm}}\, = {{13} \over 8}L\,\,\widehat x + {5 \over 8}L\widehat y$$

D

$${\overrightarrow r _{cm}}\, = {3 \over 8}L\,\,\widehat x + {{11} \over 8}L\widehat y$$

$${X_{cm}} = {{2mL + 2mL + {{5mL} \over 2}} \over {4m}} = {{13} \over 8}L$$

$${Y_{cm}} = {{2m \times L + m \times \left( {{L \over 2}} \right) + m \times 0} \over {4m}} = {{5L} \over 8}$$

2

MCQ (Single Correct Answer)

A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :

A

$$2\sqrt {{k \over p}} $$

B

$$2\sqrt {{p \over k}} $$

C

$$\sqrt {{{2p} \over 2}} $$

D

$$\sqrt {{{2k} \over p}} $$

$${{dp} \over {dt}} = F = kt$$

$$\int_P^{3P} {dP} = \int_0^T {kt\,dt} $$

$$2p = {{K{T^2}} \over 2}$$

$$T = 2\sqrt {{P \over K}} $$

$$\int_P^{3P} {dP} = \int_0^T {kt\,dt} $$

$$2p = {{K{T^2}} \over 2}$$

$$T = 2\sqrt {{P \over K}} $$

3

MCQ (Single Correct Answer)

A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its
instantaneous angular acceleration will be -

A

$${g \over {13l}}$$

B

$${g \over {2l}}$$

C

$${g \over {3l}}$$

D

$${7g \over {3l}}$$

Applying torque equation about point P.

2M

I = 2M

$$ \therefore $$ $$\alpha $$ = $$-$$ $${{{M_0}gl} \over {13{M_0}{\ell ^2}}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$-$$ $${g \over {13\ell }}$$

$$ \therefore $$ $$\alpha $$ = $${g \over {13\ell }}$$ anticlockwise

4

MCQ (Single Correct Answer)

A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms^{–1}, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is - (g = 10 ms^{–2})

A

30 m

B

40 m

C

20 m

D

10 m

Time taken for the particles to collide,

t = $${f \over {{V_{rel}}}} = {{100} \over {100}} = 1$$ sec

Speed of wood just before collision = gt = 10 m/s

& speed of bullet just before collision v-gt

= 100 $$-$$ 10 = 90 m/s

Now, conservation of linear momentum just before and after the collision -

$$-$$ (0.02) (1v) + (0.02) (9v) = (0.05)v

$$ \Rightarrow $$ 150 = 5v

$$ \Rightarrow $$ v = 30 m/s

Max. height reached by body h = $${{{v^2}} \over {2g}}$$

h = $${{30 \times 30} \over {2 \times 10}}$$ = 45m

$$ \therefore $$ Height above tower = 40 m

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Units & Measurements

Motion

Laws of Motion

Work Power & Energy

Simple Harmonic Motion

Center of Mass and Collision

Rotational Motion

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Heat and Thermodynamics

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Ray & Wave Optics

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