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1

JEE Main 2021 (Online) 27th August Evening Shift

MCQ (Single Correct Answer)
The boxes of masse 2 kg and 8 kg are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass 8 kg to strike the ground starting from rest. (use g = 10 m/s2)

A
0.34 s
B
0.2 s
C
0.25 s
D
0.4 s

Explanation


(m1g $$-$$ 2T) = m1a $$-$$ (1)

T $$-$$ m2g = m2(2a)

2T $$-$$ 2m2g = 4m2 a $$-$$ (2)

m1g $$-$$ 2m2g = (m1 + 4m2) a

a = $${{(8 - 4)g} \over {(8 + 8)}} = {4 \over {16}}g = {g \over 4}$$

a = $${{10} \over 4}$$ m/s2

S = $${1 \over 2}$$ at2

$${{0.2 \times 2 \times 4} \over {10}}$$ = t2

t = 0.4 sec
2

JEE Main 2021 (Online) 26th August Morning Shift

MCQ (Single Correct Answer)
The initial mass of a rocket is 1000 kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of 20 ms-2. The gases come out at a relative speed of 500 ms$$-$$1 with respect to the rocket : [Use g = 10 m/s2]
A
6.0 $$\times$$ 102 kg s$$-$$1
B
500 kg s$$-$$1
C
10 kg s$$-$$1
D
60 kg s$$-$$1

Explanation


$${F_{thrust}} = \left( {{{dm} \over {dt}}.{V_{rel}}} \right)$$

$$\left( {{{dm} \over {dt}}{V_{rel}} - mg} \right) = ma$$

$$ \Rightarrow \left( {{{dm} \over {dt}}} \right) \times 500 - {10^3} \times 10 = {10^3} \times 20$$

$${{dm} \over {dt}}$$ = (60 kg /s)
3

JEE Main 2021 (Online) 27th July Evening Shift

MCQ (Single Correct Answer)
A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation

$$F = {F_0}\left[ {1 - {{\left( {{{t - T} \over T}} \right)}^2}} \right]$$

Where F0 and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is :
A
2F0T/M
B
F0T/2M
C
4F0T/3M
D
F0T/3M

Explanation

At t = 0, u = 0

$$a = {{{F_0}} \over M} - {{{F_0}} \over {M{T^2}}}{(t - T)^2} = {{dv} \over {dt}}$$

$$\int\limits_0^v {dv = \int\limits_{t = 0}^{2T} {\left( {{{{F_0}} \over M} - {{{F_0}} \over {M{T^2}}}{{(t - T)}^2}} \right)dt} } $$

$$V = \left[ {{{{F_0}} \over M}t} \right]_0^{2T} - {{{F_0}} \over {M{T^2}}}\left[ {{{{t^3}} \over 3} - {t^2}T + {T^2}t} \right]_0^{2T}$$

$$ \Rightarrow $$ $$V = {{4{F_0}T} \over {3M}}$$
4

JEE Main 2021 (Online) 27th July Morning Shift

MCQ (Single Correct Answer)
Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are m, 2m and 2m respectively. A moves towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. All motions occur along same straight line. The final speed of C is :

A
6 m/s
B
9 m/s
C
4 m/s
D
3 m/s

Explanation

Collision between A and B


m $$\times$$ 9 = mv1 + 2mv2 (from momentum conservation)

$$e = 1 = {{{v_2} - {v_1}} \over 9}$$

$$\Rightarrow$$ v2 = 6 m/sec. v1 = $$-$$3 m/sec.

collision between B and C


2m $$\times$$ 6 = 4mv (from momentum conservation)

v = 3 m/s

Questions Asked from Laws of Motion

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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JEE Main 2021 (Online) 27th August Evening Shift (1)
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JEE Main 2021 (Online) 27th July Evening Shift (1)
JEE Main 2021 (Online) 27th July Morning Shift (1)
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