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1

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R (t) is :
A
B
C
D

Explanation

At any instant 't'

Total energy of charge distribution is constant

i.e.     $${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$$

$$ \therefore $$  $${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$$

$$ \therefore $$  V = $$\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$

$$ \therefore $$  V = $$\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$ = C$$\sqrt {{1 \over {{R_0}}} - {1 \over R}} $$

Also the slope of v-s curve will go on decreasing
2

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
For the given cyclic process CAB as shown for a gas, the work done is :

A
1 J
B
10 J
C
5 J
D
30 J

Explanation

Since P$$-$$V indicator diagram is given, so work done by gas is area under the cyclic diagram.

$$ \therefore $$  $$\Delta $$W = Work done by gas = $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ 5 J

= 10 J
3

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)
A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25 $$ \times $$ 106 N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms–2 , the value of x will be close to :
A
8 cm
B
4 cm
C
40 cm
D
80 cm

Explanation

velocity of 1 kg block just before it collides with 3kg block

= $$\sqrt {2gh} = \sqrt {2000} $$ m/s

Applying momentum conversation just before and just after collision.

1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s



initial compression of spring

1.25 $$ \times $$ 106 x0 = 30 $$ \Rightarrow $$ x0 $$ \approx $$ 0

applying work energy theorem,

Wg + Wsp = $$\Delta $$KE

$$ \Rightarrow $$   40 $$ \times $$ x + $${1 \over 2}$$ $$ \times $$ 1.25 $$ \times $$ 106 (02 $$-$$ x2)

= 0 $$-$$ $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ v2

solving x $$ \approx $$ 4 cm
4

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)
A particle which is experiencing a force, given by $$\overrightarrow F = 3\widehat i - 12\widehat j,$$ undergoes a displacement of $$\overrightarrow d = 4\overrightarrow i $$ particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?
A
9 J
B
10 J
C
12 J
D
15 J

Explanation

Work done = $$\overrightarrow F \cdot \overrightarrow d $$  

                    $$=$$ 12 J

work energy theorem

wnet $$=$$ $$\Delta $$K.E.

12 $$=$$ Kf $$-$$ 3

Kf = 15 J

Questions Asked from Work Power & Energy

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