 ### JEE Mains Previous Years Questions with Solutions

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### JEE Main 2021 (Online) 1st September Evening Shift

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 ____________ $$\times$$ 105 N/m.

## Explanation

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, vf = 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 $$\times$$ 105 N/m

Comparing the spring constant, k = x $$\times$$ 105

The value of the x = 16.
2

### JEE Main 2021 (Online) 31st August Morning Shift

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 = ___________.

## Explanation Pi = Pf

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

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

$$\Rightarrow$$ v = 20

(K. E.)I = $${1 \over 2}$$m $$\times$$ (40)2 = 800 m

(K. E.)f = $${1 \over 2}$$$${m \over 2}$$ . (20)2 + $${1 \over 2}$$ . $${m \over 2}$$ (60)2 = 1000 m

| $$\Delta$$ 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

### JEE Main 2021 (Online) 27th August Morning Shift

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 .................... .

## Explanation

Given WA = WB

FAd cos45$$^\circ$$ = FBd 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
4

### JEE Main 2021 (Online) 26th August Morning Shift

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/s2)

## Explanation From energy conservation

Ki + Ui = kf + Uf

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

$$-$$5 = kf $$-$$ 45

kf = 40 J

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