Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 40,000 kg is moving with a speed of 72 kmh^{$$-$$1} when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, the spring constant is ____________ $$\times$$ 10^{5} N/m.

Your Input ________

Correct Answer is **16**

Given, the length of the shock absorber, l = 1.5 m

The total mass of the system, M = 40000 kg

The speed of the wagon, v = 72 km/h

When brakes are applied, the final velocity, v_{f} = 0

The compressed spring of the shock absorber, x = 1 m

Applying the work-energy theorem,

Work done by the system = Change in kinetic energy

$$W = \Delta KE$$

$${W_{friction}} + {W_{spring}} = {1 \over 2}mv_f^2 + {1 \over 2}mv_i^2$$

$$ - {{90} \over {100}}\left( {{1 \over 2}m{v^2}} \right) + {W_{spring}} = 0 - {1 \over 2}mv_i^2$$ ($$\because$$ 90% energy lost due to friction)

$${W_{spring}} = - {{10} \over {100}} \times {1 \over 2}m{v^2}$$

$$ - {1 \over 2}k{x^2} = {1 \over {20}}m{v^2}$$

$$k = {{m{v^2}} \over {10 \times {x^2}}}$$

Substituting the values in the above equation, we get

$$k = {{40000 \times {{\left( {72 \times {5 \over {18}}} \right)}^2}} \over {10{{(1)}^2}}}$$

= 16 $$\times$$ 10^{5} N/m

Comparing the spring constant, k = x $$\times$$ 10^{5}

The value of the x = 16.

The total mass of the system, M = 40000 kg

The speed of the wagon, v = 72 km/h

When brakes are applied, the final velocity, v

The compressed spring of the shock absorber, x = 1 m

Applying the work-energy theorem,

Work done by the system = Change in kinetic energy

$$W = \Delta KE$$

$${W_{friction}} + {W_{spring}} = {1 \over 2}mv_f^2 + {1 \over 2}mv_i^2$$

$$ - {{90} \over {100}}\left( {{1 \over 2}m{v^2}} \right) + {W_{spring}} = 0 - {1 \over 2}mv_i^2$$ ($$\because$$ 90% energy lost due to friction)

$${W_{spring}} = - {{10} \over {100}} \times {1 \over 2}m{v^2}$$

$$ - {1 \over 2}k{x^2} = {1 \over {20}}m{v^2}$$

$$k = {{m{v^2}} \over {10 \times {x^2}}}$$

Substituting the values in the above equation, we get

$$k = {{40000 \times {{\left( {72 \times {5 \over {18}}} \right)}^2}} \over {10{{(1)}^2}}}$$

= 16 $$\times$$ 10

Comparing the spring constant, k = x $$\times$$ 10

The value of the x = 16.

2

Numerical

A block moving horizontally on a smooth surface with a speed of 40 ms^{$$-$$1} splits into two equal parts. If one of the parts moves at 60 ms^{$$-$$1} in the same direction, then the fractional change in the kinetic energy will be x : 4 where x = ___________.

Your Input ________

Correct Answer is **1**

P

m $$\times$$ 40 = $${m \over 2}$$ $$\times$$ v + $${m \over 2}$$ $$\times$$ 60

40 = $${v \over 2}$$ + 30

$$\Rightarrow$$ v = 20

(K. E.)

(K. E.)

| $$\Delta$$ K. E. | = | 1000m $$-$$ 800 m | = 200 m

$${{\Delta K.E.} \over {{{(K.E.)}_i}}} = {{200m} \over {800m}} = {1 \over 4} = {x \over 4}$$

x = 1

3

Numerical

Two persons A and B perform same amount of work in moving a body through a certain distance d with application of forces acting at angle 45$$^\circ$$ and 60$$^\circ$$ with the direction of displacement respectively. The ratio of force applied by person A to the force applied by person B is $${1 \over {\sqrt x }}$$. The value of x is .................... .

Your Input ________

Correct Answer is **2**

Given W_{A} = W_{B}

F_{A}d cos45$$^\circ$$ = F_{B}d cos60$$^\circ$$

$${F_A} \times {1 \over {\sqrt 2 }} = {F_B} \times {1 \over 2}$$

$${{{F_A}} \over {{F_B}}} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}$$

x = 2

F

$${F_A} \times {1 \over {\sqrt 2 }} = {F_B} \times {1 \over 2}$$

$${{{F_A}} \over {{F_B}}} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}$$

x = 2

4

Numerical

A uniform chain of length 3 meter and mass 3 kg overhangs a smooth table with 2 meter lying on the table. If k is the kinetic energy of the chain in joule as it completely slips off the table, then the value of k is ................. . (Take g = 10 m/s^{2})

Your Input ________

Correct Answer is **40**

From energy conservation

K

$$0 + \left( { - 1 \times 10 \times {1 \over 2}} \right) = {k_f} + \left( { - 3 \times 10 \times {3 \over 2}} \right)$$

$$-$$5 = k

k

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